Acid Base AP Past Exam

iiiContents 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Marian De Wane 2. Properties of Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Arden P. Zipp 3. Weak Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Valerie Ferguson 4. pH and the Properties of [H + ] in Aqueous Solution . . . . . . . . . . . . . . . . . . . 29 Lew Acampora 5. Acid–Base Titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Cheri Smith 6. Polyprotic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Lew Acampora 7. Salts and Buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Adele Mouakad 8. Acidic and Basic Reactions of Oxides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Marian DeWane 9. Misconceptions in Chemistry Demonstrated on AP ® Chemistry Examinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 George E. Miller 10. Acid–Base Chemistry Beyond AP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Arden P. Zipp 11. Acids–Bases Lesson Plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Heather Hattori 12. About the Editor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 13. About the Authors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 The Properties of Acids and Bases Arden P. Zipp SUNY Cortland Cortland, New York Acids and bases represent two of the most important classes of chemical compounds. Some of the properties of acids (such as their sour taste) and bases (bitterness) have been known to humans for hundreds of years even if their chemical explanations have not. The reactions of these substances are signifcant in many atmospheric and geological phenomena. In addition, acids and bases control many of the physiological processes in the human body and can catalyze a variety of chemical reactions. This chapter will address some of the fundamental aspects of acids and bases in order to set the stage for subsequent discussions. In addition to taste, which was mentioned above but is no longer used in the chemistry lab, several other properties of acids and bases that can be observed readily are given below. Table 1 shows that they have some properties in common, while they do not share others. Table 1 Properties of Acids and Bases Acids Bases 1 conduct electricity in solution 1 conduct electricity in solution 2 react with product(s) formed 2 react with product(s) formed many metals H 2 metal oxide and water H 2 carbonates CO 2 , H 2 O, a salt metal ions precipitates bicarbonates CO 2 , H 2 O bicarbonates CO 3 2- , H 2 O a base H 2 O, a salt an acid H 2 O, a salt dyes colors, often reddish dyes colors, often bluish SPECIAL FOCuS: Acids and Bases 4 Teaching Tip Students will appreciate these properties more if they are demonstrated by the teacher, or better yet, investigated by students themselves. Inexpensive conductivity meters can be purchased or constructed to investigate the frst property on the list while the others can be studied using well plates and solutions dispensed from dropper bottles. Early History: Arrhenius Theory Some of the properties in Table 1 were known to alchemists hundreds of years ago, and classifcation schemes were developed by many noted scientists during the 1600s and 1700s including Robert Boyle, Joseph Black, and Antoine Lavoisier. However, the beginning of the modern understanding of acids is usually attributed to Svante Arrhenius, who, in his 1884 Ph.D. thesis, explained the electrical conductivity of aqueous solutions in terms of the presence of ions and attributed the unique behavior of acids to H + ions. According to Arrhenius, an acid is a substance that, when added to H 2 O, increases the [H + ], while a base is a substance that increases the [OH - ]. Many of the reactions in Table 1 can be explained with the Arrhenius acid–base theory by the presence of H + or OH - ions as shown in the equations below: Acids Bases 2H + + M (e.g. Mg) → H 2 + M 2+ 2OH - + M 2 O 3 (e.g. Al) + 2H 2 O→ H 2 + 2M(OH) 4 – xOH - + M x+ → M(OH) x (s) 2H + + CO 3 2 - → H 2 O + CO 2 OH - + HCO3 - →H 2 O + CO 3 2- H + + HCO 3 - → H 2 O + CO 2 H + + OH - → H 2 O H + + OH - → H 2 O State of H + in H 2 O. In the above equations the acid is represented as H + , a bare proton. However, it is generally agreed that the charge density of a proton is too large for it to exist in that form in aqueous solutions. Many texts represent protons as hydronium ions, H 3 O + , as if each proton is bonded to a single H 2 O molecule. Current evidence suggests, however, that the actual species has three additional H 2 O molecules attached via hydrogen bonds to form H 9 O 4 + . H + will be used in this volume but it should be recognized that the actual species is more complex. Interestingly, the same texts that make a point of using H 3 O + usually show the hydroxide ion as OH - even though it also appears to be hydrated, existing in H 2 O as H 7 O 4 - . The Properties of Acids and Bases 5 Strong and Weak Acids Arrhenius related the degree of electrical conductivity to the strength of acids, thereby introducing the concept of strong and weak acids. Strong acids are those that ionize completely (or nearly so) in solution, while weak acids ionize only partially, i.e., have solutions that contain signifcant amounts of the parent acid and lower concentrations of H + . In terms of equations, the reaction of a strong acid goes completely to the right, so is written with a simple arrow while that for a weak acid is reversible and is depicted with a double-headed arrow. Strong acid: HA(aq) → H + (aq) + A - (aq) Weak acid: HA(aq) ↔ H + (aq) + A - (aq) Among the standard acids commonly encountered in aqueous solutions, six are strong. HCl hydrochloric acid HNO 3 nitric acid HBr hydrobromic acid HClO 4 perchloric acid HI hydroiodic acid H 2 SO 4 sulfuric acid All other acids behave as weak acids in water. Although strong and weak acids produce the same types of reactions, the rates at which these reactions occur vary depending on the concentration of free H + ions in solution. It should be noted that in the above table, only the frst proton from H 2 SO 4 can really be described as “completely ionized,” the second proton has a measurable equilibrium constant. Teaching Tip The difference between strong and weak acids can be made more concrete for students by demonstrations (or experiments) in which they compare the behavior of 1.0 M solutions of strong (e.g., HCl) and weak (HC 2 H 3 O 2 ) acids. Properties of interest include electrical conductivity and the rate of reaction with Mg metal (using a balloon attached to the mouth of a conical fask to capture the evolved H 2 ). Equal volumes of the weak and strong acids will produce the same volume of H 2 but the rates at which they do so will differ substantially. Reactions with calcium carbonate, capturing the CO 2 , work similarly. Students can measure how much time it takes for a short measured length of Mg ribbon or a measured scoop of calcium carbonate to dissolve completely. SPECIAL FOCuS: Acids and Bases 6 Strong and Weak Bases There are also six common strong bases that ionize completely in H 2 O, though some have limited solubility, which should not be confused with amount of ionization. LiOH lithium hydroxide Ca(OH) 2 calcium hydroxide NaOH sodium hydroxide Sr(OH) 2 strontium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide A solution of a strong base in H 2 O may be obtained by dissolving the appropriate solid hydroxide but may also be formed by reacting a solid metallic oxide with H 2 O. For example, sodium oxide dissolves in H 2 O according to the following equation: Na 2 O + H 2 O → 2 Na + + 2 OH - Similar reactions to produce strongly basic solutions occur with several other anions but these will not be considered here. There are a variety of compounds that behave as weak bases, although few of these contain OH - ions. These will be discussed further below. Concentrations Water itself is a weak acid, existing in equilibrium with its ions, H + and OH -, as shown in the reversible equation: H 2 O ↔ H + + OH - for which K = [H + ][OH – ] =1.8 x 10 -16 [H 2 O] where the square brackets, [], represent concentrations in moles per liter. When the above value of K is multiplied by [H 2 O], (55.5 M), the expression is K w = [H + ][OH - ] with a value of K w = 1.0 x 10 -14 at 25 ˚C. Using the K w expression it is possible to calculate [H + ] for any [OH - ] and vice versa. In a 1.0 M solution of a strong acid, [H + ] =1.0, while in a 1.0 M solution of a strong base, [OH - ] = 1.0 and [H + ] = 1.0 x 10 -14 . Because of this extremely wide range of [H + ], an alternative method of comparing [H + ] concentrations, called the pH scale, has been devised, where pH is defned as -log [H + ]. Thus, a solution with [H + ] = 1.0 has a pH = 0.00 and one with [H + ] = 1.0 x 10 -14 has a pH = 14.00. In pure H 2 O, [H + ] = [OH - ] = 1.0 x 10 -7 and pH = pOH = 7.00. Acid solutions are those in which [H + ] > 1.0 x 10 -7 or pH < 7.00 while basic (alkaline) solutions have [OH - ] > 1.0 x 10 -7 , [H + ] < 1.0 x 10 -7 or pH > 7.00. A more extensive discussion of pH and further calculations are provided in a later chapter. The Properties of Acids and Bases 7 Teaching Tip Students typically enjoy determining the pH values of common household materials (with indicator paper or drops of universal indicator). This activity can be useful for AP students if it is coupled with pH calculations and practice in pH ↔ [H + ] or pOH ↔ [OH-] conversions. Formation of Acidic or Basic Solutions It is easy to see how adding a strong acid like hydrochloric acid, HCl, or even a weak acid such as ethanoic (acetic) acid, HC 2 H 3 O 2 , to water will increase the concentration of [H + ] above 1.0 x 10 -7 as these species ionize completely (HCl, a strong acid) or partially (HC 2 H 3 O 2 , a weak acid) to release H + ions. Similarly, the release of OH - ions can be envisioned from the addition of one of the strong bases to H 2 O. However, the production of acidic solutions by substances that contain no H + (or basic ones by substances that do not possess OH - ) is less apparent. Clearly, an acid–base theory other than that proposed by Arrhenius is required. Brønsted-Lowry Acid–Base Theory As outlined above, Arrhenius defned acids as substances that produce H + ions in H 2 O and bases as species that produce OH - ions. In 1923 Johannes Brønsted and Thomas Lowry independently described acids as proton donors and bases as proton acceptors, thereby broadening the concept of acids and bases signifcantly. In addition, they introduced the notion that acids and bases exist in pairs (called conjugate pairs) that differ by the presence (acid) or absence (base) of a proton. Further, they suggested that when acid–base reactions occur, stronger acids and bases react to form weaker acids and bases as in the equation below, where HA and A - are a conjugate pair, and B and HB + are a conjugate pair: HA + B ↔ HB + + A - Such reactions are reversible and the relative amounts of the various substances depend on the strengths of the acids (bases) involved. Where HA is a stronger acid than HB + and B is a stronger base than A - , the equilibrium constant will be >1. The strengths of Brønsted-Lowry acids and their conjugate bases are inversely related. This is because stronger acids donate protons more readily and, hence, their conjugate bases have lower attractions for protons (i.e., are weaker). The Brønsted-Lowry theory includes as acids all the substances identifed by Arrhenius as acids because they donate H + ions to water and would also classify the SPECIAL FOCuS: Acids and Bases 8 OH - ion as a base because it is a proton acceptor. For example, HCl acts as an acid according to both defnitions in the equation: HCl + H 2 O → H 3 O + + Cl - because it releases an H + ion (Arrhenius) or donates a proton (Brønsted-Lowry). In this reaction, H 2 O behaves as the base. However, the Brønsted-Lowry theory can also account for the acidic behavior of metal ions and nonmetallic oxides, neither of which contain H + ions. In addition, it can explain the basic behavior of weak bases such as ammonia. Acidities of Metal Ions and Nonmetallic Oxides Just as protons are hydrated (described above), most metal ions are also hydrated, usually with six H 2 O molecules. In both instances lone pairs of electrons in H 2 O molecules are attracted to the positive charge of the ion. For metal ions with large charge densities (due to high charges and/or small radii), the attraction for the oxygen of the H 2 O may be strong enough to polarize the O-H bond and cause an H + ion to be released. This process is depicted in the equation: M(H 2 O) 6 x+ + H 2 O ↔ M(HO) 5 OH (x-1)+ + H 3 O + The effect of charge and radius on the acidic behavior of selected metal ions is shown in the Table 2 below. Table 2 Li + Be 2+ slightly acidic weakly acidic Na + Mg 2+ Al 3+ neutral weakly acidic acidic K + Ca 2+ Sc 3+ Ti 4+ neutral slightly acidic acidic very acidic Teaching Tip The relative acidities of these (and other) metal ions can be shown by adding a few drops of universal indicator to solutions containing these ions and comparing the solution colors with those of pH standards. The Properties of Acids and Bases 9 The acidic behavior of nonmetallic oxides can be explained on the basis of their reaction with water to form oxyacids as represented in the equation for SO 2 : SO 2 + H 2 O → H 2 SO 3 Such oxyacids ionize to release H + ions as: H 2 SO 3 + H 2 O ↔ H 3 O + + HSO 3 - Thus, the Brønsted-Lowry theory offers an explanation for the acidic behavior of both metal ions and nonmetallic oxides, neither of which contain H + . A more extensive discussion of oxides (both nonmetallic and metallic) is undertaken in a later section as is the relative acidities of oxyacids. Behavior of Weak Bases As mentioned above, there are six strong bases, each of which contains hydroxide ions. In contrast to the small number of strong bases, there are many substances that behave as weak bases. These substances produce an excess of OH - ions in water even though they do not contain OH - ions themselves. Most of these materials contain a nitrogen atom that can act as a Brønsted-Lowry base by accepting a proton, although some contain other atoms that can behave in this same manner. When a proton is abstracted from a water molecule, an excess of OH - ions remains in the solution as shown in the equation: H 2 O + B ↔ BH + + OH - The [OH - ] produced in such a reaction depends on the position of the above equilibrium, which is a function of the strength of B as a Brønsted-Lowry base. (H 2 O is acting as the acid in this reaction.) For ammonia, NH 3 , the most common weak base, [OH - ] = 4.2 x 10 -3 M in a 1.0 M NH 3 solution. Weak acids and bases will be treated in more detail in another section of this book. Lewis Acid–Base Theory By emphasizing the transfer of protons, the Brønsted-Lowry theory is able to account for several acid–base reactions that can not be treated by the Arrhenius theory. The same year that Brønsted and Lowry proposed their theory, G.N. Lewis noted that the reaction of a proton with a base involved the donation of a pair of electrons from the base to form a bond with the proton. He defned bases as electron pair donors, and acids as electron pair acceptors. Hence, Lewis acid–base reactions encompass SPECIAL FOCuS: Acids and Bases 10 all Brønsted-Lowry reactions but also cover many processes that do not involve proton transfer. Examples of these include all cases in which a pair of electrons is donated, such as the hydration of metal ions and the reaction of ammonia with boron compounds: M x+ + 6H 2 O → M(H 2 O) 6 x+ NH 3 + BCl 3 → H 3 N-BCl 3 This brief discussion of various aspects of acids and bases is intended to introduce the more extensive treatments in later chapters. Because of the importance of acids and bases in modern chemistry, this material has been tested frequently on the AP Chemistry Exam, both Part I and Part II. Sample Part I Released Exam 2002 Questions An analysis of this released exam revealed a total of 9 questions (out of 75) related to acids and bases. Of these, the ones below are based solely on the above material. 21. In the laboratory, H 2 (g) can be produced by adding which of the following to 1.0 M HCl(aq)? I. 1 M NH 3 (aq) II. Zn(s) III. NaHCO 3 (s) (A) I only * (B) II only (C) III only (D) I and II only (E) I, II, and III 22. In liquid ammonia, the reaction represented below occurs. 2NH 3 ↔ NH 4 + + NH 2 - In the reaction, NH 4 + acts as (A) a catalyst (B) both an acid and a base * (C) the conjugate acid of NH 3 The Properties of Acids and Bases 11 (D) the reducing agent (E) the oxidizing agent 30. At 25˚C, aqueous solutions with a pH of 8 have hydroxide concentration, [OH - ], of (A) 1 x 10 -14 M (B) 1 x 10 -8 M * (C) 1 x 10 -6 M (D) 1.0 M (E) 8.0 M Sample Part II Released Exam 2004 Questions Question 4 (c) A 0.1 M nitrous acid solution is added to the same volume of a 0.1 M sodium hydroxide solution HNO 2 + OH - → NO 2 - + H 2 O (d) Hydrogen iodide gas is bubbled into a solution of lithium carbonate: 2 HI + CO 3 2- → 2 I - + H 2 O + CO 2 (h) Concentrated hydrochloric acid is added to a solution of sodium sulfde: 2 H + + S 2- → H 2 S Released Exam 2005 Questions Question 4 (d) Solid calcium carbonate is added to a solution of ethanoic (acetic) acid: Na 2 CO 3 + 2 HC 2 H 3 O 2 → 2 C 2 H 3 O 2 - + H 2 O + CO 2 + 2 Na + (f) Boron trifuoride gas is added to ammonia gas: BF 3 + NH 3 → F 3 B-NH 3 (g) Sulfur trioxide gas is bubbled into a solution of sodium hydroxide: SO 3 + OH - → HSO 4 - SPECIAL FOCuS: Acids and Bases 12 Released Exam 2006 Questions Question 4 (c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide: HC 2 H 3 O 2 + OH - → H 2 O + C 2 H 3 O 2 - (d) Ammonia gas is bubbled into a solution of hydrofuoric acid: NH 3 + HF → NH 4 + + F - (f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas: PH 3 + BCl 3 → H 3 P-BCl 3 Released Exam 2007 Questions Question 4 (b) (i) Excess nitric acid is added to solid calcium carbonate. 2 H + + CaCO 3 → Ca 2+ + H 2 O + CO 2 (ii) Briefy explain why statues made of marble (calcium carbonate) displayed outdoors in urban areas are deteriorating. The H + ions in acid rain react with the marble statues and the soluble compounds of calcium that are formed are washed away. Bibliography Brown, T. L., H. E. LeMay Jr., and B. E. Bursten. Chemistry: The Central Science, 5th ed. Englewood Cliffs, NJ: Prentice Hall, 1991. Huheey, J. E., E. A. Keiter, and R. L. Keiter. Inorganic Chemistry, 4th ed. New York: Harper Collins, 1993. Rayner-Canham, G., and T. Overton. Descriptive Inorganic Chemistry, 4th ed. New York: W.H. Freeman, 2006. Salzberg, H. W. From Caveman to Chemist. Washington, DC: American Chemistry Society, 1991. Weak Acids and Bases Valerie Ferguson Moore High School Moore, Oklahoma Many substances dissolve in water to cause a change in the pH of pure water. Some of the substances ionize completely and others only partially ionize. Strong acids and bases are strong electrolytes in water solutions, and are almost totally ionized. The reaction depicted below of a strong acid in water occurs with all of the HCl molecules in water. Since all of the HCl molecules are converted to H 3 O + and Cl - , the concentration of the hydronium ion is equal to the initial concentration of HCl in the solution. Strong acids and bases are not thought of as establishing equilibria since the reaction essentially goes to completion. However, equilibria never “turn off.” The equilibrium autodissociation of water is always present and operative in any aqueous solution even though the conjugate ion concentration (OH - in acid, H 3 O + in base) is extremely small. Weak acid and base solutions are partially ionized. The Brønsted-Lowry defnition describes many acid and base interactions as a proton transfer from one particle to another. The particle that donates a proton is the acid and the particle accepting the proton is the base. The products of these reactions are called conjugate acids and conjugate bases. The acid and its product, the conjugate base, are a conjugate acid–base pair. SPECIAL FOCuS: Acids and Bases 14 Conjugate acid–base pairs are two substances whose formulas differ by a proton. The example below illustrates a Brønsted-Lowry reaction with acetic acid in water. Examples from Prior AP Chemistry Examinations (2005 Form B) K a = [H 3 O + ][OCl - ]/[HOCl] = 3.2 x 10 -8 1. Hypochlorous acid, HOCl, is a weak acid in water. The K a expression for HOCl is shown above. (a) Write a chemical equation showing how HOCl behaves as an acid in water. HOCl(aq) + H 2 O(l) ↔ OCl - (aq) + H 3 O + (aq) Teaching Tips Students need to practice writing equations for the reactions of weak acids and bases in water. Opportunities for such practice occur several times during the course. Questions on the AP Chemistry Exam commonly ask students to demonstrate their knowledge of writing these equations. The illustration below shows the Brønsted-Lowry reaction of a weak base, ammonia, in water. Brønsted-Lowry bases are proton acceptors. Weak Acids and Bases 15 (2002) 2 NH 3 ↔ NH 4 + + NH 2 - 22. In liquid ammonia, the reaction represented above occurs. In the reaction, NH 4 + acts as (A) a catalyst (B) both an acid and a base * (C) the conjugate acid of NH 3 (D) the reducing agent (E) the oxidizing agent (1994) HSO 4 - + H 2 O ↔ H 3 O + + SO 4 2- 22. In the equilibrium represented above, the species that act as bases include which of the following? I. HSO 4 - II. H 2 O III. SO 4 2 - (A) II only (B) III only (C) I and II (D) I and III * (E) II and III SPECIAL FOCuS: Acids and Bases 16 (1989) 34. All of the following species can function as Brønsted-Lowry bases in solution EXCEPT (A) H 2 O (B) NH 3 (C) S 2- * (D) NH 4 + (E) HCO 3 - * Proton acceptors must have a lone pair of electrons. Brønsted-Lowry reactions are reversible and will establish a chemical equilibrium between the forward and reverse reactions. The position of the equilibrium depends on the strength of the acid and/or base, in aqueous solutions, in comparison to water. The relative strengths of the various weak acids can be described in different ways: the value of the acid dissociation, K a , the [H + ] compared to the initial [HA], strength of the conjugate base compared to water, or the position of the dissociation. The weaker acids have stronger conjugate bases. Stronger acids and bases tend to react with each other to produce their weaker conjugates. Since the stronger species are more likely to react, the conjugates are less likely to react and are thus weaker. The strong acids and bases are essentially totally ionized. The conjugates are very weak and less likely to react with water. Since the weaker species is less likely to react, it is in higher concentrations in the solution. The weaker species will be in greater concentration when equilibrium is established. (1999) HC 2 H 3 O 2 (aq) + CN - (aq) ↔ HCN(aq) + C 2 H 3 O 2 - (aq) 62 . The reaction represented above has an equilibrium constant equal to 3.7 x 10 4 . Which of the following can be concluded from this information? * (A) CN - (aq) is a stronger base than C 2 H 3 O 2 - (aq). (B) HCN(aq) is a stronger acid than HC 2 H 3 O 2 (aq). (C) The conjugate base of CN - (aq) is C 2 H 3 O 2 - (aq). (D) The equilibrium constant will increase with an increase in temperature. (E) The pH of a solution equimolar amounts of CN - (aq) and HC 2 H 3 O 2 (aq) is 7.0. Weak Acids and Bases 17 (1989) H 2 PO 4 - + HBO 3 2- ↔ HPO 4 2- + H 2 BO 3 - 55. The equilibrium constant for the reaction represented by the equation above is greater than 1.0. Which of the following gives the correct relative strengths of the acids and bases in the reaction? Acids Bases * (A) H 2 PO 4 - > H 2 BO 3 - and HBO 3 2 - > HPO 4 2- (B) H 2 BO 3 - > H 2 PO 4 - and HBO 3 2 - > HPO 4 2- (C) H 2 PO 4 - > H 2 BO 3 - and HPO 4 2 - > HBO 3 2- (D) H 2 BO 3 - > H 2 PO 4 - and HPO 4 2 - > HBO 3 2- (E) H 2 PO 4 - = H 2 BO 3 - and HPO 4 2 - = HBO 3 2- (1984) 49. Each of the following can act as both a Brønsted acid and a Brønsted base EXCEPT (A) HCO 3 - (B) H 2 PO 4 - * (C) NH 4 + (D) H 2 O (E) HS - Periodic trends are observed in the relative strength of acids. Binary hydro- acids—those that contain only hydrogen and a nonmetal—tend to increase in strength from left to right across the periodic table within the same period. The strength of a binary acid increases from top to bottom within the same group. The graphic below represents the reaction of a binary acid in water. SPECIAL FOCuS: Acids and Bases 18 Electronegativity of E increases 2.5 < 2.8 < 3.0 HOI HOBr HOC Acid strength increases K a =2.3 × 10 -11 < 2.5 × 10 -9 < 2.9 × 10 -8 Acids composed of hydrogen, oxygen, and some other element are called oxoacids or oxyacids. Oxoacids are composed of a central atom with one or more O-H groups attached to it. The strength of an oxoacid depends upon the ease with which the hydrogen ion is released from the O-H group. The other atoms in the molecule affect the polarity of the O-H bond. As the electronegativity of the central atom increases, the electrons are pulled to the central atom and away from the O-H bond. The probability that H + will be released is increased. The more readily the hydrogen ion is released, the stronger the acid. When the number of oxygens is held constant, the strength of the oxoacid increases from top to bottom within a group and from left to right across a period. For the same central atom, the stronger acids contain more oxygen atoms. Weak Acids and Bases 19 (2002) 1(e). HOBr is a weaker acid than HBrO 3 . Account for this fact in terms of molecular structure. The H-O bond is weakened or increasingly polarized by the additional oxygen atoms bonded to the central bromine atom in HBrO 3 . 1 point earned for a correct explanation. Teaching Tips Students need to demonstrate an understanding of the strength of the O-H bond as related to the strength of the acid. The greater the number of electronegative atoms in the molecule, the weaker the O-H bond as the electrons are pulled away from the O-H bond. Many students found it helpful to draw a structural formula to help answer this question. An important group of oxoacids is the carboxylic acids. These organic compounds are characterized by containing a chain of carbon atoms with one or more carboxyl groups, -COOH, attached. These compounds are weak acids called carboxylic acids. Acetic acid is an example of a simple carboxylic acid. Its reaction with water is shown on page 1. Most carboxylic acids react in a similar fashion. Only the O-H bond is polar enough so the hydrogen is readily donated as a hydrogen ion. Acetic acid SPECIAL FOCuS: Acids and Bases 20 (2007 Form B) 5. Answer the following questions about laboratory solutions involving acids, bases and buffer solutions. (a) Lactic acid, HC 3 H 5 O 3 , reacts with water to produce an acidic solution. Show below are the complete Lewis structures of the reactants. In the space provided above, complete the equation by drawing the complete Lewis structures of the reaction products. (2001) 4(d). Write the net ionic equation for the following reaction: Solutions of potassium hydroxide and propanoic acid are mixed. C 2 H 5 COOH + OH - → C 2 H 5 COO - + H 2 O (2005 Form B) 4(d). Write the net ionic equation for the following reaction: Ethanoic acid (acetic acid) is added to a solution of barium hydroxide. OH - + HC 2 H 3 O 2 → C 2 H 5 O 2 - + H 2 O The strength of the carboxylic acids is affected by the substitution of one or more hydrogen atoms with electronegative atoms. More electronegative atoms pull the electron density away from the O-H bond and cause an increase in acid strength. The closer the electronegative atoms are to the O-H bond, the greater the effect on acid strength by weakening the O-H bond. Weak Acids and Bases 21 I—CH 2 CH 2 COOH Cl—CH 2 CH 2 COOH CH 3 CHCLOOH CH 3 CCL 2 COOH 3-ledopropanoic acid 3-Chloropropanoic acid 2-Chloropropanoic acid 2,2-Dichloropropanoic acid K a = 8.3 x 10 -5 < K a = 1.0 x 10 -4 < K a = 1.4 x 10 -3 < K a = 8.7 x 10 -3 Increasing acid strength Amines are organic compounds that act as weak bases. The structure is an ammonia molecule, NH 3 , with one hydrogen substituted by a hydrocarbon group. Amines react as bases similar to reactions of NH 3 . The amines act as Brønsted-Lowry bases and are proton acceptors. The proton attaches to the nitrogen in the amine group. NH 3 (aq) + H 2 O(aq) ↔ NH 4 + (aq) + OH - (aq) H 3 CNH 2 (aq) + H 2 O(l) ↔ H 3 CNH 3 + (aq) + OH - (aq) Teaching Tips The equation for the reaction of ammonia with water is compared to the equation for the reaction of methylamine with water. In each reaction the water acts as a Brønsted- Lowry acid and ammonia or methylamine as a Brønsted-Lowry base. Examples of other amines. (1999) 4(f). Write the net ionic equation for the following reaction: Methylamine gas is bubbled into distilled water. CH 3 NH 2 + H 2 O → CH 3 NH 3 + + OH - (2003) C 6 H 4 NH 2 (aq) + H 2 O(l) ↔ C 6 H 5 NH 3 + (aq) + OH - (aq) 1. Aniline, a weak base, reacts with water according to the reaction represented above. SPECIAL FOCuS: Acids and Bases 22 Teaching Tips The examples above demonstrate the inclusion of amines on past AP Exams. Students need to have a background of the behavior of amines as weak bases. In question 1 from 2003, students were asked to demonstrate an understanding of the characteristics of an amine equilibrium. Students should be encouraged to develop an understanding of the similarity of amine reactions to those of ammonia. The discussion of weak acids and bases is a good time to include some organic chemistry. Lewis acids can be described as electron-pair acceptors and Lewis bases as electron-pair donors. Sometimes the Lewis acid has fewer than eight electrons, as in compounds such as BF 3 . (2005) 4(f). Write the net ionic equation for the following reaction: Boron trifuoride gas is added to ammonia gas. BF 3 + NH 3 → BF 3 NH 3 (2006) 4(f). Write the net ionic equation for the following reaction: Hydrogen phosphide (phosphine) gas is added to boron trichloride gas. PH 3 + BCl 3 → H 3 PBCl 3 CO 2 (g) + OH - (aq) → HC0 3 - (aq) Lewis acid-base theory represents the movement of electrons in this reaction as follows. Weak Acids and Bases 23 (1999) 4(g). Write the net ionic equation for the following reaction: carbon dioxide gas is passed over hot, solid sodium oxide. CO 2 + Na 2 O → Na 2 CO 3 (2001) 4(a). Write the net ionic equation for the following reaction: Sulfur dioxide gas is bubbled into distilled water. SO 2 + H 2 O → H 2 SO 3 Hydrated metal ions can act as acids in water. The smaller metal ions with greater charge have a greater charge density and produce more acidic solutions. SPECIAL FOCuS: Acids and Bases 24 (2004 Form B) 4(g). Write the net ionic equation for the following reaction: Excess saturated sodium fuoride solution is added to a solution of aluminum sulfate. F – + Al 3 + → [AlF 6 ] 3- (2003 Form B) 4(g). Write the net ionic equation for the following reaction: Excess concentrated potassium hydroxide solution is added to a solution of nickel(II) chloride. OH – + Ni 2+ → [Ni(OH) 4 ] 2- (some other products were accepted) (2003) 4(e). Write the net ionic equation for the following reaction: Excess concentrated aqueous ammonia is added to solid silver chloride. AgCl + NH 3 → Ag(NH 3 ) 2 + + Cl - (1984) 48. Which of the following ions is the strongest Lewis acid? (A) Na + (B) Cl - (C) CH 3 COO - (D) Mg 2+ * (E) Al 3+ I highly suggest using the following Web sites with simulations of weak acid and weak base behavior: • The frst is the work of Thomas Greenbowe of Iowa State University, located at: www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ animationsindex.htm. • The second is part of a collaboration of John I. Gelder, Michael Abraham, and Kirk Haines, which can be found at: genchem1.chem.okstate.edu/CCLI/ Startup.html Weak Acids and Bases 25 Bibliography Brady, James E., and Fred Senese. Chemistry: Matter and Its Changes, 4th ed. Hoboken, NJ: Wiley, 2003. Brown, Theodore E., Eugene H . LeMay, and Bruce E. Bursten. Chemistry: The Central Science, 10th ed. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2006. Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. Silberberg, Martin S. Chemistry: The Molecular Nature of Matter and Change, 2nd ed. New York: McGraw-Hill Higher Education, 2000. Whitten, Kenneth W., Raymond E. Davis, Larry M. Peck, and George G. Stanley. Chemistry, 8th ed. Belmont, CA: Thomson/Brooks/Cole, 2007. Zumdahl, Steven S., and Susan A. Zumdahl. Chemistry, 7th ed. Boston: Houghton Miffin, 2007. SPECIAL FOCuS: Acids and Bases 26 Graphics Index (Picture 1) Brady, James E., and Fred Senese. Chemistry: Matter and Its Changes, 4th ed., 738. Hoboken, NJ: Wiley, 2003. (Picture 2) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 617. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 3) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 617. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 4) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 621. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 5) Brown, Theodore E., Eugene H. LeMay, and Bruce E. Bursten. Chemistry: The Central Science, 10th ed., 703. Upper Saddle River, NJ: Pearson/Prentice- Hall, 2006. (Picture 6) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 622. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 7) Brady, James E., and Fred Senese. Chemistry: Matter and Its Changes, 4th ed., 748. Hoboken, NJ: Wiley, 2003. (Picture 8) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 622. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 9) Brown, Theodore E., Eugene H. LeMay, and Bruce E. Bursten. Chemistry: The Central Science, 10th ed., 705. Upper Saddle River, NJ: Pearson/Prentice- Hall, 2006. (Picture 10) Brown, Theodore E., Eugene H. LeMay, and Bruce E. Bursten. Chemistry: The Central Science, 10th ed., 706. Upper Saddle River, NJ: Pearson/Prentice- Hall, 2006. Weak Acids and Bases 27 (Picture 11) Hill, John W., Ralph H. Petrucci, Terry W. McCreary, and Scott S. Perry. General Chemistry, 4th ed., 623. Upper Saddle River, NJ: Pearson/Prentice-Hall, 2005. (Picture 12) Silberberg, Martin S. Chemistry: The Molecular Nature of Matter and Change, 2nd ed., 780. New York: McGraw-Hill Higher Education, 2000. (Picture 13) Brady, James E., and Fred Senese. Chemistry: Matter and Its Changes, 4th ed., 751. Hoboken, NJ: Wiley, 2003. (Picture 14) Brady, James E., and Fred Senese. Chemistry: Matter and Its Changes, 4th ed., 752. Hoboken, NJ: Wiley, 2003. (Picture 15) Silberberg, Martin S. Chemistry: The Molecular Nature of Matter and Change, 2nd ed., 792. New York: McGraw-Hill Higher Education, 2000. (Picture 16) Silberberg, Martin S. Chemistry: The Molecular Nature of Matter and Change, 2nd ed., 786. New York: McGraw-Hill Higher Education, 2000. pH and the Properties of [H + ] in Aqueous Solution Lew Acampora St. Francis School Louisville, Kentucky The neutralization of a strong acid with a strong base occurs according to the net ionic equation 1 : Equation 1: H + + OH - → H 2 O ; ΔHº = -55.6 kJ ∕ mol ; ΔGº 298K = -79.8 kJ ∕ mol The negative sign of ΔGº 298K , and experience tell us that the reaction proceeds spontaneously from standard initial concentrations of 1.0 M H + and 1.0 M OH - . Rather than proceeding to completion, where the concentration of one or both of the reactants is completely exhausted, the reaction does attain a state of equilibrium in which small but measurable concentrations of both H + and OH - are present. The state of equilibrium is quantitatively described by the equilibrium constant expression. Because the concentration of the pure water is unaffected by the progress of the reaction, it is omitted from the equilibrium constant expression, K eq : K eq 1 H OH = [ ] ⋅ + − The value of K eq can be readily calculated as: K e e eq nG RT o = = - - - kJ mol 0.00 79.8 8831 kJ mol K 298 K ⋅ ⋅ = = × e 1.0 32.2 10 14 1. The symbol “H + (aq)” is a shorthand representation of the hydrated proton. In aqueous solutions, the acidic proton is undoubtedly coordinated to several water molecules through and is often written as the hydronium ion, “H 3 O + (aq).” Because the number of water molecules coordinated to a single proton varies with time, it is probably most accurately represented as “H 2n+1 O n + (aq),” where “n” represents the number of water molecules. In terms of calculations, when used correctly, the symbol H + (aq)” is a convenient and suitable representation. SPECIAL FOCuS: Acids and Bases 30 While the magnitude of the value of K eq refects the observation that the equilibrium state of this reaction lies far to the right, the equilibrium mixture will maintain small but measurable concentrations of both hydrogen ion and hydroxide ion. Commonly, the reverse reaction of Equation 1, which represents the auto- ionization of water, is used to determine the [H + ] and [OH - ] in aqueous solutions. In this case: Equation 2: H 2 O  H + + OH - ; ΔGº 298K = +79.8 kJ ∕ mol ; K w = [ H+ ] ∙ [ OH – ] = 1.0 × 10 -14 The implication of this calculation is that even in the purest sample of water at 25 o C, there are small but measurable concentrations of H + and OH - ion. In pure water 2 : [ H + ] = [ 0Hˉ ] Η + = = × = × OH 1.0 10 1.0 10 - -14 -7 M While the [H + ] = [OH - ] in pure water, many solutes (acids or bases) will change the relative concentrations of these ions. In any case, the equilibrium relation, [H + ] [OH - ] = 1.0 x 10 -14 , will remain applicable in any aqueous solution at 25 o C. Therefore, any substance that increases the [H + ] must also decrease the [OH - ] in order to satisfy the equilibrium relation. Solutes that have this effect on the relative concentrations of H + and OH - are called acids. Conversely, bases are substances that will increase the [OH - ] and will decrease the [H + ]. In determining the [H + ] and the [OH - ] in an acid, base, or buffer solution containing solutes of reasonable concentration, the contribution from the auto- ionization of water can be neglected. In a solution of an acid, the [H + ] can be determined directly from the concentration and strength of the acid, and the [OH - ] is subsequently calculated using Equation 2. For example, the [H + ] ion in a 0.020 M solution of the strong acid hydrochloric acid is readily determined as follows: Equation 3: HCl(g) → H + (aq) + Cl - (aq) [ ] o 0.020 ~0 0 Δ[ ] -0.020 +0.020 +0.020 [ ] f 0 0.020 0.020 The [H + ] = 0.020 M. The [OH - ] ion determined from Equation 2: H OH 1.0 10 OH 1.0 -14 + − − ⋅ = × = ×× = × = × + 10 H 1.0 10 2.0 1 -14 -14 0 020 . 00 M -13 2. The symbol K w represents the specifc equilibrium constant that refers to the auto-ionization of water. Like all equilibrium constants, its value is a function of the temperature. In terms of calculations expected on the AP Chemistry Examination, the value of K w at 25 o C is commonly used. pH and the Properties of [H+] in Aqueous Solution 31 For an aqueous solution of a base, the [OH-] is directly calculated, and subsequently the [H+] can be determined. For example, the concentrations of each ion in a 0.050 M solution of barium hydroxide are determined as: Equation 4: Ba(OH) 2 (s) → Ba 2+ (aq) + 2 OH - (aq) [ ] o 0.050 0 ~0 Δ[ ] -0.050 +0.050 +0.10 [ ] f 0 0.050 0.10 The [OH-] = 0.10 M. The [H+] ion determined from Equation 2: H OH 1.0 10 H 1.0 -14 + − + ⋅ = × = × 10 OH 1.0 10 1.0 10 -14 -14 − = × = × 0 10 . --13 M The preceding examples illustrate that both the [H + ] and the [OH - ] in aqueous solutions vary over a range that is greater than 14 orders of magnitude. The pH scale is commonly used as an alternative expression to the [H + ], where pH is defned as pH = -log 10 [H + ]. Algebraically, this is equivalent to [H + ] = 10 -pH , and shows that the pH value is the negative of the exponent of 10 when the concentration is expressed as a power of 10. The range of values for [H + ] in aqueous solutions, approximately 1.0 x 10 -14 M < [H + ] < 1.0 M , translates to a pH range from 14.00 to 0.00. Referring to the previous examples, the pH of the 0.020 M HCl(aq) solution is pH = -log 10 0.020 = 1.70, the pH of the 0.050 M Ba(OH) 2 (aq) solution is pH = -log 10 1.0 x 10 -13 = 13.00, and the pH of pure, neutral water is then pH = -log 10 1.0 x 10 -7 =7.00. Analogously, the pOH of an aqueous solution is defned as pOH = -log 10 [OH – ], and [OH – ] = 10 -pOH . The logarithmic pH scale conveniently represents the [H + ] over the 14 orders of magnitude, but no additional information is conveyed by the pH of the solution that is not already indicated by the [H + ]. In fact, the above relationships among [H + ], [OH - ], pH, and pOH imply that the value of any one of these will fx the values of the others. 3 Calculation of these values for aqueous solutions will depend on the type of solute, and fve types of calculations can be identifed: • Solutions of strong acids only. • Solutions of strong bases only. • Solutions of weak acids only. • Solutions of weak bases only. 3. The value of K w , like all equilibrium constants, is temperature dependent. While the relation: K w = [H + ] [OH - ] is generally true, the value of K w = 1.0 x 10 -14 at standard temperature of 25 o C. SPECIAL FOCuS: Acids and Bases 32 • Buffer solutions (containing appreciable concentrations of a weak acid and its conjugate base). Examples of these calculations are described elsewhere in this manual. The laboratory determination of the [H + ] of an aqueous solution, or equivalently, its pH, can be done in two ways. The use of indicators, either in solution or adsorbed onto pH paper, affords a crude determination of the pH of a sample. An indicator is itself a weak acid whose color is different from that of its conjugate base. The indicator is intensely colored, so that a negligible concentration of the indicator is added to the solution – suffciently dilute so that it does not have an appreciable effect on the [H + ] of the solution. The indicator will then ionize according to the equation: H-In ← → H + + In - The acid dissociation constant of the indicator is written as K a,In = ⋅ − [ ] + − H In H In . This equation can be rearranged as In H In H − + − [ ] = K a,In , showing that the ratio of the concentration of the conjugate base form to the acid form depends on the value of K a , of the indicator, and the [H + ] in the solution. For K a, In >> [H + ], [In - ] >> [H-In], and the solution will have the color of the base form of the indicator. Conversely, for K a, In << [H + ], [In - ] << [H-In] and the solution will have the color of the acid form of the indicator. In practice, the pK a of the indicator, defned as –log 10 K a, In , is compared to the pH of the solution, and the color of the solution changes from that of the acid form to the color of the base form as the pH of the solution changes from (pK a – 1) to (pK a + 1). Indicators with different pK a ’s can be used to measure different ranges of pH values. A chart of commonly used indicators and their transition point (pK a ’s) can be found in most textbooks. More precise pH measurements are made with a pH meter. Though the technical details vary by instrument, the fundamental operation of a pH meter is to measure the electrochemical potential difference between a glass membrane electrode, whose electrochemical potential is sensitive to the [H + ] in the solution being measured, and an internal reference electrode. 4 The measured potential depends on the [H + ] in the external solution, and the pH meter is simply a high impedance voltmeter that is calibrated to refect the pH of the solution. With careful calibration using standard buffer solutions, laboratory pH meters are routinely capable of making accurate pH measurements to ±0.01 pH unit. 4. In practice, the test electrode and the reference electrode are combined into the same body, referred to as a combination electrode. pH and the Properties of [H+] in Aqueous Solution 33 In closing, the equilibrium that exists between H + ions and OH - ions in any aqueous solution determines much of the chemistry of the solutions. Because chemical properties of the solution are sensitive to variations in the concentrations of these ions, the concentrations of these ions in such solutions are often critically important. Though these equilibrium concepts are sometimes diffcult for students to master, the role of pH as a measure of the concentration of not only the H + ion but also the OH - ion is an important step that will allow students to go further in their mastery of the material. Exercises Complete the following table: Solute and Concentration [H + ] [OH-] pH pOH ? M HCl 0.050 M 0.10 M CH 3 COOH [K a = 1.8 x 10 -5 ] 0.10 M CH 3 COOH [K a = 1.8 x 10 -5 ] & ? M CH 3 COO - 5.12 Pure Water 0.10 M NH 3 [K b = 1.8 x 10 -5 ] & 0.20 M NH4 + ? M NH 3 [K b = 1.8 x 10 -5 ] 1.0 x 10 -3 M ? M Ba(OH) 2 13.70 Answer: Solute and Concentration [H + ] [OH - ] pH pOH 0.050 M HCl 0.050 M 2.0 x 10 -13 M 1.30 13.70 0.10 M CH 3 COOH [K a = 1.8 x 10 -5 ] 1.3 x 10 -3 M 7.5 x 10 -12 M 2.87 11.13 0.10 M CH 3 COOH [K a = 1.8 x 10 -5 ] & 0.24 M CH 3 COO - 1.2 x 10 -5 M 8.8 x 10 -8 M 5.12 8.88 Pure Water 1.0 x 10 -7 M 1.0 x 10 -7 M 7.00 7.00 0.10 M NH 3 [K b = 1.8 x 10 -5 ] & 0.20 M NH 4 + 1.1 x 10 -9 M 9.0 x 10 - 6 M 8.95 5.05 0.056 M NH3 [Kb = 1.8 x 10 -5 ] 1.0 x 10 -11 M 1.0 x 10 -3 M 11.00 3.00 0.25 M Ba(OH) 2 2.0 x 10 -14 M 0.50 M 13.70 0.30 SPECIAL FOCuS: Acids and Bases 34 AP Exam Questions – Past Exams (1984) 2. The pH of 0.1-molar ammonia is approximately (A) 1 (B) 4 (C) 7 * (D) 11 (E) 14 Students should recognize that ammonia is a weak base and that a solution of a weak base will have a pH > 7. Because a 0.1-molar solution of a strong base would have a [OH - ] = 0.1 M, a [H + ] ion = 1 x 10 -13 M and pH = 13, the same concentration of a weak base would have to have a lower pH value. Choice (D) is the only choice that matches these constraints. 3. Find the pH of each of the following: a. 0.10 M HCl. b. A solution prepared by mixing 10.0 mL of a solution of 0.10 M HCl with 990.0 mL of distilled water. c. A solution prepared by mixing 45.0 mL of 0.10 M HCl with 55.0 mL of 0.10 M NaOH. d. A solution prepared by mixing 50.0 mL of a buffer of pH 4.00 with 50.0 mL of distilled water. 3. Answers: a. 0.10 M HCl is a solution of a strong acid: H HCl 0.10M pH log 0.10 1.00 f o 10 + =[ ] = = − = b. Solution of a strong acid: H 0.10 M 10 mL 990 mL 10 mL f + = × + = = − = 0.0010 M pH log 0.0010 3.00 10 c. The strong acid and strong base will neutralize, and there will be excess base remaining: pH and the Properties of [H+] in Aqueous Solution 35 Rxn: H + + OH - → H 2 O mol o 0.0045 0.0055 0 ∆mol -0.0045 -0.0045 +0.0045 mol f 0 0.0010 0.0045 OH 0.0010 M 0.10 L 0.010 M pOH log f 1 − = = = − 00 0.010 2.00 pH 14.00 - 2.00 12.00 = = = d. Because solution (d) is a buffer solution, the ratio A HA − [ ] will not be affected by dilution and the pH will remain unchanged. pH = 4.00 (assuming that after dilution, [A - ], [HA] >> [H + ], [OH - ]. (2006B, Q1ai, Q1aii) C 6 H 5 COOH(s) ← → C 6 H 5 COO - (aq) + H + (aq) K a = 6.46 x 10 -5 2. Benzoic acid, C 6 H 5 COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH. (a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.47. Calculate each of the following. (i) [H + ] in the solution (ii) [OH - ] in the solution 2. Answers: Students should be able to convert among [H + ], [OH - ], pH, and pOH easily. (a)(i) H 0 10 3.4 10 M -pH -4.47 -5 + = = = × 1 (a)(ii) OH H 1.0 10 10 -14 -5 − + = = × × = K w 3 4 . 3.0 10 M -10 × or pOH p - pH 14.00 - 4.47 9.53 OH = = = − K w == = = × 0 10 3.0 10 M -pOH -9.53 -10 1 SPECIAL FOCuS: Acids and Bases 36 (2007, Q1c - e) HF(aq) + H 2 O(l) ← → H 3 O + (aq) + F – (aq) K a = 7.2 x 10 -4 Hydrofuoric acid, HF(aq), dissociates in water as represented by the equation above. HF(aq) reacts with NaOH(aq) according to the reaction represented below. HF(aq) + OH-(aq) ← → H2O(l) + F – (aq) 2. A volume of 15.0 mL of 0.40 M NaOH(aq) is added to 25.0 mL of 0.40 M HF(aq) solution. Assume that volumes are additive. (c) Calculate the number of moles of HF(aq) remaining in the solution. (d) Calculate the molar concentration of F – (aq) in the solution. (e) Calculate the pH of the solution. 2. Answers: (c) mol HF(aq) = initial mol HF(aq) − mol OH − (aq) added = (0.025 L)(0.40 mol L −1 ) − (0.015 L)(0.40mol L −1 ) = 0.010 mol − 0.0060 mol = 0.004 mol (d) [HF] .004 mol HF .040L 0.10MHF = = (e) K a K a = + − + − [ ] = [ ]× H O F HF H O HF F 3 3 = × × ( ) = × 0.10 M 7.2 10 0.15 M 4.8 -4 110 M pH -log 4.8 10 3.32 -4 10 -4 = × ( ) = OR pH pK a F HF M M = + − = + = [ ] log . log . . . 3 14 15 10 3 32 pH and the Properties of [H+] in Aqueous Solution 37 Teaching Strategies Introduce the auto-ionization of pure water as the reverse of the neutralization reaction of a strong acid and a strong base. Emphasize that both H + and OH - are present in any aqueous solution, and that the concentration of either particle will fx the concentration of the other. Emphasize the mathematical relationships among [H + ], [OH - ], pH, and pOH for aqueous solutions and that, given the value of any one of these data, values for the remaining are fxed and can easily be determined. Remind students that the relevant equations are provided on the insert for the free-response section of the AP Chemistry Examination, but they are not available for the multiple-choice questions. In determining the [H + ] in an aqueous solution, be sure that students can identify the type of solution that they are dealing with: • strong acid, only • strong base, only • weak acid, only • weak base, only • buffer solution containing both a weak acid and its conjugate base Students should understand which principles/equations are applicable to each of these instances. Acid–Base Titrations Cheri Smith Yale Secondary School Abbotsford, British Columbia Titration is an important type of volumetric analysis usually used to determine the number of moles of solute in a certain amount of aqueous solution. Once the amount is known, other quantities may be discerned, such as the concentration of the solution or the molar mass of the solute. The titration procedure generally involves the addition of carefully measured portions of a “standard” solution of known concentration to a measured mass or volume of an unknown. In many acid–base titrations, a chemical indicator is added to signal the point at which the two reactants have been combined in just the right ratio to give a complete reaction. This is sometimes called the stoichiometric point or, more commonly, the equivalence point. Alternatively, the equivalence point may be determined using a pH meter to follow the progress of the reaction. Approaching Titration Problems with an AP ® Chemistry Class AP students may or may not be already familiar with introductory titration problems such as those involving the determination of an unknown solution’s concentration or fnding the volume of a solution required to react with a standard solution, from their frst course in chemistry. In any event, the AP Chemistry teacher would be wise to briefy review some simple problems before using a more sophisticated lecture and laboratory approach. Simple problems should involve the idea of “matching” equivalent amounts of strong acids and strong bases in solutions, for example, how many mL of 0.25 M sodium hydroxide will it take to be equivalent to 20 mL of 0.50 M hydrochloric acid. For students who have done little lab, a good introduction may be to titrate HCl with NaOH by drop counting in a well tray using phenolphthalein indicator. (For SPECIAL FOCuS: Acids and Bases 40 example, how many drops of 0.1M NaOH are equivalent to 10 drops of 0.2 M HCl?) Examples of more sophisticated lecture topics and labs that should be done in AP Chemistry include: • Determination of the molar mass of a unknown monoprotic acid. (KHP, KHSO 4 , and NaHSO 3 are good unknowns to use in lab as their masses vary signifcantly from one another.) • Determination of the K a value for an unknown monoprotic acid. (Approximately 0.1 mol/L acetic acid, formic acid and propanoic acids are good unknowns for lab use as their K a values differ by a full order of magnitude.) • Back titrations to determine the strength of antacid products. Once the teacher has reviewed and reinforced the students’ basic understanding of titration, he or she can move on to the determination of pH experimentally and mathematically at different points during a titration. Included should be titrations of a strong acid with a strong base and, more important for AP Chemistry, a strong acid or base with a weak acid or base. The teacher may elect to include diprotic titrations and should certainly consider the selection of appropriate acid–base indicators, as well as the use of pH meters or probeware. Calculating the pH at various points during a theoretical titration is an extremely valuable exercise for AP Chemistry students, as it requires the application of virtually every type of pH problem encountered in the acid–base unit. An AP Chemistry teacher would fnd it benefcial to begin with strong–strong titration curves, thus reviewing the determination of the pH of strong acids and bases alone and in combination with one or the other in excess, including the applications of K w and/ or pK w . Weak–strong curves are the source of a wealth of review, including weak acid and base pH problems (ICE table calculations), buffer calculations, hydrolysis problems, and strong acids or bases in excess. These should be reinforced by either student run pH mater/probe use, or at least teacher demonstration of actual variation of pH during titrations, buffer use, and hydrolysis reactions. Calculation of pH in a Theoretical Titration Curve— Use of the ICF Table Most AP Chemistry students are able to handle the calculation of pH in a solution containing only one dissolved species. When two solutions are combined, however, they fnd the problem considerably more diffcult. One of the biggest issues is the Acid–Base Titrations 41 organization of the increased amount of information given in such problems. The ICF table is a valuable tool for AP Chemistry teachers and students facing problems that involve chemical reactions such as neutralization with limiting and excess reactants. It is similar to the ICE chart used for equilibrium problems in that the “I” represents “initial” concentrations or numbers of moles and the “C” is the “change” in the initial quantities during the reaction. The traditional “E,” for “equilibrium” in an ICE chart, is replaced by an “F” that stands for “fnal” quantities following a reaction that goes to completion in one direction. Such a reaction in a titration curve calculation will involve neutralization by a strong acid or base. Because one (or both) of the reacting species is strong, the reaction goes virtually to completion, hence “F” replaces “E.” Such a reaction is said to be quantitative and is useful for analytic calculations. Strong Acid–Strong Base Titrations Though the pH at various points during the titration of a strong acid with a strong base (or vice versa) is relatively easy to determine, pre-AP and AP Chemistry teachers alike would be wise to apply the ICF technique early as it will build student confdence for its application to more diffcult problems later. Consider the curve that describes the titration of 25.0 mL of 0.100 mol/L HCl with NaOH. FIGuRE 1 pH Providing such a curve for the titration of 25 mL of 0.100 mol/L HCl, with no further information, the AP Chemistry teacher might ask the class members to take several minutes, working alone or collaboratively, to list any information they can SPECIAL FOCuS: Acids and Bases 42 determine about the titration from the pH profle. Sample student responses might include: • Base is being added to acid. • The reaction is represented by: HCl + NaOH → NaCl + H 2 O • The initial pH = 1.0 • The initial [HCl] = [H + ] = 10 -pH = 10 -1.0 = 0.1 mol/L • The acid is strong as the initial [H + ] = the initial [HCl] indicates 100% ionization. • The pH at equivalence = 7.0 • The volume of base required to reach equivalence = 23.5 mL. • The moles NaOH = moles OH - = 25.0 mL x 0.100 mol HCl x 1 mol NaOH = 1000 mL 1 mol HCl 0.00250 mol NaOH are required to reach equivalence. • The [NaOH] = 0.00250 mol x 1000 mL = 0.106 mol/L 23.5 mL 1 L AP Chemistry teachers should reinforce the importance of beginning titration calculations with a balanced chemical equation, ideally in molecular form. This leads to a proper mole ratio that is most likely to be correct when units include the formula of the species they refer to. Any AP Chemistry text will contain a sample curve to which teachers can refer; however, the teacher might prefer to produce a curve using probeware (such as that available from Vernier or Pasco) or to use a variety of simulations available commercially or on the Internet (see suggestions in the laboratory section at the end of this chapter). Production of the curve could be done after, during, or even before the calculations. Once students have examined a pH profle, or a titration curve for a strong monoprotic acid being titrated with a strong base, the AP Chemistry teacher can discuss the calculation of pH at various points throughout the curve. Example 1 Calculate the pH of the 25.0 mL sample of 0.100 mol/L solution of HCl being titrated in the profle above. Acid–Base Titrations 43 Strategy: The student should recognize that pH = - log [H + ] and that the concentration of a solution is independent of its volume. Hence, 25.0 mL is irrelevant to the problem at this point. Strong acids such as HCl, ionize completely, hence, [H + ] = [HCl]. (AP Chemistry students must be aware of the “BIG 6” strong acids: HCl, HBr, HI, HClO 4 , HNO 3 , and H 2 SO 4 (HClO 3 is included in some texts, although it ionizes less than 100%); H 2 SO 4 is only nearly completely ionized for its frst ionization) as there is no table of acid ionization constants included with the AP Chemistry Exam.) Solution: pH = - log [H + ] = - log [HCl] = - log (0.100 mol/L) = 1.000 Example 2 Calculate the pH of the 25.0 mL sample of HCl following the addition of 10.0 mL of 0.106 mol/L NaOH. Note that a “rounder” concentration of base might be used for introduction; however, this is what we’ve determined for our example. Strategy: The ICF table is very useful for helping students keep track of everything necessary to solve this problem. If concentrations are to be used (which may be preferable because pH depends upon concentration rather than moles), students must recognize that the concentration of a solution is affected by the addition of water or another solution as both of these cause dilution. Step one is to dilute. The dilution factors are 25/35ths and 10/35ths,respectively. Step two is to neutralize by applying the ICF table. Step three is to examine the bottom line of the ICF table to determine what affects the pH. Solution: [HCl] in = 0.100 mol/L (25.0 mL) [NaOH] in = 0.106 mol/L (10.0 mL) (35.0 mL) (35.0 mL) = 0.0714 mol/L = 0.0303 mol/L Neutralization in ICF shows: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) I: 0.0714 M 0.0303 M 0 M C: -0.0303 -0.0303 +0.0303 F: 0.0411 0 0.0303 SPECIAL FOCuS: Acids and Bases 44 The bottom line indicates a [HCl] = [H + ] = 0.0411 mol/L, hence, pH = - log [H + ] = - log (0.0411) = 1.386 (Note that moles might have been used by multiplying the molarity of each reactant by its volume and using the moles in the “I” line. The changes would then have been 0.00106 moles. The fnal moles of acid divided by a total volume of 0.0350 L would give the same concentration and consequently, the same pH as shown.) One fnal note that should be mentioned is that the ICF table indicates that the [NaOH] fnal = 0 mol/L, following the neutralization. This is, of course, impossible, as [H + ][OH - ] must always = 10 -14 = K w (at 25 o C). Hence, [OH - ] = [NaOH] at equivalence is negligible and approaches 0, though it can never actually equal 0. If [H + ] = [OH - ], then each = √ 10 -14 = 10 -7 , which is very small compared to the molarity numbers on the problem, and so ≈ 0. Example 3 Calculate the pH of the 25.0 mL sample of HCl following the addition of 23.5 mL of 0.106 mol/L NaOH. Strategy: As in example 2, dilute and neutralize, then examine the bottom line of the ICF table. Solution: [HCl] in = 0.100 mol/L (25.0 mL) [NaOH] in = 0.10638 mol/L (23.5 mL) (48.5 mL) (48.5 mL) = 0.0515 mol/L = 0.0515 mol/L Neutralization in ICF shows: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) I: 0.0515 M 0.0515 M 0 M- C: -0.0515 -0.0515 +0.0515 F: 0 0 0.0515 This process shows that the pH of 7 at the equivalence point is due to the formation of 48.5 mL of a 0.0515 mol/L solution of NaCl, which dissociates to form the neutral cation, Na + and an anion, Cl - . As the conjugate base of the strong acid, HCl, Cl - is too weak to accept any appreciable quantity of protons from the water it is dissolved in and hence, it too is neutral. Determination of the pH at some point beyond Acid–Base Titrations 45 equivalence in the same fashion is helpful as it involves the application of Kw and/or pKw along with the ICF table. Example 4 Calculate the pH of the 25.0 mL sample of HCl following the addition of 30.0 mL of 0.106 mol/L NaOH solution. Strategy: As in example 2, dilute and neutralize, then examine the bottom line of the ICF table. The application of K w and/or pK w will be required to convert the [OH - ]. Solution: [HCl]in = 0.100 mol/L (25.0 mL) [NaOH]in = 0.10638 mol/L (30.0 mL) (55.0 mL) (55.0 mL) = 0.0455 mol/L = 0.0580 mol/L Neutralization in ICF shows: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) I: 0.0455 M 0.0580 M 0 M- C: -0.0455 -0.0455 +0.0455 F: 0 0.0125 0.0455 As explained in example 3, the salt, NaCl, is completely neutral and does not affect the pH at all. However, the [NaOH] excess is very signifcant. From this number, it becomes evident that the pOH = - log [OH - ] = -log[NaOH] = - log (0.0125) = 1.90. Consequently, as pK w = pH + pOH = 14.00 at 25 o C, the pH = 14.00 – 1.90 = 12.10. The alternative approach would be to recognize that [H + ] = K w /[OH - ] = 10 -14 /0.0125 = 8.0 x 10 -13 M. Hence, pH = - log (8.0 x 10 -13 ) = 12.10. Weak–Strong Titrations A classic weak–strong titration involves the titration of a weak carboxylic acid, such as ethanoic (acetic) acid or methanoic (formic) acid, with a strong base. Once students have examined a pH profle, or titration curve for a weak monoprotic acid being titrated with a strong base, the AP Chemistry teacher can compare the calculation of pH at various points throughout the curve with that of a strong–strong titration. SPECIAL FOCuS: Acids and Bases 46 FIGuRE 2 ml NaOH added pH Providing such a curve, for the titration of 25 mL of 0.100 mol/L acetic acid, with no further information, the AP Chemistry teacher might ask the class to take several minutes, working collaboratively with a partner, to list any information they can determine about the titration from the pH profle, with emphasis on any obvious differences they notice between this curve and the earlier strong–strong one. Sample student responses might include: • Base is being added to acid. • The reaction is represented by: CH 3 COOH + NaOH→ NaCl + H 2 O. • The net ionic reaction is CH 3 COOH + OH - → H 2 O + CH 3 COO - . • The initial pH approximates 2.9 (higher than the strong acid). • The initial [H + ] = 10 -pH = 10 -2.9 = 0.0013 mol/L. • The acid is not as strong as the initial [H + ] << the initial [CH 3 COOH] indicates less than 100% ionization. • The % ionization = 0.0013 x 100% = 1.3%. This is typical of a weak acid. 0.100 • The pH rises more quickly during a weak–strong titration. • The pH at equivalence is between 8 and 9 (above the neutral value of 7). • The volume of base required to reach equivalence = 23.5 mL (same value). • This is a particularly signifcant point for students to recognize and it should be emphasized that the strength of an acid has no bearing on the volume Acid–Base Titrations 47 of base required to neutralize it. One mole of monoprotic acid requires one mole of base to neutralize it completely whether the acid is strong or weak. • The moles NaOH = 25.0 mL x 0.100 mol HA x 1 mol NaOH = 0.00250 mol 1000 mL 1 mol HA required to reach equivalence. • The [NaOH] = 0.00250 mol x 1000 mL = 0.106 mol/L. 23.5 mL 1 L The same sources suggested for strong–strong curves may be relied upon for weak–strong curves. Once students have examined a pH profle, or titration curve for a weak monoprotic acid being titrated with a strong base, the AP Chemistry teacher can discuss the calculation of pH at various points throughout the curve. The ICF table is particularly useful for students attempting to deal with the additional complexity of these problems. AP Chemistry teachers should stress the review of weak acid, weak base, buffer, and hydrolysis calculations as they arise. It is important for students to consider the chemical composition of the solution being titrated as well as the process for calculating its pH at any particular point. Example 5 Calculate the pH of the 25.0 mL sample of a 0.100 mol/L solution of CH 3 COOH being titrated in the profle above. (Given the K a of acetic acid = 1.8 x 10 -5 .) Strategy: The student should recognize that acetic acid is a weak acid, hence an ICE table type of calculation is required to determine the [H 3 O + ]. Solution: According to the Brønsted-Lowry defnition, an acid will donate a proton to the base it is dissolved in. Because this acid is weak, only a small percentage of the molecules will donate protons, resulting in the following equilibrium: CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO - I: 0.100 M - 0 M 0 M C: - x + x + x [H + ] formed is defned as x E: ~ 0.100 x x For weak acids that ionize less than 5%, x may be considered negligible compared with the initial concentration. SPECIAL FOCuS: Acids and Bases 48 K a = [H 3 O + ] [CH 3 COO - ] = x 2 = 1.8 x 10 -5 [CH 3 COOH] 0.100 Simple algebra reveals x = [H 3 O + ] = 1.3 x 10 -3 mol/L and pH = - log [H 3 O + ] = - log (1.3 x 10 -3 ) = 2.87. Example 6 Calculate the pH of the 25.0 mL sample of CH 3 COOH following the addition of 10.0 mL of 0.106 mol/L NaOH. Strategy: The ICF table may be used in an identical series of steps to those used for example 2 for the strong–strong titration shown previously. Solution: [CH 3 COOH] in = 0.100 M (25.0 mL) [NaOH] in = 0.106 M (10.0 mL) (35.0 mL) (35.0 mL) = 0.0714 mol/L = 0.0303 mol/L Neutralization in ICF shows: CH 3 COOH(aq) + NaOH(aq) → NaCH 3 COO(aq) + H 2 O(l) I: 0.0714 M 0.0303 M 0 M - C: -0.0303 -0.0303 +0.0303 F: 0.0411 0 0.0303 The bottom line indicates a [CH 3 COOH] = 0.0411 M and a [CH 3 COO - ] = [CH 3 COO - ] = 0.0303 M. The student should recognize the presence of a weak conjugate pair with signifcant quantities of both the conjugate acid and base present. The AP Chemistry teacher should point out that such a solution constitutes a buffer. The pH of this solution may be calculated as a weak acid with conjugate base present as a common ion: K a = [H 3 O + ] [CH 3 COO - ] = [H 3 O + ] (0.0303) = 1.8 x 10 -5 [CH 3 COOH] 0.0411 Since [H 3 O + ] = 2.4 x 10 -5 mol/L, pH = - log (2.4 x 10 -5 ) = 4.61. The pH of a buffer may also be calculated using the Henderson-Hasselbalch equation: Acid–Base Titrations 49 pH = pK a + log [CH 3 COO - ] = 4.74 + log (0.0303) = 4.61 [CH 3 COOH] (0.0411) This portion of the curve is referred to as the “buffer zone.” It is particularly instructive for the AP Chemistry teacher to point out that exactly half way to equivalence, the quantity of excess weak acid exactly equals the quantity of conjugate base formed by the neutralization. Substitution into each of the equations above, in turn, shows: K a = [H 3 O + ] [CH 3 COO - ] = [H 3 O + ] and pH = pK a + log [CH 3 COO - ] = pK a + log(1) = pK a [CH 3 COOH] [CH 3 COOH] The teacher should encourage AP Chemistry students to use either or both of these equations to recognize and to verify the changes that occur during the buffer zone portion of a weak–strong titration curve: • halfway to equivalence, [H + ] = K a and pH = pK a . • prior to halfway, [H + ] exceeds K a and so pH < pK a . • past halfway and before equivalence, [H + ] < K a and so pH exceeds pK a . Example 7 Calculate the pH of the 25.0 mL sample of CH 3 COOH following the addition of 23.5 mL of 0.106 mol/L NaOH. Strategy: As in example 6, dilute and neutralize, then examine the bottom line of the ICF table. Solution: [CH 3 COOH] in = 0.100 M (25.0 mL) [NaOH] in = 0.10638 M (23.5 mL) (48.5 mL) (48.5 mL) =0.0515 mol/L =0.0515 mol/L Neutralization in ICF shows: CH 3 COOH(aq) + NaOH(aq) → NaCH 3 COO(aq) + H 2 O(l) I: 0.0515 M 0.0515 M 0 M - C: -0.0515 -0.0515 +0.0515 F: 0 0 0.0515 The bottom line shows that the basic pH at the equivalence point is due to the formation of 48.5 mL of a 0.0515 mol/L solution of the salt, NaCH 3 COO, SPECIAL FOCuS: Acids and Bases 50 which dissociates to form the neutral cation, Na + and the weakly basic anion, CH 3 COO - . As the conjugate base of the weak acid, CH 3 COOH, CH 3 COO - is strong enough to accept an appreciable quantity of protons from the water it is dissolved in and hence, it forms hydroxide ions by hydrolysis. CH 3 COO - (aq) + H 2 O(l) → CH 3 COOH(aq) + OH - (aq) I: 0.0515 M - 0 M 0 M C: -x + x + x [OH - ] formed is E: ~ 0.0515 x x defned as x K b = [CH 3 COOH][OH - ] = K w = 10 -14 = 5.6 x 10 -10 = x 2 hence, [CH 3 COO - ] K a 1.8 x 10 -5 0.0515 x = [OH - ] = 5.3 x 10 -6 mol/L and pOH = - log(5.3 x 10 -6 ) = 5.27, and pH = 14 – 5.27 = 8.73 Example 8 Calculate the pH of the 25.0 mL sample of CH 3 COOH following the addition of 30.0 mL of 0.106 mol/L NaOH solution. Strategy: As in example 7, dilute and neutralize, then examine the bottom line of the ICF table. As in the strong–strong example 4 shown previously, the application of K w and/or pKw will be required to convert the [OH - ]. Solution: [CH 3 COOH] in = 0.100 M (25.0 mL) [NaOH] in = 0.10638 M (30.0 mL) (55.0 mL) (55.0 mL) = 0.0455 mol/L = 0.0580 mol/L Neutralization in ICF shows: CH 3 COOH(aq) + NaOH(aq) ↔ NaCH 3 COO(aq) + H 2 O(l) I: 0.0455 M 0.0580 M 0 M - C: -0.0455 -0.0455 +0.0455 F: 0 0.0125 0.0455 As explained in example 7, the salt, NaCH 3 COO, is weakly basic due to hydrolysis. However, the [NaOH] excess is much more signifcant. In fact, the hydroxide Acid–Base Titrations 51 ion from the NaOH forces the hydrolysis reaction so far left that its contribution to the [OH - ] may be completely ignored. The [OH - ] = [NaOH] excess for all intents and purposes. Consequently, the pOH = - log [OH - ] = - log (0.0125) = 1.90. Hence, as pK w = pH + pOH = 14.00 at 25 o C, the pH = 14.00 – 1.90 = 12.10. The alternative approach would be to recognize that [H 3 O + ] = K w /[OH - ] = 10 -14 /0.0125 = 8.0 x 10 -13 M. Hence, pH = - log (8.0 x 10 -13 ) = 12.10. The alternative to titrating a weak acid with a strong base is, of course, the titration of a weak base with a strong acid. A classic example might be the titration of ammonia with hydrochloric acid. FIGuRE 3 Such a titration would include the calculation of pH by: • The use of the ionization constant of a weak base, K b . Initially one is determining the pH of a solution of ammonia. • The use of a K b expression or Henderson-Hasselbalch during the buffer zone. In this portion of the curve the solution contains the weak conjugate pair ammonia and ammonium ion, therefore it is a basic buffer. SPECIAL FOCuS: Acids and Bases 52 • The use of the acid ionization constant of a weak acid, K a at equivalence due to the formation of an acidic salt, in this case, ammonium chloride, which undergoes cationic hydrolysis. • The application of the standard defnition of pH applied to the excess [H 3 O + ] present in a solution of ammonium chloride once excess strong acid has been added and the titration proceeds beyond equivalence. Selection of an Appropriate Acid-Base Indicator An acid–base indicator is a chemical species that has an acid and a base form with one or both forms being visibly colored. Many indicators were originally isolated from organic plants or foods. Some have been used as dyes. The general form for the equation that represents the equilibrium for an indicator is: HIn(aq) + H 2 O(l) ↔ H 3 O + (aq) + In - (aq) acid form base form (color one) (color two) If the indicator is placed in a stronger acid, the acid will donate protons to the indicator, producing the HIn form (and color). A stronger base will pull protons from the indicator, producing the In - form (and color). When indicators are in the middle of their color change, the [HIn] = [In - ] and the color shown will be a combination of the acid and base form colors. This is called the endpoint or transition point for the indicator. Since indicators are actually weak acids, they do have K a and consequently pK a values. Substitution into familiar equations shows: K a = [H 3 O + ] [In - ] = [H 3 O + ] and pH = pK a + log [ In - ] = pK a + log(1) = pK a [HIn] [HIn] Hence, an indicator will reach its endpoint and change color when the [H 3 O + ] in the system equals the indicator’s K a . Similarly, during the color change (at the indicator’s endpoint), the pH of the system will equal the indicator’s pK a . Ideally, endpoints should occur exactly at equivalence. Thus, different types of titrations will require different indicators. Phenolphthalein changes color between pH 8.2 and 10.0. It has a pK a of approximately 9.1, hence it would be quite appropriate for the acetic acid titration because example 7’s calculated equivalence point was pH 8.73. On the other hand, it would be absolutely wrong for use in the titration of ammonia with hydrochloric acid because the pH at equivalence for that titration Acid–Base Titrations 53 would be around 5.4, as shown in the third curve (Figure 3). One signifcant point demonstrated by each of the curves we’ve studied is the rapid change in pH that suddenly occurs at equivalence. AP Chemistry teachers can demonstrate this dramatically through the use of a pH meter or probe ware (such as that available from Vernier or Pasco) or a variety of simulations available commercially or on the Internet (see suggestions in the laboratory section at the end of this chapter). Because of this rapid change, an indicator’s pK a need not exactly match the pH expected at equivalence. Rather, the indicator must simply complete its entire color change during the vertical part of the titration curve. This means that phenolphthalein might be selected ahead of an indicator having a pK a closer to the pH at equivalence simply because its change from colorless to pink is so clearly visible and easy to detect. The titration curve below shows that phenolphthalein or methyl red along with many other indicators would be appropriate for determining the equivalence point for a strong–strong titration. FIGuRE 4 Sample Questions from Previous AP Chemistry Exams The topic of titration appears on virtually every AP Chemistry exam. It is often a signifcant topic in the free-response portion of the examination. Problem number SPECIAL FOCuS: Acids and Bases 54 one of the three required problems is always an equilibrium problem. This problem centers on an acid–base equilibrium nearly every other year. Such problems require student confdence with acid–base calculations of the sort discussed in this chapter. 2002 Operational Exam HOBr(aq) ↔ H + (aq) + OBr – (aq) K a = 2.3 x 10 -9 1. Hypobromous acid, HOBr, is a weak acid that dissoci ates in water, as represented by the equation above. (a) Calculate the value of [H + ] in an HOBr solution that has a pH of 4.95. (b) Write the equilibrium constant expression for the ionization of HOBr in water, then calculate the concentration of HOBr(aq) in an HOBr solution that has [H + ] equal to 1.8 x 10 -5 M. (c) A solution of Ba(OH) 2 is titrated into a solution of HOBr. (i) Calculate the volume of 0.115 M Ba(OH) 2 (aq) needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr(aq). (ii) Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7. Explain. (d) Calculate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160 M HOBr to produce a buffer solution with [H + ] = 5.00 ´ 10 –9 M. Assume that volume change is negligible. (e) HOBr is a weaker acid than HBrO 3 . Account for this fact in terms of molecular structure. Solution: (a) pH = –log[H + ] ; [H + ] = 10 -4.95 = 1.1 x 10 -5 mol/L (b) K a = [H+][OBr] = 2.3 x 10 -9 [HOBr] [H + ] = [OBr – ] = 1.8 x 10 -5 M (1.8 x 10 -5 ) 2 = 2.3 x 10 -9 X x = [HOBr] = 0.14 M Acid–Base Titrations 55 (c) (i) 65.0 mL x 0.146 mol HOBr x 1 mol H+ x 1 mol OHˉ x 1 mol Ba(OH) 2 1000 mL 1 mol HOBr 1mol H+ 2 mol OHˉ x 1000 mL = 41.3 mL 0.115 mol Ba(OH) 2 (ii) pH > 7; salt of a weak acid is a weak base (hydrolysis of OBr - ) (d) pH = pK a + log [A – ] or [A – ] = [HA]K a [HA] [H+] [OBr – ] = (0.160)(2.3 ′ 10 9 ) = 0.0736 M 5.00 ′ 109 125 mL x 0.736 mol OBr – x 1 mol NaOBr = 0.00920 mol NaOBr 1000 mL 1 mol OBr – (e) very electronegative oxygen is able to draw elec trons away from the bromine and weaken the O–H bond, making it easier for the hydrogen ion to be donated. O Br — O – H O 2003 Operational Exam C 6 H 5 NH 2 (aq) + H 2 O(l) ↔ C 6 H 5 NH 3 + (aq) + OH – (aq) 1. Aniline, a weak base, reacts with water according to the reaction represented above. (a) Write the equilibrium constant expression, K b , for the reaction represented above. (b) A sample of aniline is dissolved in water to produce 25.0 mL of 0.10 M solution. The pH of the solution is 8.82. Calculate the equilibrium constant, K b , for this reaction. (c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calculate the pH of the solution when 5.0 mL of the acid has been titrated. (d) Calculate the pH at the equivalence point of the titration in part (c). (e) The pK a values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer. SPECIAL FOCuS: Acids and Bases 56 Indicator pK a Erythrosine 3 Litmus 7 Thymolphthalein 10 Solution (a) K b = C H NH OH C H NH 6 5 3 6 5 2 + − [ ] (b) pOH = 14 – pH = 14 – 8.82 = 5.18 –log[OH – ] = 5.18; [OH – ] = 6.61×10 –6 M [OH - ] = [C 6 H 5 NH 3 + ] K b = 6 61 10 0 10 6 61 10 6 2 6 . . – . × ( ) × − − = 4.4×10 –10 (c) 25 mL x 0 1 1 . mol L = 2.5 mmol C 6 H 5 NH 2 5 mL × 0 1 1 . mol L = 0.5 mmol H + added 2.0 mmol base remains in 30.0 mL solution (may determine from ICF) 4 4 10 0 50 30 0 20 0 3 10 . . . . × [ ] +       − X X mmol ml mmol 00 0 . mL       X = 1.80×10 –9 = [OH – ] [H + ] = 1 10 1 8 10 14 9 × × − − . = 5.6×10 –6 ; pH = 5.26 (d) When neutralized, there are 2.5 mmol of C 6 H 5 NH 3 + in 50.0 mL of solution, giving a [C 6 H 5 NH 3 + ] = 0.050 M (may determine from ICF.) This cation will partially ionize according to the following equilibrium: C 6 H 5 NH 3 + (aq) ↔ C 6 H 5 NH 2 (aq) + H + (aq) at equilibrium, [C 6 H 5 NH 2 ] = [H + ] = X [C 6 H 5 NH 3 + ] = (0.050–X) X X 2 0 050 ( . ) − = K a = 2.3×10 –5 X = 1.06×10 –3 = [H + ] pH = –log[H + ] = 2.98 Acid–Base Titrations 57 (e) Erythrosine; the indicator will change color when the pH is near its pK a , since the equivalence point is near pH 3, the indicator must have a pK a near this value. The lab-based essay question has focused on acid-base titrations. Occasionally, the core concepts of titrimetry will be tested in an essay question that is not lab based. The following is an example of a lab-based acid–base essay question: 1998 Operational Exam Lab-Based Question An approximately 0.1–molar solution of NaOH is to be standardized by titration. Assume that the following materials are available. • Clean, dry 50 mL buret • 250 mL Erlenmeyer fask • Wash bottle flled with distilled water • Analytical balance • Phenolphthalein indicator solution • Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard) (a) Briefy describe the steps you would take, using the materials listed above, to standardize the NaOH solution. (b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution. (c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve. SPECIAL FOCuS: Acids and Bases 58 FIGuRE 5 pH Volume of NaOH Solution Added (mL) (d) Describe how the value of the acid–dissociation constant, K a , for the weak acid HX could be determined from the titration curve in part (c). (e) The graph below shows the results obtained by titrating a different weak acid, H 2 Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve. FIGuRE 6 Volume of NaOH Solution Added pH Acid–Base Titrations 59 Solution: (a) • Exactly mass a sample of KHP in the Erlenmeyer fask and add distilled water to dissolve the solid. • Add a few drops of phenolphthalein to the fask. • Rinse the buret with the NaOH solution and fll. • Record starting volume of base in buret. • With mixing, titrate the KHP with the NaOH solution until it just turns slightly pink. • Record end volume of buret. • Repeat to check your results. (b) mass of KHP = moles of KHP; molar mass KHP since KHP is monoprotic, this is the number of moles of NaOH. moles of NaOH = molarity of NaOH L of titrant FIGuRE 7 equivalence point pH Volume of NaOH Solution Added (mL) (d) From the titration curve, at the 12.5 mL volume point, the acid is half- neutralized and the pH = pK a . K a = 10 -pKa (e) Y 2– SPECIAL FOCuS: Acids and Bases 60 Recommended Laboratory Exercises Related to Acid–Base Titration Wet Labs Of the 22 recommended laboratory experiments for AP Chemistry, three involve acid–base titrimetry. As suggested earlier, the frst experiment should involve the preparation of a primary standard that could be used to standardize a base that might be used to determine a molar mass, a K a value, the moles of water of hydration, the effciency of an antacid or the concentration of a diprotic acid, to list a few possibilities. An even simpler experiment (titration by drops) may be useful for students who have not encountered any form of titration in a previous class. Such an experiment will help bridge students’ pre-AP or general chemistry skills to those required for AP Chemistry. The second experiment should involve the development of experimental titration curves to be graphed and interpreted by the students. The classic two curves are, of course, the titration of frst, HCl and second, CH 3 COOH with NaOH as examples of strong acid–strong base and weak acid–strong base curves. Other possibilities include the titration of a weak base such as NH 3 with HCl or a polyprotic acid, such as H 3 PO 4 with NaOH. A potentiometric titration, based on conductivity of, for example, Ba(OH) 2 with H 2 SO 4 is another possibility. The third experiment involves the determination of appropriate indicators for different acid–base titrations. There are a number of excellent resources for laboratories such as these. Essential Experiments in Chemistry by Morrison and Scodellaro, SMG Lab Books, 2005, includes all three of these labs (14A, 14G and 14 H). Laboratory Investigations: AP Chemistry by Hostage and Fossett, Peoples Publishing Group, 2005, contains two of them (#8 and #10). Additionally Chemistry and Computers by Holmquist and Volz, 2003, contains fve of these labs (#23–#27) each with specifc direction for use of probe ware. Virtual Labs There are a number of laboratory exercises that may be done virtually. Some of them are commercially available on DVD or CD. One of the more popular is the Virtual Chemlab disc provided with the text, Chemistry, the Central Science, 10th edition, by Brown, LeMay and Bursten (Prentice Hall, 2006). This disc includes several lab exercises that students may perform virtually, on the computer, including a titration Acid–Base Titrations 61 lab that allows them to select various reagents to use in a variety of titrations. The data they collect may be presented in tabular and/or graphic form. Bridging to the Lab by Jones and Tasker, W.H. Freeman and Company, 2002, includes 11 different virtual lab exercises, one of which uses an acid-base titration to determine the pK a of a food preservative. Students produce a titration curve and use it to determine the pK a of a weak acid as well as its concentration and fnally its molar mass. In addition to the commercial lab exercises, there are some Web sites whose authors have graciously provided their lab exercises free of charge. A fantastic source of Flash-based chemistry simulations has been prepared by Tom Greenbowe from Iowa State. Start at www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ fashfles/to select a titration simulation that is very easy to use. Students may select from several weak or strong acids, one of two possible indicators, and four different bases that may be placed in the buret. Another great Web site for curves was prepared by Yue-Wing Long from Wake Forest University. Visit www.wfu.edu/~ylwong/chem/titrationsimulator/index.html to actually plot a titration curve that may be used to determine the concentration and/or the pK a of a variety of weak acids. This site is particularly useful as it allows students to alter the strength of the acid they use by selecting various K a or concentration values for the acids they titrate. Polyprotic Acids Lew Acampora St. Francis School Louisville, Kentucky According to Brønsted-Lowry formalism, an acid is defned as a proton donor. 5 The common and unique reactivity of the proton, also represented as the H + ion, imbues acids with their familiar properties. Though all Brønsted-Lowry acids contain hydrogen, the converse statement—that all hydrogen-containing compounds act as acids—is not generally true. The extent to which a species donates a proton is a measure of the acidity of the species and is ultimately determined by the electronic structure of the particle. For example, because gaseous hydrogen chloride dissolves and ionizes completely in aqueous solutions, it is classifed as a strong acid: HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl - (aq) or HCl(aq) → H + (aq) + Cl - (aq) Hydrogen fuoride, however, dissolves into water and reaches a state of dynamic equilibrium in which a fraction of the HF molecules will be ionized at any instant. This state of equilibrium is indicated by the double arrows and leads to the classifcation of hydrofuoric acid as a weak acid: HF(aq) + H 2 O(l)  H 3 O + (aq) + F - (aq) or HF(aq)  H + (aq) + F - (aq) The difference in the behavior of HCl and HF can be traced to such factors as the relative strength of the H-X bond, the strength of the intermolecular forces between the HX molecule and the water molecules, and the energy of hydration of the X - ion. 5. The more general defnition of acids, attributed to G. N. Lewis, defnes acids and bases in terms of their capacity to accept or donate pairs of electron. A Lewis acid is an electron-pair acceptor. Though this defnition is more general, most quantitative problems in AP Chemistry can be approached using the Brønsted-Lowry formalism. SPECIAL FOCuS: Acids and Bases 64 Hydrochloric acid and hydrofuoric acid are examples of monoprotic acids because each molecule contains only one ionizable hydrogen. Acetic Acid, CH 3 COOH, is another example of a monoprotic acid because, though each molecule contains four hydrogen atoms, it is only the carboxylic hydrogen that is ionized in aqueous solution: CH 3 COOH(aq)  H + (aq) + CH 3 COO - (aq) Because of the covalent character of the C – H bond, the hydrogen atoms that are directly bonded to carbon are not ionized in aqueous solution. Many substances contain more than one ionizable hydrogen atom, and such substances are generally referred to as polyprotic acids, with subsets including diprotic, triprotic, etc. In these substances, each acidic hydrogen can be donated. For example, one of the most common diprotic acids is sulfuric acid, H 2 SO 4 . A structural diagram for sulfuric acid shows that both hydrogen atoms are directly bonded to oxygen atoms, and the character of these O-H bonds leads to the acidity of both hydrogen atoms. Thus, the sulfuric acid molecule can lose both protons sequentially to become, frst, the hydrogen sulfate ion, HSO 4 - , and fnally the sulfate ion, SO 4 2- . FIGuRE 1 The underlying principles of reactivity that govern the behavior of monoprotic acids are generally true for polyprotic acids; however, there are two additional factors or considerations that should be noted. The presence of more than one acidic hydrogen per molecule will affect the stoichiometry of reactions of those polyprotic acids. Additionally, the tendency to donate each proton will be different, and will have ramifcations in terms of the acid strength. The Brønsted-Lowry formalism considers an acid–base reaction as simply a proton transfer reaction between a proton donor (the acid) and a proton acceptor (the base). A general schema for such a reaction with a monoprotic acid is: HA + B →A - + HB + Polyprotic Acids 65 The stoichiometry of the reaction shows that acid and base are consumed in equimolar quantities. If the reactant is a diprotic acid, then each mole of acid will contribute two moles of protons. In order to neutralize a given quantity of base, then, only half as many moles of the diprotic acid are required: H 2 A + 2 B → A 2- + 2 HB + or ½ H 2 A + B →½ A 2- + HB + While 1 mole of hydrochloric acid will completely neutralize 1 mole of sodium hydroxide, the same neutralization would require only ½ mole of sulfuric acid. 6 The tendency of a particular species to donate a proton—the acid strength—is ultimately traced to the electronic structure of the acid and its conjugate base. For this reason, successive ionizations of a polyprotic acid are generally weaker. As protons are removed from the substance, any remaining ionizable protons are more strongly bonded because a positively charged proton will be more strongly attracted to the more negatively charged particle. The resulting species is a weaker acid than the original particle. This is refected in the values of the acid dissociation constants, 7 K an , for polyprotic acids, which decrease in value for successive ionizations. Consider, for example, the triprotic weak phosphoric acid, H 3 PO 4 . Steps in the complete ionization of phosphoric acid and the values of the acid dissociation constants are shown here: Reaction Equation Equilibrium Constant Expression Equilibrium Constant Value H 3 PO 4  H + + H 2 PO 4 - K H H PO H PO a1 2 4 3 4 =     ⋅     [ ] + − 7.3 x 10 -3 H 2 PO 4 -  H + + HPO 4 2- K H HPO H PO a2 4 2 2 4 =     ⋅         + − − 6.2 x 10 -8 HPO 4 2-  H + + PO 4 3- K H PO HPO a3 4 3 4 2 =     ⋅         + − − 4.8 x 10 -13 6. The concept of “normality” has fallen somewhat out of favor and can be omitted from the AP Chemistry course, but it still may be useful in describing the stoichiometry of acid–base reactions. Normality in this context is defned as “moles of ionizable protons per liter of solution.” For a diprotic acid, the normality is simply twice the molarity. Thus, a 0.10 M H 2 SO 4 solution would be considered 0.20 N, or 0.20 normal. Similarly, a 0.10 M solution of the triprotic phosphoric acid, H 3 PO 4 , would be 0.30 N. 7. The acid dissociation constants for polyprotic acids are generally denoted as K an , where the “n” refers to the nth proton donated by the molecular species. A diprotic acid would have different values of K a1 and K a2 , a triprotic acid would have K a1 , K a2 , and K a3 . SPECIAL FOCuS: Acids and Bases 66 The phosphoric acid molecule H 3 PO 4 is the strongest acid among the substances listed. Its conjugate base, the dihydrogen phosphate ion H 2 PO 4 - , still contains two ionizable protons and is itself an acid. 8 Because of its overall negative charge and the bond electron density, the H-O bonds in H 2 PO 4 - are stronger than those in H 3 PO 4 , and H 2 PO 4 - is a weaker acid. The observation that successive protons are increasingly diffcult to remove from polyprotic acids implies that these polyprotic acids ionize stepwise when reacting with strong bases. For phosphoric acids, the following net ionic equations apply: Reaction Description Net Ionic Equation Equal volumes of 0.10 M phosphoric acid and 0.10 M sodium hydroxide are mixed. H 3 PO 4 + OH - → H 2 PO 4 - + H 2 O Equal volumes of 0.10 M phosphoric acid and 0.20 M sodium hydroxide are mixed. H 3 PO 4 + 2 OH - → HPO 4 2- + 2 H 2 O Equal volumes of 0.10 M phosphoric acid and 0.30 M sodium hydroxide are mixed. H 3 PO 4 + 3 OH - → PO 4 3- + 3 H 2 O Equal volumes of 0.10 M sodium dihydrogen phosphate and 0.20 M sodium hydroxide are mixed. H 2 PO 4 - + 2 OH - → PO 4 3- + 2 H 2 O Since phosphoric acid, dihydrogen phosphate ion, and monohydrogen phosphate ion can each act as acids, their respective conjugate bases are dihydrogen phosphate, monohydrogen phosphate, and phosphate ion. These bases are listed in order of increasing base strength, and the values of the respective base dissociation constants refect this: Reaction Equation Equilibrium Constant Expression Equilibrium Constant Value PO 4 3- + H 2 O  HPO 4 2- + OH - K HPO OH PO b3 4 2 4 3 =     ⋅         − − − 2.1 x 10 -2 HPO 4 2- + H 2 O  H 2 PO 4 - + OH - K H PO OH HPO b2 2 4 4 2 =     ⋅         − − − 1.6 x 10 -7 H 2 PO 4 - + H 2 O  H 3 PO 4 + OH - K H PO OH H PO b1 3 4 2 4 = [ ] ⋅         − − 1.3 x 10 -12 Again, since the bases will accept protons stepwise, the following net ionic equations can be written: 8. A species such as H 2 PO 4 - , which can either donate a proton, acting as an acid, or accept a proton, acting as a base, is termed amphiprotic or amphoteric. Though these terms are not strictly synonymous, for the purposes of the AP Chemistry curriculum, they can be considered interchangeable. Other examples of such species include the monohydrogen phosphate ion, HPO 4 2- , and the hydrogen carbonate or bicarbonate ion, HCO 3 - . Polyprotic Acids 67 Reaction Description Net Ionic Equation Equal volumes of 0.10 M hydrochloric acid and 0.10 M sodium phosphate are mixed. PO 4 3- + H + → HPO 4 2- Equal volumes of 0.20 M hydrochloric acid and 0.10 M sodium phosphate are mixed. PO 4 3- + 2 H + → H 2 PO 4 - Equal volumes of 0.30 M hydrochloric acid and 0.10 M sodium phosphate are mixed. PO 4 3- + 3 H + → H 3 PO 4 Equal volumes of 0.10 M hydrochloric acid and 0.10 M sodium monohydrogen phosphate are mixed. HPO 4 2- + H + → H 2 PO 4 - Because a solution of a polyprotic acid or its conjugate base(s) may contain several species related by different equilibrium reactions and constants, calculations to determine the concentration of each species may seem intractable. Fortunately, the values of the relevant acid dissociation constants are suffciently diverse that the solution chemistry is usually dominated by one ionization, and the concentrations of each species can be determined from straightforward calculation. Consider, for example, a solution of 0.10 M phosphoric acid. In principle, the solution contains the species H 3 PO 4 , H 2 PO 4 - , HPO 4 2- , and PO 4 3- , in addition to the H + and OH - ions. In practice, the chemistry of the solution is dominated by the ionization reaction of the frst proton: H 3 PO 4  H + + H 2 PO 4 - Ionization of subsequent protons: H 2 PO 4 -  H + + HPO 4 2- and HPO 4 2 -  H + + PO 4 3- will have a negligible effect on the [H + ] in the solution. The [H + ] of the resulting solution can be determined from the ionization of the frst proton: H 3 PO 4  H + + H 2 PO 4 - [ ] o 0.10 M ~0 0 ∆[ ] -x +x +x [ ] eq 0.10 – x x x K a1 = [H + ] ∙ [H 2 PO 4 – ] [H 3 PO 4 ] 7.5 × 10 –3 = x ∙ x 0.10 – x [H + ] = [H 2 PO 4 – ] = x = 0.024 M SPECIAL FOCuS: Acids and Bases 68 The resulting [H + ] and [H 2 PO 4 - ] can be used to determine the [HPO 4 2- ]: K a2 2 2 = ⋅ × = + − − − H HPO H PO 6 10 0.0 4 2 4 8 . 224 M x 0.024 M HPO x 6 10 M 4 8 ( )⋅ ( ) = = × − − 2 2 . x x As expected, the relatively small [HPO 4 2- ] implies that the second ionization occurs to a negligible extent and has no measurable effect on the [H + ] or the [H 2 PO 4 - ]. Similarly, the [PO 4 3- ] can be determined using previously calculated concentrations: K a3 3 2 3 4 8 = ⋅ × = + − − − H PO HPO 10 0.0 4 4 1 . 224 M x 6 10 M PO x 1 10 8 4 8 ( )⋅ × ( ) = = × − − − . . 2 2 3 1 MM x x Not surprisingly, the calculated [PO 4 3- ] is extremely small. FIGuRE 2 The behavior of polyprotic acids in acid–base titrations refects the same considerations. Each of the ionizable protons is lost from the acid sequentially with increasing pK a values. The graph of pH vs. volume base added, above, illustrates Polyprotic Acids 69 what happens when a solution of a weak, monoprotic acid is titrated with a strong base. Important points on the graph are the equivalence point, where the moles of added OH - are equal to the moles of HA initially present, and the half-equivalence point. At the half-equivalence point, half of the initial moles of weak acid remain, and half have been converted to the conjugate base. At this point in the titration, [HA] = [A - ], and K a = ⋅ [ ] = + − + H A HA H , so p pH 1 2 Eq. Pt. K a = and K a K a 10 -p = . FIGuRE 3 The graph of pH vs. volume base shows two buffering regions corresponding to the sequential ionization of each proton from the acid when a solution of a weak diprotic acid is titrated with a strong base. As in the case of the monoprotic acid, the equilibrium constants for each ionization may be determined from the pH values at each half-equivalence point. At the frst half-equivalence point in the titration curve, [H 2 A] = [HA - ] and: K a1 = ⋅ [ ] = + − + H HA H A H 2 , so p pH 1 2 Eq. Pt. 1 K a1 = , and K a K a 1 1 10 -p = . At the half-equivalence point corresponding to the second ionization, [HA - ] = [A 2- ] and K a2 = ⋅ = + + H A HA H 2- - , so p pH 1 2 Eq. Pt. 2 K a2 = , and K a K a 2 2 10 -p = . The fundamental principles of acid–base chemistry remain valid for acids that have many ionizable protons, and a solid understanding of the chemistry of monoprotic acids and bases is essential for students before they can grasp the details of the chemistry of polyprotic species. SPECIAL FOCuS: Acids and Bases 70 Examples from Prior AP Chemistry Examinations (1994) H 2 C 2 O 4 + 2 H 2 O  2 H 3 O + + C 2 O 4 2- 31. Oxalic acid, H 2 C 2 O 4 , is a diprotic acid with K 1 = 5.36 x 10 -2 and K 2 = 5.3 x 10 -5 . For the reaction above, what is the equilibrium constant? (A) 5.36 x 10 -2 (B) 5.3 x 10 -5 (C) 2.8 x 10 -6 (D) 1.9 x 10 -10 (E) 1.9 x 10 -13 31. Answer: (C) From Hess’s Law H 2 C 2 O 4 + H 2 O  H 3 O + + HC 2 O 4 - K 1 = 5.36 x 10 -2 HC 2 O 4 - + H 2 O  H 3 O + + C 2 O 4 2- K 2 = 5.3 x 10 -5 H 2 C 2 O 4 + 2 H 2 O  2 H 3 O + + C 2 O 4 2- K = K 1 x K 2 = 2.8 x 10 -6 (1998) 5(e). The graph below shows the results obtained by titrating a different weak acid, H 2 Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter “A” on the curve. FIGuRE 4 5(e). Answer: A 2- Polyprotic Acids 71 (1997) 1. The overall dissociation of oxalic acid, H 2 C 2 O 4, is represented below. The overall dissociation constant is also indicated. H 2 C 2 O 4  2 H + + C 2 O 4 2- K = 3.78 x 10 -6 (a) What volume of 0.400-molar NaOH is required to neutralize completely a 5.00 x 10 -3 -mole sample of pure oxalic acid? (b) Give the equations representing the frst and second dissociations of oxalic acid. Calculate the value of the frst dissociation constant, K 1 , for oxalic acid if the value of the second dissociation constant, K 2 , is 6.40 x 10 -5 . (c) To a 0.015-molar solution of oxalic acid, a strong acid is added until the pH is 0.5. Calculate the [C 2 O 4 2- ] in the resulting solution. (Assume the change in volume is negligible.) (d) Calculate the value of the equilibrium constant, K b , for the reaction that occurs when solid Na 2 C 2 O 4 is dissolved in water. Answers: (a) 5.00 10 mol H C O 2 5.00 10 mol H 1. 3 2 2 4 3 × = × × ( ) = − − + 000 10 mol H 1.00 10 mol OH 1.00 10 mol 2 2 2 × = × × − + − − − OH 0.400M 0.0250 L 25.0 mL − = = (b) There are two successive dissociations: H 2 C 2 O 4  H + + HC 2 O 4 - (equilibrium constant = K 1 ) HC 2 O 4 -  H + + C 2 O 4 2- (equilibrium constant = K 2 ) K K K K K K = × = = × × 1 2 1 2 3.78 10 6.40 10 -4 -5 5.91 10 -2 = × (c) pH 0.5 H 10 0.32 M -0.5 = = = + Because K << 1, [H 2 C 2 O 4 ] eq ≈ 0.015 M K H C O H C O C O 2 4 2 2 2 4 2 4 2 = ⋅ [ ] + − − 2 = ⋅[ ] = × ( ) × + H C O H 3.78 10 2 2 4 -6 K 2 0.015 0.32 6.0 10 M -7 ( ) ( ) = × 2 (d) C 2 O 4 2- + H 2 O  HC 2 O 4 - + OH - K K K b w a = = × × = 2 1.00 10 6.40 10 1.56 -14 -5 10 -10 × SPECIAL FOCuS: Acids and Bases 72 (1994) 7. A chemical reaction occurs when 100. milliliters of 0.2000-molar HCl is added dropwise to 100. milliliters of 0.100-molar Na3PO4 solution. (a) Write the two net ionic equations for the formation of the major species. (b) Identify the species that acts as both a Brønsted acid and as a Brønsted base in the equations in (a). Draw the Lewis electron-dot diagram for this species. (c) Sketch a graph using the axis provided, showing the shape of the titration curve that results when 100. milliliters of the HCl solution is added slowly from a buret to the Na 3 PO 4 solution. Account for the shape of the curve. (d) Write the equation for the reaction that occurs if a few additional milliliters of HCl solution are added to the solution resulting from the titration in (c). 7. Answers: (a) H + + PO 4 3-  HPO 4 2- H + + HPO 4 2-  H 2 PO 4 - (b) HPO 4 2- Polyprotic Acids 73 (c) (d) H + + H 2 PO 4 -  H 3 PO 4 Salts and Buffers Adele Mouakad St. John’s School San Juan, Puerto Rico Acid–Base Properties of Salt Solutions A salt is an ionic compound that can be considered to be the product of the reaction between an acid and a base. The salt that is produced is sometimes neutral, but in many instances it dissolves in water to form a solution that is acidic or basic. This can be clearly explained from the Brønsted-Lowry theory, which shows that some ions can act as acids or bases. These ions can act as acids or bases because they react with water. This reaction with water is called hydrolysis. Hydrolysis is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. Notable examples are: (i) The reaction of the acetate ion with water. CH 3 COO - + H 2 O → CH 3 COOH + OH - This resulting solution would be basic due to the presence of the hydroxide ion. Since this solution has the form of a base ionization, you can write an expression for K b . K b = [OH - ][CH 3 COOH] [CH 3 COO - ] (ii) The reaction of the ammonium ion with water. NH 4 + + H 2 O → NH 3 + H 3 O + This resulting solution would be acidic due to the presence of the hydronium ion. Since the resulting solution has the form of an acid ionization, the expression for K a is: K a = [H 3 O + ][NH 3 ] [NH 4 + ] SPECIAL FOCuS: Acids and Bases 76 How can you predict whether a salt will be acidic, basic or neutral? When the acetate ion hydrolyzes, it produces acetic acid, which is a weak acid. This means that the acetate ion holds fairly strongly to the proton (i.e., it does not ionize readily). So we can say that the acetate ion acts as a base. In general, we can say that anions of weak acids will be basic. On the other hand, the anions of strong acids (e.g., Cl - from HCl) have very little basic character. In other words, these anions do not hydrolyze. Let us examine the cation conjugate to a weak base, such as the ammonium (NH 4 + ) ion, which we used in our example above. It is clearly seen that it behaves like an acid. So we can generalize that the cations of weak bases are acidic. On the other hand, the cations of strong bases (alkali metal cations and alkaline earth metal cations) have hardly any acidic character. Aqueous metal ions, that are not those of the alkali or alkaline earth metals, usually hydrolyze by acting as acids. Many of these ions form hydrated metal ions. For example, copper (II) ion forms the Cu(H 2 O) 4 2+ ion. The Cu 2+ ion acts as a Lewis acid, forming bonds with a lone pair of electrons in the O atoms of H 2 O. Because the electrons are drawn away from the oxygen atoms by the Cu atom, the O atoms, in turn, tend to draw electrons from the O – H bonds, making them increasingly polar and weaker. As a result, the H 2 O molecules in Cu(H 2 O) 4 2+ are acidic. Copper (II) salts tend to have a pH around 3.5. Cu(H 2 O) 4 2+ hydrolyzes by donating a proton to a free water molecule. One of the water molecules loses a proton, reducing the charge of the ion by 1, and the water molecule becomes a hydroxide as shown in the reaction below. Cu(H 2 O) 4 2+ (aq) + H 2 O (l) ↔ Cu(H 2 O) 3 (OH) + (aq) + H 3 O + (aq) We can summarize this in the following tables: IONS OF NEuTRAL SALTS Cations Na + , K + , Rb + , Cs + Mg 2+ , Ca 2+ , Sr 2+ , Ba 2+ Anions Cl - , Br - , I - , ClO 4 - , BrO 4 - , ClO 3 - , NO 3 - Some Common Basic Ions F - CH 3 COO - NO 2 - CN - HCO 3 - CO3 2- S 2- HS - SO 4 2- HPO 4 2- PO 4 3- ClO - ClO 2 - Salts and Buffers 77 Some Common Acidic Ions NH 4 + CH 3 NH 3 + Al 3+ Pb 2+ Transition metal ions (e.g., Cu 2+ ) HSO 4 - H 2 PO 4 - To predict whether the salt solution is acidic, basic, or neutral you need to know whether the ions comprising the salt are acidic or basic or do not hydrolyze. In other words, you have to determine the acidity or basicity of the ions comprising the salt. Consider sodium cyanide, NaCN. This salt is composed of the potassium ion, Na + and the cyanide ion, CN - . Sodium is an alkali metal, so Na + does not hydrolyze. However, the CN - is the conjugate base of a weak acid, hydrocyanic acid, a weak acid. Therefore, the cyanide ion, CN - , is basic. A salt of sodium cyanide is expected to be basic. Conclusions regarding acid or basic hydrolysis can be made by examining the K a or K b values in the literature for the species involved. Let us examine the following exercise from the 1981 AP Exam. Exercise 1 Al(NO 3 ) 3 K 2 CO 3 NaHSO 4 NH 4 Cl (a) Predict whether a 0.10 molar solution of each of the salts above is acidic, neutral, or basic. (b) For each of the solutions that is not neutral, write a bal anced chemical equation for a reaction oc curring with wa ter that supports your prediction. Solution: (a) Al(NO 3 ) 3 = acidic K 2 CO 3 = basic NaHSO 4 = acidic NH 4 Cl = acidic (b) Al(NO 3 ) 3 : Al(H 2 O) 6 3+ + H 2 O ↔ H 3 O + + Al(H 2 O) 5 (OH) 2+ NO 3 -- + H 2 O →No reaction (conjugate base of a strong acid; it does not hydrolyze) K 2 CO 3 : K + + H 2 O → No reaction. CO 3 2- + H 2 O ↔ HCO 3 - + OH - NaHSO 4 : SPECIAL FOCuS: Acids and Bases 78 Na + + H 2 O → No reaction HSO 4 - + H 2 O ↔ H 3 O + + SO 4 2- NH 4 Cl: Cl - + H 2 O → No reaction (conjugate base of strong acid; it does not hydrolyze) NH 4 + + H 2 O ↔ NH 3 + H 3 O + One of the things to note from the above exercise is that students need to be aware that the transition metal ions and other metallic ions mentioned above do form complex ions with water. Students need to know the coordination number for the more common ions, such as Al 3+ , Cu 2+ , Zn 2+ , etc. A few basic rules would be: 1. A salt of a strong acid and a strong base. (e.g. NaCl). The salt does not have any ions that will hydrolyze, so it will be neutral. 2. A salt of a strong base and a weak acid. The anion is the conjugate base of a weak acid, so it hydrolyzes giving a basic solution. 3. A salt of a weak acid and a strong base. The cation of this salt is the conjugate of a weak base, so it hydrolyzes producing an acidic solution. 4. A salt of a weak acid and weak base. In this case both ions hydrolyze, the acidity or basicity depends on the K a and K b of the conjugate acids and bases respectively. If the K a > K b , the solution will be acidic. If the K b > K a , the solution will be basic. Buffers Defnition: A buffer solution is one that resists changes in pH when small quantities of an acid or a base are added to it. If 0.01 mol of HCl is added to 1.0 L of sodium chloride solution, the pH changes from 7.0 to 2.0. On the other hand, if this same amount of hydrochloric acid is added to 1.0 L of ocean or sea water, the pH changes by only 0.1 unit. Obviously, sea water has other substances present that maintain the pH of sea water at a fairly constant level. The pH of sea water ranges from 7.5 to 8.4. This is very important because many aquatic organisms are very sensitive to changes in pH. The pH of mammalian blood is also maintained at a pH of close to 7.38. Composition: A buffer contains a weak acid and its conjugate base or a weak base and its conjugate acid. Thus a buffer solution contains both an acid species and a Salts and Buffers 79 base species. The buffer in sea water is mostly due to the presence of carbonic acid, H 2 CO 3, and its conjugate base, the bicarbonate ion HCO 3 - . In blood, both the carbonate and phosphate buffers are present. Adding an acid to the buffer: The buffer solution must remove most of the new hydrogen ions otherwise the pH would drop markedly. The hydrogen ions will combine with the bicarbonate ion to make carbonic acid: HCO 3 - (aq) + H + (aq) ↔ H 2 CO 3(aq) Even though the reaction is reversible, most of the new hydrogen ions have been removed, so the pH won’t change very much. But because the reaction is reversible, the pH will decrease a very small amount. Adding a base to the buffer solution: Bases contain hydroxide ions or react to form them, so the buffer solution must remove most of these ions. There are two ways in which the carbonic/bicarbonate buffer can remove the hydroxide ion. Either carbonic acid or bicarbonate ion reacts: H 2 CO 3(aq) + OH - (aq) ↔ HCO 3 - (aq) + H 2 O (l) HCO 3 - (aq) + OH - (aq) ↔ CO 3 2- (aq) + H 2 O (l) Since most of the hydroxide has been removed, the pH of the buffer changes very little. The pH of a buffer: The pH of buffers can vary from being in the acid range of the pH scale all the way to the basic range. The pH of a buffer is determined by the relative amounts of acid and conjugate base that are present in the solution and the K a of the acid. Let us examine a buffer made of acetic acid and sodium acetate. This is similar to one of citric acid and sodium citrate that is often found in commercial fruit juices, to maintain the pH of the juice. Let us compare the pH of two different buffer solutions. 1.(a) A buffer made of 50.0 mL of a 0.10 M solution of acetic acid, CH 3 COOH, and 50.0 mL of a 0.10 M sodium acetate, CH 3 COONa. • The K a of acetic acid is 1.7 x 10 -5 . • The equilibrium reaction is CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - (aq) + H 3 O + (aq) . • The total volume of the buffer is 100.0 mL. • The concentration of acetic acid is 0.0050 moles of acid/0.100L = 0.050 M. • The concentration of the acetate ion is 0.0050 moles of acetate ion/0.100 L = 0.050 M. SPECIAL FOCuS: Acids and Bases 80 Filling in the ICE concentration table: CH 3 COOH H 2 O CH 3 COO - H 3 O + Initial 0.050 M 0.050 M 0 Change -X +X +X Equilibrium 0.050 M – X 0.050 M + X X In this solution K a = [H 3 O + ] [CH 3 COO - ] / [CH 3 COOH]. We substitute the equilibrium concentrations in the above equation. In doing so we can assume that X is small compared with 0.050 M. So the net equation is: 1.7 x 10 -5 = [X] [0.050 M]/ [0.050 M] After rearranging, we get [H 3 O + ] = (1.7 x 10 -5 ) (0.050 M)/ (0.050 M) [H 3 O + ] = 1.7 x 10 -5 M pH = - log [1.7 x 10 -5 M] = 4.76 which is in the acid range Now in this example the concentrations of the acid and the conjugate base are equal and the concentration of the hydronium ion is equal to K a . The pH of this buffer can be altered by adjusting the relative concentrations of the acid and the conjugate base. (b) For example, if the equilibrium concentration of the acetate ion is 0.070 M and the equilibrium concentration of the acetic acid is 0.025 M, the concentration of hydronium ion is [H 3 O + ] = K a [HA]/ [A - ]. Substituting in the above equation we can see that the ratio [HA]/[A - ] = 0.025/0.070, is less than 1, and the resulting [H 3 O + ] = 5.67 x 10 -5 has a pH of 5.25. It is slightly more basic than the previous solution, but still well within the acid range. A conclusion from the above problem: If the ratio [HA]/[A - ] = 1 then the pH of the resulting solution will be equal to pK a If the ratio [HA]/[A - ] > 1 then the resulting solution will be more acidic than pK a . If the ratio [HA]/[A - ] < 1 then the resulting solution will be more basic than pK a . Salts and Buffers 81 2. A buffer solution is made up of 50.0 mL of 0.10 M NH 4 + and 50.0 mL of 0.10 M NH 3 . The K a of ammonium ion is K w / K b K a = 1 x 10 -14 / 1.8 x 10 -5 K a = 5.56 x 10 -10 The equilibrium reaction is NH 4 + (aq) + H 2 O (l) ↔ NH 3(aq) + H 3 O + . The fnal volume is 100 mL, so the fnal concentration of [NH 4 + ] = 0.050 M. The fnal concentration of [NH 3 ] = 0.050 M . Filling in the ICE concentration table: NH 4 + H 2 O NH 3 H 3 O + Initial 0.050 M 0.050 M 0 Change -X +X +X Equilibrium 0.050 M – X 0.050 M + X X Now K a = [H 3 O + ] [NH 3 ]/ [NH 4 + ]. We can assume that X is small compared to 0.050 M. Substituting we get: • 5.56 x 10 -10 = (X) (0.050 M) / (0.050 M) • We can see that X = [H 3 O + ] = 5.56 x 10 -10 M • pH = -log [5.56 x 10 -10 ] This gives a pH of 9.3, which is in the basic range. 3. From studying the two different buffers, we can clearly see that the pH of the buffer will be determined mainly by the value of K a . a. For a buffer with pH 3, we choose an acid/conjugate base combination that has a pK a close to 3. b. For a buffer with a pH of 9, we choose an acid/conjugate base combination that has a pK a close to 9. 4. The addition of water affects a buffer solution (dilutes of a buffer). When water is added to a buffer solution, the pH remains unchanged. Let us examine this. HA ↔ H + + A - K a = [H + ] [A - ] / [HA] SPECIAL FOCuS: Acids and Bases 82 Rearranging this, we can see that: [H + ] = K a [HA] / [A - ]. The addition of more water to the buffer does not change the ratio [HA]/[A - ], since the same amount of water is being added to both. If this ratio remains unchanged, the pH stays the same. This analysis does not consider new effects that appear if the solution becomes extremely dilute. 5. A buffer resists change to pH when small amounts of acid or base are added. We are now going to examine this quantitatively. Suppose we have a buffer consisting of 0.10 M nitrous acid and 0.20 M sodium nitrite. The K a of nitrous acid is 4.5 x 10 -4 . To 95.0 mL of this buffer solution we add 5.0 mL of 0.10 M hydrochloric acid. This type of problem becomes a two-part problem. This frst part is a stoichiometric calculation where we assume that all the H 3 O + from the strong acid reacts with the nitrite ion forming nitrous acid. The second part is an equilibrium problem where the concentrations from the frst part are used. Solution Part 1—Stoichiometric Calculation When a hydronium ion is added to the buffer it reacts with the nitrite ion: H 3 O + (aq) + NO 2 - (aq) → HNO 2 (aq) + H 2 O (l) Because the nitrous acid is a weak acid, we assume that the reaction goes to completion. First, we must calculate amounts of hydrogen ions, nitrous acid and nitrite ions in the solution before the reaction. Because HCl is a strong acid, the hydrogen ion concentration is equal to the molarity of the acid. • HCl is 0.10 M, therefore the concentration of H 3 O + is 0.10 M. • The moles of H 3 O + = (0.10 moles/L) (0.005L) = 0.00050 moles. • The moles of HNO 2 = (0.10 moles/L) (0.095L) = 0.0095 moles. • The moles of NO 2 - = (0.20 moles/L) (0.095L ) = 0.019 moles. We assume that all the H 3 O + added reacts with the nitrite ion. Therefore, 0.00050 moles of acid is produced and 0.00050 moles of nitrite ion is used up. • Moles of nitrite ion remaining = (0.019 – 0.00050) mol = 0.0185 mol. Salts and Buffers 83 • Moles of nitrous acid after reaction = (0.0095 + 0.00050) mol = 0.010 mol. • [HNO 2 ] = 0.010 mol/0.10 L = 0.10 M. • [NO 2 - ] = 0.0185 moles/ 0.10L = 0.185M . Part 2—Equilibrium Problem Using the concentration found in Part 1, you construct the following ICE table: HNO 2 (mol/L) H 3 O + (mol/L) NO 2 - (mol/L) Initial 0.10 M 0 0.185 M Change - X + X + X Equilibrium 0.10 – X X 0.185 + X The equilibrium constant equation is: K a = [H 3 O + ] [NO 2 - ] / [HNO 2 ] Substituting, you get: 4.5 x 10 -4 = [ X ] [0.185 + X] / [0.10 – X] Assuming that X is small enough that 0.1855 + x ≈ 0.185 and 0.10 – X ≈ 0.10, x 10 -4 = (X) (0.185) /0.10 [H 3 O + ] = X = 2.43 x 10 -4 The pH of the original buffer was 3.65, the pH after the addition of the 5.0 mL of 0.10 M HCl (aq) is 3.61. So the pH changed by only 0.04 units The buffer pH changed much less than 0.5 units, which is considered the maximum change that a buffer should have and continue to be effective at pH control. Examples from Prior AP Chemistry Examinations (1983) 2(a). Specify the properties of a buffer solution. De scribe the components and the composition of ef fective buffer solu tions. 2(b). An employer is interviewing four applicants for a job as a laboratory technician and asks each how to pre pare a buffer solution with a pH close to 9. Archie A. says he would mix acetic acid and sodium acetate solutions. Beula B. says she would mix NH 4 Cl and HCl so lutions. Carla C. says she would mix NH 4 Cl and NH 3 so lutions. Dexter D. says he would mix NH 3 and NaOH so lutions. SPECIAL FOCuS: Acids and Bases 84 Which of these applicants has given an appropri ate pro ce dure? Explain your answer, referring to your dis cussion in part (a). Explain what is wrong with the er roneous proce dures. (No calculations are necessary, but the following acid ity constants may be helpful: acetic acid, K a = 1.8 x10 –5 ; NH 4 + , K a = 5.6 x 10 –10 ) Answer: 2(a). A buffer solution resists changes in pH when small amounts of an acid or a base are added. Preparation of a buffer: Mix a weak acid with the conjugate base of the acid. 2(b). Archie has a buffer solution: he has a weak acid and its conjugate base, but it does not have the correct pH since the pK a is close to 5. So he does not have the correct buffer. • Beula does not have a buffer, since she has a weak acid (NH 4 + ) and a strong acid (HCl). • Carla has the correct buffer. She has a weak acid, (NH 4 + ) and its conjugate base (NH 3 ). She also has the correct pH, since the pK a is close to 9. • Dexter does not have a buffer since he has mixed a weak base and a strong base. (1977A) 3. The value of the ionization constant, K a , for hypochlor ous acid, HOCl, is 3.1 X 10 –8 . (a) Calculate the hydronium ion concentration of a 0.050 molar solution of HOCl. (b) Calculate the concentration of hydronium ion in a so lu tion prepared by mixing equal volumes of 0.050 molar HOCl and 0.020 molar sodium hypochlorite, NaOCl. Answer: (a) HOCl + H 2 O ↔ H 3 O + + OCl - X << 0.050 X = [H 3 O + ] = 4.0 x 10 –5 M (b) HOCl + H 2 O ↔ H 3 O + + OCl - X ‹‹ 0.010 X = [H 3 O + ] = 8.0 x 10 –8 M Salts and Buffers 85 (2005A) HC 3 H 5 O 2 (aq) ↔ C 3 H 5 O 2 – (aq) + H + (aq) K a = 1.34 × 10 –5 Propanoic acid, HC 3 H 5 O 2 , ionizes in water ac cording to the equation above. (a) Write the equilibrium constant expression for the reaction. (b) Calculate the pH of a 0.265 M solution of propanoic acid. (c) A 0.496 g sample of sodium propanoate, NaC 3 H 5 O 2 , is added to a 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following. (i) The concentration of the propanoate ion, C 3 H 5 O 2 – (aq) in the solution (ii) The concentration of the H + (aq) ion in the so lution Answer: (a) [C 3 H 5 O 2- ][H + ] = K a [HC 3 H 5 O 2 ] (b) let X be the amount of acid that ionizes, then X = [C 3 H 5 O 2 – ] = [H + ] 0.265 – X = [HC 3 H 5 O 2 ] X 2 = K a = 1.34 x10 –5 0.265 – X X = 0.00188 M = [H + ] [you can assume that 0.265 – X ≈ 0.265 in order to simplify your calculations] pH = –log[H + ] = 2.73 (c) (i) 0 496 0 50 . . g 1mol 96.0g L × = 0.103 M since each sodium propanoate dissociates com pletely when dissolved, producing 1 propanoate ion for every sodium propanoate, and this is more than thousands of times larger than the propanoate ions from the acid, then [C 3 H 5 O 2 – ] = 0.103 M (ii) let X be the amount that ionizes, then: X = [H + ] X + 0.103 = [C 3 H 5 O 2 – ] 0.265 – X = [HC 3 H 5 O 2 ] SPECIAL FOCuS: Acids and Bases 86 (X) (0.103 + X) = K a = 1.34 x 10 –5 0.265 – X X = 3.43 x 10 –5 M = [H + ] [We can assume that 0.265 – X ≈ 0.265 and X + 0.103 ≈ 0.103, in order to simplify calculations.] Buffer capacity is the amount of acid or base that can be added to a buffer solution before a major pH change (more than +- 0.5 units) occurs. All buffers have a buffer capacity. They will resist change to pH when only small amounts of acid or base added. If large amounts of acid or base are added, then the buffer’s capacity is exceeded and a major pH change will occur. The buffer’s capacity is determined by the amount of acid and conjugate base. A buffer with a large capacity has a large concentration of HA and of A - , so that it can absorb a relatively large amount of H 3 O + or OH - . The pH of a buffered solution is determined by the ratio [HA]/[A - ]. The capacity of a buffer is determined by the magnitude of [HA] and [A - ]. A good laboratory activity to do with the students is to study the buffering capacity of seawater as compared to that of a 0.50 M NaCl solution. (This concentration ≈ the concentration of NaCl in seawater.) The students can see the importance of a natural buffer such as is present in seawater and also study the buffering capacity of seawater. Bibliography Cyberspace Chemistry. “Hydrolysis of Metal Salts.” University of Waterloo. http://www. science.uwaterloo.ca/~cchieh/cact/c123/salts.html (accessed January 2008). Mitchell, Philip, and Norris Rakestraw. “The Buffer Capacity of Sea Water”, The Biological Bulletin 65 (December 1933): 437–451. Ebbing, Darrell, and Steve Gammon. General Chemistry, 8th ed. Boston: Houghton Miffin Company, 2005. Acidic and Basic Reactions of Oxides Marian DeWane Centennial High School Boise, Idaho The oxides of the elements form an interesting group of acidic and basic compounds. The combination of element X with the highly electronegative oxygen results in a polarized bond (in every instance except two oxygen atoms bonding to form O 2 ), which then can be a reaction site when the oxide is added to water to form an aqueous solution, or when the oxide is reacted directly with an acid or a base. Two types of reactions can result: (1) M – O - + H 2 O → M—O—H + OH - Here the oxide is a proton acceptor from water leaving an excess of hydroxide ion in solution, hence forming a basic solution. In this case the oxides are classifed as basic oxides. Oxides behaving in this way include those of lithium, sodium, potassium, cesium, calcium, strontium, and barium. These are among the least electronegative elements or metals group. (2) X = O + H 2 O → - O—X—O - + 2H + Here the oxide is an oxygen acceptor from water, which releases hydrogen ions into the solution, hence forming an acidic solution and in effect acting as a proton donor. Such oxides are classifed as acidic oxides. Oxides behaving in this way include those of carbon, nitrogen, sulfur, phosphorous, and the halogens. These are among the most electronegative elements or nonmetals group. These reactions are obviously oversimplifed, and can be studied more closely as follows: (1) An alternative view of the basic oxides is that these are mostly ionic compounds (with a high degree of ionic character) and can be thought of as SPECIAL FOCuS: Acids and Bases 88 containing ions in the solid state. When added to water, it is these ions that act as proton acceptors as follows: O 2- + H 2 O → 2 OH - , each O 2- ion reacting to form two hydroxide ions, hence the highly basic nature of the resulting solution. Thus it is the ionic structure of the metal oxides that best explains their reaction in aqueous solution. The combination of the highly electronegative oxygen atom with the low electronegativity atom of a typical metal produces the ionic bond in metal oxides. The electronegative oxygen is easily able to attract the most weakly bonded electron away from a metal atom to form the ionic bond. Two electrons can be taken from an alkaline earth element such as magnesium to form Mg 2+ O 2- . Typical oxides are Na 2 O, K 2 O, MgO, CaO, etc. (2) In the acidic oxides, the nonmetal atom readily forms covalent bonding with the oxygen, as in common examples such as CO 2 , SO 3 , NO 2 , Cl 2 O, for example. While this uses some electron density of the central atom, the more electronegative atoms in the nonmetals still have suffcient remaining electron density to form additional bonding with oxygen atoms in water molecules. This weakens the O-H bond in water, releasing H + ion. The complete reaction may take place in two steps to release two H + ions. For example: SO 3 + H 2 O → HSO 4 - + H + HSO 4 - → SO 4 2- + H + Direct reactions of oxides can occur without water being an obvious reagent. So, for example, sodium oxide can react with hydrochloric acid to form sodium chloride and water in a typical acid–base reaction forming a salt and water: Na 2 O + 2HCl → 2 NaCl + H 2 O However, when an acidic oxide reacts with some bases water must be present to provide the oxygen transferred to the central atom. For example, sulfur trioxide can react with ammonia to form ammonium sulfate. Water has to be included as a reagent to provide the oxygen transferred. SO 3 + 2NH 3 + H 2 O → (NH 4 ) 2 SO 4 Acidic and Basic Reactions of Oxides 89 Amphoteric Oxides As in all other areas of chemistry, there are chemical behaviors that fall in between two extremes, and oxides are no exception. Amphoteric oxides can act as acids or bases, depending on the circumstances. With hydrogen as an exception (water is clearly an example of an amphoteric oxide), metals in the center of periods and semimetals tend to form such oxides. For example aluminum oxide, Al 2 O 3 , can act as an acid or a base, as can SiO 2. Aluminum oxide is a compact strong structure relatively insoluble in water, but it can be dissolved by reacting in strong HCl. Here the aluminum oxide is acting as a base, neutralizing the hydrochloric acid, as represented in the reaction below: Al 2 O 3 + 6HCl → Al 2 Cl 6 + 3H 2 O However, aluminum oxide also dissolves in strong sodium hydroxide solution acting as an acid neutralizing the sodium hydroxide. This reaction can be represented by the equation: Al 2 O 3 + 2NaOH + 3H 2 O → 2Al(OH) 4 - + 2Na + AP Exam Examples of Free-Response Questions: Free-response questions about oxides are usually in the equations section, question 4. The question now requires students to provide a balanced net equation and answer a question about the reaction. The year these questions were given, no additional questions were asked. Metal oxide plus water examples: 2004 Form B Question 4 (f): Powdered barium oxide is mixed with water. BaO + H 2 O → Ba 2+ + 2OH - 2002 Question 4 (c): Solid cesium oxide is added to water. Cs 2 O + H 2 O → 2Cs + +2OH - 2000 Question 4 (g): Powdered strontium oxide is added to distilled water. SrO + H 2 O → Sr 2+ +2OH - 1999 Question 4 (a) Calcium oxide powder is added to distilled water. CaO + H 2 O → Ca 2+ +2OH - Nonmetal oxide plus water examples: 2003 Question 4 (h): Solid dinitrogen pentoxide is added to water. N 2 O 5 + H 2 O → 2H + + 2NO 3 - 2003 Form B Question 4 (f): Sulfur trioxide gas is bubbled into water. SO 3 + H 2 O → H + + HSO 4 - 2002 Form B Question 4 (e): Sulfur dioxide gas is bubbled into a beaker of water. SO 2 + H 2 O → H + + HSO 3 - SO 2 + H 2 O → H 2 SO 3 was also accepted. 2001 Question 4 (a): Sulfur dioxide gas is bubbled into distilled water. SO 2 + H 2 O → H + + HSO 3 - SO 2 + H 2 O → H 2 SO 3 was also accepted. Bibliography Brady, J., and F. Senese. Chemistry: Matter and Its Changes, 4th ed. New York: John Wiley and Sons, 2004. Cotton, F. A., G. Wilkinson, Carlos Murillo, and M. Bochmann. Advanced Inorganic Chemistry, 6th ed. New York: John Wiley and Sons, 1999. Oxtoby, D. W., and N. H. Nachtrieb. Principles of Modern Chemistry, 2nd ed. New York: Saunders, 1990. Rayner-Canham, G. Descriptive Inorganic Chemistry, 2nd ed. New York: Freeman, 1998. Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations George E. Miller University of California, Irvine Irvine, California Misconceptions and misunderstandings are often demonstrated by students on both multiple-choice and free-response portions of the AP Chemistry Examination. Many may refect prior thinking brought to the AP class. These are often termed “preconceptions.” Others refect misunderstanding of material presented within the AP class or read from the textbook. Frequently, misunderstandings arise from a lack of depth and/or accuracy of thinking about a topic or question. On multiple-choice questions, frequent misconceptions are often included by the item writers as possible wrong answers, or distracters. Students who select these answers are showing they have that misconception. Teachers should always ask students to explain why they selected a particular distracter. The teacher can then proceed to help the student correct the misconception in their thinking, which will help the student make the correct selection. In free-response questions, perhaps the most common misconception demonstrated is a lack of understanding of the words “explain,” or “justify.” This persists even when the examination question tries to provide guidance as to what is expected. Use of familiar examples is helpful. If a sportswriter or commentator explained why team X won the game against team Y by stating, “It was because team X scored more points,” they would not survive long in their jobs. Yet that is the format a considerable number of students exhibit when providing an explanation on the AP Chemistry Exam. Restating the question also does not answer the question. SPECIAL FOCuS: Acids and Bases 92 For example, the following question and answer might have appeared in AP: Using the information given in the following table, explain the difference in K a values for HClO and HBrO. Acid K a Electronegativity of halogen HClO 3.0 x 10 -8 3.0 HBrO 2.8 x 10 -9 2.8 Incorrect response (1): The table shows there is a difference between the K a values of HClO and HBrO. Incorrect response (2): HBrO has a larger K a because 2.8 x 10 -9 is larger than 3 x 10 -8 . Mathematical errors even creep in as a result of carelessness or real misunderstandings of conceptual algebra, or exponential notation (as seen above). For example, students may transcribe K = [H][A] / [HA] ≅ [H]2 /[HA] incorrectly as [H] = √ K/[HA]. Refective conceptual thinking would always look for responses where numerator and denominator values moved in the same direction. To foster correct thinking, teachers should encourage students to always answer the question qualitatively before plugging in any numbers. The remaining document addresses selected topics where frequent misunderstandings are demonstrated, together with some hints for instructors. Misconceptions Regarding Acids, Bases, and Aqueous Equilibria There is confusion on a number of issues relating to acids and bases, as noted below. Student and teacher dialogue (almost more than demonstrations, which also can easily be misconstrued) is key. The answer is to start with basic principles, arguing somewhat as follows: 1. Reactions of acids and bases occur in water. (No nonaqueous systems are dealt with in AP Chemistry.) Lewis acids are barely mentioned, so master Brønsted-Lowry or proton transfer systems before even thinking about anything else. 2. The largest numbers of molecules present in solutions are therefore water molecules. Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations 93 3. The obvious species for anything to react with is, therefore, water. If you jump into a crowded swimming pool, the chances are that you will hit water molecules frst! 4. Thus, frst-cut reactions are water reactions. 5. Water is amphiprotic and amphoteric, so its reactions are to accept a proton or donate a proton. Free protons do not exist in water solutions—they must be attached to something. Most will be attached to water molecules in the form of H(H 2 O) n + where n is 1-4. In most texts, this is abbreviated to H + . 6. Thus, acid–base chemistry is almost exclusively (>99%!) the study of proton transfer reactions all of which are equilibrium reactions and each of which has a well-defned equilibrium constant that only varies with temperature. 7. So we simply must list all possible reactions of any given system in water and ask: “Is the system at equilibrium?” If not, then consider reactions that happen as they go to equilibrium; if it is at equilibrium, then what do we know about all the possible equilibrium constants? 8. In one system at equilibrium, all possible equilibria and the corresponding equilibrium constant expressions must be simultaneously satisfed. One cannot have some OK, and some not OK. 9. Correspondingly, there can only be ONE value of a species concentration (e.g., one value of [H + ] or [OH - ]. The solution cannot track what H + came from what source to create separate values. So what are the misconceptions, and why do they exist? The following examples are given mostly for an acid, but the corresponding idea for a base also exists. 1. Acid–base “strength” is related to acid–base concentration. This is a semantic problem caused by the unfortunate choice of terms by chemists. To the lay public, putting a lot of any chemical in water makes it “strong.” Strong coffee or strong lemonade has a high concentration of those chemicals. However, acids and bases are defned as strong or weak solely on the basis of their equilibrium constants in reaction with water. An acid with a high K value (K value > 10 -1 ) for the reaction HA + H 2 O ↔ A - + H 3 O + is a strong acid. This is often stated in terminology regarding the “degree of ionization.” Conversely, strong bases have a high K value for the reaction B + H 2 O ↔ B + + OH - . Some texts still may use the term “dissociation” based on the idea that HA becomes H + and A - by simple dissociation. Generally, it is now recognized that proton SPECIAL FOCuS: Acids and Bases 94 transfer reactions to water are what actually takes place. It is the K value for the transfer to water that determines whether an acid is classifed as strong or weak. The “concentration” of a solution of an acid in water strictly refers to the formal concentration which is the same as [HA] + [A - ] , or the corresponding values for a base. This concentration expression is the same for either a strong or a weak acid. Thus, one can have a high concentration of a weak acid, or a small concentration of a strong acid, and vice versa. Weaker students fnd this quite diffcult. A class of one strong student is a strong class, but a class of 40 weak students is still a weak class. 2. “Reaction to completion can only occur as the K value will allow.” So a weak acid cannot react to neutralize a strong base. For example: 5 mL of 0.1 M NaOH is added to 10 mL of 0.1 M acetic acid. What is the end result? Beginning students who have learned to correctly write an equilibrium constant for the weak acid in water as HAc + H 2 O ↔ H + + Ac - with K a = 1.8 x 10 -5 frequently will argue that since [H + ] = [Ac - ], now 1.8 x 10 -5 = [H + ]2 / (0.1) and the concentration of hydrogen ion is now only √ (1.8 x 10 -6 ) or 1.3 x 10 -3 M. Thus the reaction can only consume 1.3 x 10 -3 moles of OH-, leaving the remainder in solution, so the result is strongly basic. When the solution is mixed, it is not at equilibrium. All reactions must be allowed to proceed to equilibrium before the equilibrium mixture is analyzed. With this in mind, the ICE tabulation method can be used to obtain a result. 3. “Acids are the only source of protons in water.” In many problems with high or medium concentrations of acids, this is approximately true, so often neglecting water as a source works. The full treatment of this, which leads to a third order equation for [H+], is usually given in college analytical chemistry courses. However, there are circumstances where the contribution of the “auto- ionization” or “autoprotolysis” of water according to H 2 O + H 2 O ↔ H(H 2 O) + + OH - cannot be ignored. The classic catch question is: What is the [H+] of a 1 x 10 -10 M solution of HCl in water? Since HCl is a strong acid, we assume complete transfer, so [H+] from HCl is 1 x 10 -10 M. (pH = 10). A good student will recognize this as nonsense since this is a basic solution. The water is contribution 1 x 10 -7 , so the true value is close to 1 x 10 -7 + 1 x 10 -10 or 1 x 10 -7 M, a “neutral” solution of pH = 7. Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations 95 Usually the examination committee for AP has tried to set problems on the free- response section where the water contribution can be ignored and approximations are simple. 4. [H + ] and [OH - ] values are independent of each other. A small but signifcant number of students fail to apply the water equilibrium in acid–base solutions. Thus they are comfortable to report a pH of 4 and a pOH of 4 (or the equivalent concentrations values) in the same solution. Perhaps a focus during discussions on the need to satisfy the water equilibrium, or at least a frequent repetition of the mantra pH + pOH ALWAYS = 14 (at normal room temperature) could help. 5. Acids are compounds containing H atoms, bases are compounds containing OH. Unfortunately, this defnes NH 3 as an acid. Acids and bases are defned according to their ability to donate or accept protons, almost exclusively (for AP) in reaction with water. Accepting a proton from water creates a basic solution (larger [OH - ] than [H + ]). This enables reactions such as NH 3 + H 2 O ↔ NH 4 + + OH - to be written to correctly identify ammonia as a (weak) base. Much practice in writing reaction equations with water as reagent and seeking the K values for the reactions in tables will help students learn how to classify more correctly. Of course there are nonaqueous systems, but these are rarely dealt with in any signifcant way in AP or frst-year college chemistry courses. Acid–Base Chemistry Beyond AP ® Arden P. Zipp SUNY Cortland Cortland, New York Most of the acid–base behavior discussed in AP Chemistry occurs in water. However, the use of water as a solvent limits the species that can act as acids or bases and restricts the range of reactions that can be legitimately considered as acid–base reactions. This article will address acid–base chemistry beyond water and, thus, beyond the usual considerations of AP. A further extension is provided by acidity and basicity measurements in the gas phase, where the relative tendency of a molecule to donate or accept a proton can be measured. Trends in acidity as a function of molecular structure in the gas phase can often be more easily understood, as many trends in solution depend on solvation energy considerations, which are absent in the gas phase (Barntmess, Scott, McIver 1979). Leveling Effect of Water It is well known that H 2 O ionizes into H + (or H 3 O + ) and OH - . Acids that are stronger than H 3 O + will donate protons to H 2 O to form H 3 O + ions, while bases that are stronger than OH - will abstract protons from H 2 O to leave OH - ions. As a consequence, all stronger acids (or bases) are made to appear as if they are of equal strength. This is the situation with HCl, HBr, HI, HNO 3 , HClO 4 , and H 2 SO 4 (frst proton), which appear equally strong in water. Although the K a of these substances actually varies from 10 1 to 10 11 , these differences are not apparent in water because H 2 O molecules attract the H + from the various anions so strongly that all the acid is converted into H 3 O + and the corresponding anion. This occurs because H 2 O molecules are stronger bases toward H + than any of the anions in these acids, so the differences in anion attraction for the H + are not apparent. (Recall from the discussion of Brønsted-Lowry acids and bases that reactions with H + ions involve competitions SPECIAL FOCuS: Acids and Bases 98 between two bases.) The same “leveling” phenomenon occurs with bases that are stronger than OH - (e.g., H - , C 2 H 5 O - or S 2- ) because they react with an H + from water to form H 2 , C 2 H 5 OH, or HS - , respectively, leaving OH - as the predominant anion in solution. The opposite effect occurs for very weak acids (or bases), which are not strong enough to exhibit their acid or base properties in water. As an example of this behavior, alcohols (ROH) contain hydroxyl (-OH) groups but exhibit neither acidic nor basic properties in H 2 O. The K a for a typical alcohol is about 10 -18 (pK a = 18). Note: Saying that alcohols are very weak acids is equivalent to stating that their conjugate bases (RO - ) are very strong, which would lead to them being leveled in water as discussed above. (There is virtually no tendency for an alcohol to behave as a base by either gaining an H + or losing an OH - in water.) Nominally, acids with K a values that range from 10 0 to 10 -14 (pK a s from 0 to 14) can be studied in water and substances with values outside this range must be investigated in other solvents. One way to compare the useful ranges of different solvents is to graph their effective pK a ranges relative to H 2 O as depicted below (Atkins, Overton, Rourke, Weller, and Armstrong 2006, 117). pK a Ranges of pure Solvents Relative to Water H 2 SO 4 HF HCOOH H 2 O (CH 3 ) 2 SO Effective pK a in water NH 3 -20 -10 0 10 20 30 40 In order to discriminate among the acidities of the six strong acids mentioned above, a less basic solvent than water (one with a more negative pK a value) can be used. Two examples of such solvents are liquid methanoic (formic) acid, HCOOH, and liquid HF. Because these solvent molecules have less attraction for H + than does H 2 O, the acids do not ionize completely. Thus, it is possible to measure the relative amounts of the ionized and un-ionized acids, to calculate their ionization constants, and to relate these values to aqueous solutions. In this way, the K a values of HNO 3 , H 2 SO 4 , HCl, HBr, HClO 4 and HI have been found to be 10 1 , 10 2 , 10 7 , 10 9 , 10 10 , and 10 11 , respectively. 99 Acid–Base Chemistry Beyond AP On the other hand, bases stronger than OH - can be distinguished by using solvents such as liquid dimethyl sulfoxide, (CH 3 ) 2 SO, or liquid NH 3 , which have pK a values greater than 14. The opposite strategy applies to the task of distinguishing among very weak acids (or bases). In the case of very weak bases a more acidic solvent than H 2 O is needed in order to protonate the basic site more effectively and for very weak acids a more basic solvent must be employed to help remove H + . For a series of alcohols their relative strengths could potentially be determined by studying them in either dimethylsulfoxide or ammonia. Once the constraints of an aqueous solution have been eliminated, a variety of other acid–base reactions can be observed. Chemists have developed “super acids,” i.e., acids that are stronger than 100% sulfuric acid, which can even protonate hydrocarbons! At the other extreme are super bases, such as butyllithium—LiC 4 H 9 , which can remove H + ions from virtually anything including many hydrocarbons, which do not show acidic properties under other circumstances. Relative Acid Strengths Although it is common to classify acids as either strong or weak, there is a great deal of variation among the species in each category. These variations can be accounted for on the basis of the structures of the molecules involved. Although the specifc structural factors depend on the acid category (hydroacids vs. oxoacids), the fact that acid strengths can be correlated with structures for a given category of acids provides a very powerful organizational tool. Consideration of acid strength from gas phase species, which can be done both theoretically and experimentally, can lead to interestingly different results as solvation effects are removed (Bartmess and Hinde 2005). Hydroacids These acids can be represented by the general formula H x Y and include species in which an atom of a nonmetal is attached directly to one or more hydrogen atoms. Such substances ionize according to the equation: H x Y <=> H + + H (x-1) Y - , with the extent of this reaction dependent on the ease with which the bond between H + and H (x-1) Y - can be broken. The K a values for the compounds in groups 14 to 17, SPECIAL FOCuS: Acids and Bases 100 most of which had to be measured in solvents other than water, cover a range of 10 60 (from 10 -49 to 10 11 ) as refected in the pK a s given in the table below (Bowser 1993, 314). pKa for Hydroacids CH 4 49 NH 3 39 H 2 O 15.74 HF 3.45 SiH 4 35 PH 3 27 H 2 S 7.05 HCl -7 GeH 4 25 AsH 3 23 H 2 Se 3.8 HBr -9 H 2 Te 2.6 HI -11 The pK a values of these substances decrease down the periodic table and from left to right, refecting an increase in K a values and, therefore, acid strength in the same directions. There are two factors that infuence the ease with which the H-Y bond can be broken. These are the electronegativity (EN) and the size of the atom Y. The increase in acidity going across the periodic table can be accounted for by the change in EN. As the EN of Y increases across the period, the electron pair in the H-Y bond will be drawn closer to Y, increasing the polarity of the bond. Because the atomic size does not vary appreciably across a period, the change in EN is the major factor affecting the acidity. The increase in the acidity down the table cannot be explained in terms of the EN, however, because the EN changes in the wrong direction, decreasing down the table. However, there is a large increase in atomic size descending in any group. As the size of an anion increases, its electrostatic attraction for H + decreases, which would increase the acidity. Oxoacids This group of acids is more extensive and diverse than the hydroacids discussed above, with changes in both the central atom and the number of oxygen and/or hydrogen atoms as shown in the list of several of the more common oxoacids: HClO, HClO 2 , HClO 3 , HClO 4 , HNO 2 , HNO 3 , H 2 CO 3 , H 2 SO 3 , H 2 SO 4 , and H 3 PO 4 . Despite this variability in formula, all of the acids in this group can be represented by the general formula (HO) m YO n , which emphasizes the fact that the ionizable protons are bonded to oxygen atoms, which, in turn, are bonded to the central atom Y. Thus, the structure of HClO is actually HOCl, while that of HClO 2 is HOClO and that of H 2 SO 4 is (HO) 2 SO 2 . When these species are compared in this manner, their strengths as represented by their K a (or pK a ) values are found to fall into different categories depending on the number of unprotonated oxygen atoms, regardless of the number of ionizable H’s in the molecule. This is shown in the table below (Bowser 1993, 314). 101 pK a Table for Oxoacids (HO)mY Very weak pK a > 7 (H0)mYO Weak pK a 2 – 5 (H0)mYO 2 Strong pK a < 1 (H0)mYO 3 Very Strong pK a < 5 HOCl 7.54 HOClO 1.94 HOClO 2 -2.0 HOClO 3 -7.3 HOBr 8.49 HONO 3.20 HONO 2 -1.44 (HO) 3 As 9.18 (HO) 2 CO 3.60 (HO) 2 SO 2 -3 (HO) 4 Si 9.77 (HO) 2 SO 1.89 (HO) 2 SeO 2 1.66 (HO) 2 SeO 2.62 (HO) 3 PO 2.15 (HO) 3 AsO 2.21 Several points are notable about the data in this table. First, the acid strength increases with the number of unprotonated oxygen atoms in the molecule. An increase in the number of such electron-attracting oxygen atoms will withdraw electrons more effectively from the H in the O-H bond. This increases the polarity of the bond and makes the H atom more H + -like. Second, the acid strength changes little with variations in atomic size or electronegativity in contrast to the behavior observed for the hydroacids discussed above. This can be accounted for by the fact that the ionizable H + is always attached to an oxygen atom whose size doesn’t change and, further, any change in the size of the central atom has little effect on the size of the anion as a whole. Similarly, changes in the electronegativity of the central atom represent a small effect on the attraction for the H + compared to the infuence of the greater EN effects of one or more (more electronegative) oxygen atoms. Acidity of Metal Ions Metal ions behave as Lewis acids, accepting a share of a pair of electrons from appropriate donors. One of the most common electron pair donors is the H 2 O molecule, especially in aqueous solutions, where most metal ions exist as octahedral species as shown by the equation: M x+ + 6H 2 O --> M(H 2 O) 6 x+ The interaction of some metal ions (most notably alkali metal ions and larger alkaline earth metal ions such as Ca 2+ , Sr 2+ and Ba 2+ ) with H 2 O goes no further. However, many metal ions attract the electron pairs of the coordinated H 2 O molecules so strongly that one or more molecules may actually be torn apart, releasing H + and making the solution acidic as represented in the equation: M(H 2 O) 6 x+ --> M(H 2 O) 5 OH x-1 + H + Acid–Base Chemistry Beyond AP SPECIAL FOCuS: Acids and Bases 102 The tendency of metal ions to behave in this manner depends primarily on two factors: charge and size, as shown in the table below where the pK a values for several metal ions participating in this reaction are given along with their radii (Wulfsberg 1991, 25). pKa values and Radii of Common Metal Ions Ion Na + Ca 2+ Pu 3+ Mg 2+ Ca 2+ Ba 2+ pK 14.2 12.8 7.0 11.4 12.8 13.5 Radius (nm) 0.116 0.114 0.114 0.086 0.114 0.149 The radius and charge can be combined into a single quantity called the charge density. (It is not necessary to consider the details of calculating this quantity but only to deal with it qualitatively, where it increases with an increase in charge and/or a decrease in ionic radius.) As a frst approximation, solutions containing metal ions become more acidic as the charge density of the metal ion increases. As the charge density of metal ions increases, this reaction may proceed further, to the precipitation of metal hydroxides: M(H 2 O) 6 x+ --> M(OH) x + xH + + (6-x) H 2 O The pH at which this reaction is initiated decreases as the charge density of the metal ion increases, as shown by the pH values for the formation of the following hydroxides, which are characterized by a decrease in size and/or an increase in the charge of the cation (Masterton and Hurley 2006, 427). Ca(OH) 2 pH 12.3 Fe(OH) 2 pH 8.67 Fe(OH) 3 pH 4.49 For species with extremely high charge densities, the potential cations may exist as oxoanions. This is illustrated by the ions VO 4 3- , CrO 4 2- and MnO 4 - , in which the charges on the nonexistent cations would be +5, +6, and +7, respectively. Hard and Soft Acids and Bases (HSAB) This concept represents one of the most signifcant developments in acid–base theory in recent years. It represents an extension of the Lewis theory and was developed to account for the results obtained when a variety of electron-pair acceptors were reacted with several electron pair donors. Specifcally, it was found that a cation such as Ag + reacts with halide ions in the order I - > Br - > Cl - > F - , whereas (as was pointed out above) H + reacts with this same series of ions in the opposite order: F - > Cl - > Br - > I - . 103 To account for this difference in behavior, the electron-pair acceptors (Lewis acids) H + and Ag + were labeled as hard and soft, respectively, while the halide ions F - , Cl - , Br - , and I - were assumed to progress from hard to soft. Hard acids and bases are species that interact in an electrostatic manner, whereas soft acids and bases exhibit covalent interactions. A fundamental principle of this theory is that hard acids interact preferentially with hard bases. Thus, hard H + reacts more strongly with hard F - (to form the weak acid HF) than with the soft I - (since HI is a strong acid) and the soft Ag + interacts more strongly with soft I - (to form the less soluble AgI) than with harder F - (AgF is more soluble). The HSAB theory has provided the basis to account for a wide range of chemical phenomena. For example, it offers explanations for the strengths of the hydrohalic acids and the solubilities of the silver halides as described above. Further, however, it provides the basis to understand the distribution of metal ions in nature between sulfdes (soft species such as Ag + , Hg 2+ , Pt 2+ ) and oxides (hard species Fe 2+/3+ , Cu 2+ ). Activities (1) Determine the pK a value of an acid with a K a = 1.2 x 10 2 . (2) Give the formula for the acidic and basic species in CH 3 COOH. (3) Organic amines behave as weak bases in aqueous solutions when an H + ion reacts with the lone pair of electrons on the N atom. a. Predict the relative strength of an organic amide (RCONH 2 ) behaving as a weak base. b. Give the formula of an appropriate solvent in which an amide would behave this way. (4) Give an approximate pK for the hydroacid stibine (SbH 3 ). (5) The oxoacid pyrosulfuric acid (H 2 S 2 O 7 ) is produced when SO 3 is bubbled into H 2 SO 4 . Write the formula in the (HO) m EO n style, determine the number of free O atoms per S atom and predict its strength relative to that of H 2 SO 4 . (6) Use the charge density to a. Account for the fact that the reaction of alkali metal ions and H 2 O goes only to M(H 2 O) 6 + . b. Identify the alkali metal ion that would be most likely to go beyond this stage. Acid–Base Chemistry Beyond AP SPECIAL FOCuS: Acids and Bases 104 (7) Copper can form the ions Cu + and Cu 2+ . State and explain which of these is more likely to react with the hard base F - . Conclusion This essay has attempted to present several aspects of acid–base theory beyond those discussed in earlier selections (or in the average AP Chemistry course). The topics covered here were chosen in the hope that they would further the understanding and appreciation of the topics usually found in AP courses. Bibliography Atkins, P. W., T.L. Overton, J.P. Rourke, M.T. Weller, and F.A. Armstrong. Inorganic Chemistry, 4th ed., Oxford, UK: Oxford University Press, 2006, 117. Bartmess, J. E., J. A. Scott, and R. T. McIver Jr. “Scale Of Acidities in the Gas Phase from Methanol to Phenol.” Journal of American Chemistry Society 101 (1979): 6046–56. Bartmess, J. E., and Robert J. Hinde. “The Gas-Phase Acidities of the Elemental Hydrides are Functions of Electronegativity and Bond Length.” Canadian Journal of Chemistry 83(11) (2005): 2005–12. Bowser, J. R. Inorganic Chemistry. Pacifc Grove, CA: Brooks/Cole Publishing, 1993, 312–14. Masterton, W. L., and C. N. Hurley. Chemistry: Principles and Reactions, 5th ed. Pacifc Grove, CA: Brooks/Cole Publishing, 2006, 427. Wulfsberg, G. Principles of Descriptive Inorganic Chemistry. Mill Valley, CA: University Science Books, 1991, 25. Acid–Base Lesson Plans Heather Hattori John H. Guyer High School Denton, Texas This is one of my favorite units of the year. It brings back familiar concepts and calculations to second-year students, as well as a wealth of laboratory activities to keep them engaged. I teach on an A/B day 90-minute block schedule so some activities are done during class while others are done outside (either after school or on Saturdays). Tenth-, eleventh-, and twelfth-grade students enroll in AP Chemistry with an Algebra II minimum math co-requisite, since we teach the frst year of chemistry to freshmen. The text provided to students is Zumdahl’s Chemistry, 5th edition, from the Houghton Miffin Company. You may fnd my problem selections in the study guide odd, but that would be dependent on when I teach acids and bases. This past year, we as a district moved equilibrium into the frst semester with the acid–base unit following directly after it. This may be disconcerting to some who would prefer to follow the textbook in the order it is presented. It works for us, but may not work for you. Be that as it may, here is a basic outline of what I would cover in my acid–base unit in years where I did not teach the pre-AP class and was unsure of the students’ background. General Outline Day 1: • Review characteristics and examples of acids and bases • Review acid/base theories (Arrhenius, Brønsted-Lowry, and Lewis) and associated concepts • Review strong/weak acids and bases as well as additional classifcations of acids (monoprotic, diprotic, polyprotic, ternary, oxy acids, etc.) SPECIAL FOCuS: Acids and Bases 106 • Review concentration terms (concentrated, dilute) and calculations (molarity) • Review acid–base reactions including acid–base formation from acid–base anhydrides • Review pH scale, logarithms, and associated strong acid–base calculations ([H 3 O + ], [OH - ], pH, and pOH) • Pre-lab: Determining the pH of a Substance (fts AP suggested lab topic: Determination of Appropriate Indicators for Various Acid–Base Titrations; pH Determination) • Assign Study Guides for Chapters 14 and 15. Day 2: • Determining the pH of a substance lab • Discussion of titration curves • Pre-lab: Titration Curves of Weak and Strong Acids and Bases Day 3: • 7ittotiot Cutvcs ot !co| oto Sttotj Acios oto Boscs Lab • Review titration problems • Pre-lab: % Acetic Acid in Vinegar (fts AP suggested lab topic: Determination of Concentration by Acid–Base Titration, Including a Weak Acid or Weak Base) Day 4: (this lab has been done outside the school day) • ° Acctic Acio it \itcjot Lab Day 5: • Weak acid–base equilibria—revisiting K • Calculating the pH of a weak acid or base • Use past AP free-response questions for practice Day 6: • Common-ion Effect • Buffers Acid–Base Lesson Plans 107 • Pre-lab: Preparation of a Buffer Solution at a Given pH or A Resistance to Change (fts AP suggested lab topic: Preparation and Properties of Buffer Solutions) Day 7: (this lab has been done outside the school day) • Ptcµototiot ot o Buttct So|utiot ot o Givct µH Lab Day 8: (If time permits. If not, then skip to Day 9) • Complex Ion Equilibria Day 9: • Unit Assessment Additional Attachments: • Study Guides to Accompany Zumdahl’s Chemistry, 5th edition. Hattori, Heather R. 2001. • Acids and Bases Introduction-Review Notes • Titration Notes • Calculating the pH of a Weak Acid or Base Notes • Common Ion Effect and Buffers Notes • Dctcttititj tlc µH ot o Su|stotcc. Hattori, Heather R. 1995. • AcioBoscs, Pcoctiots, oto 7ittotiot Co|cu|otiot Pcvicu !ot|slcct. Hattori, Heather R. 2007. • Dissociotiot oto µH Ptocticc Hattori, Heather R. 2005. Bibliography Holmquist Dan D., Jack Randall, and Don Volz. In Chemistry with Calculators, Beaverton, OR: Vernier Software & Technology, 2000. Hostage, David, and Martin Fossett. “Finding the Mass Percent of Acetic Acid in Vinegar,” in Laboratory Investigations: AP Chemistry, Saddle Brook, NJ: The Peoples Publishing Group, Inc., 2006. SPECIAL FOCuS: Acids and Bases 108 Hostage, David, and Martin Fossett. Laboratory Investigations: AP Chemistry, Saddle Brook, NJ: The Peoples Publishing Group, 2006. Moore, John T., and Richard Langley. 5 Steps to a 5 AP Chemistry, McGraw-Hill, 2004. Pasco Scientifc. “A Resistance to Change,” in PasPort Explorations in Chemistry. Roseville, CA: PASCO Scientifc, 2003. Zumdahl, Steven S., and Zumdahl. Chemistry, 5th ed. Houghton Miffin Company, 2000. Appendix A Review Notes on Acids and Bases Introduction Common Acids and Bases Where do I fnd acids? • Fruits (lemons, oranges, etc.) contain citric acid. • Vinegar contains acetic acid. • Rain water contains sulfuric acid. • Soft drinks contain carbonic and phosphoric acids. Where do I fnd bases? • Cleaners contain ammonia. • Milk of magnesia contains magnesium hydroxide. • Drano contains sodium hydroxide. Characteristics of Acids and Bases • Acids • Taste sour • Turn blue litmus paper red • Have a pH < 7.0 • Are corrosive • Bases • Taste bitter • Turn red litmus paper blue • Have a pH > 7.0 • Are caustic SPECIAL FOCuS: Acids and Bases 110 Three Acid–Base Theories The following theories are where we get our defnitions of what acids and bases are: • Arrhenius Concept • Brønsted-Lowry Model • Lewis Theory Arrhenius Concept Acids produce H + (hydrogen ions) in solution. HCl(aq) → H 1+ (aq) + Cl 1- (aq) H 3 PO 4 (aq) → 3 H 1+ (aq) + PO 4 3- (aq) Bases produce OH - (hydroxide ions) in solution. NaOH (aq) → Na 1+ (aq) + OH 1- (aq) Ca(OH) 2 (aq) → Ca 2+ (aq) + 2OH 1- (aq) Brønsted-Lowry Model Acids are proton donors. In other words, acids donate H + (hydrogen ions) in solution. HCl(aq) + H 2 O (l) → H 3 O 1+ (aq) + Cl 1- (aq) H 3 PO 4 (aq) + 3H 2 O (l) → 3 H 3 O 1+ (aq) + PO 4 3- (aq) Bases are proton acceptors. In other words, bases accept H + (hydrogen ions) in solution. NH 3 (aq) + H 2 O (l) → NH 4 1+ (aq) + OH 1- (aq) CaO (s) + H 2 O (l) → Ca(OH) 2 (aq) The frst equation sets up two interesting relationships. Not only does the ammonia accept a hydrogen ion to become the ammonium ion, the hydroxide ion could accept a hydrogen ion from the ammonium and become water. These two relationships make the ammonia/ammonium ion and the water/hydroxide ion conjugate acid/base pairs. Lewis Theory Acids are electron-pair acceptors in solution. Bases are electron-pair donors in solution. BF 3 (aq) + :NH 3 (aq) → F 3 B:NH 3 Appendix A 111 The pH Concept pH represents the relative acidity of a solution based on hydrogen ion or hydronium ion concentration and can be found using either of the following equations: pH = -log[H + ] or pH = -log[H 3 O + ] . Since concentrations can be large or small, such as 12 M or 4.2 X 10 -13 M, sometimes it is hard to compare the [H 3 O 1+ ]. The pH scale was developed to make it easier to recognize if a substance was acidic or basic. The amphoteric nature of pure water contributes to its neutral nature. This means that water can act as an acid or a base due to the concept known as auto- ionization shown in the following two examples. Remember that: H 2 O (l) ↔ H +1 (aq) + OH -1 (aq) (Arrhenius concept) According to the Arrhenius concept water (in the equation above) is acting as an acid by producing hydrogen ions as well as a base by producing hydroxide ions. 2H 2 O (l) ↔ H 3 O +1 (aq) + OH -1 (aq) (Brønsted-Lowry model) According to the Brønsted-Lowry model one water molecule (in the equation above) is acting as an acid by donating a hydrogen ion to the other molecule of water forming a hydronium ion. The water molecule that is accepting the hydrogen ion is acting as the base. If we write the equilibrium expression for this we get: K w = [H 3 O 1+ ][OH 1- ] = 1.0 X 10 -14 always at 25°C We can solve for the [H 3 O 1+ ] or [OH 1- ] based on the above equation. For every H 3 O 1+ produced, one OH 1- is made, so: [H 3 O 1+ ][OH 1- ] = 1.0 X 10 -14 becomes [x][x] = 1.0 X 10 -14 x 2 = 1.0 X 10 -14 x = 1.0 X 10 -7 ∴ [H 3 O 1+ ] = 1.0 X 10 -7 M = [OH 1- ] In solutions when the: [H 1+ ] > [OH 1- ], then the solution is acidic [H 1+ ] < [OH 1- ], then the solution is basic [H 1+ ] = [OH 1- ], then the solution is neutral SPECIAL FOCuS: Acids and Bases 112 Based on this and the fact that [H 1+ ] and [OH 1- ] can be calculated if you know one or the other, it’s time to solve some problems. Examples 1. Given the [OH 1- ] = 1.0 X 10 -5 M, solve for the [H 3 O 1+ ] and determine the solution’s nature (acidic, basic, or neutral). 2. Given the [H 3 O 1+ ] = 1.0 X 10 -2 M, solve for the [OH 1- ] and determine the solution’s nature (acidic, basic, or neutral). Substances other than water can also be amphoteric. Let’s look at a hydrated aluminum hydroxide ion as an example. [Al(H 2 O) 4 (OH) 2 ] 1+ + NaOH ↔ Al(H 2 O) 3 (OH) 3 + OH 1- When a water of hydration from [Al(H 2 O) 4 (OH) 2 ] 1+ donates the hydrogen ion to becomes the third hydroxide ion, it is acting as a Brønsted-Lowry acid. [Al(H 2 O) 4 (OH) 2 ] 1+ + HCl ↔ [Al(H 2 O) 5 OH] 2+ + Cl 1- When a hydroxide from [Al(H 2 O) 4 (OH) 2 ] 1+ accepts the hydrogen ion to create the ffth water of hydration, it is acting as a Brønsted-Lowry base. pH scale Based on logarithms, the pH scale was developed. 0 7 14 Acidic Neutral Basic pH = -log[H + ] or pH = -log [H 3 O + ] 3. What is the pH of a solution with a [H + ] = 1.0 X 10 -9 M? A similar formula can be used to calculate the pOH (relative basicity) if you know the [OH 1- ]. pOH = -log [OH 1- ] 4. What is the pOH of a solution with a [OH 1- ] = 2.4 X 10 -7 M? By combining all three of the previous formulas: K w = [H 3 O 1+ ][OH 1- ] = 1.0 X 10 -14 pH = -log [H 3 O 1+ ] Appendix A 113 pOH = -log [OH 1- ], we can derive the formula: pH + pOH = 14. 5. What is the pOH of a solution that has a pH of 10.0? Vinegar has a pH of 2.8. What are the pOH, [H 3 O 1+ ], and [OH 1- ] of a vinegar solution? Relative Strengths of Acids and Bases Acids and bases can be concentrated or dilute, weak or strong. Do not confuse the frst two, which describe the solution concentration, with the last two, which describe the amount of dissociation. A strong acid or base is one that completely dissociates in water or has 100% ionization. A weak acid or base is one in which only a small portion of the particles dissociate or one that only partially ionizes. • Strong acids include: HCl, HBr, HI, HNO 3 , H 2 SO 4 , and HClO 4 . All others are considered weak. • Strong bases include: LiOH, NaOH, KOH, Mg(OH) 2 , Ca(OH) 2 , Sr(OH) 2, and Ba(OH) 2 . All others are considered weak. Appendix B Review Notes on Acids and Bases— Teacher’s Edition Common Acids and Bases Where do I fnd acids? • Fruits (lemons, oranges, etc.) contain citric acid. • Vinegar contains acetic acid. • Rain water contains sulfuric acid. • Soft drinks contain carbonic and phosphoric acids. Where do I fnd bases? • Cleaners contain ammonia. • Milk of magnesia contains magnesium hydroxide. • Drano contains sodium hydroxide. Characteristics of Acids and Bases Acids: • taste sour • turn blue litmus paper red • have a pH < 7.0 • are corrosive Bases: • taste bitter • turn red litmus paper blue • have a pH > 7.0 • are caustic SPECIAL FOCuS: Acids and Bases 116 Three Acid–Base Theories The following theories are where we get our defnitions of what acids and bases are: • Arrhenius Concept • Brønsted-Lowry Model • Lewis Theory Arrhenius Concept Acids produce H + (hydrogen ions) in solution. HCl(aq) → H + (aq) + Cl - (aq) H 3 PO 4 (aq) → H + (aq) + H 2 PO 4 - (aq) Bases produce OH - (hydroxide ions) in solution. NaOH (aq) → Na + (aq) + OH - (aq) Ca(OH) 2 (aq) → Ca 2+ (aq) + 2OH - (aq) Brønsted-Lowry Model Acids are proton donors. In other words, acids donate H + (hydrogen ions) in solution, usually to water. These are hydrated to form H(H 2 O) n + , where n may be 1–6. For convenience, this species is usually abbreviated to H+ except when a balanced equation is desired, as shown in some of the examples below. HCl(aq) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) H 3 PO 4 (aq) + H 2 O (l) → H 3 O + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) + H 2 O (l) → H 3 O + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) + H 2 O (l) → H 3 O + (aq) + PO 4 3- (aq) Bases are proton acceptors. In other words, bases accept H + (hydrogen ions) in solution. NH 3 (g) + H 2 O (l) → NH 4 + (aq) + OH - (aq) CaO (s) + H 2 O (l) → Ca(OH) 2 (aq) The frst equation sets up two interesting relationships. Not only does the ammonia accept a hydrogen ion to become the ammonium ion, the hydroxide ion could accept a hydrogen ion from the ammonium ion and become water. These two relationships make the ammonia/ammonium ion and the water/hydroxide ion conjugate acid–base pairs. The reaction goes both ways and so reaches equilibrium. Appendix B 117 Lewis Theory Acids are electron pair acceptors in solution.Bases are electron pair donors in solution. BF 3 (aq) + :NH 3 (aq) → F 3 B:NH 3 The pH Concept Conventionally, the symbol [H + ] is used to represent the concentration of H + ions in moles per liter. Since concentrations can be large or small, such as 12 M or 4.2 X 10 -13 M, sometimes it is hard to compare the [H + ]. The pH scale was developed to make it easier to recognize if a substance was acidic or basic. pH represents the relative acidity of a solution based on hydrogen ion or hydronium ion concentration and can be found using either of the following equations: pH = -log[H + ] or pH + -log[H + ] . The amphoteric nature of pure water contributes to its neutral nature. This means that water can act as an acid or a base by accepting or donating protons. Water also undergoes a process called auto-ionization, which is shown in the following two examples. Remember that: H 2 O (l) ↔ H + (aq) + OH - (aq) (Arrhenius concept) According to the Arrhenius concept, water, in the equation above, is acting as an acid by producing hydrogen ions as well as a base by producing hydroxide ions. H 2 O (l) + H 2 O (l) ↔ H 3 O + (aq) + OH - (aq) (Brønsted-Lowry model) According to the Brønsted-Lowry model, one water molecule in the equation above is acting as an acid by donating a hydrogen ion to the other molecule of water forming a hydronium ion. The water molecule that is accepting the hydrogen ion is acting as the base. If we write the equilibrium expression for this we get: K w = [H 3 O + ][OH - ] = 1.0 X 10 -14 at 25°C We can solve for the [H 3 O + ] or [OH - ] based on the above equation. In pure water, with no other substances present, for every H 3 O + produced, one OH - is made so [H 3 O + ] = [OH - ] = x [H 3 O 1+ ][OH 1- ] = 1.0 X 10 -14 becomes [x][x] = 1.0 X 10 -14 SPECIAL FOCuS: Acids and Bases 118 x 2 = 1.0 X 10 -14 x = 1.0 X 10 -7 ∴ [H 3 O + ] = 1.0 X 10 -7 M = [OH - ] In solutions when the: [H + ] > [OH - ], then the solution is acidic. [H + ] < [OH - ], then the solution is basic. [H + ] = [OH - ], then the solution is neutral. Based on this and the fact that [H + ] and [OH - ] can be calculated using K w if you know one or the other, it’s time to solve some problems. Examples 1. Given the [OH - ] = 1.0 X 10 -5 M, solve for the [H 3 O + ] and determine the solution’s nature (acidic, basic, or neutral). K w = [H 3 O + ][OH - ] = 1.0 X 10 -14 [H 3 O + ] [1.0X 10 -5 ] = 1.0 X 10 -14 [H 3 O + ] = 1.0 X 10 -14 M 2 = 1.0 X 10 -9 M which is less than the [OH - ] so the 1.0X 10 -5 M solution is basic. 2. Given the [H 3 O + ] = 1.0 X 10 -2 M, solve for the [OH - ] and determine the solution’s nature (acidic, basic, or neutral). K w = [H 3 O + ][OH - ] = 1.0 X 10 -14 [1.0X 10 -2 M][OH - ] = 1.0 X 10 -14 [OH - ] = 1.0 X 10 -14 M 2 = 1.0 X 10 -12 M which is less than the [H 3 O + ] so 1.0X 10 -2 M the solution is basic. Substances other than water can also be amphoteric. Let’s look at a hydrated aluminum hydroxide ion as an example. [Al(H 2 O) 4 (OH) 2 ] + + NaOH ↔ Al(H 2 O) 3 (OH) 3 + OH - When a water of hydration from [Al(H 2 O) 4 (OH) 2 ] + donates the hydrogen ion to becomes the third hydroxide ion it is acting as a Brønsted-Lowry acid. [Al(H 2 O) 4 (OH) 2 ] + + HCl ↔ [Al(H 2 O) 5 OH] 2+ + Cl - When a hydroxide from [Al(H 2 O) 4 (OH) 2 ] + accepts the hydrogen ion to create the ffth water of hydration, it is acting as a Brønsted-Lowry base. Appendix B 119 pH scale Based on logarithms, the pH scale was developed. 0 7 14 Acidic Neutral Basic pH = -log[H + ] or pH = -log [H 3 O + ] 3. What is the pH of a solution with a [H + ] = 1.0 X 10 -9 M? pH = -log[H + ] pH = -log(1.0X10 -9 M) pH = 9.00 A similar formula can be used to calculate the pOH (relative basicity) if you know the [OH - ]. pOH = -log [OH - ] What is the pOH of a solution with a [OH - ] = 2.4 X 10 -7 M? pOH = -log [OH - ] pOH = -log(2.4 X 10 -7 M) pOH = 6.62 By combining all three of the previous formulas: K w = [H + ][OH - ] = 1.0 X 10 -14 pH = -log [H + ] pOH = -log [OH - ], we can derive the formula: pH + pOH = 14 4. What is the pOH of a solution that has a pH of 10.0? pH + pOH = 14 10.0 + pOH = 14 pOH = 4.0 Vinegar has a pH = 2.8. What are the pOH, [H + ], and [OH - ] of a vinegar solution? pH + pOH = 14 2.8 + pOH = 14 pOH = 11.2 pH = -log[H + ] 2.8 = -log[H + ] [H + ] = 2 X 10 -3 M pOH = -log[OH - ] 11.2 = -log[OH - ] [H + ] = 6 X 10 -12 M SPECIAL FOCuS: Acids and Bases 120 Relative Strengths of Acids and Bases Acids and bases can be concentrated or dilute, weak or strong. Do not confuse the frst two, which describe the solution concentration, with the last two, which describe the amount of dissociation. A strong acid or base is one that completely dissociates in water or has 100% ionization. A weak acid or base is one in which only a small portion of the particles dissociate, or one that only partially ionizes. Strong acids include: HCl, HBr, HI, HNO 3 , H 2 SO 4 , H 3 PO 4 , and HClO 4 . All others are considered weak. Strong bases include: LiOH, NaOH, KOH, Mg(OH) 2 , Ca(OH) 2 , Sr(OH) 2, and Ba(OH) 2 . All others are considered weak. Appendix C Titration Notes A titration is the experimental procedure used to determine the concentration of an unknown acid or base. The necessary chemical components of a titration include a standard solution, and solution of unknown concentration, and an indicator. Titration curves are a way to visually represent the change in pH during a titration. Titrations and titration curves tell you many things about a reaction. During a titration, the indicator experiences a color change at the endpoint of the titration. If the best indicator is chosen, the endpoint and the equivalence point (when there are equal amounts of acid and base) occur at nearly the same time. Common indicators used in acid–base titrations include phenolphthalein (clear to pink pH = 8-10) and bromothymol blue (yellow to blue pH = 6-8). The equivalence point can be visually determined from the titration curve. It is the area in which the slope of the line approaches infnity (or is nearly vertical). Acid–Base Reactions When an acid reacts with a base a neutralization reaction occurs. The same thing happens when a base reacts with an acid. An acid–base reaction is a double displacement reaction. AB + CD → AD + CB The products of an acid–base reaction are a salt (metal/nonmetal compound) and water. Examples HCl + NaOH → NaCl + HOH H 2 SO 4 + KOH → SPECIAL FOCuS: Acids and Bases 122 H 3 PO 4 + LiOH → Hydrobromic acid + calcium hydroxide HBr + Ca(OH) 2 → CaBr 2 + 2H 2 O Hydrofuoric acid + aluminum hydroxide HF + Al(OH) 3 → AlF 3 + 3H 2 O Titration Problems Often, the goal of a titration is to calculate the concentration of one of the reactants. This can be done using a balanced equation and stoichiometry. 1. A 25.0 mL sample of HCl was titrated to the endpoint with 15.0 mL of 2.0 M NaOH. What was the molarity of the acid? 2. A 10.0 mL sample of H 2 SO 4 was neutralized by 13.5 mL of 1.0 M KOH. What was the molarity of the acid? 3. What volume of 1.50 M NaOH is necessary to neutralize 20.0 mL of 2.5M H 3 PO 4 ? 4. How much 0.5M HNO 3 is necessary to titrate 25.0 mL of 0.05M Ca(OH) 2 solution to the endpoint? Appendix D Titration Notes – Teacher’s Edition A titration is the experimental procedure used to determine the concentration of an unknown acid or base. The necessary chemical components of a titration include a standard solution, a solution of unknown concentration, and may include a colored indicator. Titration curves are a way to visually represent the change in pH during a titration. Titrations and titration curves tell you many things about a reaction. During a titration, the indicator experiences a color change at the endpoint of the titration. If the best indicator is chosen, the endpoint and the equivalence point (when equivalent amounts of acid and base have been added) occur at nearly the same time. Common indicators used in acid–base titrations include phenolphthalein (clear to pink pH = 8–10) and bromothymol blue (yellow to blue pH = 6–8). The equivalence point can be visually determined from the titration curve. It is the area where the slope of the line approaches infnity (or is nearly vertical). Titration curves and the progress of titrations can also be followed by using a pH meter or probe. Acid–Base Reactions When an acid reacts with a base, a neutralization reaction occurs. The same thing happens when a base reacts with an acid. An acid–base reaction is a double displacement reaction. AB + CD → AD + CB The overall products of an acid–base reaction are a salt (metal/nonmetal compound) and water. SPECIAL FOCuS: Acids and Bases 124 Examples HCl + NaOH → NaCl + HOH (which also can be written H 2 O) H 2 SO 4 + 2KOH → K 2 SO 4 + 2H 2 O H 3 PO 4 + 3LiOH → Li 3 PO 4 + 3H 2 O Hydrobromic acid + calcium hydroxide HBr + Ca(OH) 2 → CaBr 2 + 2H 2 O Hydrofuoric acid + aluminum hydroxide HF + Al(OH) 3 → AlF 3 + 3H 2 O Titration Problems Often, the goal of a titration is to calculate the concentration of one of the reactants. This can be done using a balanced equation and stoichiometry. 1. A 25.0 mL sample of HCl was titrated to the equivalence point with 15.0 mL of 2.0 M NaOH. What was the molarity of the acid? HCl + NaOH → NaCl + H 2 O 15.0 mL NaOH × 1 L NaOH × 2.0 mol NaOH × 1 mol HCl × 1 × 1000 mL = 1.2 M HCl 1000mL NaOH 1 L NaOH 1 mol NaOH 25.0 mL HCl 1 L HCl 2. A 10.0 mL sample of H 2 SO 4 was completely reacted by 13.5 mL of 1.0 M KOH. What was the molarity of the acid? H 2 SO 4 + 2KOH → K 2 SO 4 + 2H 2 O 13.5 mL KOH × 1 L KOH × 1.0 mol KOH × 1 mol H 2 SO 4 × 1000 mL H 2 SO 4 × 1 = 0.68 M H 2 SO 4 1000mL 1L 2 mol KOH 1 L H 2 SO 4 10.0 mL H 2 SO 4 3. What volume of 1.50 M NaOH is necessary to completely react with 20.0 mL of 2.5M H 3 PO 4 ? Note: M is molarity, which can be mol/L or millimoles/mL. H 3 PO 4 + 3NaOH → Na 3 PO 4 + 3H 2 O 20.0 mL H 3 PO 4 × 2.5 mmol H 3 PO 4 × 3 mmol NaOH × 1mL NaOH = 1.00 × 10 2 mL NaOH 1.0 mL H 3 PO 4 1 mmol H 3 PO 4 1.50mmol NaOH Appendix D 125 4. How much 0.5M HNO 3 is necessary to titrate 25.0 mL of 0.05M Ca(OH) 2 solution to the equivalence point? 2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O 25.0 mL Ca(OH) 2 × 0.05 mmol Ca(OH) 2 × 2 mmol HNO 3 × 1mL HNO 3 = 5.00 mL HNO 3 1 mL Ca(OH) 2 1 mmol Ca(OH) 2 0.5 mmol HNO 3 Appendix E Calculating the pH of a Weak Acid or Base When working with weak acids, weak bases, or buffer systems, one must take into account the fact that many species exist in the reaction vessel. Take, for example, the case of calculating the pH of a 0.100 M acetic acid solution whose K a = 1.8 X 10 -5 . The reaction: HC 2 H 3 O 2 (aq) + H 2 O (l) ↔ H 3 O 1+ (aq) + C 2 H 3 O 2 1- (aq). Acetic acid is a weak acid and thus, only a small percentage of it dissociates. This becomes an equilibrium problem instead of a simple stoichiometry problem (like it would if it were a strong acid). We begin by writing the equilibrium expression: Ka Ka [H O ][ C H O ] [ HC H O ] 3 1 2 3 2 1- 2 3 2 = + Next we need to look at the conditions during the reaction: HC 2 H 3 O 2 H 3 O 1+ C 2 H 3 O 2 1- Initial 0.100M 0 0 Change -x +x +x Equilibrium 0.100M - x x x Substituting into the equilibrium expression we get: Ka = [x][x] = 1.8 X 10 -5 [ 0.100M -x ] Assuming the effect of x on 0.100M is negligible because of the large difference (power difference > 10 -3 units) in concentration and K a , solving for x we get: x 2 x 2 = = = − − + 1 8 10 1 34 10 6 3 3 1 . . [ ] X x X H O SPECIAL FOCuS: Acids and Bases 128 Now we can solve for pH using the equation: pH = -log[H 3 O 1+ ] and fnd the pH to be 2.87. Example 1 Calculate the pH of a 2.00M ammonia solution with a K a = 5.71 X 10 -10 @ 25°C. A. -0.30 B. 2.22 C. 8.94 D. 11.78 Solution: The correct answer is D. The reaction is NH3(aq) + H2O(l) ↔ NH41+(aq) + OH1-(aq) Since K w = K a K b = 1.0 X 10 -14 so K b = K K w a substituting in values we get: K b = 1 0 10 5 71 10 1 8 10 14 10 5 . . . X X X M − − − = . NH 3 NH 4 1+ OH 1- Initial 2.00 M 0 0 Change - x +x +x Equilibrium 2.00 - x x x Substituting into the equilibrium expression we get: K b = [x][x] = 1.8 X 10 -5 [ 2.00M -x ] Assuming the effect of x on 2.00M is negligible because of the large difference (power difference > 10 -3 units) in concentration and K b , solving for x we get: x 2 x 2 = = = − − − 3 6 10 6 0 10 6 3 1 . . [ ] X x X OH Now we can solve for pOH using the equation: pOH = -log[OH 1- ] and fnd the pOH to be 2.22. Since pH + pOH = 14, pH = 14 – 2.22, and the pH = 11.78. If you chose answer A, you only used the initial concentration of ammonia and substituted it into the pH = -log[H 3 O 1+ ] assuming that the concentration of ammonia was equal to the hydronium ion concentration. If you chose answer B, you selected the value that was equal to the pOH. If you chose answer C, you used 5.71 X 10 -10 for your K b value instead of 1.8 X 10 -5 . Appendix F Applications of Aqueous Equilibria Notes Different Constants (to hopefully clear confusion) K b molal boiling point constant ∆t b = K b m K f molal freezing point constant ∆t f = K f m k rate constant Rate = k [reactants] K w ionization constant for water K w = K a K b = 1.0 X 10 -14 @ 25°C K a acid dissociation constant K a = [H + ][A - ]/[HA] K b base dissociation constant K b = [X + ][OH - ]/[XOH] K c equilibrium constant for concentrations K p equilibrium constant based on pressures K sp equilibrium constant for solubility product K d equilibrium constant for decomposition reactions K f equilibrium constant for formation reactions The Effect of Common Ions When a salt is added to an acid or base containing similar ions, the equilibrium shifts to relieve the stress this is an application of . If NaF is added to HF: If NH 4 Cl is added to NH 3 : HF ↔ H + + F - NH 3 + H 2 O ↔ NH 4 + + OH - NaF ↔ Na + + F - NH 4 Cl ↔ NH 4 + + Cl - More fuoride ions are added to the solution. More ammonium ions are added to the solution. The equilibrium shifts to the left. The equilibrium shifts to the left. This shifting process is called the common ion effect. This complicates the equilibrium by providing an additional source for ions. SPECIAL FOCuS: Acids and Bases 130 Example 1 Calculate the pH of a mixture containing 0.100 M HC 3 H 5 O 2 and 0.100 M NaC 3 H 5 O 2 . K a = 1.3 x 10 -5 Buffers Combinations of weak acids or weak bases with their salts form buffers. Buffers are solutions that resist changes in pH when hydrogen or hydroxide ions are added. When working with buffers, be sure to look at the stoichiometry before setting up the equilibrium. Example 2 What is the difference in pH of a solution containing 2.00 mol of ammonia and 3.00 mol of ammonium chloride in 1.00 L of solution versus the pH of the ammonia alone? K b = 1.81 X 10 -5 The Henderson-Hasselbalch equation can also be used to solve for pH. pH = pK a + log ([A - ]/[HA]) pOH = pK b + log ([X + ]/[WB]) Solution: Titrations are laboratory procedures used to determine the concentration of acids or bases. Along the way, the pH of the mixture can be calculated. Problems like this have appeared in the past on the AP Exam. A 100.0 mL sample of 0.150 M HNO 2 (pK a = 3.35) was titrated with 0.300 M NaOH. Determine the pH after the following quantities of base have been added to the acid solution and draw the resulting titration curve. a. 0.00 mL b. 25.00 mL c. 49.50 mL d. 50.00 mL e. 55.00 mL f. 75.00 mL Indicators and Titrations Common indicators used in acid base titrations are: phenolphthalein (clear to pink pH =8-10) bromothymol blue (yellow to blue pH = 6-8) universal indicator (red to purple pH = 1-13) Appendix F 131 To prepare a buffer with a specifc pH (part of pre-lab discussion): pH = pKa + pH pK A HA a = + − log [ ] [ ] 6.0 = pKa + 6 pK A HA a . log [ ] [ ] 0 = + − Acids to choose from: Their K a values H 2 SO 3 1.5 X 10 -2 1.0 X 10 -7 HOAc 1.8 X 10 -5 H 3 PO 4 7.3 X 10 -3 6.2 X 10 -8 4.8 X 10 -13 H 2 CO 3 4.3 X 10 -7 5.6 X 10 -11 HSO 4 1- 1.2 X 10 -2 H 2 SO 4 Large 1.2 X 10 -2 So if I am not mistaken, for a buffer with pH = 6, you would choose HOAc and NaOAc since the pK a of HOAc is closest to 6. 6 p(1.8 X 10 ) A HA -5 . log [ ] [ ] 0 = + − 6 0 4 74 . . log = +     [ ] − A HA 1.26 A HA = − log [ ] [ ] 10 1 26 − − = . [ ] A [HA] 0 055 . [ ] = − A [HA] If [HA] = 0.1 M, then [A - ] = 0.0055M. Solubility Equilibria When solids dissolve to form aqueous solutions, the equilibrium involved is the solubility equilibria. When a salt is added to water at frst, none is dissolved, then as the water molecules collide with the salt and interact with the intermolecular forces, the formula units of the salt begin to dissolve. Eventually the maximum amount of SPECIAL FOCuS: Acids and Bases 132 solid that can dissolve at that particular temperature does dissolve and equilibrium is reached. BaCO 3 (s) ↔ Ba 2+ (aq) + CO 3 2- (aq) K sp = [Ba 2+ ] [CO 3 2- ] = 1.6 X 10 -9 Why is the BaCO 3 not included in the K sp expression? Notice that the K value is very small. We would expect this since barium carbonate is not considered soluble in water based on what we have learned from solubility rules. Do realize, however, that solubility and the solubility product are not the same. Solubility is a position that changes with temperature and the concentration of common ions present in the solution. The solubility product is a constant value for a particular temperature (usually room temperature when you look it up in a chart). K sp problems are much like K a and K b problems. Calculating K sp from solubility Calculating solubility from K sp PbSO 4 has a measured solubility of 1.14 X 10 -4 mol/L Al(OH) 3 has a K sp of 2 X 10 -32 at 25°C. At 25°C,what is the K sp of PbSO 4 ? What is its solubility of PbSO 4 at 25°C? Relative Solubilities If salts produce the same number of ions when they dissolve, then their K sp values can be used to determine their solubilities relative to each other. The smaller the K sp , the less soluble it is. If the salts produce different numbers of ions when they dissolve, then you have to calculate the solubilities from the K sp values, as in the previous example. The Common Ion Effect on Solubility A common ion affects the solubility of a solid, generally reducing the amount that dissolves by providing an initial concentration of one of the ions. RICE diagrams are very useful in demonstrating this affect. Example 3 The K sp of silver sulfate is 1.2 X 10 -5 . Calculate the solubility of silver sulfate in a 0.20 M solution of potassium sulfate. Appendix G Applications of Aqueous Equilibria Notes – Teacher’s Edition Different Constants (to hopefully clear confusion) K b molal boiling point constant ∆t b = K b m K f molal freezing point constant ∆t f = K f m k rate constant Rate = k [reactants] K w ionization constant for water K w = K a K b = 1.0 X 10 -14 @ 25°C K a acid dissociation constant K a = [H + ][A - ]/[HA] K b base dissociation constant K b = [X + ][OH - ]/[XOH] K c equilibrium constant for concentrations K p equilibrium constant based on pressures K sp equilibrium constant for solubility product K d equilibrium constant for decomposition reactions K f equilibrium constant for formation reactions The Effect of Common Ions When a salt is added to an acid or base containing similar ions, the equilibrium shifts to relieve the stress this is an application of LeChatlier’s Principle If NaF is added to HF: If NH 4 Cl is added to NH 3 : HF ↔ H + + F - NH 3 + H 2 O ↔ NH 4 + + OH - NaF ↔ Na + + F - NH 4 Cl ↔ NH 4 + + Cl - More fuoride ions are added to the solution. More ammonium ions are added to the solution. The equilibrium shifts to the left. The equilibrium shifts to the left. This shifting process is called the common ion effect. This complicates the equilibrium by providing an additional source for ions. SPECIAL FOCuS: Acids and Bases 134 Example 1 Calculate the pH of a mixture containing 0.100 M HC 3 H 5 O 2 and 0.100 M NaC 3 H 5 O 2 . K a = 1.3 x 10 -5 HC 3 H 5 O 2 (aq)  H + (aq) + C 3 H 5 O 2 - (aq) HC 3 H 5 O 2 H 1+ C 3 H 5 O 2 1- Initial 0.100M 0 0 Change -x +x x + 0.100 [The 0.100 comes from the NaC 3 H 5 O 2 .] Equilibrium 0.100M - x x x + 0.100 K a = [x][0.100 + x] =1.3 X 10 -5 [0.100M -x] If you consider x to be negligible, then this simplifes to : K a K [0.100x] [ 0.100M] a = = − 1 3 10 5 . X and x = 1.3 X 10 -5 M = [H + ] pH = -log[H + ] pH = -log(1.3X10 -5 ) pH = 4.89, which is consistent with a weak acid. Buffers Combinations of weak acids or weak bases with their salts form buffers. Buffers are solutions that resist changes in pH when hydrogen or hydroxide ions are added. When working with buffers, be sure to look at the stoichiometry before setting up the equilibrium. Example 2 What is the difference in pH of a solution containing 2.00 mol of ammonia and 3.00 mol of ammonium chloride in 1.00 L of solution versus the pH of the ammonia alone? K b = 1.81 X 10 -5 2.00 mol NH 3 only 2.00 mole NH 3 + 3.00 mol NH 4 Cl NH 3 + H 2 O(aq)  NH 4 + (aq) + OH - (aq) NH 3 + H 2 O(aq) ↔ NH 4 + (aq) + OH - (aq) NH 3 NH 4 1+ OH 1- NH 3 NH 4 1+ OH 1- Initial 2.00 M 0 0 2.00 M 0 0 Change -x +x +x -x 3.00 + x x Equilibrium 2.00M - x x x 2.00M - x x x Appendix G 135 K a = [x][x] = 1.81 X 10 -5 [2.00M -x] K a = [3.00M + x][x] = 1.81 X 10 -5 [2.00M -x] Considering x to be negligible: Considering x to be negligible: x 2 = 2.00 M × (1.81 X 10 -5 ) 3x = 2.00 M × 1.81 X 10 -5 x= 6.01 × 10 -3 M = [OH - ] x = 1.21 × 10 -5 M = [OH - ] pOH = -log[OH - ] pOH = -log[OH - ] pOH = 2.221 pOH = 4.917 pH + pOH = 14 pH + pOH = 14 pH + 2.221 = 14 pH + 4.917 = 14 pH = 11.779 pH = 9.083 The Henderson-Hasselbalch equation can also be used to solve for pH. pH = pK a + log ([A - ]/[HA]) pOH = pK b + log ([X + ]/[WB]) Titrations are laboratory procedures used to determine the concentration of acids or bases. Along the way, the pH of the mixture can be calculated. Problems like this have appeared in the past on the AP Exam. A 50.0 mL sample of 0.300 M HNO 2 (pK a = 3.35) was titrated with 0.450 M NaOH. Determine the pH after the following quantities of base have been added to the acid solution and draw the resulting titration curve: a. 0.00 mL b. 16.5 mL c. 32.0 mL d. 33.3 mL e. 35.0 mL f. 50.0 mL a. HNO 2 (aq)  H + (aq) + NO 2 - (aq) pKa = -log(K a ) 3.35 = -log(K a ) K a = 4.5 × 10 -4 K a K [H ][NO ] [HNO ] a 2 - 2 = = + − 4 5 10 4 . X =4.5 X 10 -4 SPECIAL FOCuS: Acids and Bases 136 [HNO 2 ] [H 1+ ] [NO 2 1- ] Initial 0.300 M 0 0 Change -x +x +x Equilibrium 0.300 M - x x x K a = [x][x] =4.5 × 10 [0.300M -x] x = 1.2 × 10 -2 M = [H + ] pH = -log[H + ] pH = 1.92 b. HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O 50.0 mL HNO 2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO 2 16.5 mL NaOH × 1L/1000mL × 0.450 M = 0.00743 mol NaOH 0.0150 mol – 0.007425 mol = 0.07575 mol HNO 2 left after reaction 50.0mL + 16.5mL = 66.5 mL or 0.0665 L HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O [HNO 2 ] [NaOH] [Na 1+ ] [NO 2 1- ] Initial 0.0150 mol 0.0743 mol 0 0 Change -0.0743mol -0.0743 mol + 0.0743 mol + 0.0743mol Equilibrium 0.0757 mol 0 0.0743 mol 0.0743 mol pH = pK a + log ([A - ]/[HA]) pH = 3.35 + log(0.0743 mol/0.0665 L ÷ 0.0757 mol/0.0665 L) pH = 3.34 c. HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O 50.0 mL HNO 2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO 2 32.0 mL NaOH × 1L/1000mL × 0.450 M = 0.0144 mol NaOH 0.015 mol – 0.014 mol = 0.001 mol HNO 2 left after reaction 50.0mL + 32.0mL = 82.0 mL or 0.0820 L HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O [HNO 2 ] [NaOH] [Na 1+ ] [NO 2 1- ] Initial 0.0150 mol 0.0144 mol 0 0 Change -0.0144 mol -0.0144 mol + 0.0144 mol + 0.0144 mol Equilibrium 0.0006 mol 0 0.0144 mol 0.0144 mol Appendix G 137 pH = pK a + log ([A - ]/[HA]) pH = 3.35 + log(0.0144 mol/0.0820 L ÷ 0.0006 mol/0.0820 L) pH = 4.73 d. HNO 2 (aq) + NaOH(aq) Na + (aq) + NO 2 - (aq) + H 2 O 50.0 mL HNO 2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO 2 33.3 mL NaOH × 1L/1000mL × 0.450 M = 0.0150 mol NaOH 0.0150 mol – 0.0150 mol = 0.00 mol HNO 2 left after reaction 50.0mL + 33.3mL = 83.3 mL or 0.0833 L HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O [HNO 2 ] [NaOH] [Na 1+ ] [NO 2 1- ] Initial 0.0150 mol 0.0150 mol 0 0 Change 0.0150 mol 0.0150 mol + 0.0150 mol + 0.0150 mol Equilibrium 0 0 0.0150 mol 0.0150 mol 0.0150 mol NO 2 1- = 0.180 M 0.0833 L NO 2 - is the conjugate base of a weak acid so pK a + pK b = 14.0 pK b = 14.0 – 3.35 = 10.65 NO 2 - (aq) + H 2 O (l)  OH - (aq) + HNO 2 (aq) [NO 2 - ] [H 2 O] [OH - ] [HNO 2 ] Initial 0.180M 0 0 Change -x +x +x Equilibrium 0.180M-x x x pK b = -log(K b ) K b = 2.24 × 10 -11 K b K [OH ][HNO ] [NO ] b 2 2 - = = − − 2 24 10 11 . X x 2 = (0.180M) × 2.24 × 10 -11 x = 2.01 ×10 -6 = [OH - ] pOH = -log[OH - ] pOH = 5.697 14 - pOH = pH pH = 8.303 e. HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O 50.0 mL HNO 2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO 2 35.0 mL NaOH × 1L/1000mL × 0.450 M = 0.0158 mol NaOH SPECIAL FOCuS: Acids and Bases 138 0.0158 mol – 0.0150 mol = 0.0008 mol NaOH left after reaction 50.0 mL + 35.0 mL = 85.0 mL or 0.0850 L HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O [HNO 2 ] [NaOH] [Na 1+ ] [NO 2 1- ] Initial 0.0150 mol 0.0158mol 0 0 Change 0.0150 mol 0.0150 mol + 0.0150 mol + 0.0150 mol Equilibrium 0 0.0008 mol 0.0150 mol 0.0150 mol Here the concentration of NaOH controls the pH since it is a strong base. pOH = -log[OH - ] pOH = -log(0.0008 M) pOH = 3 pH + pOH = 14 pH = 14 – pOH pH = 11 f. HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O 50.0 mL HNO 2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO 2 50.0 mL NaOH × 1L/1000mL × 0.450 M = 0.0225 mol NaOH 0.0225 mol – 0.0150 mol = 0.0075 mol NaOH left after reaction 50.0mL + 50.0mL = 100.0 mL or 0.1000 L HNO 2 (aq) + NaOH(aq)  Na + (aq) + NO 2 - (aq) + H 2 O [HNO 2 ] [NaOH] [Na 1+ ] [NO 2 1- ] Initial 0.0150 mol 0.0225 mol 0 0 Change 0.0150 mol 0.0150 mol + 0.0150 mol + 0.0150 mol Equilibrium 0 0.0075 mol 0.0150 mol 0.0150 mol Here the concentration of NaOH controls the pH since it is a strong base. pOH = -log[OH - ] pOH = -log(0.0075M) pOH = 2.125 pH + pOH = 14 pH = 14 – pOH pH = 11.875 Indicators and Titrations Common indicators used in acid base titrations are: • phenolphthalein (clear to pink pH = 8-10) • bromothymol blue (yellow to blue pH = 6-8) • universal indicator (red to purple pH = 1-13) To prepare a buffer with a specifc pH (part of pre-lab discussion) pH = pKa pH pK A HA a = + − log [ ] [ ] Appendix G 139 6.0 = pKa 6 pK A HA a . log [ ] [ ] 0 = + − Acids to choose from: Their K a values H 2 SO 3 1.5 X 10 -2 1.0 X 10 -7 HOAc 1.8 X 10 -5 H 3 PO 4 7.3 X 10 -3 6.2 X 10 -8 4.8 X 10 -13 H 2 CO 3 4.3 X 10 -7 5.6 X 10 -11 HSO 4 1- 1.2 X 10 -2 H 2 SO 4 Large 1.2 X 10 -2 So if I am not mistaken, for a buffer with a pH of 6, you would choose HOAc and NaOAc since the pK a of HOAc is closest to 6. 6 p(1.8 X 10 ) A HA -5 . log [ ] [ ] 0 = + − 6 4.74 A HA . log [ ] [ ] 0 = + − 1.26 A HA = − log [ ] [ ] 10 1 26 − − = . [ ] A [HA] 0 055 . [ ] = − A [HA] If [HA] = 0.1 M, then [A - ] = 0.0055M. Solubility Equilibria When solids dissolve to form aqueous solutions, the equilibrium involved is the solubility equilibria. When a salt is added to water at frst, none is dissolved, then as the water molecules collide with the salt and the intermolecular forces interact with the water, the formula units of the salt begin to dissolve. Eventually, the maximum amount of solid that can dissolve at that particular temperature does dissolve and equilibrium is reached. BaCO 3 (s) ↔ Ba 2+ (aq) + CO 3 2- (aq) K sp = [Ba 2+ ] [CO 3 2- ] = 1.6 X 10 -9 Why is the BaCO 3 not included in the K sp expression? Barium carbonate is a solid and solids are not found in equilibrium expressions. Notice that the K value is very small. We would expect this since barium carbonate is not considered soluble in water, based on what we have learned from solubility rules. Do realize, however, that solubility and the solubility product are not the same. Solubility is a position that changes with temperature and the concentration of common ions present in the solution. The solubility product is a constant value for a particular temperature (usually room temperature when you look it up on a chart). K sp problems are much like K a and K b problems. Calculating K sp from solubility PbSO 4 has a measured solubility of 1.14 X 10 -4 mol/L What is the K sp of PbSO 4 ? PbSO 4 (s)  Pb 2+ (aq) + SO 4 2- (aq) K sp = [Pb 2+ ][SO 4 2- ] K sp = (x)(x) K sp = (1.14 × 10 -4 ) 2 K sp = 1.30 × 10 -8 Calculating solubility from K sp Al(OH) 3 has a K sp of 2 X 10 -32 at 25°C Al(OH) 3 (s)  Al 3+ (aq) + 3OH - (aq) K sp = [Al 3+ ][OH - ] 3 2 X 10 -32 = (x)(3x) 3 x = 5 × 10 -9 M Relative Solubilities If salts produce the same number of ions when they dissolve, then their K sp values can be used to determine their solubilities relative to each other. The smaller the K sp , the less soluble it is. If the salts produce different numbers of ions when they dissolve, then you have to calculate the solubilities from the K sp values, as in the previous example. The Common Ion Effect on Solubility A common ion affects the solubility of a solid, generally reducing the amount that dissolves by providing an initial concentration of one of the ions. RICE diagrams are very useful in demonstrating this effect. Appendix G 141 Example 3 The K sp of silver sulfate is 1.2 × 10 -5 . Calculate the solubility of silver sulfate in a 0.20 M solution of potassium sulfate. Ag 2 SO 4 (s)  2Ag 1+ (aq) + SO 4 2- (aq) K sp = [Ag 1+ ] 2 [SO 4 2- ] The K 2 SO 4 solution contributes 0.20 M of SO 4 2- [Ag 1+ ] [SO 4 2- ] Initial 0 0.20 Change +2x + x Equilibrium 2x 0.20 + x 1.2 × 10 -5 = (2x) 2 (0.20 + x) x = 3.9 × 10 -3 M [Ag 1+ ] = 2x = 7.7 × 10 -3 M The rest of Chapter 15 discusses qualitative analysis, selective precipitation, and complex ion equilibria—topics that generally receive very little attention on the AP Exam. Read about it if you have time, otherwise, we will come back to it after all the other AP material has been covered or after the AP Exam. Appendix H Study Guides and Worksheets Chapter 14 Answer the following questions, in order, on notebook paper. Due date: 1. Make a chart comparing the defnitions of an acid and a base according to the Arrhenius concept, the Bronsted-Lowry model, and the Lewis acid–base model. 2. What is a hydronium ion? 3. What is a conjugate acid–base pair? Give an example. 4. What is K a ? 5. What is the difference between a strong and weak acid? 6. What is a diprotic acid? Give an example. 7. Organic acids are generally strong or weak? Why? 8. Water is amphoteric. What does that mean? 9. What is K w ? What is its value at 25 °C? 10. When calculating pH from [H + ], how do you determine the number of signifcant digits in your answer? 11. How does calculating the pH of a strong acid differ from calculating the pH of a weak acid? 12. How do you calculate percent dissociation? 13. How do you calculate the pH of a strong base? 14. Give examples of salts that produce acidic solutions, basic solutions, and neutral solutions. 15. Hydrofuoric acid is the only weak acid consisting of a halogen. Why? Read through the In-Class Discussion Questions. They may be used for warm-ups. SPECIAL FOCuS: Acids and Bases 144 Complete the following problems, in order, from pages 712–717. (19, 21–23, 27, 29, 31–32, 37–38, 43–46, 49–54, 61, 63, 81, 89, 95–96, 103–04, 109–10, 115–118, 123, 129, 138, 142–143, and 153) Chapter 15 Answer the following questions, in order, on notebook paper. Due date: 1. What is a common ion? 2. How do common ions affect equilibrium? 3. What is a buffered solution? What does it do? What substance in our body is considered a buffered solution? 4. When dealing with a strong acid or strong base being added to a buffer, what do you do before setting up the ICE chart? 5. What is the Henderson-Hasselbalch equation? When can you use it? 6. What is buffer capacity and how is it calculated? 7. Draw the pH curves for a strong acid/strong base titration, a weak acid– strong base titration, and a weak base–strong acid titration. 8. What is the solubility product? 9. What is the common ion effect? 10. How are pH and solubility related? 11. What is the ion product? How is it related to K sp and precipitation? 12. What is selective precipitation and what is it used for? 13. What is qualitative analysis? 14. What is a complex ion? 15. What is a ligand? Read through the In-Class Discussion Questions. They may be used for warm-ups. Complete the following problems, in order, from pages 781–789. (13–14, 18–19, 23–24, 33 –36, 40, 51–53, 55, 57–60, 61, 64, 69, 71–72, 79–82, 84–85, 88, 92, 97–99, 106, 119–121, and 123) Appendix H 145 Acid–Bases, Reactions, and Titration Calculation Review Worksheet 1. Complete the following table. Show all work on a separate sheet of paper. [H 3 O + ] [OH - ] pH pOH 7.23 X 10 -2 M 8.89 X 10 -7 M 10.3 9.39 3.99 X 10 -5 M 1.19 X 10 -10 M 2. The approximate pH of oranges is 3.5. What are the pOH, the [OH - ], and the [H 3 O + ]? 3. The approximate pOH of tomatoes is 9.8. What are the pH, the [H 3 O + ], and the [OH - ]? 4. Write and balance the chemical equation for the reaction between: a. barium hydroxide and sulfuric acid b. aluminum hydroxide and nitric acid c. phosphoric acid and potassium hydroxide 5. How many mL of 1.50M magnesium hydroxide are necessary to titrate 25.0 mL of 2.50 M phosphoric acid? 6. What is the molarity of a lithium hydroxide solution if 15.0 mL is neutralized by 7.5 mL of a 0.020 M acetic acid solution? 7. A student reacted a 25.00 mL sample of vinegar (acetic acid solution) with 38.25 mL of 0.500 M sodium hydroxide. a. Write and balance the chemical equation for the reaction. b. Calculate the molarity of the vinegar solution. c. How many grams of acetic acid would be in 1.00 L of the vinegar? d. What is the percent by mass of acetic acid in the vinegar solution (assume a density of 1.00 g/mL)? SPECIAL FOCuS: Acids and Bases 146 Acid–Bases, Reactions, and Titration Calculation Review Worksheet— Teacher’s Edition 1. Complete the following table. Show all work on a separate sheet of paper. [H + ] [OH - ] pH pOH 7.23 X 10 -2 M 1.38 × 10 -13 M 1.141 12.859 1.12 × 10 -8 M 8.89 X 10 -7 M 7.949 6.051 1.38 × 10 -13 M 1.38 × 10 -13 M 10.3 3.7 1.38 × 10 -13 M 1.38 × 10 -13 M 4.61 9.39 2.51 × 10 -10 M 3.99 X 10 -5 M 9.601 4.399 1.19 X 10 -10 M 8.40 × 10 -5 M 9.924 4.076 2. The approximate pH of oranges is 3.5. What are the pOH, the [OH - ], and the [H + ]? a. pOH = 10.5 b. [OH - ]=3 × 10 -11 M c. [H + ] =3 × 10 -4 M 3. The approximate pOH of tomatoes is 9.8. What are the pH, the [H + ], and the [OH - ]? a. pH = 4.2 b. [H + ] = 6× 10 -5 M c. [OH - ] = 2 × 10 -10 M 4. Write and balance the chemical equation for the complete reaction between: a. barium hydroxide and sulfuric acid Ba(OH) 2 + H 2 SO 4 → BaSO 4 + 2H 2 O b. aluminum hydroxide and nitric acid Al(OH) 3 + 3HNO 3 → Al(NO 3 ) 3 + 3H 2 O c. phosphoric acid and potassium hydroxide H 3 PO 4 + 3KOH → K 3 PO 4 + 3H 2 O 5. How many mL of 1.50M magnesium hydroxide are necessary to completely titrate 25.0 mL of 2.50 M phosphoric acid? 3Mg(OH) 2 + 2H 3 PO 4 → Mg 3 (PO 4 ) 2 + 6H 2 O 25.0 mL H 3 PO 4 × 1 L H 3 PO 4 × 2.50 mol H 3 PO 4 × 3 mol Mg(OH) 2 × 1L Mg(OH) 2 × 1000mLMg(OH) 2 = 62.5mL Mg(OH) 2 1 1000mL H 3 PO 4 1L H 3 PO 4 2 mol H 3 PO 4 1.50 mol Mg(OH) 2 1L Mg(OH) 2 Appendix H 147 6. What is the molarity of a lithium hydroxide solution if 15.0 mL is completely titrated by 7.5 mL of a 0.020 M acetic acid solution? LiOH + HC 2 H 3 O 2 → LiC 2 H 3 O 2 + H 2 O 7.5 mL HC 2 H 3 O 2 × 0.020 mmol HC 2 H 3 O 2 × 1 mmol LiOH × 1 = 0.010M LiOH 1 1mL HC 2 H 3 O 2 1mmol HC 2 H 3 O 2 15.0 mL LiOH 7. A student reacted a 25.00 mL sample of vinegar (acetic acid solution) with 38.25 mL of 0.500 M sodium hydroxide. a. Write and balance the chemical equation for the reaction. HC 2 H 3 O 2 (aq)+ NaOH(aq)→ NaC 2 H 3 O 2 (aq) + H 2 O b. Calculate the molarity of the vinegar solution. 38.25 mL NaOH × 0.500 mmol NaOH × 1 mmol HC 2 H 3 O 2 × 1 = 0.765M HC 2 H 3 O 2 1mL NaOH 1mmol NaOH 25.00 mL HC 2 H 3 O 2 1 c. How many grams of acetic acid would be in 1.00 L of the vinegar? 1.00 L HC 2 H 3 O 2 × 0.765M HC 2 H 3 O 2 × 60.05 g HC 2 H 3 O 2 = 45.9 g HC 2 H 3 O 2 1 1L HC 2 H 3 O 2 1mol HC 2 H 3 O 2 d. What is the percent by mass of acetic acid in the vinegar solution (assume a density of 1.00 g/mL)? 45.9 g HC 2 H 3 O 2 × 1L HC 2 H 3 O 2 × 1 mL HC 2 H 3 O 2 × 100 = 4.59% HC 2 H 3 O 2 1L HC 2 H 3 O 2 1000mL HC 2 H 3 O 2 1.00 g HC 2 H 3 O 2 Appendix I Dissociation and pH Practice Weak Acid or Weak Base Systems The ionization constant (or equilibrium constant) is very important here. Only ions contribute. Water, which is usually the solvent, does not participate in any major way here. The general form for a reaction containing an acid is: HA + 2H 2 O ↔ H 3 O + + A - so: K a = HA is an acid because it begins with hydrogen or because it donates a proton to form H 3 O + . The A - is the element or polyatomic ion portion of the acid with a negative charge. Write the K a for the dissociation of chlorous acid, HClO 2 , and formic acid, HCOOH. K a = K a = Weak polyprotic acids dissociate (break apart) stepwise. Each step is written as a different reaction. Generally there is a different value for K a each time an acid loses a hydrogen ion. H 3 PO 4 + H 2 O ↔H 3 O + + H 2 PO 4 1- K a1 = H 2 PO 4 1- + H 2 O ↔ H 3 O + + HPO 4 2- K a2 = HPO 4 2- + H 2 O ↔ H 3 O + + PO 4 3- K a3 = Examples: Write the dissociation reactions for sulfuric acid and carbonic acid. H 2 SO 4 H 2 CO 3 This also holds true for weak bases like ammonia. SPECIAL FOCuS: Acids and Bases 150 Write the K b for NH 3 + H 2 O ↔ NH 4 + + OH - K b = Steps for calculating the pH of a weak acid or base: 1) Write and balance the equation if it is not already done for you. 2) Determine the equilibrium expression, K a or K b . 3) Develop the ICE table. 4) Substitute the fnal values into the equilibrium expression. 5) Solve for the concentrations. 6) Solve for pH or pOH based on the [H + ] or [OH - ]. Calculate the pH of a 1.00M acetic acid solution with a K a = 1.8 X 10 -5 1) HC 2 H 3 O 2 (aq) + H 2 O (l) ↔ H 3 O 1+ (aq) + C 2 H 3 O 2 1- (aq) 2) K a = 3) HC 2 H 3 O 2 H 3 O 1+ C 2 H 3 O 2 1- Initial Change Equilibrium 4) K a = 5) pH = -log[H 3 O 1+ ] Try this one: Calculate the pH of a 0.500M ammonia solution with a K b = 1.75 X 10 -5 1) NH 3 (aq) + H 2 O (l) ↔ NH 4 1+ (aq) + OH 1- (aq) 2) K b = 3) NH 3 NH 4 1+ OH 1- Initial Change Equilibrium 4) K b = 5) pOH = -log[OH 1- ] pH + pOH = 14 Appendix J Dissociation and pH Practice—Teacher’s Edition Weak Acid or Weak Base Systems The ionization constant (or equilibrium constant) is very important here. Only ions contribute. Water, which is usually the solvent, does not participate in any major way here. The general form for a reaction containing an acid is: HA + 2H 2 O ↔ H 3 O + + A - so: K a = [H 3 O + ][A - ] [HA] HA is an acid because it begins with hydrogen or because it donates a proton to form H 3 O + . The A - is the element or polyatomic ion portion of the acid with a negative charge. Write the K a for the dissociation of chlorous acid, HClO 2 , and formic acid, HCOOH. K a = [H + ][ClO 2 - ] K a = [H + ][COOH - ] [HClO 2 ] [HCOOH] Polyprotic acids dissociate (break apart) stepwise. Each step is written as a different reaction. Generally there is a different value for K a each time an acid loses a hydrogen ion. H 3 PO 4 + H 2 O ↔ H 3 O + + H 2 PO 4 - K a1 = [H 3 O + ][ H 2 PO 4 - ] [H 3 PO 4 ] H 2 PO 4 - + H 2 O ↔ H 3 O + + HPO 4 2- K a2 = [H 3 O + ][ HPO 4 2- ] [H 2 PO 4 - ] HPO 4 2- + H 2 O ↔ H 3 O + + PO 4 3- K a3 = [H 3 O + ][ PO 4 3- ] [HPO 4 2- ] SPECIAL FOCuS: Acids and Bases 152 Practice Write the dissociation reactions for sulfuric acid and carbonic acid. H 2 SO 4 H 2 CO 3 H 2 SO 4 + H 2 O ↔ H 3 O + + HSO 4 - H 2 CO 3 + H 2 O ↔ H 3 O + + HCO 3 - HSO 4 - + H 2 O ↔ H 3 O + + SO 4 2- HCO 3 - + H 2 O ↔ H 3 O + + CO 3 2- Equilibrium expressions also can be written for weak bases like ammonia. Write the K b for NH 3 + H 2 O ↔ NH 4 + + OH - K b = [NH 4 + ][OH - ] [NH 3 ] Steps for calculating the pH of a weak acid or base: 1) Write and balance the equation if it is not already done for you. 2) Determine the equilibrium expression, K a or K b . 3) Develop the ICE table. 4) Substitute the fnal values into the equilibrium expression. 5) Solve for the concentrations. 6) Solve for pH or pOH based on the [H + ] or [OH - ]. Examples: Calculate the pH of a 1.00M acetic acid solution with a K a = 1.8 X 10 -5 1) HC 2 H 3 O 2 (aq) + H 2 O (l) ↔ H 3 O + (aq) + C 2 H 3 O 2 - (aq) 2) K a = [H 3 O + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] 3) [HC 2 H 3 O 2 ] [H + ] [C 2 H 3 O 2 - ] Initial 1.00 M 0 0 Change -x +x +x Equilibrium 1.00 M - x x x 4) K a = [x][x] = 1.8×10 -5 [1.00-x] 5) x = 4.2×10 -3 M = [H + ] 6) pH = -log[H + ] = 2.87 Calculate the pH of a 0.500M ammonia solution with a K b = 1.75 X 10 -5 Appendix J 153 1.) NH 3 (aq) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq) 2) K b = [NH 4 + ][OH - ] [NH 3 ] 3.) [NH 3 ] [NH 4 + ] [OH - ] Initial 0.500 M 0 0 Change -x +x +x Equilibrium 0.500-x x x 4) K b = [x][x] = 1.75×10-5 [0.500-x] 5) x = 2.96×10 -3 M = [OH - ] 6) pOH = -log[OH-] pH + pOH = 14 pOH = 2.529 pH = 14 – pOH pH = 11.471 Appendix K Lab: Determining the pH of a Substance Introduction Many classifcation systems for compounds exist. One way to classify substances is based on the relative amount of acid present. An acid produces hydrogen ions in an aqueous solution according to the Arrhenius concept. The pH scale was developed to measure the amount of hydrogen ions present in a solution. The use of an indicator, a weak organic acid that changes colors, is another way to measure the hydrogen ion concentration. Different indicators work in different pH ranges. Use the following table as a reference: Indicators Color at a bottom of range Color at top of range pH range Methyl red Red Yellow 4.8-6.0 Litmus Red Blue 5.2-7.5 Bromothymol blue Yellow Blue 6.0-7.6 Phenolphthalein Clear Pink 8.0-10.0 Universal indicator: changes colors at different pH values as follows 3.0 red-orange 4.0 orange 5.0 yellow 6.0 green-yellow 7.0 green 8.0 blue-green 9.0 blue-gray 10.0 violet Objectives 1. To observe typical reactions of several indicators when treated with acids and bases. 2. To estimate the pH of some common household products using these reactions. SPECIAL FOCuS: Acids and Bases 156 Procedure Part 1: 1. Obtain a set of standard solutions and a set of indicators. 2. Add 1–3 drops of each standard across a row in wells according to data table 1. 3. Add 1 drop of indicator down the wells according to data table 1. 4. Record the colors observed for each reaction in the data table. 5. Estimate the pH of each solution. Record in the data table. 6. Rinse out the spot plate with deionized water. Part 2: 1. Obtain a set of household products. 2. Add 1–3 drops of each product across a row in wells according to data table 2. 3. Add 1 drop of indicator down the wells according to data table 2. 4. Record the colors observed for each reaction in the data table. 5. Estimate the pH of each solution. Record in the data table. 6. Rinse out the spot plate with deionized water. Part 3: (To be done after you estimate the pH of each household substance based on the indicator colors.) 1. Go to a station set up for measuring the pH of the products. 2. Follow the directions for calibrating your pH probe. 3. Submerge the probe tip in a small amount of the frst solution to be measured. 4. Record the pH value in the data table. 5. Rinse the probe thoroughly with deionized water. 6. Repeat steps 3–5 with each household product to be tested. Appendix K 157 Data Table 1: Standard Solution Methyl Red Litmus Indicator Bromthy- mol Blue Phenol- phthalein universal Indicator Estimated pH 0.1 M HCl 0.1 M acetic acid distilled water 0.1 M NH3 (ammonia) 0.1 M NaOH Data Table 2: Product Methyl Red Litmus Indicator Bromthy- mol Blue Phenol- phthalein universal Indicator Estimated pH aspirin bleach detergent eggs lemon juice milk milk of magnesia clear soft drink tea SPECIAL FOCuS: Acids and Bases 158 Analysis: 1. Based on your observations, how does the estimated pH of the hydrochloric acid compare to the estimated pH of the acetic acid? 2. Based on your observations, how does the estimated pH of ammonia compare to the estimated pH of sodium hydroxide? 3. Using your textbook, fnd the strengths of the listed acids and bases. How does your estimated pH values compare to the rankings of these substances? 4. Indicators also can come from objects found in nature, your home, or even in an offce. Locate one such indicator through research and explain how you could tell the difference between an acid and a base using this indicator. Conclusions Describe your results. Where they as you expected? Does one indicator work better than another? If so, which one and in what cases? Appendix K 159 Was it diffcult to determine the color of the indicator in any particular substance? If so, which one(s)? Why do you think this might have been the case? If you start with 15.0 mL of hydrochloric acid in a beaker, add a couple drops of methyl red and a couple drops of phenolphthalein, what would you expect to happen as you add 10.0 mL of sodium hydroxide? 15.0 mL of NaOH? 20.0 mL of NaOH? 161 About the Editor Marian DeWane teaches AP Chemistry at Centennial High School in Boise, Idaho. She has a long list of accomplishments including Teacher of the Year, National Board Certifcation in Chemistry, and Presidential Scholar Teacher. She is currently a Question Leader for the AP Chemistry Examination and has also served as a Reader and Table Leader. She also serves on the ACS Exams Institute Advanced High School Committee and National Chemistry Olympiad Task Force, and is a National Leader for the College Board. About the Authors Lew Acampora currently teaches AP Chemistry at the St. Francis School in Louisville, Kentucky. For the last 16 years he has taught AP Chemistry at independent schools across the country. Since 1995, he has served as a Reader, Table Leader, and Question Leader at the AP Chemistry Exam Reading. He is a consultant for the College Board and serves on the ACS Exams Institute Advanced High School Committee. Valerie Ferguson teaches AP Chemistry at Moore High School in Moore, Oklahoma. She has served as a Reader and Table Leader for the AP Chemistry Examination. She also works as a consultant for the College Board and teaches Summer Institutes for AP Chemistry teachers. She has served as chair of the Oklahoma ACS, president of the Oklahoma Science Teachers Organization, and was co-organizer of the Teaching AP Chemistry symposium at the 14th Biennial Conference on Chemistry Education. Adele Mouakad is an award-winning teacher at St. John’s School in San Juan, Puerto Rico. She has been teaching AP Chemistry since 1983. She has won the Presidential Award for Excellence in Science Teaching, and has been a member of the Woodrow Wilson Foundation Teacher Outreach Program and a Dreyfus Master Teacher. She has also been a Reader for the AP Chemistry Examination and a member of the National Chemistry Olympiad Task Force. Cheri Smith is an award-winning teacher of AP Chemistry at Yale Secondary School in Abbotsford, British Colombia in Canada. She has been a Reader and Table Leader for the AP Chemistry Examination. She has also served as a consultant for the College Board, conducting AP workshops worldwide. Smith has contributed to numerous publications and served as a reviewer for the AP Vertical Teams ® Guide for Science and the Pre-AP ® : Topics for Vertical Teams in Science workshop. Arden Zipp is a distinguished faculty member at State University of New York College at Cortland in Cortland, New York. He has served in every capacity for the College Board in the development and scoring of the AP Examination: He has been a Reader, Table Leader, Questions Leader, Chief Reader, and Development Committee member and chair. He is also involved in the development and scoring of the IB Chemistry Program, and is chair of the National Chemistry Olympiad Examination Task Force. Heather Hattori teaches AP Chemistry, AP Environmental Science, pre-AP chemistry, and pre-AP physics at John H. Guyer High School in Denton, Texas. Hattori received her bachelor’s degree in chemistry and English from Stephen F. Austin State University in 1991 and is currently attending the University of North Texas in pursuit of a Ph.D. in chemistry. She has been teaching AP Chemistry and/or AP Environmental Science since 2000 and has been an AP Chemistry Reader since 2005. Hattori was a College Board consultant in the mid-90s, but has been taking a break to raise Arabian horses and most recently attend graduate school in her spare time. George E. Miller is a senior lecturer emeritus in the Chemistry department at the University of California Irvine, where he is also principal scientist and supervisor of the nuclear reactor facility and responsible for K–12 science teacher professional development. He has been on the chemistry faculty at UCI since 1965, and has taught freshman chemistry, analytical chemistry, radiochemistry, and instrumental analysis. He was member and chair of the AP Chemistry Development Committee and a member of the College Board Subject Test in Chemistry Committee. He has been a Reader, Table Leader, and Question Leader for AP Chemistry over a period of 20 years, and has presented both day- and weeklong workshops for AP and general chemistry teachers. He has more than 60 publications in analytical chemistry and in science education. F08CMSF131 The Properties of Acids and Bases Arden P. Zipp SUNY Cortland Cortland, New York Acids and bases represent two of the most important classes of chemical compounds. Some of the properties of acids (such as their sour taste) and bases (bitterness) have been known to humans for hundreds of years even if their chemical explanations have not. The reactions of these substances are significant in many atmospheric and geological phenomena. In addition, acids and bases control many of the physiological processes in the human body and can catalyze a variety of chemical reactions. This chapter will address some of the fundamental aspects of acids and bases in order to set the stage for subsequent discussions. In addition to taste, which was mentioned above but is no longer used in the chemistry lab, several other properties of acids and bases that can be observed readily are given below. Table 1 shows that they have some properties in common, while they do not share others. Table 1 Properties of Acids and Bases Acids 1 2 conduct electricity in solution react with many metals carbonates bicarbonates a base dyes product(s) formed H2 CO2, H2O, a salt CO2, H2O H2O, a salt colors, often reddish Bases 1 2 conduct electricity in solution react with metal oxide and water metal ions bicarbonates an acid dyes product(s) formed H2 precipitates CO32-, H2O H2O, a salt colors, often bluish SPECIAL FOCuS: Acids and Bases Teaching Tip Students will appreciate these properties more if they are demonstrated by the teacher, or better yet, investigated by students themselves. Inexpensive conductivity meters can be purchased or constructed to investigate the first property on the list while the others can be studied using well plates and solutions dispensed from dropper bottles. Early History: Arrhenius Theory Some of the properties in Table 1 were known to alchemists hundreds of years ago, and classification schemes were developed by many noted scientists during the 1600s and 1700s including Robert Boyle, Joseph Black, and Antoine Lavoisier. However, the beginning of the modern understanding of acids is usually attributed to Svante Arrhenius, who, in his 1884 Ph.D. thesis, explained the electrical conductivity of aqueous solutions in terms of the presence of ions and attributed the unique behavior of acids to H+ ions. According to Arrhenius, an acid is a substance that, when added to H2O, increases the [H+], while a base is a substance that increases the [OH-]. Many of the reactions in Table 1 can be explained with the Arrhenius acid–base theory by the presence of H+ or OH- ions as shown in the equations below: Acids 2H + M (e.g. Mg) → H2 + M 2H+ + CO32- → H2O + CO2 H+ + HCO3 - → H2O + CO2 H+ + OH - → H O 2 Bases 2+ + 2OH - + M2O3 (e.g. Al) + 2H2O→ H2 + 2M(OH)4 – xOH - + Mx+→ M(OH) (s) OH - + HCO3 - →H2O + CO32 H+ + OH - → H2O x State of H+ in H2O. In the above equations the acid is represented as H+, a bare proton. However, it is generally agreed that the charge density of a proton is too large for it to exist in that form in aqueous solutions. Many texts represent protons as hydronium ions, H3O+, as if each proton is bonded to a single H2O molecule. Current evidence suggests, however, that the actual species has three additional H2O molecules attached via hydrogen bonds to form H9O4+. H+ will be used in this volume but it should be recognized that the actual species is more complex. Interestingly, the same texts that make a point of using H3O+ usually show the hydroxide ion as OH- even though it also appears to be hydrated, existing in H2O as H7O4-. 4 the rates at which these reactions occur vary depending on the concentration of free H+ ions in solution. thereby introducing the concept of strong and weak acids. 5 . i.. work similarly.. the reaction of a strong acid goes completely to the right. have solutions that contain significant amounts of the parent acid and lower concentrations of H+. Equal volumes of the weak and strong acids will produce the same volume of H2 but the rates at which they do so will differ substantially. while weak acids ionize only partially. only the first proton from H2SO4 can really be described as “completely ionized. six are strong. so is written with a simple arrow while that for a weak acid is reversible and is depicted with a double-headed arrow. HCl HBr HI hydrochloric acid hydrobromic acid hydroiodic acid HNO3 HClO4 H2SO4 nitric acid perchloric acid sulfuric acid All other acids behave as weak acids in water. Although strong and weak acids produce the same types of reactions. Teaching Tip The difference between strong and weak acids can be made more concrete for students by demonstrations (or experiments) in which they compare the behavior of 1.0 M solutions of strong (e.” the second proton has a measurable equilibrium constant. It should be noted that in the above table. Strong acids are those that ionize completely (or nearly so) in solution. Properties of interest include electrical conductivity and the rate of reaction with Mg metal (using a balloon attached to the mouth of a conical flask to capture the evolved H2). HCl) and weak (HC2H3O2) acids. In terms of equations. Strong acid: Weak acid: HA(aq) → H+(aq) + A-(aq) HA(aq) ↔ H+(aq) + A-(aq) Among the standard acids commonly encountered in aqueous solutions.e. Students can measure how much time it takes for a short measured length of Mg ribbon or a measured scoop of calcium carbonate to dissolve completely. capturing the CO2. Reactions with calcium carbonate.The Properties of Acids and Bases Strong and Weak Acids Arrhenius related the degree of electrical conductivity to the strength of acids.g. called the pH scale.0 x 10-14.00 while basic (alkaline) solutions have [OH-] > 1. When the above value of K is multiplied by [H2O]. the expression is Kw = [H+][OH-] with a value of Kw = 1.5 M).00.0 x 10-7 and pH = pOH = 7. Acid solutions are those in which [H+] > 1. For example. Because of this extremely wide range of [H+]. represent concentrations in moles per liter. These will be discussed further below. which should not be confused with amount of ionization. as shown in the reversible equation: [H+][OH–] =1.00. while in a 1.0 x 10-7 or pH < 7. sodium oxide dissolves in H2O according to the following equation: Na2O + H2O → 2 Na+ + 2 OHSimilar reactions to produce strongly basic solutions occur with several other anions but these will not be considered here. Concentrations Water itself is a weak acid. [H+] = [OH-] = 1. In a 1. H+ and OH-. [H+] =1. A more extensive discussion of pH and further calculations are provided in a later chapter. Using the Kw expression it is possible to calculate [H+] for any [OH-] and vice versa.0 x 10-14 at 25 ˚C. a solution with [H+] = 1. existing in equilibrium with its ions.0 and [H+] = 1.8 x 10-16 [H2O] where the square brackets. has been devised.0 has a pH = 0. There are a variety of compounds that behave as weak bases.00 and one with [H+] = 1.SPECIAL FOCuS: Acids and Bases Strong and Weak Bases There are also six common strong bases that ionize completely in H2O.0 x 10-7. H2O ↔ H+ + OHfor which K = 6 . (55. where pH is defined as -log [H+].ions. Thus. although few of these contain OH. []. though some have limited solubility.0.0 x 10-7 or pH > 7. [H+] < 1. [OH-] = 1.00.0 M solution of a strong acid.0 M solution of a strong base. In pure H2O. an alternative method of comparing [H+] concentrations.0 x 10-14 has a pH = 14. LiOH lithium hydroxide NaOH sodium hydroxide KOH potassium hydroxide Ca(OH)2 Sr(OH)2 Ba(OH)2 calcium hydroxide strontium hydroxide barium hydroxide A solution of a strong base in H2O may be obtained by dissolving the appropriate solid hydroxide but may also be formed by reacting a solid metallic oxide with H2O. they suggested that when acid–base reactions occur. The Brønsted-Lowry theory includes as acids all the substances identified by Arrhenius as acids because they donate H+ ions to water and would also classify the 7 . Brønsted-Lowry Acid–Base Theory As outlined above. are weaker). the release of OH. the production of acidic solutions by substances that contain no H+ (or basic ones by substances that do not possess OH-) is less apparent. the equilibrium constant will be >1.ions can be envisioned from the addition of one of the strong bases to H2O. This activity can be useful for AP students if it is coupled with pH calculations and practice in pH ↔ [H+] or pOH ↔ [OH-] conversions. However.ions. HCl.The Properties of Acids and Bases Teaching Tip Students typically enjoy determining the pH values of common household materials (with indicator paper or drops of universal indicator). their conjugate bases have lower attractions for protons (i. Arrhenius defined acids as substances that produce H+ ions in H2O and bases as species that produce OH. a strong acid) or partially (HC2H3O2. where HA and A. thereby broadening the concept of acids and bases significantly.are a conjugate pair. In 1923 Johannes Brønsted and Thomas Lowry independently described acids as proton donors and bases as proton acceptors. or even a weak acid such as ethanoic (acetic) acid. The strengths of Brønsted-Lowry acids and their conjugate bases are inversely related. and B and HB+ are a conjugate pair: HA + B ↔ HB+ + ASuch reactions are reversible and the relative amounts of the various substances depend on the strengths of the acids (bases) involved. HC2H3O2. Where HA is a stronger acid than HB+ and B is a stronger base than A-. Similarly. In addition. they introduced the notion that acids and bases exist in pairs (called conjugate pairs) that differ by the presence (acid) or absence (base) of a proton. an acid–base theory other than that proposed by Arrhenius is required. Clearly.e. hence.. Further. to water will increase the concentration of [H+] above 1. stronger acids and bases react to form weaker acids and bases as in the equation below. a weak acid) to release H+ ions. Formation of Acidic or Basic Solutions It is easy to see how adding a strong acid like hydrochloric acid.0 x 10-7 as these species ionize completely (HCl. This is because stronger acids donate protons more readily and. the Brønsted-Lowry theory can also account for the acidic behavior of metal ions and nonmetallic oxides. However. In addition. 8 . the attraction for the oxygen of the H2O may be strong enough to polarize the O-H bond and cause an H+ ion to be released. most metal ions are also hydrated. usually with six H2O molecules. Acidities of Metal Ions and Nonmetallic Oxides Just as protons are hydrated (described above). In this reaction. it can explain the basic behavior of weak bases such as ammonia.SPECIAL FOCuS: Acids and Bases OH. For example.ion as a base because it is a proton acceptor. This process is depicted in the equation: M(H2O)6x+ + H2O ↔ M(HO)5OH(x-1)+ + H3O+ The effect of charge and radius on the acidic behavior of selected metal ions is shown in the Table 2 below. Table 2 Li+ slightly acidic Be2+ weakly acidic Na+ neutral Mg2+ weakly acidic Al3+ acidic K+ neutral Ca2+ slightly acidic Sc3+ acidic Ti4+ very acidic Teaching Tip The relative acidities of these (and other) metal ions can be shown by adding a few drops of universal indicator to solutions containing these ions and comparing the solution colors with those of pH standards. In both instances lone pairs of electrons in H2O molecules are attracted to the positive charge of the ion. For metal ions with large charge densities (due to high charges and/or small radii). H2O behaves as the base. neither of which contain H+ ions. HCl acts as an acid according to both definitions in the equation: HCl + H2O → H3O+ + Clbecause it releases an H+ ion (Arrhenius) or donates a proton (Brønsted-Lowry). A more extensive discussion of oxides (both nonmetallic and metallic) is undertaken in a later section as is the relative acidities of oxyacids. Lewis acid–base reactions encompass 9 . G.ions remains in the solution as shown in the equation: H2O + B ↔ BH+ + OHThe [OH-] produced in such a reaction depends on the position of the above equilibrium. the Brønsted-Lowry theory is able to account for several acid–base reactions that can not be treated by the Arrhenius theory. which is a function of the strength of B as a Brønsted-Lowry base. When a proton is abstracted from a water molecule. Behavior of Weak Bases As mentioned above. These substances produce an excess of OH. and acids as electron pair acceptors. (H2O is acting as the acid in this reaction. The same year that Brønsted and Lowry proposed their theory. [OH-] = 4.0 M NH3 solution. Hence. there are many substances that behave as weak bases.The Properties of Acids and Bases The acidic behavior of nonmetallic oxides can be explained on the basis of their reaction with water to form oxyacids as represented in the equation for SO2: SO2 + H2O → H2SO3 Such oxyacids ionize to release H+ ions as: H2SO3 + H2O ↔ H3O+ + HSO3Thus. the Brønsted-Lowry theory offers an explanation for the acidic behavior of both metal ions and nonmetallic oxides. neither of which contain H+.N. NH3. although some contain other atoms that can behave in this same manner.ions in water even though they do not contain OH.2 x 10-3 M in a 1. an excess of OH. Most of these materials contain a nitrogen atom that can act as a Brønsted-Lowry base by accepting a proton. Weak acids and bases will be treated in more detail in another section of this book. Lewis noted that the reaction of a proton with a base involved the donation of a pair of electrons from the base to form a bond with the proton.) For ammonia. In contrast to the small number of strong bases. Lewis Acid–Base Theory By emphasizing the transfer of protons. He defined bases as electron pair donors. each of which contains hydroxide ions. there are six strong bases. the most common weak base.ions themselves. 21. Of these. NaHCO3(s) (A) * (B) (C) (D) (E) I only II only III only I and II only I. II. 2NH3 ↔ NH4+ + NH2In the reaction. NH4+ acts as (A) a catalyst (B) both an acid and a base * (C) the conjugate acid of NH3 10 . In the laboratory.SPECIAL FOCuS: Acids and Bases all Brønsted-Lowry reactions but also cover many processes that do not involve proton transfer. Sample Part I Released Exam 2002 Questions An analysis of this released exam revealed a total of 9 questions (out of 75) related to acids and bases. the reaction represented below occurs. Examples of these include all cases in which a pair of electrons is donated.0 M HCl(aq)? I. H2(g) can be produced by adding which of the following to 1. 1 M NH3(aq) II. both Part I and Part II. Zn(s) III. such as the hydration of metal ions and the reaction of ammonia with boron compounds: Mx+ + 6H2O → M(H2O)6x+ NH3 + BCl3 → H3N-BCl3 This brief discussion of various aspects of acids and bases is intended to introduce the more extensive treatments in later chapters. Because of the importance of acids and bases in modern chemistry. this material has been tested frequently on the AP Chemistry Exam. the ones below are based solely on the above material. In liquid ammonia. and III 22. + H2O + CO2 + 2 Na+ (f) Boron trifluoride gas is added to ammonia gas: BF3 + NH3 → F3B-NH3 (g) Sulfur trioxide gas is bubbled into a solution of sodium hydroxide: SO3 + OH. [OH-].+ H2O + CO2 (h) Concentrated hydrochloric acid is added to a solution of sodium sulfide: 2 H+ + S2.+ H2O (d) Hydrogen iodide gas is bubbled into a solution of lithium carbonate: 2 HI + CO32.0 M Sample Part II Released Exam 2004 Questions Question 4 (c) A 0. aqueous solutions with a pH of 8 have hydroxide concentration.→ 2 I.→ HSO4- 11 . At 25˚C.0 M 8.→ H2S Released Exam 2005 Questions Question 4 (d) Solid calcium carbonate is added to a solution of ethanoic (acetic) acid: Na2CO3 + 2 HC2H3O2 → 2 C2H3O2.1 M sodium hydroxide solution HNO2 + OH.The Properties of Acids and Bases (D) the reducing agent (E) the oxidizing agent 30. of (A) (B) * (C) (D) (E) 1 x 10-14 M 1 x 10-8 M 1 x 10-6 M 1.→ NO2.1 M nitrous acid solution is added to the same volume of a 0. Descriptive Inorganic Chemistry. Overton. NJ: Prentice Hall. 2 H+ + CaCO3 → Ca2+ + H2O + CO2 (ii) Briefly explain why statues made of marble (calcium carbonate) displayed outdoors in urban areas are deteriorating.. Bibliography Brown. Keiter. G. W. 1991. 1993. Keiter.. Freeman. 1991. 5th ed. The H+ ions in acid rain react with the marble statues and the soluble compounds of calcium that are formed are washed away. L. E. E. E. Inorganic Chemistry. Rayner-Canham. E. New York: W. 2006. From Caveman to Chemist. Huheey. A. Bursten. and T. H. Salzberg. 12 . L.. DC: American Chemistry Society. Englewood Cliffs. 4th ed. J. T. Washington.SPECIAL FOCuS: Acids and Bases Released Exam 2006 Questions Question 4 (c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide: HC2H3O2 + OH. New York: Harper Collins. Chemistry: The Central Science. and B.H. 4th ed. H. LeMay Jr. and R.→ H2O + C2H3O2 (d) Ammonia gas is bubbled into a solution of hydrofluoric acid: NH3 + HF → NH4+ + F(f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas: PH3 + BCl3 → H3P-BCl3 Released Exam 2007 Questions Question 4 (b) (i) Excess nitric acid is added to solid calcium carbonate.. Oklahoma Many substances dissolve in water to cause a change in the pH of pure water. Strong acids and bases are strong electrolytes in water solutions. and are almost totally ionized. The acid and its product. Since all of the HCl molecules are converted to H3O+ and Cl-. The particle that donates a proton is the acid and the particle accepting the proton is the base.” The equilibrium autodissociation of water is always present and operative in any aqueous solution even though the conjugate ion concentration (OH. equilibria never “turn off. Strong acids and bases are not thought of as establishing equilibria since the reaction essentially goes to completion. The products of these reactions are called conjugate acids and conjugate bases. are a conjugate acid–base pair. The reaction depicted below of a strong acid in water occurs with all of the HCl molecules in water. H3O+ in base) is extremely small. The Brønsted-Lowry definition describes many acid and base interactions as a proton transfer from one particle to another. However.Weak Acids and Bases Valerie Ferguson Moore High School Moore. the concentration of the hydronium ion is equal to the initial concentration of HCl in the solution. . Weak acid and base solutions are partially ionized. the conjugate base. Some of the substances ionize completely and others only partially ionize.in acid. in water.2 x 10-8 1. ammonia. The example below illustrates a Brønsted-Lowry reaction with acetic acid in water.SPECIAL FOCuS: Acids and Bases Conjugate acid–base pairs are two substances whose formulas differ by a proton. is a weak acid in water. 14 . Brønsted-Lowry bases are proton acceptors. HOCl. The illustration below shows the Brønsted-Lowry reaction of a weak base. The Ka expression for HOCl is shown above. (a) Write a chemical equation showing how HOCl behaves as an acid in water. Opportunities for such practice occur several times during the course. Questions on the AP Chemistry Exam commonly ask students to demonstrate their knowledge of writing these equations. Hypochlorous acid. Examples from Prior AP Chemistry Examinations (2005 Form B) Ka = [H3O+][OCl-]/[HOCl] = 3. HOCl(aq) + H2O(l) ↔ OCl-(aq) + H3O+(aq) Teaching Tips Students need to practice writing equations for the reactions of weak acids and bases in water. H2O III. In the reaction.Weak Acids and Bases (2002) 2 NH3 ↔ NH4+ + NH222. the reaction represented above occurs. In liquid ammonia. NH4+ acts as (A) (B) * (C) (D) (E) a catalyst both an acid and a base the conjugate acid of NH3 the reducing agent the oxidizing agent (1994) HSO4. HSO4II.+ H2O ↔ H3O+ + SO4222. SO42(A) II only (B) III only (C) I and II (D) I and III * (E) II and III 15 . In the equilibrium represented above. the species that act as bases include which of the following? I. 0. Ka. (D) The equilibrium constant will increase with an increase in temperature. The position of the equilibrium depends on the strength of the acid and/or base. The strong acids and bases are essentially totally ionized.SPECIAL FOCuS: Acids and Bases (1989) 34. All of the following species can function as Brønsted-Lowry bases in solution EXCEPT (A) H2O (B) NH3 (C) S2+ * (D) NH4 (E) HCO3* Proton acceptors must have a lone pair of electrons. the conjugates are less likely to react and are thus weaker. The weaker species will be in greater concentration when equilibrium is established. or the position of the dissociation. Since the stronger species are more likely to react. it is in higher concentrations in the solution. Which of the following can be concluded from this information? * (A) CN (aq) is a stronger base than C2H3O2 (aq).The reaction represented above has an equilibrium constant equal to 3. 16 . in comparison to water. The weaker acids have stronger conjugate bases. (C) The conjugate base of CN-(aq) is C2H3O2-(aq). Since the weaker species is less likely to react. strength of the conjugate base compared to water.7 x 104. in aqueous solutions. Stronger acids and bases tend to react with each other to produce their weaker conjugates. The conjugates are very weak and less likely to react with water. Brønsted-Lowry reactions are reversible and will establish a chemical equilibrium between the forward and reverse reactions. The relative strengths of the various weak acids can be described in different ways: the value of the acid dissociation. (E) The pH of a solution equimolar amounts of CN-(aq) and HC2H3O2(aq) is 7. (1999) HC2H3O2(aq) + CN-(aq) ↔ HCN(aq) + C2H3O2-(aq) 62 . the [H+] compared to the initial [HA]. (B) HCN(aq) is a stronger acid than HC2H3O2(aq). > HPO42HPO42. Each of the following can act as both a Brønsted acid and a Brønsted base EXCEPT (A) HCO3(B) H2PO4+ * (C) NH4 (D) H2O (E) HSPeriodic trends are observed in the relative strength of acids.↔ HPO42.= H2BO3and and and and and Bases HBO32.+ HBO32. The graphic below represents the reaction of a binary acid in water.> HBO32HPO42. 17 .Weak Acids and Bases (1989) H2PO4. Binary hydroacids—those that contain only hydrogen and a nonmetal—tend to increase in strength from left to right across the periodic table within the same period.> H2BO3(D) H2BO3.> H2PO4(E) H2PO4.= HBO32- (1984) 49. Which of the following gives the correct relative strengths of the acids and bases in the reaction? Acids * (A) H2PO4 > H2BO3 (B) H2BO3. The equilibrium constant for the reaction represented by the equation above is greater than 1.+ H2BO355.0.> HBO32HPO42.> HPO42HBO32. The strength of a binary acid increases from top to bottom within the same group.> H2PO4(C) H2PO4. 0 HOC 2. the electrons are pulled to the central atom and away from the O-H bond.5 HOI Ka=2. 18 . the stronger acids contain more oxygen atoms.SPECIAL FOCuS: Acids and Bases Electronegativity of E increases Acid strength increases 2.8 HOBr 2. The other atoms in the molecule affect the polarity of the O-H bond. The more readily the hydrogen ion is released. the stronger the acid.3 × 10-11 < < 2. oxygen. As the electronegativity of the central atom increases.9 × 10-8 Acids composed of hydrogen. Oxoacids are composed of a central atom with one or more O-H groups attached to it. When the number of oxygens is held constant.5 × 10-9 < < 3. The strength of an oxoacid depends upon the ease with which the hydrogen ion is released from the O-H group. and some other element are called oxoacids or oxyacids. the strength of the oxoacid increases from top to bottom within a group and from left to right across a period. The probability that H+ will be released is increased. For the same central atom. Acetic acid is an example of a simple carboxylic acid. The greater the number of electronegative atoms in the molecule. attached. An important group of oxoacids is the carboxylic acids. Only the O-H bond is polar enough so the hydrogen is readily donated as a hydrogen ion. Acetic acid 19 . Many students found it helpful to draw a structural formula to help answer this question. 1 point earned for a correct explanation. Teaching Tips Students need to demonstrate an understanding of the strength of the O-H bond as related to the strength of the acid. HOBr is a weaker acid than HBrO3. Most carboxylic acids react in a similar fashion. These organic compounds are characterized by containing a chain of carbon atoms with one or more carboxyl groups. the weaker the O-H bond as the electrons are pulled away from the O-H bond. The H-O bond is weakened or increasingly polarized by the additional oxygen atoms bonded to the central bromine atom in HBrO3. These compounds are weak acids called carboxylic acids. Account for this fact in terms of molecular structure. Its reaction with water is shown on page 1. -COOH.Weak Acids and Bases (2002) 1(e). complete the equation by drawing the complete Lewis structures of the reaction products.+ H2O (2005 Form B) 4(d). OH.→ C2H5COO. 20 . More electronegative atoms pull the electron density away from the O-H bond and cause an increase in acid strength. (a) Lactic acid.+ HC2H3O2 → C2H5O2. The closer the electronegative atoms are to the O-H bond. (2001) 4(d).SPECIAL FOCuS: Acids and Bases (2007 Form B) 5. Write the net ionic equation for the following reaction: Ethanoic acid (acetic acid) is added to a solution of barium hydroxide. the greater the effect on acid strength by weakening the O-H bond. Write the net ionic equation for the following reaction: Solutions of potassium hydroxide and propanoic acid are mixed. C2H5COOH + OH. In the space provided above. bases and buffer solutions. HC3H5O3. Show below are the complete Lewis structures of the reactants.+ H2O The strength of the carboxylic acids is affected by the substitution of one or more hydrogen atoms with electronegative atoms. reacts with water to produce an acidic solution. Answer the following questions about laboratory solutions involving acids. In each reaction the water acts as a BrønstedLowry acid and ammonia or methylamine as a Brønsted-Lowry base. (1999) 4(f). The amines act as Brønsted-Lowry bases and are proton acceptors. reacts with water according to the reaction represented above. 21 . NH3.Weak Acids and Bases I—CH2CH2COOH 3-ledopropanoic acid Cl—CH2CH2COOH 3-Chloropropanoic acid CH3CHCLOOH 2-Chloropropanoic acid CH3CCL2COOH 2. Write the net ionic equation for the following reaction: Methylamine gas is bubbled into distilled water.0 x 10-4 < Ka = 1.3 x 10-5 Increasing acid strength < Ka = 1. NH3(aq) + H2O(aq) ↔ NH4+(aq) + OH-(aq) H3CNH2(aq) + H2O(l) ↔ H3CNH3+(aq) + OH-(aq) Teaching Tips The equation for the reaction of ammonia with water is compared to the equation for the reaction of methylamine with water. Aniline. Examples of other amines.2-Dichloropropanoic acid Ka = 8. with one hydrogen substituted by a hydrocarbon group. CH3NH2 + H2O → CH3NH3+ + OH- (2003) C6H4NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH-(aq) 1. The structure is an ammonia molecule. a weak base.4 x 10-3 < Ka = 8.7 x 10-3 Amines are organic compounds that act as weak bases. The proton attaches to the nitrogen in the amine group. Amines react as bases similar to reactions of NH3. In question 1 from 2003. The discussion of weak acids and bases is a good time to include some organic chemistry. Students need to have a background of the behavior of amines as weak bases. Sometimes the Lewis acid has fewer than eight electrons. Lewis acids can be described as electron-pair acceptors and Lewis bases as electron-pair donors. Students should be encouraged to develop an understanding of the similarity of amine reactions to those of ammonia. as in compounds such as BF3. PH3 + BCl3 → H3PBCl3 CO2(g) + OH-(aq) → HC03-(aq) Lewis acid-base theory represents the movement of electrons in this reaction as follows. students were asked to demonstrate an understanding of the characteristics of an amine equilibrium. Write the net ionic equation for the following reaction: Hydrogen phosphide (phosphine) gas is added to boron trichloride gas. Write the net ionic equation for the following reaction: Boron trifluoride gas is added to ammonia gas. 22 .SPECIAL FOCuS: Acids and Bases Teaching Tips The examples above demonstrate the inclusion of amines on past AP Exams. BF3 + NH3 → BF3NH3 (2006) 4(f). (2005) 4(f). CO2 + Na2O → Na2CO3 (2001) 4(a). Write the net ionic equation for the following reaction: carbon dioxide gas is passed over hot. Write the net ionic equation for the following reaction: Sulfur dioxide gas is bubbled into distilled water. The smaller metal ions with greater charge have a greater charge density and produce more acidic solutions.Weak Acids and Bases (1999) 4(g). SO2 + H2O → H2SO3 Hydrated metal ions can act as acids in water. solid sodium oxide. 23 . edu/CCLI/ Startup. Write the net ionic equation for the following reaction: Excess saturated sodium fluoride solution is added to a solution of aluminum sulfate. AgCl + NH3 → Ag(NH3)2+ + Cl- (1984) 48. Write the net ionic equation for the following reaction: Excess concentrated aqueous ammonia is added to solid silver chloride. which can be found at: genchem1.html 24 .SPECIAL FOCuS: Acids and Bases (2004 Form B) 4(g).chem. and Kirk Haines.edu/group/Greenbowe/sections/projectfolder/ animationsindex.htm. F– + Al3+ → [AlF6]3- (2003 Form B) 4(g).chem. Write the net ionic equation for the following reaction: Excess concentrated potassium hydroxide solution is added to a solution of nickel(II) chloride. Gelder. Which of the following ions is the strongest Lewis acid? (A) Na+ (B) Cl(C) CH3COO(D) Mg2+ 3+ * (E) Al I highly suggest using the following Web sites with simulations of weak acid and weak base behavior: • The first is the work of Thomas Greenbowe of Iowa State University. Michael Abraham. located at: www.iastate. • The second is part of a collaboration of John I.(some other products were accepted) (2003) 4(e). OH– + Ni2+ → [Ni(OH)4]2.okstate. 25 . 2005. 10th ed. and George G. Chemistry: The Molecular Nature of Matter and Change. Upper Saddle River.. LeMay. Steven S. Chemistry: The Central Science. Whitten. 7th ed. McCreary. Brown. Upper Saddle River. Zumdahl. 4th ed. Boston: Houghton Mifflin.. General Chemistry. Martin S. Ralph H. and Fred Senese. Eugene H . Larry M. Chemistry. James E. 2000. NJ: Pearson/Prentice-Hall. and Scott S. Petrucci. 8th ed. Hill. Zumdahl. Chemistry: Matter and Its Changes. 2003. and Bruce E.. Kenneth W. Hoboken. Perry. Stanley. Theodore E. NJ: Wiley. 2nd ed. New York: McGraw-Hill Higher Education. 2007. Peck. John W. 2007. and Susan A.Weak Acids and Bases Bibliography Brady. Davis. Chemistry. Belmont. Silberberg.. 4th ed. 2006. Bursten. Raymond E. Terry W.. CA: Thomson/Brooks/Cole. NJ: Pearson/Prentice-Hall. LeMay. 2005. 2006.. Perry.. Petrucci. 4th ed. 4th ed. John W.. Petrucci. 621. 705. General Chemistry. Ralph H. Chemistry: The Central Science. James E.. Hoboken. McCreary. and Scott S.. (Picture 7) Brady. Chemistry: The Central Science.. Ralph H. (Picture 3) Hill. NJ: Pearson/PrenticeHall. Hoboken. 4th ed. Terry W. Theodore E. General Chemistry. (Picture 2) Hill.... (Picture 5) Brown. 2005. McCreary. Upper Saddle River. Upper Saddle River. 706. Perry. Terry W. Chemistry: The Central Science. 2006. Terry W.SPECIAL FOCuS: Acids and Bases Graphics Index (Picture 1) Brady. John W. Ralph H. Upper Saddle River.. LeMay. Chemistry: Matter and Its Changes. 703. 748. and Scott S. Ralph H.. NJ: Pearson/Prentice-Hall. 2003. 2003. 2005. Theodore E. and Fred Senese.. 4th ed. 10th ed. 4th ed. LeMay. 617. 2005. 622. (Picture 8) Hill. (Picture 10) Brown.. (Picture 9) Brown. (Picture 6) Hill. Perry. 738. John W. Theodore E. Bursten. and Bruce E. General Chemistry. Bursten. and Scott S. Ralph H. John W. Petrucci.. Terry W. and Bruce E. Terry W. James E. NJ: Wiley... 622. John W. and Scott S. Upper Saddle River. (Picture 4) Hill. Upper Saddle River. and Bruce E. 2005. Upper Saddle River.. NJ: Pearson/Prentice-Hall. NJ: Wiley. Upper Saddle River. 10th ed. and Scott S. Eugene H.. NJ: Pearson/PrenticeHall. Chemistry: Matter and Its Changes. NJ: Pearson/Prentice-Hall. Perry. Eugene H.. NJ: Pearson/PrenticeHall. 617. Bursten. McCreary. 2006. General Chemistry. 26 . 10th ed. 4th ed. Eugene H. NJ: Pearson/Prentice-Hall. Perry. Petrucci. NJ: Pearson/Prentice-Hall. Petrucci. McCreary. McCreary. and Fred Senese.. Upper Saddle River. General Chemistry. 4th ed. New York: McGraw-Hill Higher Education. NJ: Wiley. James E. (Picture 16) Silberberg.. 752.. 4th ed. (Picture 15) Silberberg. Chemistry: The Molecular Nature of Matter and Change.. General Chemistry. 2003. New York: McGraw-Hill Higher Education. 4th ed. Perry. Martin S. 623. 2nd ed. 2nd ed. NJ: Pearson/Prentice-Hall. (Picture 12) Silberberg. 2003. Chemistry: The Molecular Nature of Matter and Change. Chemistry: Matter and Its Changes. 780.Weak Acids and Bases (Picture 11) Hill. and Scott S. 751. NJ: Wiley. Ralph H. Chemistry: The Molecular Nature of Matter and Change. Upper Saddle River. Hoboken. Terry W. Chemistry: Matter and Its Changes. Hoboken. James E. New York: McGraw-Hill Higher Education. 2nd ed.. Petrucci. (Picture 14) Brady. 27 . 2000.. 792. Martin S. John W. and Fred Senese. (Picture 13) Brady. Martin S.. 786.. 4th ed. McCreary. 2005. 2000.. and Fred Senese. 2000.. . Rather than proceeding to completion. where the concentration of one or both of the reactants is completely exhausted. In terms of calculations.-79.6 kJ ∕mol . Kentucky The neutralization of a strong acid with a strong base occurs according to the net ionic equation1: Equation 1: H+ + OH.0 M OH-.0 × 1014 1. ΔGº298K = -79. it is omitted from the equilibrium constant expression.00831 K eq = e . . when used correctly.nG o RT = e kJ ⋅298 K mol ⋅K = e32. “H3O+(aq). Because the concentration of the pure water is unaffected by the progress of the reaction. the reaction does attain a state of equilibrium in which small but measurable concentrations of both H+ and OH.→ H2O .2 = 1. The symbol “H+(aq)” is a shorthand representation of the hydrated proton.8 kJ mol 0.” Because the number of water molecules coordinated to a single proton varies with time. the symbol H+(aq)” is a convenient and suitable representation. ΔHº = -55. the acidic proton is undoubtedly coordinated to several water molecules through and is often written as the hydronium ion. Keq: K eq = 1 [ H ] ⋅ OH− + The value of Keq can be readily calculated as: .8 kJ ∕mol The negative sign of ΔGº298K . Francis School Louisville.pH and the Properties of [H+] in Aqueous Solution Lew Acampora St. and experience tell us that the reaction proceeds spontaneously from standard initial concentrations of 1. The state of equilibrium is quantitatively described by the equilibrium constant expression. it is probably most accurately represented as “H2n+1On+(aq).” where “n” represents the number of water molecules. In aqueous solutions.are present.0 M H+ and 1. 0 × 10-14 OH− = 1. Solutes that have this effect on the relative concentrations of H+ and OH.ion.0 × 10-7 M While the [H+] = [OH-] in pure water.SPECIAL FOCuS: Acids and Bases While the magnitude of the value of Keq reflects the observation that the equilibrium state of this reaction lies far to the right.0 × 10-14 = 1. In determining the [H+] and the [OH-] in an acid. the value of Kw at 25oC is commonly used.020 0. any substance that increases the [H+] must also decrease the [OH-] in order to satisfy the equilibrium relation. For example. In pure water2: [ H+ Η + = 0Hˉ. base. the equilibrium mixture will maintain small but measurable concentrations of both hydrogen ion and hydroxide ion.020 -0.020 The [H+] = 0. the [H+] can be determined directly from the concentration and strength of the acid. The symbol Kw represents the specific equilibrium constant that refers to the auto-ionization of water. the reverse reaction of Equation 1.020 0. the contribution from the autoionization of water can be neglected. Conversely.020 M solution of the strong acid hydrochloric acid is readily determined as follows: Equation 3: [ ]o Δ[ ] [ ]f HCl(g) 0. many solutes (acids or bases) will change the relative concentrations of these ions. is used to determine the [H+] and [OH-] in aqueous solutions.. Commonly.020 M.0 × 10-14 = = 2.0 × 10-13 M + H 0. Therefore. [H+] [OH-] = 1. In terms of calculations expected on the AP Chemistry Examination. In a solution of an acid. In any case. and the [OH-] is subsequently calculated using Equation 2. its value is a function of the temperature. or buffer solution containing solutes of reasonable concentration.020 2. Like all equilibrium constants.0 × 10-14 The implication of this calculation is that even in the purest sample of water at 25 oC. Kw = [ H+ ] ∙ [ OH– ] = 1.020 0 → H+(aq) ~0 +0.020 + Cl-(aq) 0 +0.8 kJ ∕mol . bases are substances that will increase the [OH-] and will decrease the [H+]. ΔGº298K = +79. the equilibrium relation.0 x 10-14.] ] = [ OH = 1. 30 .0 × 10-14 1. there are small but measurable concentrations of H+ and OH. the [H+] ion in a 0.are called acids. will remain applicable in any aqueous solution at 25 oC. The [OH-] ion determined from Equation 2: H+ ⋅ OH− = 1. which represents the autoionization of water. In this case: Equation 2: H2O  H+ + OH. 00. In fact.00 to 0. [OH-]. Solutions of weak acids only. Solutions of weak bases only. Solutions of strong bases only. but no additional information is conveyed by the pH of the solution that is not already indicated by the [H+].0 × 10-14 = = 1. The value of Kw. The range of values for [H+] in aqueous solutions.050 0 ~0 -0. the pOH of an aqueous solution is defined as pOH = -log10 [OH–].70. like all equilibrium constants. is temperature dependent. and pOH imply that the value of any one of these will fix the values of the others.050 M solution of barium hydroxide are determined as: Equation 4: [ ]o Δ[ ] [ ]f Ba(OH)2(s) → Ba2+(aq) + 2 OH-(aq) 0.00. 31 .10 The [OH-] = 0.050 +0.pH and the Properties of [H+] in Aqueous Solution For an aqueous solution of a base. the above relationships among [H+].050 0. 3. and subsequently the [H+] can be determined. pH. and five types of calculations can be identified: • • • • Solutions of strong acids only.050 +0.10 0 0.0 x 10-13 = 13. Analogously. and shows that the pH value is the negative of the exponent of 10 when the concentration is expressed as a power of 10.10 M. Referring to the previous examples.020 = 1. where pH is defined as pH = -log10[H+]. the [OH-] is directly calculated.0 x 10-14 M < [H+] < 1. approximately 1.050 M Ba(OH)2(aq) solution is pH = -log10 1. translates to a pH range from 14.020 M HCl(aq) solution is pH = -log10 0. The [H+] ion determined from Equation 2: H+ ⋅ OH− = 1.0 × 10-13 M − OH 0. the concentrations of each ion in a 0. For example.0 x 10-7 =7.0 M . and [OH–] = 10-pOH.10 The preceding examples illustrate that both the [H+] and the [OH-] in aqueous solutions vary over a range that is greater than 14 orders of magnitude. and the pH of pure. The pH scale is commonly used as an alternative expression to the [H+].3 Calculation of these values for aqueous solutions will depend on the type of solute. this is equivalent to [H+] = 10-pH. the pH of the 0. neutral water is then pH = -log10 1.0 x 10-14 at standard temperature of 25 oC. The logarithmic pH scale conveniently represents the [H+] over the 14 orders of magnitude.0 × 10-14 H+ = 1. the value of Kw = 1.0 × 10-14 1. the pH of the 0. Algebraically. While the relation: Kw = [H+] [OH-] is generally true.00. The indicator is intensely colored.01 pH unit. Indicators with different pKa’s can be used to measure different ranges of pH values. defined as –log10 Ka. In practice.SPECIAL FOCuS: Acids and Bases • Buffer solutions (containing appreciable concentrations of a weak acid and its conjugate base). The laboratory determination of the [H+] of an aqueous solution. The indicator will then ionize according to the equation: H-In ← H+ + In→ The acid dissociation constant of the indicator is written as K a. the test electrode and the reference electrode are combined into the same body. and the color of the solution changes from that of the acid form to the color of the base form as the pH of the solution changes from (pKa – 1) to (pKa + 1). Examples of these calculations are described elsewhere in this manual. and the solution will have the color of the base form of the indicator. either in solution or adsorbed onto pH paper. so that a negligible concentration of the indicator is added to the solution – sufficiently dilute so that it does not have an appreciable effect on the [H+] of the solution. [In-] >> [H-In]. and the pH meter is simply a high impedance voltmeter that is calibrated to reflect the pH of the solution. is compared to the pH of the solution. whose electrochemical potential is sensitive to the [H+] in the solution being measured. More precise pH measurements are made with a pH meter. or equivalently. In . [In-] << [H-In] and the solution will have the color of the acid form of the indicator. and an internal reference electrode. laboratory pH meters are routinely capable of making accurate pH measurements to ±0. [ H − In ] In− K a. showing that the ratio of the This equation can be rearranged as [ H − In ] H concentration of the conjugate base form to the acid form depends on the value of Ka. In practice. of the indicator.In = + . Conversely. A chart of commonly used indicators and their transition point (pKa’s) can be found in most textbooks. can be done in two ways. For Ka. affords a crude determination of the pH of a sample. In >> [H+].4 The measured potential depends on the [H+] in the external solution. 4. In << [H+]. An indicator is itself a weak acid whose color is different from that of its conjugate base. the fundamental operation of a pH meter is to measure the electrochemical potential difference between a glass membrane electrode. and the [H+] in the solution. the pKa of the indicator.In = . With careful calibration using standard buffer solutions. H+ ⋅ In− 32 . for Ka. its pH. referred to as a combination electrode. Though the technical details vary by instrument. The use of indicators. 8 x 10 -5] ? M Ba(OH)2 1. the concentrations of these ions in such solutions are often critically important.2 x 10 -5 M 1.05 3.8 x 10 -5] & 0.10 M CH3COOH [Ka = 1.30 33 .12 [H+] 0.50 M pH 1.0 x 10 -7 M 1.00 0.0 x 10 -3 M 13.1 x 10 -9 M 1.8 x 10 -8 M 1.87 5.20 M NH4+ ? M NH3 [Kb = 1.0 x 10 -14 M [OH-] 2.10 M NH3 [Kb = 1.70 5.00 13.0 x 10 -13 M 7. the role of pH as a measure of the concentration of not only the H+ ion but also the OH.050 M HCl 0.70 pOH 13.88 7.13 8.12 7.0 x 10 -3 M 0.30 2.0 x 10 -6 M 1.10 M CH3COOH [Ka = 1. Though these equilibrium concepts are sometimes difficult for students to master.0 x 10 -11 M 2.0 x 10 -7 M 9. Because chemical properties of the solution are sensitive to variations in the concentrations of these ions.10 M NH3 [Kb = 1.050 M 1.00 5.ion is an important step that will allow students to go further in their mastery of the material. the equilibrium that exists between H+ ions and OH.8 x 10 -5] & 0.5 x 10 -12 M 8.20 M NH4+ 0.8 x 10 -5] & ? M CH3COO Pure Water 0.050 M [OH-] pH pOH Answer: Solute and Concentration 0.10 M CH3COOH [Ka = 1.10 M CH3COOH [Ka = 1.ions in any aqueous solution determines much of the chemistry of the solutions.8 x 10 -5] 0.00 8.056 M NH3 [Kb = 1.8 x 10 -5] 0. Exercises Complete the following table: Solute and Concentration ? M HCl 0.95 11.25 M Ba(OH)2 [H+] 0.24 M CH3COO Pure Water 0.8 x 10 -5] 0.pH and the Properties of [H+] in Aqueous Solution In closing.3 x 10 -3 M 1.70 11.8 x 10 -5] & 0. 10 M HCl is a solution of a strong acid: H+ f = [ HCl ]o = 0.10 M NaOH. Answers: a.0 mL of a buffer of pH 4. Find the pH of each of the following: a.0 mL of 0.0 mL of a solution of 0. The strong acid and strong base will neutralize. and there will be excess base remaining: 34 .10 M × = 0. Choice (D) is the only choice that matches these constraints.10 = 1.0 mL of distilled water.10 M HCl. Because a 0. 3.0010 = 3.10M pH = −log 10 0. 3.10 M HCl with 990.1-molar solution of a strong base would have a [OH-] = 0.1-molar ammonia is approximately (A) 1 (B) 4 (C) 7 * (D) 11 (E) 14 Students should recognize that ammonia is a weak base and that a solution of a weak base will have a pH > 7. d. A solution prepared by mixing 50. 0.00 b. a [H+] ion = 1 x 10-13 M and pH = 13. Solution of a strong acid: 10 mL 990 mL + 10 mL pH = −log10 0.0010 M c.1 M.SPECIAL FOCuS: Acids and Bases AP Exam Questions – Past Exams (1984) 2.10 M HCl with 55.0 mL of 0.00 with 50. c.00 H+ f = 0. b. 0. the same concentration of a weak base would have to have a lower pH value. The pH of 0. A solution prepared by mixing 45.0 mL of distilled water. A solution prepared by mixing 10. 0 mL sample of an aqueous solution of pure benzoic acid is titrated using standardized 0.00 . (2006B.4 × 10 pOH = pK w .46x 10-5 C6H5COOH(s) ← C6H5COO-(aq) + H+(aq) → 2.0010 M = 0.0 × 10-10 M 35 .4.010 = 2.00 (assuming that after dilution.00 A− d.150 M NaOH.53 OH− = 10-pOH = 10-9. Benzoic acid. Q1ai.0 mL of the 0.150 M NaOH.00 = 12.53 = 3. C6H5COOH.0045 0. Answers: Students should be able to convert among [H+]. [OH-]. pH = 4.0045 0 OH− f = OH.47.pH = 14. (i) [H+] in the solution (ii) [OH-] in the solution 2.0010 H 2O 0 +0.0 × 10-14 = 3.47 = 3.00 . and pOH easily. [OH-]. (a) After addition of 15.10 L pOH = −log 10 0. [HA] >> [H+].0 × 10-10 M -5 3. pH. A 25. [A-]. the pH of the resulting solution is 4.0045 0.47 = 9. Because solution (d) is a buffer solution. Q1aii) Ka = 6.00 pH = 14.010 M 0. Calculate each of the following.2. the ratio [ HA ] will not be affected by dilution and the pH will remain unchanged.4 × 10-5 M (a)(ii) OH− = or Kw H+ = 1.0045 -0.→ 0. dissociates in water as shown in the equation above.0045 0. (a)(i) H+ = 10-pH = 10-4.0055 -0.pH and the Properties of [H+] in Aqueous Solution Rxn: molo ∆mol molf H+ + 0. 10 M = 3.2x 10-4 Hydrofluoric acid. A volume of 15.004 mol (d) [HF] = .40 mol L−1) − (0.15 M .015 L)(0. HF(aq) + OH-(aq) ← H2O(l) + F– (aq) → 2.40 M HF(aq) solution. Answers: (c) mol HF(aq) = initial mol HF(aq) − mol OH−(aq) added = (0.010 mol − 0.040L = 0.14 + log .32 [ HF ] 36 . (d) Calculate the molar concentration of F– (aq) in the solution. HF(aq) reacts with NaOH(aq) according to the reaction represented below.025 L)(0. (c) Calculate the number of moles of HF(aq) remaining in the solution.8 × 10-4 ) = 3.e) HF(aq) + H2O(l) ← H3O+(aq) + F– (aq) → Ka = 7. (e) Calculate the pH of the solution.32 OR pH = pK a + log F− = 3.40mol L−1) = 0.SPECIAL FOCuS: Acids and Bases (2007. dissociates in water as represented by the equation above.0 mL of 0. 2.2 × 10-4 ) 0.004 mol HF .8 × 10-4 M pH = -log10 (4.15 M = 4.10M HF K = a H3O+ F− [ HF ] [ HF ]× K a H3O (e) + = F − = 0.10 M × (7. HF(aq).0 mL of 0.40 M NaOH(aq) is added to 25. Q1c . Assume that volumes are additive.0060 mol = 0. given the value of any one of these data. values for the remaining are fixed and can easily be determined. but they are not available for the multiple-choice questions. only weak base. [OH-]. only buffer solution containing both a weak acid and its conjugate base Students should understand which principles/equations are applicable to each of these instances. pH. 37 . only weak acid.pH and the Properties of [H+] in Aqueous Solution Teaching Strategies Introduce the auto-ionization of pure water as the reverse of the neutralization reaction of a strong acid and a strong base. Emphasize the mathematical relationships among [H+]. Emphasize that both H+ and OH. Remind students that the relevant equations are provided on the insert for the free-response section of the AP Chemistry Examination. be sure that students can identify the type of solution that they are dealing with: • • • • • strong acid. and that the concentration of either particle will fix the concentration of the other. only strong base. and pOH for aqueous solutions and that. In determining the [H+] in an aqueous solution.are present in any aqueous solution. . the equivalence point. Approaching Titration Problems with an AP® Chemistry Class AP students may or may not be already familiar with introductory titration problems such as those involving the determination of an unknown solution’s concentration or finding the volume of a solution required to react with a standard solution. how many mL of 0. other quantities may be discerned. the equivalence point may be determined using a pH meter to follow the progress of the reaction. from their first course in chemistry. for example. British Columbia Titration is an important type of volumetric analysis usually used to determine the number of moles of solute in a certain amount of aqueous solution. (For . Alternatively. For students who have done little lab. In many acid–base titrations. Simple problems should involve the idea of “matching” equivalent amounts of strong acids and strong bases in solutions. a chemical indicator is added to signal the point at which the two reactants have been combined in just the right ratio to give a complete reaction. This is sometimes called the stoichiometric point or.50 M hydrochloric acid. The titration procedure generally involves the addition of carefully measured portions of a “standard” solution of known concentration to a measured mass or volume of an unknown. Once the amount is known. more commonly.Acid–Base Titrations Cheri Smith Yale Secondary School Abbotsford. the AP Chemistry teacher would be wise to briefly review some simple problems before using a more sophisticated lecture and laboratory approach. In any event.25 M sodium hydroxide will it take to be equivalent to 20 mL of 0. such as the concentration of the solution or the molar mass of the solute. a good introduction may be to titrate HCl with NaOH by drop counting in a well tray using phenolphthalein indicator. he or she can move on to the determination of pH experimentally and mathematically at different points during a titration.) • Determination of the Ka value for an unknown monoprotic acid. and strong acids or bases in excess. or at least teacher demonstration of actual variation of pH during titrations. more important for AP Chemistry. One of the biggest issues is the 40 . Once the teacher has reviewed and reinforced the students’ basic understanding of titration. including weak acid and base pH problems (ICE table calculations). (KHP. how many drops of 0. a strong acid or base with a weak acid or base. formic acid and propanoic acids are good unknowns for lab use as their Ka values differ by a full order of magnitude.1 mol/L acetic acid. KHSO4. When two solutions are combined. and hydrolysis reactions. Weak–strong curves are the source of a wealth of review. • Calculation of pH in a Theoretical Titration Curve— Use of the ICF Table Most AP Chemistry students are able to handle the calculation of pH in a solution containing only one dissolved species. An AP Chemistry teacher would find it beneficial to begin with strong–strong titration curves.1M NaOH are equivalent to 10 drops of 0. buffer use. Included should be titrations of a strong acid with a strong base and. buffer calculations.2 M HCl?) Examples of more sophisticated lecture topics and labs that should be done in AP Chemistry include: Determination of the molar mass of a unknown monoprotic acid. hydrolysis problems. The teacher may elect to include diprotic titrations and should certainly consider the selection of appropriate acid–base indicators. (Approximately 0. they find the problem considerably more difficult. These should be reinforced by either student run pH mater/probe use. and NaHSO3 are good unknowns to use in lab as their masses vary significantly from one another. however. as well as the use of pH meters or probeware. Calculating the pH at various points during a theoretical titration is an extremely valuable exercise for AP Chemistry students. including the applications of Kw and/ or pKw.SPECIAL FOCuS: Acids and Bases example. as it requires the application of virtually every type of pH problem encountered in the acid–base unit. thus reviewing the determination of the pH of strong acids and bases alone and in combination with one or the other in excess.) • Back titrations to determine the strength of antacid products. the AP Chemistry teacher might ask the class members to take several minutes. with no further information. hence “F” replaces “E. It is similar to the ICE chart used for equilibrium problems in that the “I” represents “initial” concentrations or numbers of moles and the “C” is the “change” in the initial quantities during the reaction.Acid–Base Titrations organization of the increased amount of information given in such problems. is replaced by an “F” that stands for “final” quantities following a reaction that goes to completion in one direction. to list any information they can 41 .100 mol/L HCl with NaOH. Such a reaction in a titration curve calculation will involve neutralization by a strong acid or base. FIGuRE 1 pH Providing such a curve for the titration of 25 mL of 0. working alone or collaboratively. pre-AP and AP Chemistry teachers alike would be wise to apply the ICF technique early as it will build student confidence for its application to more difficult problems later. The ICF table is a valuable tool for AP Chemistry teachers and students facing problems that involve chemical reactions such as neutralization with limiting and excess reactants. Because one (or both) of the reacting species is strong.0 mL of 0. Consider the curve that describes the titration of 25. Strong Acid–Strong Base Titrations Though the pH at various points during the titration of a strong acid with a strong base (or vice versa) is relatively easy to determine. the reaction goes virtually to completion.” Such a reaction is said to be quantitative and is useful for analytic calculations. The traditional “E.100 mol/L HCl.” for “equilibrium” in an ICE chart. the teacher might prefer to produce a curve using probeware (such as that available from Vernier or Pasco) or to use a variety of simulations available commercially or on the Internet (see suggestions in the laboratory section at the end of this chapter).100 mol HCl x 1 mol NaOH = 1000 mL 1 mol HCl 0. • The moles NaOH = moles OH.0 • The volume of base required to reach equivalence = 23. 42 .5 mL. Production of the curve could be done after. the AP Chemistry teacher can discuss the calculation of pH at various points throughout the curve.= 25. • The reaction is represented by: HCl + NaOH → NaCl + H2O • The initial pH = 1. ideally in molecular form. during.0 • The initial [HCl] = [H+] = 10-pH = 10-1.100 mol/L solution of HCl being titrated in the profile above. Sample student responses might include: • Base is being added to acid. • The pH at equivalence = 7. Once students have examined a pH profile.00250 mol x 23. or a titration curve for a strong monoprotic acid being titrated with a strong base.SPECIAL FOCuS: Acids and Bases determine about the titration from the pH profile.0 mL sample of 0. Any AP Chemistry text will contain a sample curve to which teachers can refer.1 mol/L • The acid is strong as the initial [H+] = the initial [HCl] indicates 100% ionization. however.00250 mol NaOH are required to reach equivalence. or even before the calculations. • The [NaOH] = 0.106 mol/L AP Chemistry teachers should reinforce the importance of beginning titration calculations with a balanced chemical equation.5 mL 1000 mL 1L = 0. Example 1 Calculate the pH of the 25. This leads to a proper mole ratio that is most likely to be correct when units include the formula of the species they refer to.0 = 0.0 mL x 0. HI.0714 mol/L [NaOH]in = 0.log [HCl] = . Note that a “rounder” concentration of base might be used for introduction. this is what we’ve determined for our example. H2SO4 is only nearly completely ionized for its first ionization) as there is no table of acid ionization constants included with the AP Chemistry Exam. Strategy: The ICF table is very useful for helping students keep track of everything necessary to solve this problem.0303 F: 0. and H2SO4 (HClO3 is included in some texts. If concentrations are to be used (which may be preferable because pH depends upon concentration rather than moles).0303 mol/L (10. The dilution factors are 25/35ths and 10/35ths.100 mol/L) = 1. Solution: [HCl]in = 0.0303 -0. Hence.log [H+] and that the concentration of a solution is independent of its volume. although it ionizes less than 100%).respectively.0303 +0.0714 M 0.0 mL of 0.log (0. (AP Chemistry students must be aware of the “BIG 6” strong acids: HCl.Acid–Base Titrations Strategy: The student should recognize that pH = .0 mL is irrelevant to the problem at this point.0 mL) (35. however. Step one is to dilute. HBr.0 mL) = 0. ionize completely.0303 43 .106 mol/L = 0. HNO3. HClO4.100 mol/L (25.0 mL sample of HCl following the addition of 10. students must recognize that the concentration of a solution is affected by the addition of water or another solution as both of these cause dilution.106 mol/L NaOH.0411 0 0. [H+] = [HCl].0303 M 0M C: -0.0 mL) Neutralization in ICF shows: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) I: 0.0 mL) (35.) Solution: pH = . Step two is to neutralize by applying the ICF table. 25. Step three is to examine the bottom line of the ICF table to determine what affects the pH.log [H+] = .000 Example 2 Calculate the pH of the 25. Strong acids such as HCl. hence. 0515 -0.5 mL) = 0. though it can never actually equal 0.5 mL of a 0.100 mol/L (25.0 mL sample of HCl following the addition of 23. Hence. which dissociates to form the neutral cation.5 mL of 0.0515 F: 0 0 (23.SPECIAL FOCuS: Acids and Bases The bottom line indicates a [HCl] = [H+] = 0.5 mL) (48. dilute and neutralize.log [H+] = .) One final note that should be mentioned is that the ICF table indicates that the [NaOH]final = 0 mol/L. Cl-. and so ≈ 0.0515 M C: -0. As the conjugate base of the strong acid. [OH-] = [NaOH] at equivalence is negligible and approaches 0. the same pH as shown. Solution: [HCl]in = 0. This is. then examine the bottom line of the ICF table. impossible. Strategy: As in example 2. as [H+][OH-] must always = 10-14 = Kw (at 25oC). Na+ and an anion. HCl.0515 0.0515 mol/L solution of NaCl.106 mol/L NaOH. The final moles of acid divided by a total volume of 0. Cl. Determination of the pH at some point beyond 44 . it too is neutral. of course.0 mL) [NaOH]in = 0.5 mL) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 0 M+0.386 (Note that moles might have been used by multiplying the molarity of each reactant by its volume and using the moles in the “I” line.00106 moles. Example 3 Calculate the pH of the 25.0515 mol/L = 0.0411) = 1.0515 mol/L Neutralization in ICF shows: I: 0. pH = . then each = √ 10-14 = 10-7.log (0.0411 mol/L.0515 M 0. following the neutralization. hence.is too weak to accept any appreciable quantity of protons from the water it is dissolved in and hence. which is very small compared to the molarity numbers on the problem. If [H+] = [OH-].0350 L would give the same concentration and consequently. The changes would then have been 0.10638 mol/L (48.0515 This process shows that the pH of 7 at the equivalence point is due to the formation of 48. 0 mL of 0.0 mL sample of HCl following the addition of 30.00 – 1.log (0. dilute and neutralize.90 = 12. Weak–Strong Titrations A classic weak–strong titration involves the titration of a weak carboxylic acid.0455 M 0.10.0455 +0.0455 F: 0 0. Strategy: As in example 2. is completely neutral and does not affect the pH at all.0125 0. NaCl.0125 = 8. From this number.0 x 10-13) = 12.0 mL) Neutralization in ICF shows: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) I: 0.0 mL) = 0. it becomes evident that the pOH = .0 mL) [NaOH]in = 0.106 mol/L NaOH solution.log (8. However. Solution: [HCl]in = 0. the pH = 14.0580 M 0 MC: -0.Acid–Base Titrations equivalence in the same fashion is helpful as it involves the application of Kw and/or pKw along with the ICF table. pH = . Example 4 Calculate the pH of the 25. The application of Kw and/or pKw will be required to convert the [OH-]. The alternative approach would be to recognize that [H+] = Kw/[OH-] = 10-14/0.0455 mol/L = 0. Hence.90. such as ethanoic (acetic) acid or methanoic (formic) acid. Consequently. 45 . then examine the bottom line of the ICF table. Once students have examined a pH profile. the AP Chemistry teacher can compare the calculation of pH at various points throughout the curve with that of a strong–strong titration.0 x 10-13 M.0 mL) (55.0455 As explained in example 3. the salt.0125) = 1.100 mol/L (25.0455 -0.10638 mol/L (55.0580 mol/L (30. with a strong base.log [OH-] = -log[NaOH] = . or titration curve for a weak monoprotic acid being titrated with a strong base.10.00 at 25oC. the [NaOH]excess is very significant. as pKw = pH + pOH = 14. 100 • The pH rises more quickly during a weak–strong titration.5 mL (same value).→ H2O + CH3COO-. the AP Chemistry teacher might ask the class to take several minutes.100 mol/L acetic acid. 0. • The initial pH approximates 2. working collaboratively with a partner. for the titration of 25 mL of 0. • The pH at equivalence is between 8 and 9 (above the neutral value of 7).9 (higher than the strong acid).3%. This is typical of a weak acid.0013 mol/L. • The volume of base required to reach equivalence = 23. Sample student responses might include: • • Base is being added to acid. to list any information they can determine about the titration from the pH profile. • The net ionic reaction is CH3COOH + OH. • This is a particularly significant point for students to recognize and it should be emphasized that the strength of an acid has no bearing on the volume 46 .9 = 0. with emphasis on any obvious differences they notice between this curve and the earlier strong–strong one. with no further information. • The initial [H+] = 10-pH = 10-2.0013 x 100% = 1.SPECIAL FOCuS: Acids and Bases FIGuRE 2 pH ml NaOH added Providing such a curve. • The % ionization = 0. • The acid is not as strong as the initial [H+] << the initial [CH3COOH] indicates less than 100% ionization. The reaction is represented by: CH3COOH + NaOH→ NaCl + H2O. 23.00250 mol x 1000 mL = 0. resulting in the following equilibrium: CH3COOH + H2O ↔ H3O+ + CH3COOI: 0.100 mol HA x 1 mol NaOH = 0.Acid–Base Titrations of base required to neutralize it.0 mL x 0. (Given the Ka of acetic acid = 1. • The moles NaOH = 25. weak base. Once students have examined a pH profile. only a small percentage of the molecules will donate protons.) Strategy: The student should recognize that acetic acid is a weak acid. One mole of monoprotic acid requires one mole of base to neutralize it completely whether the acid is strong or weak. or titration curve for a weak monoprotic acid being titrated with a strong base. Because this acid is weak.8 x 10-5. 47 . Example 5 Calculate the pH of the 25. AP Chemistry teachers should stress the review of weak acid. Solution: According to the Brønsted-Lowry definition. an acid will donate a proton to the base it is dissolved in. x may be considered negligible compared with the initial concentration.0 mL sample of a 0.00250 mol 1000 mL 1 mol HA required to reach equivalence. and hydrolysis calculations as they arise. the AP Chemistry teacher can discuss the calculation of pH at various points throughout the curve.100 M 0M 0M C: -x +x + x [H+] formed is defined as x E: ~ 0. The ICF table is particularly useful for students attempting to deal with the additional complexity of these problems.106 mol/L.100 mol/L solution of CH3COOH being titrated in the profile above.100 x x For weak acids that ionize less than 5%. It is important for students to consider the chemical composition of the solution being titrated as well as the process for calculating its pH at any particular point.5 mL 1L • The same sources suggested for strong–strong curves may be relied upon for weak–strong curves. buffer. hence an ICE table type of calculation is required to determine the [H3O+]. The [NaOH] = 0. Strategy: The ICF table may be used in an identical series of steps to those used for example 2 for the strong–strong titration shown previously.4 x 10-5) = 4.0303 0 0M +0.87.0303 F: 0.8 x 10-5 Since [H3O+] = 2.0 mL) = 0.0 mL sample of CH3COOH following the addition of 10.0411 0.log (2.3 x 10-3) = 2.0303 The bottom line indicates a [CH3COOH] = 0.0303 mol/L Neutralization in ICF shows: CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l) I: 0. The pH of a buffer may also be calculated using the Henderson-Hasselbalch equation: 48 .0411 = 1.100 Simple algebra reveals x = [H3O+] = 1.0 mL) (35. Example 6 Calculate the pH of the 25.61. The pH of this solution may be calculated as a weak acid with conjugate base present as a common ion: Ka = [H3O+] [CH3COO-] = [H3O+] (0.0 mL) (35.log (1.8 x 10-5 [CH3COOH] 0.3 x 10-3 mol/L and pH = .0303 M -0.log [H3O+] = .0303 0.0714 M C: -0.4 x 10-5 mol/L.0714 mol/L = 0. pH = . Solution: [CH3COOH]in = 0.0 mL) [NaOH]in = 0. The AP Chemistry teacher should point out that such a solution constitutes a buffer.0 mL of 0.106 M (10.0303 M.0303) [CH3COOH] 0.SPECIAL FOCuS: Acids and Bases Ka = [H3O+] [CH3COO-] = x2 = 1.106 mol/L NaOH.100 M (25.0411 M and a [CH3COO-] = [CH3COO-] = 0. The student should recognize the presence of a weak conjugate pair with significant quantities of both the conjugate acid and base present. 0515 F: 0 0.Acid–Base Titrations pH = pKa + log [CH3COO-] [CH3COOH] = 4.5 mL) (48.0 mL) [NaOH]in = 0. the quantity of excess weak acid exactly equals the quantity of conjugate base formed by the neutralization.0515 0 0M +0.0515 - The bottom line shows that the basic pH at the equivalence point is due to the formation of 48.5 mL) (48.0515 mol/L =0. Example 7 Calculate the pH of the 25. [H+] = Ka and pH = pKa.0 mL sample of CH3COOH following the addition of 23.” It is particularly instructive for the AP Chemistry teacher to point out that exactly half way to equivalence.0515 M C: -0. • past halfway and before equivalence.61 (0.0515 0.0515 M -0.74 + log (0. shows: Ka =[H3O+] [CH3COO-] = [H3O+] and pH = pKa + log [CH3COOH] [CH3COO-] = pKa+ log(1) = pKa [CH3COOH] The teacher should encourage AP Chemistry students to use either or both of these equations to recognize and to verify the changes that occur during the buffer zone portion of a weak–strong titration curve: • halfway to equivalence.0515 mol/L Neutralization in ICF shows: CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l) I: 0.5 mL of 0. [H+] < Ka and so pH exceeds pKa.0411) This portion of the curve is referred to as the “buffer zone.5 mL of a 0.100 M (25. [H+] exceeds Ka and so pH < pKa. Solution: [CH3COOH]in = 0. then examine the bottom line of the ICF table.10638 M (23. in turn.0515 mol/L solution of the salt. • prior to halfway.106 mol/L NaOH. Strategy: As in example 6.5 mL) =0. NaCH3COO. Substitution into each of the equations above. dilute and neutralize.0303) = 4. 49 . 27. NaCH3COO. CH3COO-. As in the strong–strong example 4 shown previously. Strategy: As in example 7.8 x 10-5 0.6 x 10-10 = x2 [CH3COO-] Ka 1. and pH = 14 – 5.73 Example 8 Calculate the pH of the 25.0 mL) [NaOH]in = 0. is weakly basic due to hydrolysis.0455 0. dilute and neutralize. CH3COO-(aq) + H2O(l) → CH3COOH(aq) + OH-(aq) I: C: E: 0. CH3COO.0515 x = [OH-] = 5.0M +x x 0M + x [OH-] formed is x defined as x hence.3 x 10-6 mol/L and pOH = .SPECIAL FOCuS: Acids and Bases which dissociates to form the neutral cation. However. Na+ and the weakly basic anion. the application of Kw and/or pKw will be required to convert the [OH-].0455 0.0 mL) (55.0515 M -x ~ 0. As the conjugate base of the weak acid.is strong enough to accept an appreciable quantity of protons from the water it is dissolved in and hence. the hydroxide 50 .0 mL) = 0.3 x 10-6) = 5. the [NaOH]excess is much more significant.0 mL sample of CH3COOH following the addition of 30.10638 M (30.106 mol/L NaOH solution. CH3COOH. it forms hydroxide ions by hydrolysis.0580 mol/L Neutralization in ICF shows: CH3COOH(aq) + NaOH(aq) ↔ NaCH3COO(aq) + H2O(l) I: C: F: 0.0455 mol/L = 0.0455 - As explained in example 7.0125 0M +0.0455 0 0. the salt. In fact.100 M (25.0 mL) (55. then examine the bottom line of the ICF table.log(5.0 mL of 0. Kb = [CH3COOH][OH-] = Kw = 10-14 = 5. Solution: [CH3COOH]in = 0.0515 .0580 M -0.0455 M -0.27 = 8. 90 = 12.Acid–Base Titrations ion from the NaOH forces the hydrolysis reaction so far left that its contribution to the [OH-] may be completely ignored. In this portion of the curve the solution contains the weak conjugate pair ammonia and ammonium ion. of course. the pOH = . The alternative approach would be to recognize that [H3O+] = Kw/[OH-] = 10-14/0. therefore it is a basic buffer.log (8. the pH = 14. • The use of a Kb expression or Henderson-Hasselbalch during the buffer zone. Initially one is determining the pH of a solution of ammonia.log [OH-] = .log (0.10. A classic example might be the titration of ammonia with hydrochloric acid. Kb. 51 .00 – 1. pH = .00 at 25oC.0125 = 8.0125) = 1. Consequently. the titration of a weak base with a strong acid.0 x 10-13 M. The [OH-] = [NaOH]excess for all intents and purposes. The alternative to titrating a weak acid with a strong base is. Hence. as pKw = pH + pOH = 14.90. FIGuRE 3 Such a titration would include the calculation of pH by: • The use of the ionization constant of a weak base.0 x 10-13) = 12.10. Hence. Since indicators are actually weak acids.73.0.2 and 10. an indicator will reach its endpoint and change color when the [H3O+] in the system equals the indicator’s Ka. in this case. Thus. The general form for the equation that represents the equilibrium for an indicator is: HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq) acid form base form (color one) (color two) If the indicator is placed in a stronger acid. Some have been used as dyes. endpoints should occur exactly at equivalence. A stronger base will pull protons from the indicator. Similarly.form (and color). ammonium chloride. On the other hand.SPECIAL FOCuS: Acids and Bases • The use of the acid ionization constant of a weak acid. which undergoes cationic hydrolysis. producing the HIn form (and color). This is called the endpoint or transition point for the indicator. different types of titrations will require different indicators. the acid will donate protons to the indicator.1. the pH of the system will equal the indicator’s pKa. When indicators are in the middle of their color change. the [HIn] = [In-] and the color shown will be a combination of the acid and base form colors. Selection of an Appropriate Acid-Base Indicator An acid–base indicator is a chemical species that has an acid and a base form with one or both forms being visibly colored. Ka at equivalence due to the formation of an acidic salt. Many indicators were originally isolated from organic plants or foods. Ideally. It has a pKa of approximately 9. • The application of the standard definition of pH applied to the excess [H3O+] present in a solution of ammonium chloride once excess strong acid has been added and the titration proceeds beyond equivalence. Substitution into familiar equations shows: Ka = [H3O+] [In-] = [H3O+] and pH = pKa + log [ In-] = pKa+ log(1) = pKa [HIn] [HIn] Hence. producing the In. hence it would be quite appropriate for the acetic acid titration because example 7’s calculated equivalence point was pH 8. it would be absolutely wrong for use in the titration of ammonia with hydrochloric acid because the pH at equivalence for that titration 52 . Phenolphthalein changes color between pH 8. during the color change (at the indicator’s endpoint). they do have Ka and consequently pKa values. It is often a significant topic in the free-response portion of the examination. as shown in the third curve (Figure 3). FIGuRE 4 Sample Questions from Previous AP Chemistry Exams The topic of titration appears on virtually every AP Chemistry exam. an indicator’s pKa need not exactly match the pH expected at equivalence. The titration curve below shows that phenolphthalein or methyl red along with many other indicators would be appropriate for determining the equivalence point for a strong–strong titration. Because of this rapid change. the indicator must simply complete its entire color change during the vertical part of the titration curve. Rather. One significant point demonstrated by each of the curves we’ve studied is the rapid change in pH that suddenly occurs at equivalence.Acid–Base Titrations would be around 5.4. AP Chemistry teachers can demonstrate this dramatically through the use of a pH meter or probe ware (such as that available from Vernier or Pasco) or a variety of simulations available commercially or on the Internet (see suggestions in the laboratory section at the end of this chapter). This means that phenolphthalein might be selected ahead of an indicator having a pKa closer to the pH at equivalence simply because its change from colorless to pink is so clearly visible and easy to detect. Problem number 53 . SPECIAL FOCuS: Acids and Bases one of the three required problems is always an equilibrium problem. or greater than 7.160 M HOBr to produce a buffer solution with [H+] = 5. equal to 7.8 x 10-5 M (1.8 x 10-5 M. (e) HOBr is a weaker acid than HBrO3.14 M 54 + .3 x 10-9 1. is a weak acid that dissociates in water. (i) Calculate the volume of 0.95. (ii) Indicate whether the pH at the equivalence point is less than 7. Hypobromous acid. Assume that volume change is negligible.146 M HOBr(aq). This problem centers on an acid–base equilibrium nearly every other year. Explain.0 mL sample of 0. 2002 Operational Exam HOBr(aq) ↔ H+(aq) + OBr–(aq) Ka = 2. Such problems require student confidence with acid–base calculations of the sort discussed in this chapter. as represented by the equation above.3 x 10-9 X x = [HOBr] = 0.1 x 10-5 mol/L (b) Ka = [H+][OBr] = 2.00 ´ 10–9 M.3 x 10-9 [HOBr] [H+] = [OBr–] = 1. (c) A solution of Ba(OH)2 is titrated into a solution of HOBr. HOBr.8 x 10-5)2 = 2. Solution: (a) pH = –log[H ] . then calculate the concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.115 M Ba(OH)2(aq) needed to reach the equivalence point when titrated into a 65. [H+] = 10-4. (a) Calculate the value of [H+] in an HOBr solution that has a pH of 4. Account for this fact in terms of molecular structure. (d) Calculate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0. (b) Write the equilibrium constant expression for the ionization of HOBr in water.95 = 1. Calculate the equilibrium constant.160)(2. (b) A sample of aniline is dissolved in water to produce 25. Which of the indicators listed is most suitable for this titration? Justify your answer. (c) The solution prepared in part (b) is titrated with 0.736 mol OBr– x 1000 mL 1 mol OBr– (e) very electronegative oxygen is able to draw electrons away from the bromine and weaken the O–H bond. The pH of the solution is 8. 55 .3 mL 0.146 mol HOBr x 1 mol H+ x 1 mol OHˉ x 1 mol Ba(OH)2 1000 mL 1 mol HOBr 1mol H+ 2 mol OHˉ 1000 mL x = 41.82.00 ′ 109 1 mol NaOBr = 0.115 mol Ba(OH)2 (ii) pH > 7.0 mL of 0.0 mL of the acid has been titrated. (d) Calculate the pH at the equivalence point of the titration in part (c).00920 mol NaOBr 125 mL x 0. (a) Write the equilibrium constant expression. for the reaction represented above.0 mL x 0. making it easier for the hydrogen ion to be donated. Aniline. Calculate the pH of the solution when 5. for this reaction. salt of a weak acid is a weak base (hydrolysis of OBr-) (d) pH = pKa + log [A ] [HA] – or [A–] = [HA]Ka [H+] [OBr–] = (0. (e) The pKa values for several indicators are given below. a weak base.Acid–Base Titrations (c) (i) 65. Kb.3 ′ 109) = 0. O Br — O – H O 2003 Operational Exam C6H5NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH–(aq) 1. reacts with water according to the reaction represented above.10 M HCl.0736 M 5.10 M solution. Kb. SPECIAL FOCuS: Acids and Bases Indicator Erythrosine Litmus Thymolphthalein pKa 3 7 10 Solution (a) Kb = C6H5 NH+ OH− 3 [ C6H5 NH2 ] (b) pOH = 14 – pH = 14 – 8.82 = 5.18 –log[OH–] = 5.18; [OH–] = 6.61×10–6 M [OH-] = [C6H5NH3+] = 4.4×10–10 0.10 – 6.61× 10−6 0.1mol (c) 25 mL x = 2.5 mmol C6H5NH2 1L 0.1mol 5 mL × = 0.5 mmol H+ added 1L 2.0 mmol base remains in 30.0 mL solution (may determine from ICF) 0.50mmol  30.0ml    4.4 × 10−10  20.0mmol   30.0mL    –9 – X = 1.80×10 = [OH ] 1 × 10−14 [H+] = = 5.6×10–6; pH = 5.26 1.8 × 10−9 Kb = (6.61×10−6 ) 2 [ X ] X +  (d) When neutralized, there are 2.5 mmol of C6H5NH3+ in 50.0 mL of solution, giving a [C6H5NH3+] = 0.050 M (may determine from ICF.) This cation will partially ionize according to the following equilibrium: C6H5NH3+(aq) ↔ C6H5NH2(aq) + H+(aq) at equilibrium, [C6H5NH2] = [H+] = X [C6H5NH3+] = (0.050–X) X2 = Ka = 2.3×10–5 (0.050 − X ) X = 1.06×10–3 = [H+] pH = –log[H+] = 2.98 56 Acid–Base Titrations (e) Erythrosine; the indicator will change color when the pH is near its pKa, since the equivalence point is near pH 3, the indicator must have a pKa near this value. The lab-based essay question has focused on acid-base titrations. Occasionally, the core concepts of titrimetry will be tested in an essay question that is not lab based. The following is an example of a lab-based acid–base essay question: 1998 Operational Exam Lab-Based Question An approximately 0.1–molar solution of NaOH is to be standardized by titration. Assume that the following materials are available. • • • • • • Clean, dry 50 mL buret 250 mL Erlenmeyer flask Wash bottle filled with distilled water Analytical balance Phenolphthalein indicator solution Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard) (a) Briefly describe the steps you would take, using the materials listed above, to standardize the NaOH solution. (b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution. (c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve. 57 SPECIAL FOCuS: Acids and Bases FIGuRE 5 pH Volume of NaOH Solution Added (mL) (d) Describe how the value of the acid–dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c). (e) The graph below shows the results obtained by titrating a different weak acid, H2Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve. FIGuRE 6 pH Volume of NaOH Solution Added 58 5 mL volume point. Ka = 10-pKa (e) Y2– 59 . • Record end volume of buret. moles of NaOH = molarity of NaOH L of titrant FIGuRE 7 equivalence point pH Volume of NaOH Solution Added (mL) (d) From the titration curve. (b) mass of KHP molar mass KHP = moles of KHP.Acid–Base Titrations Solution: (a) • Exactly mass a sample of KHP in the Erlenmeyer flask and add distilled water to dissolve the solid. • Rinse the buret with the NaOH solution and fill. • Repeat to check your results. at the 12. • Record starting volume of base in buret. since KHP is monoprotic. • With mixing. the acid is halfneutralized and the pH = pKa. • Add a few drops of phenolphthalein to the flask. this is the number of moles of NaOH. titrate the KHP with the NaOH solution until it just turns slightly pink. This disc includes several lab exercises that students may perform virtually. contains two of them (#8 and #10). 2006). 2005. for example. Peoples Publishing Group. Additionally Chemistry and Computers by Holmquist and Volz. such as H3PO4 with NaOH. by Brown. Chemistry. includes all three of these labs (14A. contains five of these labs (#23–#27) each with specific direction for use of probe ware. CH3COOH with NaOH as examples of strong acid–strong base and weak acid–strong base curves. 10th edition. the efficiency of an antacid or the concentration of a diprotic acid. of course. based on conductivity of. 2003. There are a number of excellent resources for laboratories such as these. An even simpler experiment (titration by drops) may be useful for students who have not encountered any form of titration in a previous class. 2005. to list a few possibilities. Such an experiment will help bridge students’ pre-AP or general chemistry skills to those required for AP Chemistry. As suggested earlier. Ba(OH)2 with H2SO4 is another possibility. Some of them are commercially available on DVD or CD. LeMay and Bursten (Prentice Hall. Other possibilities include the titration of a weak base such as NH3 with HCl or a polyprotic acid. The third experiment involves the determination of appropriate indicators for different acid–base titrations.SPECIAL FOCuS: Acids and Bases Recommended Laboratory Exercises Related to Acid–Base Titration Wet Labs Of the 22 recommended laboratory experiments for AP Chemistry. HCl and second. the first experiment should involve the preparation of a primary standard that could be used to standardize a base that might be used to determine a molar mass. The classic two curves are. 14G and 14 H). on the computer. the moles of water of hydration. including a titration 60 . A potentiometric titration. One of the more popular is the Virtual Chemlab disc provided with the text. Laboratory Investigations: AP Chemistry by Hostage and Fossett. Essential Experiments in Chemistry by Morrison and Scodellaro. The second experiment should involve the development of experimental titration curves to be graphed and interpreted by the students. three involve acid–base titrimetry. Virtual Labs There are a number of laboratory exercises that may be done virtually. a Ka value. the titration of first. SMG Lab Books. the Central Science. H. and four different bases that may be placed in the buret. The data they collect may be presented in tabular and/or graphic form. Students produce a titration curve and use it to determine the pKa of a weak acid as well as its concentration and finally its molar mass.edu/~ylwong/chem/titrationsimulator/index. Another great Web site for curves was prepared by Yue-Wing Long from Wake Forest University.edu/group/Greenbowe/sections/projectfolder/ flashfiles/to select a titration simulation that is very easy to use. Bridging to the Lab by Jones and Tasker. one of which uses an acid-base titration to determine the pKa of a food preservative.chem. 2002. Students may select from several weak or strong acids. This site is particularly useful as it allows students to alter the strength of the acid they use by selecting various Ka or concentration values for the acids they titrate.iastate. Visit www. Start at www.html to actually plot a titration curve that may be used to determine the concentration and/or the pKa of a variety of weak acids. 61 . Freeman and Company.wfu. one of two possible indicators. A fantastic source of Flash-based chemistry simulations has been prepared by Tom Greenbowe from Iowa State. there are some Web sites whose authors have graciously provided their lab exercises free of charge.Acid–Base Titrations lab that allows them to select various reagents to use in a variety of titrations. includes 11 different virtual lab exercises. W. In addition to the commercial lab exercises. . attributed to G. A Lewis acid is an electron-pair acceptor. an acid is defined as a proton donor.ion. . The more general definition of acids. it is classified as a strong acid: HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) or HCl(aq) → H+(aq) + Cl-(aq) Hydrogen fluoride. however. Though this definition is more general. For example. Though all Brønsted-Lowry acids contain hydrogen. the strength of the intermolecular forces between the HX molecule and the water molecules. This state of equilibrium is indicated by the double arrows and leads to the classification of hydrofluoric acid as a weak acid: HF(aq) + H2O(l)  H3O+(aq) + F-(aq) or HF(aq)  H+(aq) + F-(aq) The difference in the behavior of HCl and HF can be traced to such factors as the relative strength of the H-X bond. Francis School Louisville. The extent to which a species donates a proton is a measure of the acidity of the species and is ultimately determined by the electronic structure of the particle.5 The common and unique reactivity of the proton.Polyprotic Acids Lew Acampora St. Kentucky According to Brønsted-Lowry formalism. N. 5. defines acids and bases in terms of their capacity to accept or donate pairs of electron. and the energy of hydration of the X . most quantitative problems in AP Chemistry can be approached using the Brønsted-Lowry formalism. imbues acids with their familiar properties. dissolves into water and reaches a state of dynamic equilibrium in which a fraction of the HF molecules will be ionized at any instant. also represented as the H+ ion. because gaseous hydrogen chloride dissolves and ionizes completely in aqueous solutions. Lewis. the converse statement—that all hydrogen-containing compounds act as acids—is not generally true. first. The Brønsted-Lowry formalism considers an acid–base reaction as simply a proton transfer reaction between a proton donor (the acid) and a proton acceptor (the base). and finally the sulfate ion. A structural diagram for sulfuric acid shows that both hydrogen atoms are directly bonded to oxygen atoms. etc. the hydrogen atoms that are directly bonded to carbon are not ionized in aqueous solution. each acidic hydrogen can be donated. and such substances are generally referred to as polyprotic acids.SPECIAL FOCuS: Acids and Bases Hydrochloric acid and hydrofluoric acid are examples of monoprotic acids because each molecule contains only one ionizable hydrogen. the sulfuric acid molecule can lose both protons sequentially to become. For example. is another example of a monoprotic acid because. SO42-. In these substances. Additionally. H2SO4. there are two additional factors or considerations that should be noted.+ HB+ 64 . HSO4-. triprotic. one of the most common diprotic acids is sulfuric acid. A general schema for such a reaction with a monoprotic acid is: HA + B →A. and the character of these O-H bonds leads to the acidity of both hydrogen atoms. Acetic Acid. it is only the carboxylic hydrogen that is ionized in aqueous solution: CH3COOH(aq)  H+(aq) + CH3COO-(aq) Because of the covalent character of the C – H bond. the tendency to donate each proton will be different. the hydrogen sulfate ion. however. and will have ramifications in terms of the acid strength. FIGuRE 1 The underlying principles of reactivity that govern the behavior of monoprotic acids are generally true for polyprotic acids. Many substances contain more than one ionizable hydrogen atom. Thus. though each molecule contains four hydrogen atoms. The presence of more than one acidic hydrogen per molecule will affect the stoichiometry of reactions of those polyprotic acids. CH3COOH. with subsets including diprotic. or 0. Consider. H+ + PO4 3- 4. for polyprotic acids. As protons are removed from the substance.20 normal. If the reactant is a diprotic acid. In order to neutralize a given quantity of base. Similarly.10 M H 2SO4 solution would be considered 0. a 0. any remaining ionizable protons are more strongly bonded because a positively charged proton will be more strongly attracted to the more negatively charged particle. which decrease in value for successive ionizations.10 M solution of the triprotic phosphoric acid. Thus.2 x 10 -8 HPO42. the normality is simply twice the molarity.7 Kan. The acid dissociation constants for polyprotic acids are generally denoted as Kan.6 The tendency of a particular species to donate a proton—the acid strength—is ultimately traced to the electronic structure of the acid and its conjugate base. where the “n” refers to the nth proton donated by the molecular species.Polyprotic Acids The stoichiometry of the reaction shows that acid and base are consumed in equimolar quantities. H3PO4. the triprotic weak phosphoric acid. 7. only half as many moles of the diprotic acid are required: H2A + 2 B → A2.” For a diprotic acid. Steps in the complete ionization of phosphoric acid and the values of the acid dissociation constants are shown here: Reaction Equation H3PO4  H+ + H2PO4 - Equilibrium Constant Expression H +  ⋅ H 2PO4 −   K a1 =    [H3PO4 ] Equilibrium Constant Value 7. This is reflected in the values of the acid dissociation constants. then.+ HB+ While 1 mole of hydrochloric acid will completely neutralize 1 mole of sodium hydroxide. H+ + HPO42- H +  ⋅ HPO4 2−   Ka2 =    − H 2PO4    H +  ⋅ PO4 3 −  K a 3 =    2−  HPO4    6. then each mole of acid will contribute two moles of protons. for example. A diprotic acid would have different values of Ka1 and Ka2. the same neutralization would require only ½ mole of sulfuric acid. a 0.+ 2 HB+ or ½ H2A + B →½ A2. For this reason. The resulting species is a weaker acid than the original particle. 65 . a triprotic acid would have Ka1.3 x 10 -3 H2PO4 .20 N. Ka2. The concept of “normality” has fallen somewhat out of favor and can be omitted from the AP Chemistry course. and Ka3. would be 0. H3PO4. Normality in this context is defined as “moles of ionizable protons per liter of solution. successive ionizations of a polyprotic acid are generally weaker.8 x 10 -13 6. but it still may be useful in describing the stoichiometry of acid–base reactions.30 N. Other examples of such species include the monohydrogen phosphate ion. Equal volumes of 0.→ H2PO4 . dihydrogen phosphate ion.+ H2O H3PO4 + 2 OH . since the bases will accept protons stepwise. for the purposes of the AP Chemistry curriculum. is termed amphiprotic or amphoteric.6 x 10 -7 H2PO4 . and the hydrogen carbonate or bicarbonate ion. still contains two ionizable protons and is itself an acid. For phosphoric acids.→ PO4 3. Though these terms are not strictly synonymous.1 x 10 -2 HPO4 + H2O  H2PO4 + OH 2- H 2PO4 −  ⋅ OH −     Kb2 =  HPO4 2−    1.30 M sodium hydroxide are mixed. Net Ionic Equation H3PO4 + OH .+ 3 H2O H2PO4 .10 M sodium dihydrogen phosphate and 0. acting as a base.+ 2 H2O H3PO4 + 3 OH . and H2PO4.+ H2O  H3PO4 + OH - K b1 = [H PO ] ⋅ OH  3 4 − H 2PO4    −   1.are stronger than those in H3PO4. HPO42-. Equal volumes of 0. HCO3-. acting as an acid. the H-O bonds in H2PO4.10 M phosphoric acid and 0.20 M sodium hydroxide are mixed.→ HPO42. and the values of the respective base dissociation constants reflect this: Reaction Equation PO4 3.8 Because of its overall negative charge and the bond electron density.3 x 10 -12 Again.+ 2 OH .+ OH - Equilibrium Constant Expression HPO4 2−  ⋅ OH −     Kb3 =  PO4 3 −    - Equilibrium Constant Value 2.→ PO4 3. the following net ionic equations apply: Reaction Description Equal volumes of 0. The observation that successive protons are increasingly difficult to remove from polyprotic acids implies that these polyprotic acids ionize stepwise when reacting with strong bases. or accept a proton.10 M phosphoric acid and 0. Its conjugate base. they can be considered interchangeable.is a weaker acid. the dihydrogen phosphate ion H2PO4-. Equal volumes of 0. the following net ionic equations can be written: 8. and phosphate ion. their respective conjugate bases are dihydrogen phosphate. 66 . which can either donate a proton. A species such as H2PO4-.20 M sodium hydroxide are mixed. and monohydrogen phosphate ion can each act as acids. These bases are listed in order of increasing base strength.+ H2O  HPO 42.+ 2 H2O Since phosphoric acid.SPECIAL FOCuS: Acids and Bases The phosphoric acid molecule H3PO4 is the strongest acid among the substances listed. monohydrogen phosphate.10 M phosphoric acid and 0.10 M sodium hydroxide are mixed. ions.10 M sodium phosphate are mixed. H+ + PO43will have a negligible effect on the [H+] in the solution.10 M hydrochloric acid and 0. H+ + HPO42and HPO42.10 M sodium phosphate are mixed.20 M hydrochloric acid and 0. and PO43-.5 × 10–3 = 67 . In practice. calculations to determine the concentration of each species may seem intractable.024 M 7.10 M -x 0. the chemistry of the solution is dominated by the ionization reaction of the first proton: H3PO4  H+ + H2PO4Ionization of subsequent protons: H2PO4.+ H+ → H2PO4 - Because a solution of a polyprotic acid or its conjugate base(s) may contain several species related by different equilibrium reactions and constants. Equal volumes of 0.+ 3 H+ → H3PO4 HPO 42.10 M sodium phosphate are mixed.10 M hydrochloric acid and 0. in addition to the H+ and OH. In principle. for example. Equal volumes of 0. the solution contains the species H3PO4. a solution of 0. the values of the relevant acid dissociation constants are sufficiently diverse that the solution chemistry is usually dominated by one ionization.+ 2 H+ → H2PO4 PO4 3. Consider.Polyprotic Acids Reaction Description Equal volumes of 0.10 M phosphoric acid. HPO42-. and the concentrations of each species can be determined from straightforward calculation.30 M hydrochloric acid and 0.10 – x [H3PO4]  H+ ~0 +x x + H2PO40 +x x [H+] ∙ [H2PO4–] x∙x 0. H2PO4-.+ H+ → HPO42PO4 3. Fortunately. The [H+] of the resulting solution can be determined from the ionization of the first proton: H3PO4 [ ]o ∆[ ] [ ]eq Ka1= 0. Net Ionic Equation PO4 3. Equal volumes of 0.10 – x + [H ] = [H2PO4–] = x = 0.10 M sodium monohydrogen phosphate are mixed. the calculated [PO43-] is extremely small.2 × 10−18 M Not surprisingly.SPECIAL FOCuS: Acids and Bases The resulting [H+] and [H2PO4-] can be used to determine the [HPO42-]: K a2 = H+ ⋅ HPO 4 2− H 2 PO 4− (0. above.2 × 10−8 M As expected.024 M) 6. illustrates 68 . FIGuRE 2 The behavior of polyprotic acids in acid–base titrations reflects the same considerations. the relatively small [HPO42-] implies that the second ionization occurs to a negligible extent and has no measurable effect on the [H+] or the [H2PO4-]. Each of the ionizable protons is lost from the acid sequentially with increasing pKa values. The graph of pH vs. the [PO43-] can be determined using previously calculated concentrations: K a3 = H+ ⋅ PO 4 3− HPO 4 2− (0.8 × 10−13 = PO 4 3− = x = 1.024 M) ⋅ x (6. Similarly.2 × 10−8 = HPO 4 2− = x = 6.2 ×10−8 M) 4.024 M) ⋅ x (0. volume base added. At the half-equivalence point. and a solid understanding of the chemistry of monoprotic acids and bases is essential for students before they can grasp the details of the chemistry of polyprotic species. 69 . volume base shows two buffering regions corresponding to the sequential ionization of each proton from the acid when a solution of a weak diprotic acid is titrated with a strong base. monoprotic acid is titrated with a strong base. 2 HA The fundamental principles of acid–base chemistry remain valid for acids that have many ionizable protons. 2 [ H2 A ] At the half-equivalence point corresponding to the second ionization. 2 . Important points on the graph are the equivalence point. and K a1 = 10 a1 . At this point in the titration. [HA-] = [A2-] and H+ ⋅ A 2K a2 = = H+ . and K a = = H+ . so pK a2 = pH 1 Eq. At the first half-equivalence point in the titration curve. Pt. As in the case of the monoprotic acid. and half have been converted to the conjugate base.are equal to the moles of HA initially present. half of the initial moles of weak acid remain. [HA] = H+ ⋅ A − -pK [A-]. so pK a1 = pH 1 Eq. so pK a = pH 1 Eq. 1 . Pt. [H2A] = [HA-] and: H+ ⋅ HA− -pK K a1 = = H+ . 2 [ HA ] FIGuRE 3 The graph of pH vs. the equilibrium constants for each ionization may be determined from the pH values at each half-equivalence point. and K a = 10 a . and K a 2 = 10-pK a 2 . and the half-equivalence point. where the moles of added OH. Pt.Polyprotic Acids what happens when a solution of a weak. + H2O  H3O+ + C2O42K1 = 5. Answer: A270 . FIGuRE 4 5(e).K = K1 x K2 = 2. H2C2O4.36 x 10-2 (B) 5. Oxalic acid.8 x 10-6 (D) 1. For the reaction above.SPECIAL FOCuS: Acids and Bases Examples from Prior AP Chemistry Examinations (1994) H2C2O4 + 2 H2O  2 H3O+ + C2O4231.3 x 10-5. H2Y. The graph below shows the results obtained by titrating a different weak acid.3 x 10-5 H2C2O4 + 2 H2O  2 H3O+ + C2O42.36 x 10-2 K2 = 5. is a diprotic acid with K1 = 5.3 x 10-5 (C) 2. what is the equilibrium constant? (A) 5.36 x 10-2 and K 2 = 5. with the standardized NaOH solution. Answer: (C) From Hess’s Law H2C2O4 + H2O  H3O+ + HC2O4HC2O4.9 x 10-13 31.9 x 10-10 (E) 1. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter “A” on the curve.8 x 10-6 (1998) 5(e). 00 x 10-3-mole sample of pure oxalic acid? (b) Give the equations representing the first and second dissociations of oxalic acid.015 M K = H+ ⋅ C 2 O 4 2− 2 [ H 2C 2O 4 ] C 2 O 4 2− = K ⋅[ H 2C 2O 4 ] H + 2 = (3.+ OHKb = Kw 1.00 ×10−3 mol H2C2O4 = 2 ×(5.00 × 10-14 = 1. for the reaction that occurs when solid Na2C2O4 is dissolved in water. K1.00 ×10−2 mol H+ = 1.+ H2O  HC2O4. [H2C2O4]eq ≈ 0.5 = 0.00 ×10−2mol OH− 1.78 × 10-4 = = 5.0250 L = 25.78 x 10-6 (a) What volume of 0.0 × 10-7 M (d) C2O42.(equilibrium constant = K2) K = K1 × K 2 K1 = K 3.78 × 10-6 ) × (0.32 M (c) pH = 0. H2C2O4  2 H+ + C2O42.5 Because K << 1. H2C2O4.400-molar NaOH is required to neutralize completely a 5.91 × 10-2 K2 6.K = 3.00 × 10−2 mol OH− = 0.5.00 ×10−3mol) H+ = 1.400M (b) There are two successive dissociations: H2C2O4  H+ + HC2O4. Calculate the value of the first dissociation constant.32)2 = 6. a strong acid is added until the pH is 0.) (d) Calculate the value of the equilibrium constant. The overall dissociation constant is also indicated. The overall dissociation of oxalic acid.56 × 10-10 = -5 K a2 6. Answers: (a) 5.40 × 10-5 H+ = 10-0. K2.Polyprotic Acids (1997) 1.40 × 10 71 .40 x 10-5. (Assume the change in volume is negligible. Kb. for oxalic acid if the value of the second dissociation constant. is 6. Calculate the [C2O42-] in the resulting solution.(equilibrium constant = K1) HC2O4. is represented below.0 mL 0.015-molar solution of oxalic acid. H+ + C2O42.015) (0. (c) To a 0. Account for the shape of the curve. Draw the Lewis electron-dot diagram for this species. showing the shape of the titration curve that results when 100. (d) Write the equation for the reaction that occurs if a few additional milliliters of HCl solution are added to the solution resulting from the titration in (c). (c) Sketch a graph using the axis provided. HPO42H+ + HPO42. 7. Answers: (a) H+ + PO43.100-molar Na3PO4 solution. A chemical reaction occurs when 100.2000-molar HCl is added dropwise to 100. (b) Identify the species that acts as both a Brønsted acid and as a Brønsted base in the equations in (a).SPECIAL FOCuS: Acids and Bases (1994) 7. (a) Write the two net ionic equations for the formation of the major species. milliliters of the HCl solution is added slowly from a buret to the Na3PO4 solution. milliliters of 0. milliliters of 0. H2PO4(b) HPO42- 72 . Polyprotic Acids (c) (d) H+ + H2PO4. H3PO4 73 . . This reaction with water is called hydrolysis. These ions can act as acids or bases because they react with water. Puerto Rico Acid–Base Properties of Salt Solutions A salt is an ionic compound that can be considered to be the product of the reaction between an acid and a base. which shows that some ions can act as acids or bases. you can write an expression for Kb. Since the resulting solution has the form of an acid ionization. Since this solution has the form of a base ionization. The salt that is produced is sometimes neutral. Kb = [OH-][CH3COOH] [CH3COO-] (ii) The reaction of the ammonium ion with water. NH4+ + H2O → NH3 + H3O+ This resulting solution would be acidic due to the presence of the hydronium ion. Hydrolysis is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. Notable examples are: (i) The reaction of the acetate ion with water. This can be clearly explained from the Brønsted-Lowry theory. the expression for Ka is: Ka = [H3O+][NH3] [NH4+] . CH3COO. but in many instances it dissolves in water to form a solution that is acidic or basic. John’s School San Juan.Salts and Buffers Adele Mouakad St.+ H2O → CH3COOH + OHThis resulting solution would be basic due to the presence of the hydroxide ion. usually hydrolyze by acting as acids. ClO3 . which we used in our example above.g. Sr 2+. and the water molecule becomes a hydroxide as shown in the reaction below. reducing the charge of the ion by 1. Rb+.from HCl) have very little basic character. Aqueous metal ions. the anions of strong acids (e. I -. it does not ionize readily). the cations of strong bases (alkali metal cations and alkaline earth metal cations) have hardly any acidic character. we can say that anions of weak acids will be basic. copper (II) ion forms the Cu(H2O)42+ ion. that are not those of the alkali or alkaline earth metals. Ca2+. Because the electrons are drawn away from the oxygen atoms by the Cu atom. ClO4 -. On the other hand. tend to draw electrons from the O – H bonds. in turn. For example. the H2O molecules in Cu(H2O)42+ are acidic.5. On the other hand. Cl. these anions do not hydrolyze. Let us examine the cation conjugate to a weak base. So we can say that the acetate ion acts as a base. Cu(H2O)42+ (aq) + H2O (l) ↔ Cu(H2O)3(OH)+(aq) + H3O+(aq) We can summarize this in the following tables: IONS OF NEuTRAL SALTS Cations Na+.. Copper (II) salts tend to have a pH around 3.e. The Cu2+ ion acts as a Lewis acid. It is clearly seen that it behaves like an acid. Br-. Cu(H2O)42+ hydrolyzes by donating a proton to a free water molecule. making them increasingly polar and weaker. forming bonds with a lone pair of electrons in the O atoms of H2O. K+. In general..SPECIAL FOCuS: Acids and Bases How can you predict whether a salt will be acidic. Ba2+ Cl -. Cs+ Mg2+. As a result. Many of these ions form hydrated metal ions. it produces acetic acid. BrO4 -. NO3 - Anions Some Common Basic Ions FHCO3 SO42ClO2CH3COO CO32HPO42NO2S2PO4 3CN HS ClO - 76 . One of the water molecules loses a proton. the O atoms. In other words. such as the ammonium + (NH4 ) ion. So we can generalize that the cations of weak bases are acidic. which is a weak acid. basic or neutral? When the acetate ion hydrolyzes. This means that the acetate ion holds fairly strongly to the proton (i.. so Na+ does not hydrolyze. Therefore.g. Cu2+) To predict whether the salt solution is acidic. hydrocyanic acid.+ H2O ↔ HCO3. CN-. you have to determine the acidity or basicity of the ions comprising the salt. basic.+ H2O →No reaction (conjugate base of a strong acid. a weak acid. or neutral you need to know whether the ions comprising the salt are acidic or basic or do not hydrolyze. Sodium is an alkali metal. or basic. Conclusions regarding acid or basic hydrolysis can be made by examining the Ka or Kb values in the literature for the species involved. write a balanced chemical equation for a reaction occurring with water that supports your prediction. Na+ and the cyanide ion. This salt is composed of the potassium ion. However. neutral. CN-. Let us examine the following exercise from the 1981 AP Exam. A salt of sodium cyanide is expected to be basic. the CN.10 molar solution of each of the salts above is acidic.Salts and Buffers Some Common Acidic Ions NH4+ HSO4- CH3NH3+ H2PO4- Al3+ Pb2+ Transition metal ions (e. is basic. it does not hydrolyze) K2CO3: K+ + H2O → No reaction. NaCN. CO32. (b) For each of the solutions that is not neutral.. Consider sodium cyanide.+ OHNaHSO4: 77 . In other words. Exercise 1 Al(NO3)3 K2CO3 NaHSO4 NH4Cl (a) Predict whether a 0. the cyanide ion. Solution: (a) Al(NO3)3 = acidic NaHSO4 = acidic K2CO3 = basic NH4Cl = acidic (b) Al(NO3)3: Al(H2O)63+ + H2O ↔ H3O+ + Al(H2O)5(OH)2+ NO3-.is the conjugate base of a weak acid. etc. such as Al3+. so it hydrolyzes giving a basic solution. A salt of a strong base and a weak acid.0. The pH of sea water ranges from 7.01 mol of HCl is added to 1. Thus a buffer solution contains both an acid species and a 78 . NaCl). The pH of mammalian blood is also maintained at a pH of close to 7. If the Ka > Kb.1 unit. Zn2+. A few basic rules would be: 1. Buffers Definition: A buffer solution is one that resists changes in pH when small quantities of an acid or a base are added to it. Students need to know the coordination number for the more common ions.0 to 2. A salt of a weak acid and weak base. if this same amount of hydrochloric acid is added to 1. A salt of a weak acid and a strong base. The salt does not have any ions that will hydrolyze.g. On the other hand.0 L of sodium chloride solution. The cation of this salt is the conjugate of a weak base. the solution will be basic. the pH changes from 7. so it will be neutral. Cu2+.+ H2O → No reaction (conjugate base of strong acid. In this case both ions hydrolyze. so it hydrolyzes producing an acidic solution. If 0. A salt of a strong acid and a strong base. 2.4.5 to 8. the pH changes by only 0. Composition: A buffer contains a weak acid and its conjugate base or a weak base and its conjugate acid. If the Kb > Ka. This is very important because many aquatic organisms are very sensitive to changes in pH. Obviously. (e. the solution will be acidic. the acidity or basicity depends on the Ka and Kb of the conjugate acids and bases respectively.38. 4. The anion is the conjugate base of a weak acid.SPECIAL FOCuS: Acids and Bases Na+ + H2O → No reaction HSO4. sea water has other substances present that maintain the pH of sea water at a fairly constant level. 3. it does not hydrolyze) NH4+ + H2O ↔ NH3 + H3O+ One of the things to note from the above exercise is that students need to be aware that the transition metal ions and other metallic ions mentioned above do form complex ions with water.0 L of ocean or sea water.+ H2O ↔ H3O+ + SO42NH4Cl: Cl. 100 L = 0. • The concentration of the acetate ion is 0. The pH of a buffer: The pH of buffers can vary from being in the acid range of the pH scale all the way to the basic range. CH3COONa. so the buffer solution must remove most of these ions. • The equilibrium reaction is CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq). • The Ka of acetic acid is 1.100L = 0. In blood. the bicarbonate ion HCO3-.Salts and Buffers base species. 1. the pH of the buffer changes very little. so the pH won’t change very much.0 mL. Adding an acid to the buffer: The buffer solution must remove most of the new hydrogen ions otherwise the pH would drop markedly. There are two ways in which the carbonic/bicarbonate buffer can remove the hydroxide ion.10 M solution of acetic acid. the pH will decrease a very small amount.050 M. The buffer in sea water is mostly due to the presence of carbonic acid. 79 . both the carbonate and phosphate buffers are present. The pH of a buffer is determined by the relative amounts of acid and conjugate base that are present in the solution and the Ka of the acid. most of the new hydrogen ions have been removed.(a) A buffer made of 50. • The concentration of acetic acid is 0. The hydrogen ions will combine with the bicarbonate ion to make carbonic acid: HCO3-(aq) + H+(aq) ↔ H2CO3(aq) Even though the reaction is reversible. and its conjugate base. But because the reaction is reversible.0050 moles of acid/0.050 M. and 50. to maintain the pH of the juice.0 mL of a 0. • The total volume of the buffer is 100.10 M sodium acetate.7 x 10-5. CH3COOH. This is similar to one of citric acid and sodium citrate that is often found in commercial fruit juices. Either carbonic acid or bicarbonate ion reacts: H2CO3(aq) + OH-(aq) ↔ HCO3-(aq) + H2O(l) HCO3-(aq) + OH-(aq) ↔ CO32-(aq) + H2O(l) Since most of the hydroxide has been removed. Let us examine a buffer made of acetic acid and sodium acetate.0050 moles of acetate ion/0. H2CO3.0 mL of a 0. Let us compare the pH of two different buffer solutions. Adding a base to the buffer solution: Bases contain hydroxide ions or react to form them. The pH of this buffer can be altered by adjusting the relative concentrations of the acid and the conjugate base.050 M] After rearranging.67 x 10-5 has a pH of 5.7 x 10-5 = [X] [0. Substituting in the above equation we can see that the ratio [HA]/[A-] = 0.25.050 M]/ [0.025/0.7 x 10-5 M pH = .070.050 M.050 M -X 0. the concentration of hydronium ion is [H3O+] = Ka [HA]/ [A-].050 M)/ (0. we get [H3O+] = (1.025 M. (b) For example. It is slightly more basic than the previous solution. We substitute the equilibrium concentrations in the above equation.050 M + X H3O + 0 +X X In this solution Ka = [H3O+ ] [CH3COO-] / [CH3COOH].070 M and the equilibrium concentration of the acetic acid is 0. In doing so we can assume that X is small compared with 0. is less than 1. If the ratio [HA]/[A-] < 1 then the resulting solution will be more basic than pKa. if the equilibrium concentration of the acetate ion is 0.050 M – X H 2O CH3COO0.050 M) [H3O+] = 1. A conclusion from the above problem: If the ratio [HA]/[A-] = 1 then the pH of the resulting solution will be equal to pKa If the ratio [HA]/[A-] > 1 then the resulting solution will be more acidic than pKa. So the net equation is: 1. but still well within the acid range.7 x 10-5M] = 4.050 M +X 0.76 which is in the acid range Now in this example the concentrations of the acid and the conjugate base are equal and the concentration of the hydronium ion is equal to Ka.log [1. and the resulting [H3O+] = 5. 80 .SPECIAL FOCuS: Acids and Bases Filling in the ICE concentration table: CH3COOH Initial Change Equilibrium 0.7 x 10-5 ) (0. 050 M + X H3O + 0 +X X Now Ka = [H3O+] [NH3]/ [NH4+]. Let us examine this.56 x 10-10 The equilibrium reaction is NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+. we choose an acid/conjugate base combination that has a pKa close to 9. Filling in the ICE concentration table: NH4+ Initial Change Equilibrium 0. The addition of water affects a buffer solution (dilutes of a buffer).050 M .56 x 10-10 M • pH = -log [5.10 M NH4+ and 50.8 x 10-5 Ka = 5.050 M) / (0.3. so the final concentration of [NH4+] = 0. 4. HA ↔ H+ + AKa = [H+] [A-] / [HA] 81 .Salts and Buffers 2.56 x 10-10] This gives a pH of 9.050 M.10 M NH3. A buffer solution is made up of 50.050 M +X 0. 3. We can assume that X is small compared to 0.0 mL of 0. the pH remains unchanged. The Ka of ammonium ion is Kw/ Kb Ka = 1 x 10-14 / 1. a. From studying the two different buffers.56 x 10-10 = (X) (0. The final concentration of [NH3] = 0.050 M) • We can see that X = [H3O+] = 5. which is in the basic range. we can clearly see that the pH of the buffer will be determined mainly by the value of Ka. we choose an acid/conjugate base combination that has a pKa close to 3. When water is added to a buffer solution.050 M -X 0. Substituting we get: • 5. b. For a buffer with pH 3. The final volume is 100 mL.0 mL of 0.050 M. For a buffer with a pH of 9.050 M – X H 2O NH3 0. we must calculate amounts of hydrogen ions.0 mL of 0. 5. The second part is an equilibrium problem where the concentrations from the first part are used. Solution Part 1—Stoichiometric Calculation When a hydronium ion is added to the buffer it reacts with the nitrite ion: H3O+(aq) + NO2. • The moles of NO2. Because HCl is a strong acid.019 moles. A buffer resists change to pH when small amounts of acid or base are added.0095 moles. The Ka of nitrous acid is 4. • The moles of HNO2 = (0.10 M.10 moles/L) (0. therefore the concentration of H3O+ is 0.00050 moles of acid is produced and 0.00050 moles. • The moles of H3O+ = (0.SPECIAL FOCuS: Acids and Bases Rearranging this. To 95. This type of problem becomes a two-part problem.10 M hydrochloric acid. This analysis does not consider new effects that appear if the solution becomes extremely dilute.10 M. the pH stays the same.095L) = 0.20 M sodium nitrite. the hydrogen ion concentration is equal to the molarity of the acid.0185 mol.005L) = 0. • HCl is 0.20 moles/L) (0.00050 moles of nitrite ion is used up.019 – 0.10 M nitrous acid and 0.(aq) → HNO2 (aq) + H2O(l) Because the nitrous acid is a weak acid. since the same amount of water is being added to both.00050) mol = 0. 82 . we can see that: [H+] = Ka [HA] / [A-]. We assume that all the H3O+ added reacts with the nitrite ion.10 moles/L) (0.095L ) = 0. We are now going to examine this quantitatively. 0. First. This first part is a stoichiometric calculation where we assume that all the H3O+ from the strong acid reacts with the nitrite ion forming nitrous acid.5 x 10-4. Therefore. we assume that the reaction goes to completion.0 mL of this buffer solution we add 5. If this ratio remains unchanged. The addition of more water to the buffer does not change the ratio [HA]/[A-]. nitrous acid and nitrite ions in the solution before the reaction. • Moles of nitrite ion remaining = (0. Suppose we have a buffer consisting of 0.= (0. Part 2—Equilibrium Problem Using the concentration found in Part 1.185 + X] / [0. says he would mix NH3 and NaOH solutions.10 – X] Assuming that X is small enough that 0.185 M +X 0. 2(b). 83 .(mol/L) 0. • [HNO2] = 0. says she would mix NH4Cl and HCl solutions.10 – X H3O+ (mol/L) 0 +X X NO2. Dexter D.65.10L = 0.010 mol.1855 + x ≈ 0.10 L = 0. Beula B.10 – X ≈ 0.5 units.185M .0 mL of 0.0185 moles/ 0. says she would mix NH4Cl and NH3 solutions.185 + X The equilibrium constant equation is: Ka = [H3O+] [NO2-] / [HNO2] Substituting. x 10-4 = (X) (0. Archie A.10 M HCl(aq) is 3. • [NO2-] = 0. Carla C.10 M.185) /0. the pH after the addition of the 5. which is considered the maximum change that a buffer should have and continue to be effective at pH control. you construct the following ICE table: HNO2 (mol/L) Initial Change Equilibrium 0.5 x 10-4 = [ X ] [0. says he would mix acetic acid and sodium acetate solutions.00050) mol = 0. Describe the components and the composition of effective buffer solutions.10. So the pH changed by only 0. you get: 4.43 x 10-4 The pH of the original buffer was 3.04 units The buffer pH changed much less than 0.10 M -X 0.Salts and Buffers • Moles of nitrous acid after reaction = (0.0095 + 0. Specify the properties of a buffer solution.10 [H3O+] = X = 2. An employer is interviewing four applicants for a job as a laboratory technician and asks each how to prepare a buffer solution with a pH close to 9.185 and 0. Examples from Prior AP Chemistry Examinations (1983) 2(a).61.010 mol/0. 8 x10–5.0 x 10–8M 84 . HOCl. since the pKa is close to 9. Ka = 5. • Carla has the correct buffer. She also has the correct pH. • Beula does not have a buffer. She has a weak acid.6 x 10–10) Answer: 2(a). NaOCl. but the following acidity constants may be helpful: acetic acid. Explain what is wrong with the erroneous procedures.050 molar solution of HOCl.050 molar HOCl and 0. for hypochlorous acid. • Dexter does not have a buffer since he has mixed a weak base and a strong base. but it does not have the correct pH since the pKa is close to 5. is 3. since she has a weak acid (NH4+) and a strong acid (HCl). So he does not have the correct buffer.SPECIAL FOCuS: Acids and Bases Which of these applicants has given an appropriate procedure? Explain your answer. Ka= 1. Answer: (a) HOCl + H2O ↔ H3O+ + OClX << 0. Preparation of a buffer: Mix a weak acid with the conjugate base of the acid. (a) Calculate the hydronium ion concentration of a 0.010 X = [H3O+] = 8. referring to your discussion in part (a). (NH4+) and its conjugate base (NH3). Ka. A buffer solution resists changes in pH when small amounts of an acid or a base are added. (No calculations are necessary. Archie has a buffer solution: he has a weak acid and its conjugate base. (1977A) 3.020 molar sodium hypochlorite. 2(b). NH4+.1 X 10 –8. (b) Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of 0.0 x 10–5M (b) HOCl + H2O ↔ H3O+ + OClX ‹‹ 0.050 X = [H3O+] = 4. The value of the ionization constant. ionizes in water according to the equation above.Salts and Buffers (2005A) HC3H5O2(aq) ↔ C3H5O2–(aq) + H+(aq) Ka = 1.265 – X = [HC3H5O2] X2 = Ka = 1. then: X = [H+] X + 0.265 – X ≈ 0. (b) Calculate the pH of a 0.0 mL sample of a 0.00188 M = [H ] [you can assume that 0. then [C3H5O2–] = 0.265 – X X = 0.496 g sample of sodium propanoate. then X = [C3H5O2–] = [H+] 0.103 = [C3H5O2–] 0. C3H5O2–(aq) in the solution (ii) The concentration of the H+(aq) ion in the solution Answer: (a) [C3H5O2-][H+] = Ka [HC3H5O2] (b) let X be the amount of acid that ionizes.73 1mol 96.265 in order to simplify your calculations] pH = –log[H+] = 2. Assuming that no change in the volume of the solution occurs.34 × 10–5 Propanoic acid. (c) A 0.103 M 0.265 – X = [HC3H5O2] 85 + .103 M 0. (a) Write the equilibrium constant expression for the reaction.496g × (ii) let X be the amount that ionizes.50L since each sodium propanoate dissociates completely when dissolved.0g (c) (i) = 0.34 x10–5 0. NaC3H5O2.265 M solution of propanoic acid. (i) The concentration of the propanoate ion. HC3H5O2 .265 M solution of propanoic acid. producing 1 propanoate ion for every sodium propanoate. is added to a 50. calculate each of the following. and this is more than thousands of times larger than the propanoate ions from the acid. and Norris Rakestraw.50 M NaCl solution. All buffers have a buffer capacity. “Hydrolysis of Metal Salts.html (accessed January 2008). science. 8th ed.265 – X ≈ 0.uwaterloo.103 ≈ 0. A good laboratory activity to do with the students is to study the buffering capacity of seawater as compared to that of a 0. Philip.ca/~cchieh/cact/c123/salts.0.] Buffer capacity is the amount of acid or base that can be added to a buffer solution before a major pH change (more than +. A buffer with a large capacity has a large concentration of HA and of A-. “The Buffer Capacity of Sea Water”. 86 .5 units) occurs. General Chemistry.265 and X + 0. The pH of a buffered solution is determined by the ratio [HA]/[A-].265 – X X = 3.34 x 10–5 0. so that it can absorb a relatively large amount of H3O+ or OH-. The capacity of a buffer is determined by the magnitude of [HA] and [A-]. The Biological Bulletin 65 (December 1933): 437–451. Boston: Houghton Mifflin Company.” University of Waterloo. then the buffer’s capacity is exceeded and a major pH change will occur. Darrell. 2005.103. The buffer’s capacity is determined by the amount of acid and conjugate base. http://www. Bibliography Cyberspace Chemistry.43 x 10–5 M = [H+] [We can assume that 0. in order to simplify calculations. (This concentration ≈ the concentration of NaCl in seawater. They will resist change to pH when only small amounts of acid or base added. and Steve Gammon. Mitchell. If large amounts of acid or base are added.103 + X) = Ka = 1. Ebbing.SPECIAL FOCuS: Acids and Bases (X) (0.) The students can see the importance of a natural buffer such as is present in seawater and also study the buffering capacity of seawater. These are among the most electronegative elements or nonmetals group. which then can be a reaction site when the oxide is added to water to form an aqueous solution. phosphorous.Acidic and Basic Reactions of Oxides Marian DeWane Centennial High School Boise. (2) X = O + H2O → -O—X—O. Two types of reactions can result: (1) M – O. Idaho The oxides of the elements form an interesting group of acidic and basic compounds. and barium. In this case the oxides are classified as basic oxides. sodium. These are among the least electronegative elements or metals group. sulfur. Oxides behaving in this way include those of carbon. nitrogen.+ H2O → M—O—H + OHHere the oxide is a proton acceptor from water leaving an excess of hydroxide ion in solution. calcium. Oxides behaving in this way include those of lithium. Such oxides are classified as acidic oxides. hence forming a basic solution. cesium. hence forming an acidic solution and in effect acting as a proton donor. which releases hydrogen ions into the solution.+ 2H+ Here the oxide is an oxygen acceptor from water. The combination of element X with the highly electronegative oxygen results in a polarized bond (in every instance except two oxygen atoms bonding to form O2). strontium. and can be studied more closely as follows: (1) An alternative view of the basic oxides is that these are mostly ionic compounds (with a high degree of ionic character) and can be thought of as . potassium. or when the oxide is reacted directly with an acid or a base. and the halogens. These reactions are obviously oversimplified. K2O. Water has to be included as a reagent to provide the oxygen transferred. for example. For example.ion reacting to form two hydroxide ions. the more electronegative atoms in the nonmetals still have sufficient remaining electron density to form additional bonding with oxygen atoms in water molecules. (2) In the acidic oxides. as in common examples such as CO2. NO2.SPECIAL FOCuS: Acids and Bases containing ions in the solid state. SO3.+ H+ HSO4 .→ SO4 2.. Two electrons can be taken from an alkaline earth element such as magnesium to form Mg2+O2-. when an acidic oxide reacts with some bases water must be present to provide the oxygen transferred to the central atom. MgO. CaO. When added to water. hence the highly basic nature of the resulting solution. releasing H+ ion. the nonmetal atom readily forms covalent bonding with the oxygen. This weakens the O-H bond in water. it is these ions that act as proton acceptors as follows: O2. So. The complete reaction may take place in two steps to release two H+ ions. Cl2O. The combination of the highly electronegative oxygen atom with the low electronegativity atom of a typical metal produces the ionic bond in metal oxides.+ H+ Direct reactions of oxides can occur without water being an obvious reagent.+ H2O → 2 OH. For example: SO3 + H2O → HSO4 . etc. sodium oxide can react with hydrochloric acid to form sodium chloride and water in a typical acid–base reaction forming a salt and water: Na2O + 2HCl → 2 NaCl + H2O However. Thus it is the ionic structure of the metal oxides that best explains their reaction in aqueous solution. for example. sulfur trioxide can react with ammonia to form ammonium sulfate. The electronegative oxygen is easily able to attract the most weakly bonded electron away from a metal atom to form the ionic bond. SO3 + 2NH3 + H2O → (NH4)2SO4 88 . While this uses some electron density of the central atom. each O2. Typical oxides are Na2O. and oxides are no exception. as represented in the reaction below: Al2O3 + 6HCl → Al2Cl6 + 3H2O However. SrO + H2O → Sr2+ +2OH1999 Question 4 (a) Calcium oxide powder is added to distilled water. With hydrogen as an exception (water is clearly an example of an amphoteric oxide). as can SiO2. Aluminum oxide is a compact strong structure relatively insoluble in water. The question now requires students to provide a balanced net equation and answer a question about the reaction. no additional questions were asked. can act as an acid or a base. Here the aluminum oxide is acting as a base. Metal oxide plus water examples: 2004 Form B Question 4 (f): Powdered barium oxide is mixed with water. aluminum oxide also dissolves in strong sodium hydroxide solution acting as an acid neutralizing the sodium hydroxide. This reaction can be represented by the equation: Al2O3 + 2NaOH + 3H2O → 2Al(OH)4. For example aluminum oxide. Cs2O + H2O → 2Cs+ +2OH2000 Question 4 (g): Powdered strontium oxide is added to distilled water.+ 2Na+ AP Exam Examples of Free-Response Questions: Free-response questions about oxides are usually in the equations section. CaO + H2O → Ca2+ +2OH89 . but it can be dissolved by reacting in strong HCl. Amphoteric oxides can act as acids or bases.Acidic and Basic Reactions of Oxides Amphoteric Oxides As in all other areas of chemistry. BaO + H2O → Ba2+ + 2OH2002 Question 4 (c): Solid cesium oxide is added to water. metals in the center of periods and semimetals tend to form such oxides. neutralizing the hydrochloric acid. question 4. Al2O3. depending on the circumstances. there are chemical behaviors that fall in between two extremes. The year these questions were given. Rayner-Canham. F. and M. . Chemistry: Matter and Its Changes. 4th ed. Bibliography Brady.Nonmetal oxide plus water examples: 2003 Question 4 (h): Solid dinitrogen pentoxide is added to water. SO2 + H2O → H+ + HSO3SO2 + H2O → H2SO3 was also accepted.. Advanced Inorganic Chemistry. J. Wilkinson. 6th ed. SO3 + H2O → H+ + HSO42002 Form B Question 4 (e): Sulfur dioxide gas is bubbled into a beaker of water. SO2 + H2O → H+ + HSO3SO2 + H2O → H2SO3 was also accepted. W. 1999. 2nd ed. 2nd ed. 1990. and N. G. D. N2O5 + H2O → 2H+ + 2NO32003 Form B Question 4 (f): Sulfur trioxide gas is bubbled into water. and F. G. Descriptive Inorganic Chemistry.. New York: Saunders. Oxtoby. New York: Freeman. New York: John Wiley and Sons. 1998. New York: John Wiley and Sons. 2001 Question 4 (a): Sulfur dioxide gas is bubbled into distilled water. Nachtrieb. 2004. A. Senese.. Bochmann. Carlos Murillo. Cotton. Principles of Modern Chemistry. H. Irvine Irvine. In free-response questions. California Misconceptions and misunderstandings are often demonstrated by students on both multiple-choice and free-response portions of the AP Chemistry Examination. Use of familiar examples is helpful. misunderstandings arise from a lack of depth and/or accuracy of thinking about a topic or question. Teachers should always ask students to explain why they selected a particular distracter. perhaps the most common misconception demonstrated is a lack of understanding of the words “explain. If a sportswriter or commentator explained why team X won the game against team Y by stating.” they would not survive long in their jobs. which will help the student make the correct selection.” or “justify. The teacher can then proceed to help the student correct the misconception in their thinking. Miller University of California. or distracters.” This persists even when the examination question tries to provide guidance as to what is expected. “It was because team X scored more points.Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations George E.” Others reflect misunderstanding of material presented within the AP class or read from the textbook. Yet that is the format a considerable number of students exhibit when providing an explanation on the AP Chemistry Exam. These are often termed “preconceptions. Restating the question also does not answer the question. Students who select these answers are showing they have that misconception. On multiple-choice questions. frequent misconceptions are often included by the item writers as possible wrong answers. Many may reflect prior thinking brought to the AP class. Frequently. . as noted below.) Lewis acids are barely mentioned. students may transcribe K = [H][A] / [HA] ≅ [H]2 /[HA] incorrectly as [H] = √ K/[HA]. arguing somewhat as follows: 1. Reflective conceptual thinking would always look for responses where numerator and denominator values moved in the same direction. For example. and Aqueous Equilibria There is confusion on a number of issues relating to acids and bases. which also can easily be misconstrued) is key.8 x 10-9 is larger than 3 x 10-8 .0 x 10 -8 Electronegativity of halogen 3. 92 . Acid HClO HBrO Ka 3. Misconceptions Regarding Acids. explain the difference in K a values for HClO and HBrO.SPECIAL FOCuS: Acids and Bases For example. or exponential notation (as seen above). To foster correct thinking. The largest numbers of molecules present in solutions are therefore water molecules.0 2. Student and teacher dialogue (almost more than demonstrations. 2. Bases. together with some hints for instructors. so master Brønsted-Lowry or proton transfer systems before even thinking about anything else. the following question and answer might have appeared in AP: Using the information given in the following table. The answer is to start with basic principles. Reactions of acids and bases occur in water. Incorrect response (2): HBrO has a larger Ka because 2. teachers should encourage students to always answer the question qualitatively before plugging in any numbers.8 2.8 x 10 -9 Incorrect response (1): The table shows there is a difference between the Ka values of HClO and HBrO. Mathematical errors even creep in as a result of carelessness or real misunderstandings of conceptual algebra. The remaining document addresses selected topics where frequent misunderstandings are demonstrated. (No nonaqueous systems are dealt with in AP Chemistry. The solution cannot track what H+ came from what source to create separate values. This is a semantic problem caused by the unfortunate choice of terms by chemists. Most will be attached to water molecules in the form of H(H2O)n+ where n is 1-4.. 7. Generally. Acid–base “strength” is related to acid–base concentration. Water is amphiprotic and amphoteric. this is abbreviated to H+. One cannot have some OK. This is often stated in terminology regarding the “degree of ionization. So we simply must list all possible reactions of any given system in water and ask: “Is the system at equilibrium?” If not. it is now recognized that proton 93 .by simple dissociation. the chances are that you will hit water molecules first! 4. first-cut reactions are water reactions. all possible equilibria and the corresponding equilibrium constant expressions must be simultaneously satisfied. and some not OK. Thus. if it is at equilibrium. 5. If you jump into a crowded swimming pool. but the corresponding idea for a base also exists. putting a lot of any chemical in water makes it “strong. An acid with a high K value (K value > 10-1) for the reaction HA + H2O ↔ A.+ H3O+ is a strong acid. Free protons do not exist in water solutions—they must be attached to something. so its reactions are to accept a proton or donate a proton. So what are the misconceptions. Thus. therefore. and why do they exist? The following examples are given mostly for an acid. Correspondingly. 1. water. In most texts. To the lay public.” Strong coffee or strong lemonade has a high concentration of those chemicals. then consider reactions that happen as they go to equilibrium. acid–base chemistry is almost exclusively (>99%!) the study of proton transfer reactions all of which are equilibrium reactions and each of which has a well-defined equilibrium constant that only varies with temperature. then what do we know about all the possible equilibrium constants? 8. In one system at equilibrium. acids and bases are defined as strong or weak solely on the basis of their equilibrium constants in reaction with water.g. 6. However.Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations 3. Some texts still may use the term “dissociation” based on the idea that HA becomes H+ and A. there can only be ONE value of a species concentration (e. strong bases have a high K value for the reaction B + H2O ↔ B+ + OH-. The obvious species for anything to react with is.” Conversely. one value of [H+] or [OH-]. 9. a “neutral” solution of pH = 7. 2. leaving the remainder in solution. A class of one strong student is a strong class.” So a weak acid cannot react to neutralize a strong base. it is not at equilibrium. or the corresponding values for a base. or a small concentration of a strong acid. one can have a high concentration of a weak acid. For example: 5 mL of 0.3 x 10-3 M. It is the K value for the transfer to water that determines whether an acid is classified as strong or weak. so the true value is close to 1 x 10-7 + 1 x 10-10 or 1 x 10-7 M. so often neglecting water as a source works.1 M NaOH is added to 10 mL of 0. Weaker students find this quite difficult. which leads to a third order equation for [H+]. so the result is strongly basic. “Reaction to completion can only occur as the K value will allow.8 x 10-6) or 1. 94 . we assume complete transfer. This concentration expression is the same for either a strong or a weak acid. (pH = 10). The full treatment of this. there are circumstances where the contribution of the “autoionization” or “autoprotolysis” of water according to H2O + H2O ↔ H(H2O)+ + OHcannot be ignored. so [H+] from HCl is 1 x 10-10 M.1 M acetic acid. What is the end result? Beginning students who have learned to correctly write an equilibrium constant for the weak acid in water as HAc + H2O ↔ H+ + Ac. A good student will recognize this as nonsense since this is a basic solution. When the solution is mixed. but a class of 40 weak students is still a weak class.1) and the concentration of hydrogen ion is now only √ (1. is usually given in college analytical chemistry courses. All reactions must be allowed to proceed to equilibrium before the equilibrium mixture is analyzed. and vice versa. The “concentration” of a solution of an acid in water strictly refers to the formal concentration which is the same as [HA] + [A-] .with Ka = 1. The water is contribution 1 x 10-7.SPECIAL FOCuS: Acids and Bases transfer reactions to water are what actually takes place.3 x 10-3 moles of OH-. this is approximately true. the ICE tabulation method can be used to obtain a result. 3.” In many problems with high or medium concentrations of acids. With this in mind. Thus. now 1.8 x 10-5 = [H+]2 / (0. “Acids are the only source of protons in water. However. The classic catch question is: What is the [H+] of a 1 x 10-10 M solution of HCl in water? Since HCl is a strong acid.8 x 10-5 frequently will argue that since [H+] = [Ac-]. Thus the reaction can only consume 1. bases are compounds containing OH. Acids and bases are defined according to their ability to donate or accept protons. 4. 5. [H+] and [OH-] values are independent of each other. or at least a frequent repetition of the mantra pH + pOH ALWAYS = 14 (at normal room temperature) could help. 95 .Misconceptions in Chemistry Demonstrated on AP Chemistry Examinations Usually the examination committee for AP has tried to set problems on the freeresponse section where the water contribution can be ignored and approximations are simple. A small but significant number of students fail to apply the water equilibrium in acid–base solutions. This enables reactions such as NH3 + H2O ↔ NH4+ + OH. Thus they are comfortable to report a pH of 4 and a pOH of 4 (or the equivalent concentrations values) in the same solution. Perhaps a focus during discussions on the need to satisfy the water equilibrium. this defines NH3 as an acid. Acids are compounds containing H atoms. but these are rarely dealt with in any significant way in AP or first-year college chemistry courses. almost exclusively (for AP) in reaction with water. Much practice in writing reaction equations with water as reagent and seeking the K values for the reactions in tables will help students learn how to classify more correctly. Unfortunately. Of course there are nonaqueous systems. Accepting a proton from water creates a basic solution (larger [OH-] than [H+]).to be written to correctly identify ammonia as a (weak) base. . Acid–Base Chemistry Beyond AP® Arden P. New York Most of the acid–base behavior discussed in AP Chemistry occurs in water. However. HNO3. Acids that are stronger than H3O+ will donate protons to H2O to form H3O+ ions. all stronger acids (or bases) are made to appear as if they are of equal strength. HI. as many trends in solution depend on solvation energy considerations. Zipp SUNY Cortland Cortland. Trends in acidity as a function of molecular structure in the gas phase can often be more easily understood. This is the situation with HCl. (Recall from the discussion of Brønsted-Lowry acids and bases that reactions with H+ ions involve competitions . A further extension is provided by acidity and basicity measurements in the gas phase. the use of water as a solvent limits the species that can act as acids or bases and restricts the range of reactions that can be legitimately considered as acid–base reactions. As a consequence. which are absent in the gas phase (Barntmess. and H2SO4 (first proton).will abstract protons from H2O to leave OH. HBr. HClO4. these differences are not apparent in water because H2O molecules attract the H+ from the various anions so strongly that all the acid is converted into H3O+ and the corresponding anion. This article will address acid–base chemistry beyond water and. while bases that are stronger than OH. thus. where the relative tendency of a molecule to donate or accept a proton can be measured. which appear equally strong in water.ions. This occurs because H2O molecules are stronger bases toward H+ than any of the anions in these acids. so the differences in anion attraction for the H+ are not apparent. McIver 1979). Leveling Effect of Water It is well known that H2O ionizes into H+ (or H3O+) and OH-. beyond the usual considerations of AP. Although the Ka of these substances actually varies from 101 to 1011. Scott. C2H5OH.SPECIAL FOCuS: Acids and Bases between two bases.in water. H-. Overton. alcohols (ROH) contain hydroxyl (-OH) groups but exhibit neither acidic nor basic properties in H2O. HBr. the acids do not ionize completely.g. (There is virtually no tendency for an alcohol to behave as a base by either gaining an H+ or losing an OH. C2H5O. 98 . or HS-. 107. 102. Because these solvent molecules have less attraction for H+ than does H2O. The opposite effect occurs for very weak acids (or bases). Note: Saying that alcohols are very weak acids is equivalent to stating that their conjugate bases (RO-) are very strong. HCl. and liquid HF. H2SO4. As an example of this behavior. which are not strong enough to exhibit their acid or base properties in water. Thus. In this way. respectively. 1010. Weller. and Armstrong 2006.as the predominant anion in solution. and 1011.) Nominally. a less basic solvent than water (one with a more negative pKa value) can be used. to calculate their ionization constants. Two examples of such solvents are liquid methanoic (formic) acid. respectively. it is possible to measure the relative amounts of the ionized and un-ionized acids. and to relate these values to aqueous solutions. which would lead to them being leveled in water as discussed above. leaving OH. acids with Ka values that range from 100 to 10-14 (pKas from 0 to 14) can be studied in water and substances with values outside this range must be investigated in other solvents.(e.) The same “leveling” phenomenon occurs with bases that are stronger than OH. 117). Rourke. pKa Ranges of pure Solvents Relative to Water H2SO4 HF HCOOH H 2O (CH3)2SO NH3 -20 -10 0 10 20 Effective pKa in water 30 40 In order to discriminate among the acidities of the six strong acids mentioned above. The Ka for a typical alcohol is about 10-18 (pKa = 18). the Ka values of HNO3. HCOOH.or S2-) because they react with an H+ from water to form H2. HClO4 and HI have been found to be 101. 109. One way to compare the useful ranges of different solvents is to graph their effective pKa ranges relative to H2O as depicted below (Atkins.. with the extent of this reaction dependent on the ease with which the bond between H+ and H(x-1)Y.e. acids that are stronger than 100% sulfuric acid. 99 . (CH3)2SO. oxoacids). there is a great deal of variation among the species in each category. Such substances ionize according to the equation: HxY <=> H+ + H(x-1)Y. Relative Acid Strengths Although it is common to classify acids as either strong or weak. In the case of very weak bases a more acidic solvent than H2O is needed in order to protonate the basic site more effectively and for very weak acids a more basic solvent must be employed to help remove H+. These variations can be accounted for on the basis of the structures of the molecules involved. which can remove H+ ions from virtually anything including many hydrocarbons. bases stronger than OH. or liquid NH3. which have pKa values greater than 14.” i. Hydroacids These acids can be represented by the general formula H xY and include species in which an atom of a nonmetal is attached directly to one or more hydrogen atoms. which can be done both theoretically and experimentally.can be distinguished by using solvents such as liquid dimethyl sulfoxide. such as butyllithium—LiC4H9. Once the constraints of an aqueous solution have been eliminated. The opposite strategy applies to the task of distinguishing among very weak acids (or bases). a variety of other acid–base reactions can be observed. which do not show acidic properties under other circumstances. can lead to interestingly different results as solvation effects are removed (Bartmess and Hinde 2005).. For a series of alcohols their relative strengths could potentially be determined by studying them in either dimethylsulfoxide or ammonia. The Ka values for the compounds in groups 14 to 17. Although the specific structural factors depend on the acid category (hydroacids vs. Chemists have developed “super acids.Acid–Base Chemistry Beyond AP On the other hand.can be broken. the fact that acid strengths can be correlated with structures for a given category of acids provides a very powerful organizational tool. which can even protonate hydrocarbons! At the other extreme are super bases.. Consideration of acid strength from gas phase species. 74 7. and H3PO4. their strengths as represented by their Ka (or pKa) values are found to fall into different categories depending on the number of unprotonated oxygen atoms. HNO2. These are the electronegativity (EN) and the size of the atom Y.05 3. regardless of the number of ionizable H’s in the molecule. 100 . decreasing down the table. which emphasizes the fact that the ionizable protons are bonded to oxygen atoms. the structure of HClO is actually HOCl. however. As the size of an anion increases. However. are bonded to the central atom Y. which. with changes in both the central atom and the number of oxygen and/or hydrogen atoms as shown in the list of several of the more common oxoacids: HClO.6 HF HCl HBr HI 3. Despite this variability in formula. its electrostatic attraction for H+ decreases. When these species are compared in this manner. acid strength in the same directions. which would increase the acidity. the change in EN is the major factor affecting the acidity. pKa for Hydroacids CH4 SiH4 GeH4 49 35 25 NH3 39 PH3 27 AsH3 23 H 2O H2S H2Se H2Te 15. there is a large increase in atomic size descending in any group. reflecting an increase in Ka values and. HClO2. because the EN changes in the wrong direction.8 2. Because the atomic size does not vary appreciably across a period. HNO3. This is shown in the table below (Bowser 1993. As the EN of Y increases across the period. There are two factors that influence the ease with which the H-Y bond can be broken. The increase in acidity going across the periodic table can be accounted for by the change in EN. HClO4.SPECIAL FOCuS: Acids and Bases most of which had to be measured in solvents other than water. H2SO4. HClO3. increasing the polarity of the bond. The increase in the acidity down the table cannot be explained in terms of the EN. H2CO3. while that of HClO2 is HOClO and that of H2SO4 is (HO)2SO2. the electron pair in the H-Y bond will be drawn closer to Y.45 -7 -9 -11 The pKa values of these substances decrease down the periodic table and from left to right. H2SO3. in turn. Thus. cover a range of 10 60 (from 10-49 to 1011) as reflected in the pKas given in the table below (Bowser 1993. 314). all of the acids in this group can be represented by the general formula (HO)mYOn. therefore. 314). Oxoacids This group of acids is more extensive and diverse than the hydroacids discussed above. This increases the polarity of the bond and makes the H atom more H+-like.21 (H0)mYO2 Strong pKa < 1 HOClO2 HONO2 (HO)2SO2 (HO)2SeO2 -2. Second. releasing H+ and making the solution acidic as represented in the equation: M(H2O)6x+ --> M(H2O)5OHx-1 + H+ 101 .3 Several points are notable about the data in this table. An increase in the number of such electron-attracting oxygen atoms will withdraw electrons more effectively from the H in the O-H bond.Acid–Base Chemistry Beyond AP pKa Table for Oxoacids (HO)mY Very weak pKa > 7 HOCl HOBr (HO)3As (HO)4Si 7.18 9.66 (H0)mYO3 Very Strong pKa < 5 HOClO3 -7. where most metal ions exist as octahedral species as shown by the equation: Mx+ + 6H2O --> M(H2O)6x+ The interaction of some metal ions (most notably alkali metal ions and larger alkaline earth metal ions such as Ca2+.94 3.15 2.44 -3 1. Acidity of Metal Ions Metal ions behave as Lewis acids. Similarly.20 3.0 -1. the acid strength changes little with variations in atomic size or electronegativity in contrast to the behavior observed for the hydroacids discussed above. One of the most common electron pair donors is the H2O molecule.62 2. many metal ions attract the electron pairs of the coordinated H2O molecules so strongly that one or more molecules may actually be torn apart.49 9. First. the acid strength increases with the number of unprotonated oxygen atoms in the molecule. any change in the size of the central atom has little effect on the size of the anion as a whole. especially in aqueous solutions. However.60 1.54 8. further. accepting a share of a pair of electrons from appropriate donors.89 2. Sr2+ and Ba2+) with H2O goes no further. This can be accounted for by the fact that the ionizable H+ is always attached to an oxygen atom whose size doesn’t change and.77 (H0)mYO Weak pKa 2 – 5 HOClO HONO (HO)2CO (HO)2SO (HO)2SeO (HO)3PO (HO)3AsO 1. changes in the electronegativity of the central atom represent a small effect on the attraction for the H+ compared to the influence of the greater EN effects of one or more (more electronegative) oxygen atoms. SPECIAL FOCuS: Acids and Bases The tendency of metal ions to behave in this manner depends primarily on two factors: charge and size, as shown in the table below where the pKa values for several metal ions participating in this reaction are given along with their radii (Wulfsberg 1991, 25). pKa values and Radii of Common Metal Ions Ion pK Radius (nm) Na+ 14.2 0.116 Ca2+ 12.8 0.114 Pu3+ 7.0 0.114 Mg2+ 11.4 0.086 Ca2+ 12.8 0.114 Ba2+ 13.5 0.149 The radius and charge can be combined into a single quantity called the charge density. (It is not necessary to consider the details of calculating this quantity but only to deal with it qualitatively, where it increases with an increase in charge and/or a decrease in ionic radius.) As a first approximation, solutions containing metal ions become more acidic as the charge density of the metal ion increases. As the charge density of metal ions increases, this reaction may proceed further, to the precipitation of metal hydroxides: M(H2O)6x+ --> M(OH)x + xH+ + (6-x) H2O The pH at which this reaction is initiated decreases as the charge density of the metal ion increases, as shown by the pH values for the formation of the following hydroxides, which are characterized by a decrease in size and/or an increase in the charge of the cation (Masterton and Hurley 2006, 427). Ca(OH)2 pH 12.3 Fe(OH)2 pH 8.67 Fe(OH)3 pH 4.49 For species with extremely high charge densities, the potential cations may exist as oxoanions. This is illustrated by the ions VO43-, CrO42- and MnO4-, in which the charges on the nonexistent cations would be +5, +6, and +7, respectively. Hard and Soft Acids and Bases (HSAB) This concept represents one of the most significant developments in acid–base theory in recent years. It represents an extension of the Lewis theory and was developed to account for the results obtained when a variety of electron-pair acceptors were reacted with several electron pair donors. Specifically, it was found that a cation such as Ag+ reacts with halide ions in the order I- > Br- > Cl- > F-, whereas (as was pointed out above) H+ reacts with this same series of ions in the opposite order: F- > Cl- > Br- > I- . 102 Acid–Base Chemistry Beyond AP To account for this difference in behavior, the electron-pair acceptors (Lewis acids) H+ and Ag+ were labeled as hard and soft, respectively, while the halide ions F-, Cl-, Br-, and I- were assumed to progress from hard to soft. Hard acids and bases are species that interact in an electrostatic manner, whereas soft acids and bases exhibit covalent interactions. A fundamental principle of this theory is that hard acids interact preferentially with hard bases. Thus, hard H+ reacts more strongly with hard F- (to form the weak acid HF) than with the soft I- (since HI is a strong acid) and the soft Ag+ interacts more strongly with soft I- (to form the less soluble AgI) than with harder F- (AgF is more soluble). The HSAB theory has provided the basis to account for a wide range of chemical phenomena. For example, it offers explanations for the strengths of the hydrohalic acids and the solubilities of the silver halides as described above. Further, however, it provides the basis to understand the distribution of metal ions in nature between sulfides (soft species such as Ag+, Hg2+, Pt2+) and oxides (hard species Fe2+/3+, Cu2+). Activities (1) Determine the pKa value of an acid with a Ka = 1.2 x 102. (2) Give the formula for the acidic and basic species in CH3COOH. (3) Organic amines behave as weak bases in aqueous solutions when an H+ ion reacts with the lone pair of electrons on the N atom. a. Predict the relative strength of an organic amide (RCONH2) behaving as a weak base. b. Give the formula of an appropriate solvent in which an amide would behave this way. (4) Give an approximate pK for the hydroacid stibine (SbH3). (5) The oxoacid pyrosulfuric acid (H2S2O7) is produced when SO3 is bubbled into H2SO4. Write the formula in the (HO)mEOn style, determine the number of free O atoms per S atom and predict its strength relative to that of H2SO4. (6) Use the charge density to a. Account for the fact that the reaction of alkali metal ions and H2O goes only to M(H2O)6+. b. Identify the alkali metal ion that would be most likely to go beyond this stage. 103 SPECIAL FOCuS: Acids and Bases (7) Copper can form the ions Cu+ and Cu2+. State and explain which of these is more likely to react with the hard base F-. Conclusion This essay has attempted to present several aspects of acid–base theory beyond those discussed in earlier selections (or in the average AP Chemistry course). The topics covered here were chosen in the hope that they would further the understanding and appreciation of the topics usually found in AP courses. Bibliography Atkins, P. W., T.L. Overton, J.P. Rourke, M.T. Weller, and F.A. Armstrong. Inorganic Chemistry, 4th ed., Oxford, UK: Oxford University Press, 2006, 117. Bartmess, J. E., J. A. Scott, and R. T. McIver Jr. “Scale Of Acidities in the Gas Phase from Methanol to Phenol.” Journal of American Chemistry Society 101 (1979): 6046–56. Bartmess, J. E., and Robert J. Hinde. “The Gas-Phase Acidities of the Elemental Hydrides are Functions of Electronegativity and Bond Length.” Canadian Journal of Chemistry 83(11) (2005): 2005–12. Bowser, J. R. Inorganic Chemistry. Pacific Grove, CA: Brooks/Cole Publishing, 1993, 312–14. Masterton, W. L., and C. N. Hurley. Chemistry: Principles and Reactions, 5th ed. Pacific Grove, CA: Brooks/Cole Publishing, 2006, 427. Wulfsberg, G. Principles of Descriptive Inorganic Chemistry. Mill Valley, CA: University Science Books, 1991, 25. 104 ) . from the Houghton Mifflin Company.Acid–Base Lesson Plans Heather Hattori John H. here is a basic outline of what I would cover in my acid–base unit in years where I did not teach the pre-AP class and was unsure of the students’ background. General Outline Day 1: • • • Review characteristics and examples of acids and bases Review acid/base theories (Arrhenius. polyprotic. but that would be dependent on when I teach acids and bases. diprotic. Tenth-. 5th edition. This past year. and twelfth-grade students enroll in AP Chemistry with an Algebra II minimum math co-requisite. oxy acids. It brings back familiar concepts and calculations to second-year students. Guyer High School Denton. etc. as well as a wealth of laboratory activities to keep them engaged. Texas This is one of my favorite units of the year. Be that as it may. Brønsted-Lowry. ternary. You may find my problem selections in the study guide odd. we as a district moved equilibrium into the first semester with the acid–base unit following directly after it. since we teach the first year of chemistry to freshmen. This may be disconcerting to some who would prefer to follow the textbook in the order it is presented. eleventh-. It works for us. but may not work for you. I teach on an A/B day 90-minute block schedule so some activities are done during class while others are done outside (either after school or on Saturdays). and Lewis) and associated concepts Review strong/weak acids and bases as well as additional classifications of acids (monoprotic. The text provided to students is Zumdahl’s Chemistry. • • Day 2: • • • Determining the pH of a substance lab Discussion of titration curves Pre-lab: Titration Curves of Weak and Strong Acids and Bases Day 3: • Titration Curves of Weak and Strong Acids and Bases Lab • Review titration problems • Pre-lab: % Acetic Acid in Vinegar (fits AP suggested lab topic: Determination of Concentration by Acid–Base Titration. dilute) and calculations (molarity) Review acid–base reactions including acid–base formation from acid–base anhydrides • Review pH scale. pH. pH Determination) • Assign Study Guides for Chapters 14 and 15. and pOH) • Pre-lab: Determining the pH of a Substance (fits AP suggested lab topic: Determination of Appropriate Indicators for Various Acid–Base Titrations. Including a Weak Acid or Weak Base) Day 4: (this lab has been done outside the school day) • % Acetic Acid in Vinegar Lab Day 5: • • • Weak acid–base equilibria—revisiting K Calculating the pH of a weak acid or base Use past AP free-response questions for practice Day 6: • • 106 Common-ion Effect Buffers . logarithms. and associated strong acid–base calculations ([H3O+].SPECIAL FOCuS: Acids and Bases Review concentration terms (concentrated. [OH-]. Bibliography Holmquist Dan D. David. Heather R. and Don Volz. 2006. and Titration Calculation Review Worksheet. OR: Vernier Software & Technology. 1995. Beaverton.Acid–Base Lesson Plans • Pre-lab: Preparation of a Buffer Solution at a Given pH or A Resistance to Change (fits AP suggested lab topic: Preparation and Properties of Buffer Solutions) Day 7: (this lab has been done outside the school day) • Preparation of a Buffer Solution at a Given pH Lab Day 8: (If time permits. Heather R. Heather R. Hattori. 2005. In Chemistry with Calculators.. 2007. • Acids and Bases Introduction-Review Notes • Titration Notes • Calculating the pH of a Weak Acid or Base Notes • Common Ion Effect and Buffers Notes • Determining the pH of a Substance. • Dissociation and pH Practice. then skip to Day 9) • Complex Ion Equilibria Day 9: • Unit Assessment Additional Attachments: • Study Guides to Accompany Zumdahl’s Chemistry. Heather R. NJ: The Peoples Publishing Group. Hostage. Inc. Hattori.” in Laboratory Investigations: AP Chemistry. “Finding the Mass Percent of Acetic Acid in Vinegar. and Martin Fossett. If not. 5th edition. Hattori.. • Acid-Bases. Hattori. Jack Randall. 2000. Saddle Brook. 2001. 107 . Reactions. CA: PASCO Scientific. 5th ed. Chemistry. and Martin Fossett. Zumdahl. Saddle Brook..SPECIAL FOCuS: Acids and Bases Hostage. David. 2004. Roseville.” in PasPort Explorations in Chemistry. 108 . Laboratory Investigations: AP Chemistry. 2006. Houghton Mifflin Company. NJ: The Peoples Publishing Group. Moore. 2000. “A Resistance to Change. Steven S. McGraw-Hill. and Richard Langley.. 2003. Pasco Scientific. and Zumdahl. John T. 5 Steps to a 5 AP Chemistry. • Milk of magnesia contains magnesium hydroxide.0 • Are corrosive Bases • Taste bitter • Turn red litmus paper blue • Have a pH > 7. Soft drinks contain carbonic and phosphoric acids.) contain citric acid. • Drano contains sodium hydroxide. Characteristics of Acids and Bases • Acids • Taste sour • Turn blue litmus paper red • Have a pH < 7. Vinegar contains acetic acid.Appendix A Review Notes on Acids and Bases Introduction Common Acids and Bases Where do I find acids? • • • • Fruits (lemons. Where do I find bases? • Cleaners contain ammonia.0 • Are caustic • . Rain water contains sulfuric acid. oranges. etc. (aq) Bases produce OH. the hydroxide ion could accept a hydrogen ion from the ammonium and become water.(hydroxide ions) in solution.(aq) CaO (s) + H2O (l) → Ca(OH)2 (aq) The first equation sets up two interesting relationships. HCl(aq) + H2O (l) → H3O1+ (aq) + Cl1. HCl(aq) → H1+ (aq) + Cl1. NH3 (aq) + H2O (l) → NH41+ (aq) + OH1.(aq) Brønsted-Lowry Model Acids are proton donors. Not only does the ammonia accept a hydrogen ion to become the ammonium ion. In other words. Bases are electron-pair donors in solution. These two relationships make the ammonia/ammonium ion and the water/hydroxide ion conjugate acid/base pairs.(aq) H3PO4 (aq) + 3H2O (l) → 3 H3O1+ (aq) + PO43.SPECIAL FOCuS: Acids and Bases Three Acid–Base Theories The following theories are where we get our definitions of what acids and bases are: • • • Arrhenius Concept Brønsted-Lowry Model Lewis Theory Arrhenius Concept Acids produce H+ (hydrogen ions) in solution. NaOH (aq) → Na1+ (aq) + OH1. BF3(aq) + :NH3 (aq) → F3B:NH3 110 . bases accept H+ (hydrogen ions) in solution.(aq) Bases are proton acceptors. acids donate H+ (hydrogen ions) in solution. In other words.(aq) H3PO4 (aq) → 3 H1+ (aq) + PO43. Lewis Theory Acids are electron-pair acceptors in solution.(aq) Ca(OH)2 (aq) → Ca2+ (aq) + 2OH1. 0 X 10-14 x2 = 1.0 X 10-14 always at 25°C We can solve for the [H3O1+] or [OH1-] based on the above equation. then the solution is acidic [H1+] < [OH1-].0 X 10-14 becomes [x][x] = 1.0 X 10-7 ∴ [H3O1+] = 1. one OH1. Since concentrations can be large or small. such as 12 M or 4.0 X 10-14 x = 1. 2H2O (l) ↔ H3O+1 (aq) + OH-1 (aq) (Brønsted-Lowry model) According to the Brønsted-Lowry model one water molecule (in the equation above) is acting as an acid by donating a hydrogen ion to the other molecule of water forming a hydronium ion.is made. so: [H3O1+][OH1-] = 1. Remember that: H2O (l) ↔ H+1 (aq) + OH-1 (aq) (Arrhenius concept) According to the Arrhenius concept water (in the equation above) is acting as an acid by producing hydrogen ions as well as a base by producing hydroxide ions. then the solution is neutral 111 . If we write the equilibrium expression for this we get: Kw = [H3O1+][OH1-] = 1. then the solution is basic [H1+] = [OH1-]. The pH scale was developed to make it easier to recognize if a substance was acidic or basic. sometimes it is hard to compare the [H3O1+]. The water molecule that is accepting the hydrogen ion is acting as the base.Appendix A The pH Concept pH represents the relative acidity of a solution based on hydrogen ion or hydronium ion concentration and can be found using either of the following equations: pH = -log[H+] or pH = -log[H3O+]. The amphoteric nature of pure water contributes to its neutral nature. For every H3O1+ produced.0 X 10-7 M = [OH1-] In solutions when the: [H1+] > [OH1-].2 X 10-13 M. This means that water can act as an acid or a base due to the concept known as autoionization shown in the following two examples. What is the pH of a solution with a [H+] = 1. Let’s look at a hydrated aluminum hydroxide ion as an example. solve for the [OH1-] and determine the solution’s nature (acidic. [Al(H2O)4(OH)2]1+ + HCl ↔ [Al(H2O)5OH]2+ + Cl1When a hydroxide from [Al(H2O)4(OH)2]1+ accepts the hydrogen ion to create the fifth water of hydration. Given the [H3O1+] = 1. pH scale Based on logarithms. Substances other than water can also be amphoteric. Examples 1. or neutral).0 X 10-9 M? A similar formula can be used to calculate the pOH (relative basicity) if you know the [OH1-].4 X 10-7M? By combining all three of the previous formulas: Kw = [H3O1+][OH1-] = 1. 2. solve for the [H3O1+] and determine the solution’s nature (acidic. Given the [OH1-] = 1.0 X 10-14 pH = -log [H3O1+] 112 . basic. the pH scale was developed.0 X 10-2M. What is the pOH of a solution with a [OH1-] = 2. it is acting as a Brønsted-Lowry base.SPECIAL FOCuS: Acids and Bases Based on this and the fact that [H1+] and [OH1-] can be calculated if you know one or the other. [Al(H2O)4(OH)2]1++ NaOH ↔ Al(H2O)3(OH)3 + OH1When a water of hydration from [Al(H2O)4(OH)2]1+ donates the hydrogen ion to becomes the third hydroxide ion. basic. pOH = -log [OH1-] 4. or neutral). it’s time to solve some problems.0 X 10-5M. it is acting as a Brønsted-Lowry acid. 0 Acidic pH = -log[H+] or 7 Neutral pH = -log [H3O+] 14 Basic 3. A weak acid or base is one in which only a small portion of the particles dissociate or one that only partially ionizes. Mg(OH)2. A strong acid or base is one that completely dissociates in water or has 100% ionization. Sr(OH)2. [H3O1+]. All others are considered weak. HI. and HClO4. with the last two. and Ba(OH)2. HBr. Strong bases include: LiOH.8. 113 . What is the pOH of a solution that has a pH of 10. All others are considered weak. Ca(OH)2. What are the pOH. Do not confuse the first two. weak or strong. which describe the solution concentration. and [OH1-] of a vinegar solution? Relative Strengths of Acids and Bases Acids and bases can be concentrated or dilute.Appendix A pOH = -log [OH1-]. H2SO4.0? Vinegar has a pH of 2. which describe the amount of dissociation. KOH. NaOH. HNO3. • • Strong acids include: HCl. 5. we can derive the formula: pH + pOH = 14. . ) contain citric acid. • Soft drinks contain carbonic and phosphoric acids.Appendix B Review Notes on Acids and Bases— Teacher’s Edition Common Acids and Bases Where do I find acids? • Fruits (lemons. Where do I find bases? • Cleaners contain ammonia. • Vinegar contains acetic acid. • Rain water contains sulfuric acid. etc. oranges. • Drano contains sodium hydroxide. Characteristics of Acids and Bases Acids: • • • • taste sour turn blue litmus paper red have a pH < 7.0 are corrosive Bases: • taste bitter • turn red litmus paper blue • have a pH > 7.0 • are caustic . • Milk of magnesia contains magnesium hydroxide. (aq) Bases produce OH.(aq) + H2O (l) → H3O+ (aq) + PO43.(aq) + H2O (l) → H3O+ (aq) + HPO42.(aq) H2PO4.(aq) H3PO4 (aq) → H+ (aq) + H2PO4.(aq) Brønsted-Lowry Model Acids are proton donors.(aq) H3PO4 (aq) + H2O (l) → H3O+ (aq) + H2PO4.(hydroxide ions) in solution.(aq) HPO42. NH3 (g) + H2O (l) → NH4+ (aq) + OH. HCl(aq) + H2O (l) → H3O+ (aq) + Cl. acids donate H+ (hydrogen ions) in solution. These are hydrated to form H(H2O)n+. the hydroxide ion could accept a hydrogen ion from the ammonium ion and become water.(aq) Ca(OH)2 (aq) → Ca2+ (aq) + 2OH. usually to water.SPECIAL FOCuS: Acids and Bases Three Acid–Base Theories The following theories are where we get our definitions of what acids and bases are: • Arrhenius Concept • Brønsted-Lowry Model • Lewis Theory Arrhenius Concept Acids produce H+ (hydrogen ions) in solution. The reaction goes both ways and so reaches equilibrium. Not only does the ammonia accept a hydrogen ion to become the ammonium ion. bases accept H+ (hydrogen ions) in solution. In other words.(aq) Bases are proton acceptors. this species is usually abbreviated to H+ except when a balanced equation is desired. 116 . as shown in some of the examples below. In other words. These two relationships make the ammonia/ammonium ion and the water/hydroxide ion conjugate acid–base pairs. For convenience. HCl(aq) → H+ (aq) + Cl.(aq) CaO (s) + H2O (l) → Ca(OH)2 (aq) The first equation sets up two interesting relationships. NaOH (aq) → Na+ (aq) + OH. where n may be 1–6. The water molecule that is accepting the hydrogen ion is acting as the base.is made so [H3O+] = [OH-] = x [H3O1+][OH1-] = 1.(aq) (Brønsted-Lowry model) According to the Brønsted-Lowry model. Remember that: H2O (l) ↔ H+ (aq) + OH. pH represents the relative acidity of a solution based on hydrogen ion or hydronium ion concentration and can be found using either of the following equations: pH = -log[H+] or pH + -log[H+] . The pH scale was developed to make it easier to recognize if a substance was acidic or basic. This means that water can act as an acid or a base by accepting or donating protons. which is shown in the following two examples. Since concentrations can be large or small.(aq) (Arrhenius concept) According to the Arrhenius concept. If we write the equilibrium expression for this we get: Kw = [H3O+][OH-] = 1. one OH. water.2 X 10-13 M. sometimes it is hard to compare the [H+].0 X 10-14 at 25°C We can solve for the [H3O+] or [OH-] based on the above equation. The amphoteric nature of pure water contributes to its neutral nature.0 X 10-14 becomes [x][x] = 1. BF3(aq) + :NH3 (aq) → F3B:NH3 The pH Concept Conventionally. for every H3O+ produced. with no other substances present. the symbol [H+] is used to represent the concentration of H+ ions in moles per liter. in the equation above.Appendix B Lewis Theory Acids are electron pair acceptors in solution. is acting as an acid by producing hydrogen ions as well as a base by producing hydroxide ions. In pure water.0 X 10-14 117 .Bases are electron pair donors in solution. one water molecule in the equation above is acting as an acid by donating a hydrogen ion to the other molecule of water forming a hydronium ion. Water also undergoes a process called auto-ionization. such as 12 M or 4. H2O (l) + H2O (l) ↔ H3O+ (aq) + OH. 118 .0 X 10-12 M which is less than the [H3O+] so 1.0 X 10-9 M which is less than the [OH-] so the 1. then the solution is neutral.0 X 10-14 [H3O+] = 1. Let’s look at a hydrated aluminum hydroxide ion as an example. solve for the [H3O+] and determine the solution’s nature (acidic. solve for the [OH-] and determine the solution’s nature (acidic.0 X 10-14 [1.0 X 10-14 [H3O+] [1. basic. [H+] < [OH-]. [Al(H2O)4(OH)2]++ NaOH ↔ Al(H2O)3(OH)3 + OHWhen a water of hydration from [Al(H2O)4(OH)2]+ donates the hydrogen ion to becomes the third hydroxide ion it is acting as a Brønsted-Lowry acid. then the solution is basic. [H+] = [OH-]. Kw = [H3O+][OH-] = 1. it’s time to solve some problems.SPECIAL FOCuS: Acids and Bases x2 = 1. [Al(H2O)4(OH)2]+ + HCl ↔ [Al(H2O)5OH]2+ + ClWhen a hydroxide from [Al(H2O)4(OH)2]+ accepts the hydrogen ion to create the fifth water of hydration. or neutral). Given the [H3O+] = 1. then the solution is acidic. or neutral).0 X 10-5M. Based on this and the fact that [H+] and [OH-] can be calculated using Kw if you know one or the other.0 X 10-7 ∴ [H3O+] = 1.0X 10-5] = 1.0X 10-5 M solution is basic.0X 10-2 M the solution is basic. 2.0 X 10-14 x = 1. basic. Given the [OH-] = 1. Kw = [H3O+][OH-] = 1.0 X 10-7 M = [OH-] In solutions when the: [H+] > [OH-]. it is acting as a Brønsted-Lowry base.0 X 10-14 [OH-] = 1. Substances other than water can also be amphoteric.0 X 10-14 M2 = 1. Examples 1.0X 10-2 M][OH-] = 1.0 X 10-14 M2 = 1.0 X 10-2M. we can derive the formula: pH + pOH = 14 4.8.0 X 10-9 M? pH = -log[H+] pH = -log(1.0 pOH = 6. What is the pOH of a solution that has a pH of 10. the pH scale was developed.2 pH = -log[H+] 2. [H+]. pOH = -log [OH-] What is the pOH of a solution with a [OH-] = 2.00 A similar formula can be used to calculate the pOH (relative basicity) if you know the [OH-].0 + pOH = 14 pOH = 4. What are the pOH. What is the pH of a solution with a [H+] = 1.4 X 10-7M? pOH = -log [OH-] pOH = -log(2.62 Vinegar has a pH = 2.4 X 10-7 M) By combining all three of the previous formulas: Kw = [H+][OH-] = 1.0X10-9 M) pH = 9.2 = -log[OH-] [H+] = 6 X 10-12 M 119 .Appendix B pH scale Based on logarithms.0? pH + pOH = 14 10.8 = -log[H+] [H+] = 2 X 10-3 M pOH = -log[OH-] 11.8 + pOH = 14 pOH = 11. and [OH-] of a vinegar solution? pH + pOH = 14 2. 0 Acidic pH = -log[H+] or 7 Neutral pH = -log [H3O+] 14 Basic 3.0 X 10-14 pH = -log [H+] pOH = -log [OH-]. A weak acid or base is one in which only a small portion of the particles dissociate. H2SO4. HI.SPECIAL FOCuS: Acids and Bases Relative Strengths of Acids and Bases Acids and bases can be concentrated or dilute. HBr. Ca(OH)2. weak or strong. or one that only partially ionizes. 120 . which describe the amount of dissociation. HNO3. Do not confuse the first two. H3PO4 . All others are considered weak. Strong bases include: LiOH. with the last two. Strong acids include: HCl. A strong acid or base is one that completely dissociates in water or has 100% ionization. All others are considered weak. Mg(OH)2. KOH. Sr(OH)2. and HClO4. which describe the solution concentration. and Ba(OH)2. NaOH. Acid–Base Reactions When an acid reacts with a base a neutralization reaction occurs. the indicator experiences a color change at the endpoint of the titration. The necessary chemical components of a titration include a standard solution. and solution of unknown concentration.Appendix C Titration Notes A titration is the experimental procedure used to determine the concentration of an unknown acid or base. AB + CD → AD + CB The products of an acid–base reaction are a salt (metal/nonmetal compound) and water. An acid–base reaction is a double displacement reaction. It is the area in which the slope of the line approaches infinity (or is nearly vertical). Titrations and titration curves tell you many things about a reaction. Examples HCl + NaOH → NaCl + HOH H2SO4 + KOH → . the endpoint and the equivalence point (when there are equal amounts of acid and base) occur at nearly the same time. Titration curves are a way to visually represent the change in pH during a titration. If the best indicator is chosen. Common indicators used in acid–base titrations include phenolphthalein (clear to pink pH = 8-10) and bromothymol blue (yellow to blue pH = 6-8). The same thing happens when a base reacts with an acid. During a titration. The equivalence point can be visually determined from the titration curve. and an indicator. 0 mL of 2.0 mL sample of HCl was titrated to the endpoint with 15. What was the molarity of the acid? 2. What volume of 1.5M HNO3 is necessary to titrate 25. A 10.0 M KOH.0 M NaOH.5 mL of 1. How much 0. This can be done using a balanced equation and stoichiometry. A 25.0 mL sample of H2SO4 was neutralized by 13.50 M NaOH is necessary to neutralize 20.SPECIAL FOCuS: Acids and Bases H3PO4 + LiOH → Hydrobromic acid + calcium hydroxide HBr + Ca(OH)2 → CaBr2 + 2H2O Hydrofluoric acid + aluminum hydroxide HF + Al(OH)3→ AlF3 + 3H2O Titration Problems Often.5M H3PO4? 4. What was the molarity of the acid? 3.0 mL of 0.0 mL of 2. the goal of a titration is to calculate the concentration of one of the reactants.05M Ca(OH)2 solution to the endpoint? 122 . 1. a neutralization reaction occurs. The same thing happens when a base reacts with an acid. If the best indicator is chosen. Titration curves and the progress of titrations can also be followed by using a pH meter or probe. The equivalence point can be visually determined from the titration curve. the endpoint and the equivalence point (when equivalent amounts of acid and base have been added) occur at nearly the same time. Common indicators used in acid–base titrations include phenolphthalein (clear to pink pH = 8–10) and bromothymol blue (yellow to blue pH = 6–8). During a titration. Titration curves are a way to visually represent the change in pH during a titration. It is the area where the slope of the line approaches infinity (or is nearly vertical). the indicator experiences a color change at the endpoint of the titration. An acid–base reaction is a double displacement reaction. AB + CD → AD + CB The overall products of an acid–base reaction are a salt (metal/nonmetal compound) and water. The necessary chemical components of a titration include a standard solution. and may include a colored indicator.Appendix D Titration Notes – Teacher’s Edition A titration is the experimental procedure used to determine the concentration of an unknown acid or base. Acid–Base Reactions When an acid reacts with a base. a solution of unknown concentration. Titrations and titration curves tell you many things about a reaction. . 0 mL H2SO4 3.5 mmol H3PO4 × 1.SPECIAL FOCuS: Acids and Bases Examples HCl + NaOH → NaCl + HOH (which also can be written H2O) H2SO4 + 2KOH → K2SO4 + 2H2O H3PO4 + 3LiOH → Li3PO4 + 3H2O Hydrobromic acid + calcium hydroxide HBr + Ca(OH)2 → CaBr2 + 2H2O Hydrofluoric acid + aluminum hydroxide HF + Al(OH)3 → AlF3 + 3H2O Titration Problems Often.0 mL NaOH × 1 L NaOH × 1000mL NaOH 2.0 mL H3PO4 3 mmol NaOH 1 mmol H3PO4 × 1mL NaOH 1. What was the molarity of the acid? HCl + NaOH → NaCl + H2O 15.0 mL of 2. This can be done using a balanced equation and stoichiometry. What was the molarity of the acid? H2SO4 + 2KOH → K2SO4 + 2H2O 13.0 mL sample of HCl was titrated to the equivalence point with 15.50mmol NaOH = 1.50 M NaOH is necessary to completely react with 20.0 mL HCl × 1000 mL 1 L HCl = 1. A 25.0 mL of 2.0 mL sample of H2SO4 was completely reacted by 13.0 M KOH.00 × 102mL NaOH 124 . H3PO4 + 3NaOH → Na3PO4 + 3H2O 20. What volume of 1. 1.0 M NaOH.0 mol KOH × 1 mol H2 SO4 × 1000 mL H2 SO4 1000mL 1L 2 mol KOH 1 L H2SO4 × 1 = 0.0 mL H3PO4 × 2.68 M H2SO4 10.5M H3PO4? Note: M is molarity.5 mL KOH × 1 L KOH × 1.5 mL of 1. A 10.2 M HCl 2. the goal of a titration is to calculate the concentration of one of the reactants. which can be mol/L or millimoles/mL.0 mol NaOH × 1 L NaOH 1 mol HCl × 1 mol NaOH 1 25. 0 mL of 0.5 mmol HNO3 = 5.5M HNO3 is necessary to titrate 25.0 mL Ca(OH)2 × 0.05 mmol Ca(OH)2 × 1 mL Ca(OH)2 2 mmol HNO3 1 mmol Ca(OH)2 × 1mL HNO3 0.Appendix D 4.00 mL HNO3 125 .05M Ca(OH)2 solution to the equivalence point? 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O 25. How much 0. . 34 X 10−3 = [ H 3O 1+ ] . one must take into account the fact that many species exist in the reaction vessel.100M -x ] Assuming the effect of x on 0. only a small percentage of it dissociates. Acetic acid is a weak acid and thus. solving for x we get: x2 x 2 = 1.100 M acetic acid solution whose Ka = 1.100M -x 0. The reaction: HC2H3O2 (aq) + H2O (l) ↔ H3O1+ (aq) + C2H3O21. the case of calculating the pH of a 0. for example. weak bases.100M .] [ HC2H3O 2 ] Next we need to look at the conditions during the reaction: HC2H3O2 Initial Change Equilibrium 0.8 X 10-5.8 X 10-5 [ 0.Appendix E Calculating the pH of a Weak Acid or Base When working with weak acids.8 X 10−6 x = 1. This becomes an equilibrium problem instead of a simple stoichiometry problem (like it would if it were a strong acid). or buffer systems.x H3O1+ 0 +x x C2H3O210 +x x Substituting into the equilibrium expression we get: Ka = [x][x] = 1. Take.100M is negligible because of the large difference (power difference > 10-3 units) in concentration and Ka.(aq). We begin by writing the equilibrium expression: Ka Ka = [H3O1+ ][ C2H3O 21. 00M ammonia solution with a K a = 5.0 X 10-14 so Kb = substituting in values we get: Ka 1. A. and the pH = 11. you used 5. pH = 14 – 2.00 . you selected the value that was equal to the pOH. 2.8 X 10-5 [ 2. solving for x we get: x 22 = 3.87.22 C. If you chose answer A.8 X 10−5 M .SPECIAL FOCuS: Acids and Bases Now we can solve for pH using the equation: pH = -log[H3O1+] and find the pH to be 2. 8.71 X 10-10 for your Kb value instead of 1.00 M -x 2.0 X 10−3 = [OH 1− ] Now we can solve for pOH using the equation: pOH = -log[OH1-] and find the pOH to be 2.78 Solution: The correct answer is D.6 X 10−6 x x = 6. you only used the initial concentration of ammonia and substituted it into the pH = -log[H3O1+] assuming that the concentration of ammonia was equal to the hydronium ion concentration.71 X 10 NH3 Initial Change Equilibrium 2.22.00M is negligible because of the large difference (power difference > 10-3 units) in concentration and Kb.x NH41+ 0 +x x OH10 +x x Substituting into the equilibrium expression we get: [x][x] Kb = = 1. Since pH + pOH = 14. Kb = −10 5.94 D.00M -x ] Assuming the effect of x on 2.8 X 10-5. 128 .30 B. If you chose answer B.71 X 10-10 @ 25°C.78. Example 1 Calculate the pH of a 2. The reaction is NH3(aq) + H2O(l) ↔ NH41+(aq) + OH1-(aq) Kw Since Kw = Ka Kb = 1. If you chose answer C. 11. -0.0 X 10−14 = 1.22. Appendix F Applications of Aqueous Equilibria Notes Different Constants (to hopefully clear confusion) Kb Kf k Kw Ka Kb Kc Kp Ksp Kd Kf molal boiling point constant molal freezing point constant rate constant ionization constant for water acid dissociation constant base dissociation constant equilibrium constant for concentrations equilibrium constant based on pressures equilibrium constant for solubility product equilibrium constant for decomposition reactions equilibrium constant for formation reactions ∆tb = K bm ∆tf = Kfm Rate = k [reactants] Kw = Ka Kb = 1. If NaF is added to HF: HF ↔ H+ + F NaF ↔ Na+ + F More fluoride ions are added to the solution. This complicates the equilibrium by providing an additional source for ions.0 X 10 -14 @ 25°C Ka = [H+][A -]/[HA] Kb = [X+][OH -]/[XOH] The Effect of Common Ions When a salt is added to an acid or base containing similar ions. the equilibrium shifts to relieve the stress this is an application of . . This shifting process is called the common ion effect. The equilibrium shifts to the left. If NH4Cl is added to NH3: NH3 + H2O ↔ NH4+ + OH NH4Cl ↔ NH4+ + Cl More ammonium ions are added to the solution. The equilibrium shifts to the left. When working with buffers. Ka = 1. A 100. 0.300 M NaOH. e. Determine the pH after the following quantities of base have been added to the acid solution and draw the resulting titration curve. d. a. Problems like this have appeared in the past on the AP Exam. Along the way.00 mL Indicators and Titrations Common indicators used in acid base titrations are: phenolphthalein (clear to pink pH =8-10) bromothymol blue (yellow to blue pH = 6-8) universal indicator (red to purple pH = 1-13) 130 .00 mol of ammonium chloride in 1. c.SPECIAL FOCuS: Acids and Bases Example 1 Calculate the pH of a mixture containing 0.3 x 10-5 Buffers Combinations of weak acids or weak bases with their salts form buffers.00 L of solution versus the pH of the ammonia alone? Kb = 1. f.0 mL sample of 0.35) was titrated with 0.100 M HC3H5O2 and 0. the pH of the mixture can be calculated.81 X 10-5 The Henderson-Hasselbalch equation can also be used to solve for pH. be sure to look at the stoichiometry before setting up the equilibrium. pH = pKa + log ([A-]/[HA]) pOH = pKb + log ([X+]/[WB]) Solution: Titrations are laboratory procedures used to determine the concentration of acids or bases.00 mL 55.150 M HNO2 (pKa = 3.00 mL 25.00 mL 75.50 mL 50.00 mL 49. Example 2 What is the difference in pH of a solution containing 2.00 mol of ammonia and 3.100 M NaC3H5O2. Buffers are solutions that resist changes in pH when hydrogen or hydroxide ions are added. b. Solubility Equilibria When solids dissolve to form aqueous solutions.26 = log 10−1. none is dissolved.1 M.8 X 10-13 So if I am not mistaken.0055M.2 X 10-2 4.2 X 10-2 Large 1.2 X 10-8 4. When a salt is added to water at first.5 X 10-2 1.0 = pK a + 6.0 = pKa + log [HA] Acids to choose from: H2SO3 HOAc H3PO4 H2CO3 HSO41H2SO4 Their Ka values 1. the formula units of the salt begin to dissolve.0 X 10-7 1.3 X 10-3 6.8 X 10-5 7.6 X 10-11 1.Appendix F To prepare a buffer with a specific pH (part of pre-lab discussion): = pK a + pH = pKa + log [ A− ] [HA] [ A− ] 6. [ A− ] 6.3 X 10-7 5.055 = [HA] If [HA] = 0. then as the water molecules collide with the salt and interact with the intermolecular forces.74 + log   [HA ] 1.0 = p(1. Eventually the maximum amount of 131 . for a buffer with pH = 6.8 X 10 ) + log [HA] -5 A−  6. you would choose HOAc and NaOAc since the pKa of HOAc is closest to 6. the equilibrium involved is the solubility equilibria.0 = 4. then [A-] = 0.26 = [ A− ] [HA] [ A− ] [HA] [ A− ] 0. 132 .2 X 10-5. If the salts produce different numbers of ions when they dissolve.(aq) Why is the BaCO3 not included in the K sp expression? Notice that the K value is very small. Ksp problems are much like Ka and Kb problems.20 M solution of potassium sulfate. Calculate the solubility of silver sulfate in a 0.SPECIAL FOCuS: Acids and Bases solid that can dissolve at that particular temperature does dissolve and equilibrium is reached. generally reducing the amount that dissolves by providing an initial concentration of one of the ions. The Common Ion Effect on Solubility A common ion affects the solubility of a solid. Calculating Ksp from solubility Calculating solubility from Ksp PbSO4 has a measured solubility of 1. RICE diagrams are very useful in demonstrating this affect. At 25°C. as in the previous example. then their Ksp values can be used to determine their solubilities relative to each other.6 X 10-9 + CO32. The solubility product is a constant value for a particular temperature (usually room temperature when you look it up in a chart). Do realize. BaCO3 (s) ↔ Ba2+ (aq) Ksp = [Ba2+] [CO32-] = 1. however. We would expect this since barium carbonate is not considered soluble in water based on what we have learned from solubility rules. Example 3 The Ksp of silver sulfate is 1. The smaller the Ksp. that solubility and the solubility product are not the same.14 X 10-4 mol/L Al(OH)3 has a Ksp of 2 X 10-32 at 25°C.what is the Ksp of PbSO4? What is its solubility of PbSO4 at 25°C? Relative Solubilities If salts produce the same number of ions when they dissolve. the less soluble it is. then you have to calculate the solubilities from the Ksp values. Solubility is a position that changes with temperature and the concentration of common ions present in the solution. 0 X 10 -14 @ 25°C Ka = [H+][A -]/[HA] Kb = [X+][OH -]/[XOH] The Effect of Common Ions When a salt is added to an acid or base containing similar ions. This complicates the equilibrium by providing an additional source for ions. . This shifting process is called the common ion effect. the equilibrium shifts to relieve the stress this is an application of LeChatlier’s Principle If NaF is added to HF: HF ↔ H+ + F NaF ↔ Na+ + F More fluoride ions are added to the solution. If NH4Cl is added to NH3: NH3 + H2O ↔ NH4+ + OH NH4Cl ↔ NH4+ + Cl More ammonium ions are added to the solution. The equilibrium shifts to the left.Appendix G Applications of Aqueous Equilibria Notes – Teacher’s Edition Different Constants (to hopefully clear confusion) Kb Kf k Kw Ka Kb Kc Kp Ksp Kd Kf molal boiling point constant molal freezing point constant rate constant ionization constant for water acid dissociation constant base dissociation constant equilibrium constant for concentrations equilibrium constant based on pressures equilibrium constant for solubility product equilibrium constant for decomposition reactions equilibrium constant for formation reactions ∆tb = Kbm ∆tf = Kfm Rate = k [reactants] Kw = Ka Kb = 1. The equilibrium shifts to the left. x NH41+ 0 3. Buffers are solutions that resist changes in pH when hydrogen or hydroxide ions are added.00 mole NH3 + 3.100M] x = 1.100 x + 0.] Ka = [x][0.00 L of solution versus the pH of the ammonia alone? Kb = 1.00M .00 + x x OH1- 0 x x 134 . be sure to look at the stoichiometry before setting up the equilibrium.100 M HC3H5O2 and 0.100 comes from the NaC3H5O2.89. Example 2 What is the difference in pH of a solution containing 2.3 X 10-5M = [H+] pH = -log[H+] pH = -log(1.00 mol of ammonium chloride in 1.00 mol of ammonia and 3.00 mol NH3 only NH3 + H2O(aq)  NH4+(aq) + OH .100 M NaC3H5O2.81 X 10-5 2.100 + x] [0.x H1+ 0 +x x C3H5O210 x + 0.00 mol NH4Cl NH3 + H2O(aq) ↔ NH4+(aq) + OH .3 X 10−5 and [ 0.3X10-5) pH = 4. Buffers Combinations of weak acids or weak bases with their salts form buffers.3 X 10-5 If you consider x to be negligible.(aq) 2. which is consistent with a weak acid.(aq) NH3 Initial Change Equilibrium 2.100 [The 0. When working with buffers.3 x 10-5 HC3H5O2(aq)  H+(aq) + C3H5O2-(aq) HC3H5O2 Initial Change Equilibrium 0. then this simplifies to : K K aa = [0.00 M -x 2.100x] = 1.100M -x 0.x NH41+ 0 +x x OH10 +x x NH3 2.00 M -x 2.100M . Ka = 1.SPECIAL FOCuS: Acids and Bases Example 1 Calculate the pH of a mixture containing 0.00M .100M -x] =1. Appendix G [x][x] [2.00M -x] [3.00M + x][x] [2.00M -x] Ka = = 1.81 X 10 -5 Ka = = 1.81 X 10 -5 Considering x to be negligible: x = 2.00 M × (1.81 X 10 ) 2 -5 Considering x to be negligible: 3x = 2.00 M × 1.81 X 10 -5 x = 1.21 × 10 -5 M = [OH -] pOH = -log[OH -] pOH = 4.917 pH + pOH = 14 pH + 4.917 = 14 pH = 9.083 x= 6.01 × 10 -3 M = [OH -] pOH = -log[OH -] pOH = 2.221 pH + pOH = 14 pH + 2.221 = 14 pH = 11.779 The Henderson-Hasselbalch equation can also be used to solve for pH. pH = pKa + log ([A-]/[HA]) pOH = pKb + log ([X+]/[WB]) Titrations are laboratory procedures used to determine the concentration of acids or bases. Along the way, the pH of the mixture can be calculated. Problems like this have appeared in the past on the AP Exam. A 50.0 mL sample of 0.300 M HNO2 (pKa = 3.35) was titrated with 0.450 M NaOH. Determine the pH after the following quantities of base have been added to the acid solution and draw the resulting titration curve: a. b. c. d. e. f. 0.00 mL 16.5 mL 32.0 mL 33.3 mL 35.0 mL 50.0 mL pKa = -log(Ka) 3.35 = -log(Ka) a. HNO2(aq)  H+(aq) + NO2-(aq) Ka = 4.5 × 10-4 KKa = a [H+ ][NO 2- ] =4.55X 10-4−4 = 4. X 10 [HNO 2 ] 135 SPECIAL FOCuS: Acids and Bases [HNO2] Initial Change Equilibrium 0.300 M -x 0.300 M - x [H1+] 0 +x x [NO21-] 0 +x x Ka = [x][x] =4.5 × 10 [0.300M -x] x = 1.2 × 10-2M = [H+] pH = -log[H+] pH = 1.92 b. HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O 50.0 mL HNO2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO2 16.5 mL NaOH × 1L/1000mL × 0.450 M = 0.00743 mol NaOH 0.0150 mol – 0.007425 mol = 0.07575 mol HNO2 left after reaction 50.0mL + 16.5mL = 66.5 mL or 0.0665 L HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O [HNO2] Initial Change Equilibrium 0.0150 mol -0.0743mol 0.0757 mol [NaOH] 0.0743 mol -0.0743 mol 0 [Na1+] 0 + 0.0743 mol 0.0743 mol [NO21-] 0 + 0.0743mol 0.0743 mol pH = pKa + log ([A-]/[HA]) pH = 3.35 + log(0.0743 mol/0.0665 L ÷ 0.0757 mol/0.0665 L) pH = 3.34 c. HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O 50.0 mL HNO2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO2 32.0 mL NaOH × 1L/1000mL × 0.450 M = 0.0144 mol NaOH 0.015 mol – 0.014 mol = 0.001 mol HNO2 left after reaction 50.0mL + 32.0mL = 82.0 mL or 0.0820 L HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O [HNO2] Initial Change Equilibrium 0.0150 mol -0.0144 mol 0.0006 mol [NaOH] 0.0144 mol -0.0144 mol 0 [Na1+] 0 + 0.0144 mol 0.0144 mol [NO21-] 0 + 0.0144 mol 0.0144 mol 136 Appendix G pH = pKa + log ([A-]/[HA]) pH = 3.35 + log(0.0144 mol/0.0820 L ÷ 0.0006 mol/0.0820 L) pH = 4.73 d. HNO2(aq) + NaOH(aq) Na+(aq) + NO2-(aq) + H2O 50.0 mL HNO2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO2 33.3 mL NaOH × 1L/1000mL × 0.450 M = 0.0150 mol NaOH 0.0150 mol – 0.0150 mol = 0.00 mol HNO2 left after reaction 50.0mL + 33.3mL = 83.3 mL or 0.0833 L HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O [HNO2] Initial Change Equilibrium 0.0150 mol 0.0150 mol 0 [NaOH] 0.0150 mol 0.0150 mol 0 [Na1+] 0 + 0.0150 mol 0.0150 mol [NO21-] 0 + 0.0150 mol 0.0150 mol 0.0150 mol NO21- = 0.180 M 0.0833 L NO2- is the conjugate base of a weak acid so pKa + pKb = 14.0 pKb = 14.0 – 3.35 = 10.65 NO2- (aq) + H2O (l)  OH- (aq) + HNO2(aq) [NO2-] Initial Change Equilibrium 0.180M -x 0.180M-x [H2O] [OH -] 0 +x x [HNO2] 0 +x x pKb = -log(Kb) Kb = 2.24 × 10-11 [OH− ][HNO 2 ] KKb = = 2.24 X 10−11 b [NO 2 ] x2 = (0.180M) × 2.24 × 10-11 x = 2.01 ×10-6 = [OH-] pOH = -log[OH-] pOH = 5.697 14 - pOH = pH pH = 8.303 e. HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O 50.0 mL HNO2 × 1L/1000mL × 0.300 M = 0.0150 mol HNO2 35.0 mL NaOH × 1L/1000mL × 0.450 M = 0.0158 mol NaOH 137 0150 mol = 0.125 pH + pOH = 14 pH = 14 – pOH pH = 11.0225 mol NaOH 0.0150 mol = 0.0150 mol 0.0150 mol 0.0150 mol [NO21-] 0 + 0.0 mL + 35.0150 mol Here the concentration of NaOH controls the pH since it is a strong base.SPECIAL FOCuS: Acids and Bases 0.0158mol 0.0150 mol Here the concentration of NaOH controls the pH since it is a strong base.0150 mol 0 [NaOH] 0.0150 mol [NO21-] 0 + 0. pOH = -log[OH-] pOH = -log(0.875 Indicators and Titrations Common indicators used in acid base titrations are: • • • phenolphthalein (clear to pink pH = 8-10) bromothymol blue (yellow to blue pH = 6-8) universal indicator (red to purple pH = 1-13) To prepare a buffer with a specific pH (part of pre-lab discussion) pH = pKa + log pK a 138 [ A− ] [HA] .0 mL or 0.0225 mol – 0.0008 M) pOH = 3 pH + pOH = 14 pH = 14 – pOH pH = 11 f.0150 mol 0. pOH = -log[OH-] pOH = -log(0.0150 mol 0.0075M) pOH = 2.0150 mol 0 [NaOH] 0.0225 mol 0.0008 mol NaOH left after reaction 50.0 mL HNO2 × 1L/1000mL × 0.0 mL = 85.0mL = 100.0 mL NaOH × 1L/1000mL × 0.0075 mol [Na1+] 0 + 0.0 mL or 0.0mL + 50. HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O 50.450 M = 0.1000 L HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O [HNO2] Initial Change Equilibrium 0.0150 mol 0.300 M = 0.0150 mol HNO2 50.0150 mol 0.0150 mol 0.0850 L HNO2(aq) + NaOH(aq)  Na+(aq) + NO2-(aq) + H2O [HNO2] Initial Change Equilibrium 0.0150 mol 0.0158 mol – 0.0075 mol NaOH left after reaction 50.0008 mol [Na1+] 0 + 0. 0 pK a 6. When a salt is added to water at first.0055M.3 X 10-7 HSO411.74 + log [ A− ] [HA] [ A− ] −1. for a buffer with a pH of 6.5 X 10-2 HOAc 1.0 = 4.055 = [HA] 1.8 X 10-5 ) + log 6.26 10 = [HA] [ A− ] 0.3 X 10-3 H2CO3 4. the equilibrium involved is the solubility equilibria. none is dissolved.(aq) Ksp = [Ba2+] [CO32-] = 1. 139 .0 = pKa + log [ A− ] [HA] Their Ka values 1.8 X 10-13 Acids to choose from: H2SO3 1.1 M. Solubility Equilibria When solids dissolve to form aqueous solutions. then as the water molecules collide with the salt and the intermolecular forces interact with the water. you would choose HOAc and NaOAc since the pKa of HOAc is closest to 6.26 = log [ A− ] [HA] [ A− ] [HA] If [HA] = 0.0 X 10-7 6. the maximum amount of solid that can dissolve at that particular temperature does dissolve and equilibrium is reached.2 X 10-2 H2SO4 Large So if I am not mistaken. 6.6 X 10-9 Why is the BaCO3 not included in the K sp expression? Barium carbonate is a solid and solids are not found in equilibrium expressions. Eventually. the formula units of the salt begin to dissolve.2 X 10-2 4.8 X 10-5 H3PO4 7. then [A-] = 0.2 X 10-8 5. BaCO3 (s) ↔ Ba2+ (aq) + CO32.0 = p(1.Appendix G 6.6 X 10-11 1. The Common Ion Effect on Solubility A common ion affects the solubility of a solid. Solubility is a position that changes with temperature and the concentration of common ions present in the solution. however.14 × 10-4)2 Ksp = 1. as in the previous example. Ksp problems are much like Ka and Kb problems. Calculating Ksp from solubility PbSO4 has a measured solubility of 1.Notice that the K value is very small. . that solubility and the solubility product are not the same. then you have to calculate the solubilities from the Ksp values. RICE diagrams are very useful in demonstrating this effect. then their Ksp values can be used to determine their solubilities relative to each other.30 × 10-8 Calculating solubility from Ksp Al(OH)3 has a Ksp of 2 X 10-32 at 25°C Al(OH)3(s)  Al3+(aq) + 3OH-(aq) Ksp = [Al3+][OH-]3 2 X 10-32 = (x)(3x)3 x = 5 × 10-9 M Relative Solubilities If salts produce the same number of ions when they dissolve. based on what we have learned from solubility rules. The smaller the Ksp.14 X 10-4 mol/L What is the Ksp of PbSO4? PbSO4(s)  Pb2+(aq) + SO42-(aq) Ksp = [Pb2+][SO42-] Ksp = (x)(x) Ksp = (1. We would expect this since barium carbonate is not considered soluble in water. Do realize. the less soluble it is. generally reducing the amount that dissolves by providing an initial concentration of one of the ions. If the salts produce different numbers of ions when they dissolve. The solubility product is a constant value for a particular temperature (usually room temperature when you look it up on a chart). 141 .20 +x 0.20 M of SO42[Ag1+] Initial Change Equilibrium 0 +2x 2x [SO42-] 0. we will come back to it after all the other AP material has been covered or after the AP Exam. selective precipitation. and complex ion equilibria—topics that generally receive very little attention on the AP Exam.9 × 10-3 M [Ag1+] = 2x = 7.Appendix G Example 3 The Ksp of silver sulfate is 1. Read about it if you have time. otherwise.20 + x) x = 3.20 + x 1.20 M solution of potassium sulfate.2 × 10-5. Calculate the solubility of silver sulfate in a 0.2 × 10-5 = (2x)2(0.7 × 10-3 M The rest of Chapter 15 discusses qualitative analysis. Ag2SO4 (s)  2Ag1+(aq) + SO42-(aq) Ksp = [Ag1+]2[SO42-] The K2SO4 solution contributes 0. . on notebook paper. 7. Why? Read through the In-Class Discussion Questions. They may be used for warm-ups. What is a diprotic acid? Give an example. What is Ka? 5. Organic acids are generally strong or weak? Why? 8. How do you calculate percent dissociation? 13. and the Lewis acid–base model. . in order. What is Kw? What is its value at 25 °C? 10.Appendix H Study Guides and Worksheets Chapter 14 Answer the following questions. Due date: 1. 4. 15. Water is amphoteric. Make a chart comparing the definitions of an acid and a base according to the Arrhenius concept. how do you determine the number of significant digits in your answer? 11. How does calculating the pH of a strong acid differ from calculating the pH of a weak acid? 12. Give examples of salts that produce acidic solutions. Hydrofluoric acid is the only weak acid consisting of a halogen. and neutral solutions. What is a hydronium ion? 3. 2. the Bronsted-Lowry model. How do you calculate the pH of a strong base? 14. What is a conjugate acid–base pair? Give an example. What is the difference between a strong and weak acid? 6. basic solutions. When calculating pH from [H+]. What does that mean? 9. and 153) Chapter 15 Answer the following questions. 81. 61. 95–96. 138. in order. 49–54. 18–19. Draw the pH curves for a strong acid/strong base titration. 79–82. (19. 37–38. What is qualitative analysis? 14. 109–10. 29. What is the ion product? How is it related to Ksp and precipitation? 12. 129. 40. 61. 8. in order. They may be used for warm-ups. 64. from pages 781–789. What is the common ion effect? 10. 31–32. What is selective precipitation and what is it used for? 13. What is the solubility product? 9. a weak acid– strong base titration. what do you do before setting up the ICE chart? 5. 97–99. on notebook paper.SPECIAL FOCuS: Acids and Bases Complete the following problems. What is buffer capacity and how is it calculated? 7. and a weak base–strong acid titration. What is a common ion? 2. 84–85. How are pH and solubility related? 11. Due date: 1. 123. 43–46. 142–143. 23–24. 103–04. How do common ions affect equilibrium? 3. 63. 57–60. 27. 89. What is a complex ion? 15. 119–121. 33 –36. 51–53. and 123) 144 . from pages 712–717. What is a ligand? Read through the In-Class Discussion Questions. 106. Complete the following problems. 21–23. What is the Henderson-Hasselbalch equation? When can you use it? 6. 55. When dealing with a strong acid or strong base being added to a buffer. 69. 92. 115–118. 71–72. in order. What is a buffered solution? What does it do? What substance in our body is considered a buffered solution? 4. (13–14. 88. 500 M sodium hydroxide.19 X 10 -10 M [OH-] pH pOH 2. Reactions. The approximate pOH of tomatoes is 9. What is the molarity of a lithium hydroxide solution if 15.Appendix H Acid–Bases. Write and balance the chemical equation for the reaction between: a. barium hydroxide and sulfuric acid b.020 M acetic acid solution? 7. aluminum hydroxide and nitric acid c. [H3O+] 7. d.00 mL sample of vinegar (acetic acid solution) with 38. the [H3O+].50 M phosphoric acid? 6.89 X 10 -7 M 10. and the [H3O+]? 3. How many grams of acetic acid would be in 1.00 g/mL)? 145 .5.0 mL is neutralized by 7.00 L of the vinegar? What is the percent by mass of acetic acid in the vinegar solution (assume a density of 1.5 mL of a 0.99 X 10 -5 M 1.23 X 10 -2 M 8. What are the pH.39 3. Write and balance the chemical equation for the reaction. the [OH-]. c. and Titration Calculation Review Worksheet 1. How many mL of 1. a.8.25 mL of 0. Calculate the molarity of the vinegar solution. b. A student reacted a 25. What are the pOH.3 9. phosphoric acid and potassium hydroxide 5. Show all work on a separate sheet of paper.50M magnesium hydroxide are necessary to titrate 25. Complete the following table. and the [OH-]? 4.0 mL of 2. The approximate pH of oranges is 3. 141 7.99 X 10 -5 M 8.38 × 10 -13M 8. and the [H+]? a.23 X 10 -2 M 1.8. barium hydroxide and sulfuric acid Ba(OH)2 + H2SO4 → BaSO4 + 2H2O b. The approximate pOH of tomatoes is 9.50 mol H3PO4 × 3 mol Mg(OH)2 × 1L Mg(OH)2 × 1000mLMg(OH)2 11000mL H3PO4 1L H3PO4 2 mol H3PO4 1. the [H+]. [H+] = 6× 10-5M c.SPECIAL FOCuS: Acids and Bases Acid–Bases.5mL Mg(OH)2 146 .3 4.40 × 10 M -5 M 2.859 6.5. pOH = 10.399 4.5 b.50M magnesium hydroxide are necessary to completely titrate 25. and the [OH-]? a.949 10.601 9.38 × 10 -13M 1. the [OH-].61 9. [OH-]=3 × 10-11M c. aluminum hydroxide and nitric acid Al(OH)3 + 3HNO3 → Al(NO3)3 + 3H2O c. Write and balance the chemical equation for the complete reaction between: a. pH = 4. The approximate pH of oranges is 3.50 M phosphoric acid? 3Mg(OH)2 + 2H3PO4 → Mg3(PO4)2 + 6H2O 25. What are the pOH.89 X 10 M -7 pH 1. How many mL of 1. [OH-] = 2 × 10-10M 4.12 × 10 M -8 [OH-] 1.7 9.19 X 10 -10 3. phosphoric acid and potassium hydroxide H3PO4 + 3KOH → K3PO4 + 3H2O 5.38 × 10 -13M 1.38 × 10 M -13 2. and Titration Calculation Review Worksheet— Teacher’s Edition 1. Show all work on a separate sheet of paper. [H+] =3 × 10-4M 3. Reactions.0 mL of 2.50 mol Mg(OH)2 1L Mg(OH)2 = 62.51 × 10 -10M 1. What are the pH. Complete the following table.2 b.051 3.0 mL H3PO4 × 1 L H3PO4 × 2.39 4.076 1.924 pOH 12.38 × 10 M -13 1. [H+] 7. Appendix H 6. What is the molarity of a lithium hydroxide solution if 15.0 mL is completely titrated by 7.5 mL of a 0.020 M acetic acid solution? LiOH + HC2H3O2 → LiC2H3O2 + H2O 7.5 mL HC2H3O2 × 0.020 mmol HC2H3O2 × 1 mmol LiOH × 1 1mL HC2H3O2 1mmol HC2H3O2 1 = 0.010M LiOH 15.0 mL LiOH 7. A student reacted a 25.00 mL sample of vinegar (acetic acid solution) with 38.25 mL of 0.500 M sodium hydroxide. a. Write and balance the chemical equation for the reaction. HC2H3O2 (aq)+ NaOH(aq)→ NaC2H3O2 (aq) + H2O b. Calculate the molarity of the vinegar solution. 38.25 mL NaOH × 1mL NaOH 0.500 mmol NaOH 1mmol NaOH × 1 mmol HC2H3O2 25.00 mL HC2H3O2 × 1 1 = 0.765M HC2H3O2 c. How many grams of acetic acid would be in 1.00 L of the vinegar? 1.00 L HC2H3O2 1 × 0.765M HC2H3O2 1L HC2H3O2 × 60.05 g HC2H3O2 1mol HC2H3O2 = 45.9 g HC2H3O2 d. What is the percent by mass of acetic acid in the vinegar solution (assume a density of 1.00 g/mL)? 45.9 g HC2H3O2 1L HC2H3O2 × 1L HC2H3O2 × 1000mL HC2H3O2 1 mL HC2H3O2 1.00 g HC2H3O2 × 100 = 4.59% HC2H3O2 147 Appendix I Dissociation and pH Practice Weak Acid or Weak Base Systems The ionization constant (or equilibrium constant) is very important here. Only ions contribute. Water, which is usually the solvent, does not participate in any major way here. The general form for a reaction containing an acid is: HA + 2H2O ↔ H3O+ + A- so: Ka = HA is an acid because it begins with hydrogen or because it donates a proton to form H3O+. The A- is the element or polyatomic ion portion of the acid with a negative charge. Write the Ka for the dissociation of chlorous acid, HClO2, and formic acid, HCOOH. Ka = Ka = Weak polyprotic acids dissociate (break apart) stepwise. Each step is written as a different reaction. Generally there is a different value for K a each time an acid loses a hydrogen ion. H3PO4 + H2O ↔H3O+ + H2PO41H2PO41- + H2O ↔ H3O+ + HPO42HPO42- + H2O ↔ H3O+ + PO43Ka1 = Ka2 = Ka3 = Examples: Write the dissociation reactions for sulfuric acid and carbonic acid. H2SO4 H2CO3 This also holds true for weak bases like ammonia. (aq) 2) Kb = 3) NH3 Initial Change Equilibrium NH41+ OH1- 4) Kb = 5) pOH = -log[OH1-] pH + pOH = 14 150 .75 X 10-5 1) NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1. Determine the equilibrium expression. Develop the ICE table.8 X 10-5 1) HC2H3O2 (aq) + H2O (l) ↔ H3O1+ (aq) + C2H3O21. Solve for the concentrations.00M acetic acid solution with a K a = 1. Solve for pH or pOH based on the [H+] or [OH-]. Ka or Kb. Substitute the final values into the equilibrium expression.500M ammonia solution with a K b = 1.SPECIAL FOCuS: Acids and Bases Write the K b for NH3 + H2O ↔ NH4+ + OHKb = Steps for calculating the pH of a weak acid or base: 1) 2) 3) 4) 5) 6) Write and balance the equation if it is not already done for you. Calculate the pH of a 1.(aq) 2) Ka = 3) HC2H3O2 Initial Change Equilibrium H3O1+ C2H3O21- 4) Ka = 5) pH = -log[H3O1+] Try this one: Calculate the pH of a 0. H3PO4 + H2O ↔ H3O+ + H2PO4H2PO4. HCOOH. Water. Generally there is a different value for Ka each time an acid loses a hydrogen ion. which is usually the solvent. Ka = [H+][ClO2-] Ka = [H+][COOH-] [HClO2] [HCOOH] Polyprotic acids dissociate (break apart) stepwise.Appendix J Dissociation and pH Practice—Teacher’s Edition Weak Acid or Weak Base Systems The ionization constant (or equilibrium constant) is very important here. The A. and formic acid. does not participate in any major way here. Write the Ka for the dissociation of chlorous acid. Each step is written as a different reaction.+ H2O ↔ H3O+ + HPO42HPO42.so: Ka = [H3O+][A-] [HA] HA is an acid because it begins with hydrogen or because it donates a proton to form H3O+. HClO2.is the element or polyatomic ion portion of the acid with a negative charge. The general form for a reaction containing an acid is: HA + 2H2O ↔ H3O+ + A. Only ions contribute.+ H2O ↔ H3O+ + PO43Ka1 = [H3O+][ H2PO4-] [H3PO4] Ka2 = [H3O+][ HPO42-] [H2PO4-] Ka3 = [H3O+][ PO43-] [HPO42-] . 8×10-5 [1.87 Calculate the pH of a 0.SPECIAL FOCuS: Acids and Bases Practice Write the dissociation reactions for sulfuric acid and carbonic acid.00-x] 5) x = 4.+ H2O ↔ H3O+ + SO42HCO3.00M acetic acid solution with a K a = 1. Solve for the concentrations.00 M -x 1. Determine the equilibrium expression.(aq) 2) Ka = 3) [HC2H3O2] Initial Change Equilibrium 1.500M ammonia solution with a K b = 1.00 M . Develop the ICE table. Substitute the final values into the equilibrium expression. H2SO4 H2CO3 H2SO4 + H2O ↔ H3O+ + HSO4H2CO3 + H2O ↔ H3O+ + HCO3HSO4.75 X 10-5 152 . Write the Kb for NH3 + H2O ↔ NH4+ + OHKb = [NH4+][OH-] [NH3] Steps for calculating the pH of a weak acid or base: 1) 2) 3) 4) 5) 6) Write and balance the equation if it is not already done for you.2×10-3M = [H+] 6) pH = -log[H+] = 2.8 X 10-5 1) HC2H3O2 (aq) + H2O (l) ↔ H3O+ (aq) + C2H3O2.+ H2O ↔ H3O+ + CO32Equilibrium expressions also can be written for weak bases like ammonia. Examples: Calculate the pH of a 1. Solve for pH or pOH based on the [H+] or [OH-].x [H3O+][C2H3O2-] [HC2H3O2] [H+] 0 +x x [C2H3O2-] 0 +x x 4) Ka = [x][x] = 1. Ka or Kb. ) NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH.500 M -x 0.75×10-5 5) x = 2.500-x] = 1.) [NH3] Initial Change Equilibrium 0.500-x [NH4+] 0 +x x [OH-] 0 +x x 4) Kb = [x][x] [0.529 pH + pOH = 14 pH = 14 – pOH pH = 11.(aq) 2) K b = [NH4+][OH-] [NH3] 3.471 153 .96×10-3M = [OH-] 6) pOH = -log[OH-] pOH = 2.Appendix J 1. . 0 Bromothymol blue Phenolphthalein Universal indicator: changes colors at different pH values as follows 3. To observe typical reactions of several indicators when treated with acids and bases.0 green 8.0 5. . One way to classify substances is based on the relative amount of acid present. Use the following table as a reference: Indicators Methyl red Litmus Color at a bottom of range Red Red Yellow Clear Color at top of range Yellow Blue Blue Pink pH range 4.2-7. is another way to measure the hydrogen ion concentration.8-6. The use of an indicator. An acid produces hydrogen ions in an aqueous solution according to the Arrhenius concept.0 blue-gray 10.Appendix K Lab: Determining the pH of a Substance Introduction Many classification systems for compounds exist.0-7.0 blue-green 9. a weak organic acid that changes colors.6 8. Different indicators work in different pH ranges. To estimate the pH of some common household products using these reactions. The pH scale was developed to measure the amount of hydrogen ions present in a solution.0 violet Objectives 1.0 yellow 6. 2.0 orange 5.5 6.0-10.0 green-yellow 7.0 red-orange 4. Rinse out the spot plate with deionized water. 2. 2. 4. 4. Add 1–3 drops of each product across a row in wells according to data table 2. Estimate the pH of each solution. Add 1–3 drops of each standard across a row in wells according to data table 1. Add 1 drop of indicator down the wells according to data table 1. 4. 6. Record in the data table. Go to a station set up for measuring the pH of the products. Submerge the probe tip in a small amount of the first solution to be measured. Estimate the pH of each solution. Rinse out the spot plate with deionized water. Add 1 drop of indicator down the wells according to data table 2. Record the pH value in the data table. Record the colors observed for each reaction in the data table. 6. 3. Repeat steps 3–5 with each household product to be tested. 5. 3. 6. Part 2: 1. Part 3: (To be done after you estimate the pH of each household substance based on the indicator colors. 156 .) 1. 2. Record in the data table. 3. 5.SPECIAL FOCuS: Acids and Bases Procedure Part 1: 1. Follow the directions for calibrating your pH probe. 5. Rinse the probe thoroughly with deionized water. Obtain a set of standard solutions and a set of indicators. Obtain a set of household products. Record the colors observed for each reaction in the data table. 1 M NH3 (ammonia) 0.1 M NaOH Data Table 2: Product aspirin bleach detergent eggs lemon juice milk milk of magnesia clear soft drink tea Methyl Red Litmus Indicator Bromthymol Blue Phenolphthalein universal Indicator Estimated pH 157 .1 M HCl Methyl Red Litmus Indicator Bromthymol Blue Phenolphthalein universal Indicator Estimated pH 0.Appendix K Data Table 1: Standard Solution 0.1 M acetic acid distilled water 0. your home. How does your estimated pH values compare to the rankings of these substances? 4. Based on your observations. Indicators also can come from objects found in nature. Locate one such indicator through research and explain how you could tell the difference between an acid and a base using this indicator. Where they as you expected? Does one indicator work better than another? If so. find the strengths of the listed acids and bases. how does the estimated pH of ammonia compare to the estimated pH of sodium hydroxide? 3. Using your textbook. Conclusions Describe your results. which one and in what cases? 158 . how does the estimated pH of the hydrochloric acid compare to the estimated pH of the acetic acid? 2. or even in an office.SPECIAL FOCuS: Acids and Bases Analysis: 1. Based on your observations. 0 mL of sodium hydroxide? 15.0 mL of NaOH? 20.Appendix K Was it difficult to determine the color of the indicator in any particular substance? If so. add a couple drops of methyl red and a couple drops of phenolphthalein.0 mL of hydrochloric acid in a beaker. which one(s)? Why do you think this might have been the case? If you start with 15.0 mL of NaOH? 159 . what would you expect to happen as you add 10. . Idaho. He is a consultant for the College Board and serves on the ACS Exams Institute Advanced High School Committee. She has also been a Reader for the AP Chemistry Examination and a member of the National Chemistry Olympiad Task Force. She has served as a Reader and Table Leader for the AP Chemistry Examination. She also works as a consultant for the College Board and teaches Summer Institutes for AP Chemistry teachers. conducting AP workshops worldwide. She also serves on the ACS Exams Institute Advanced High School Committee and National Chemistry Olympiad Task Force. Francis School in Louisville. president of the Oklahoma Science Teachers Organization. She has been teaching AP Chemistry since 1983. Oklahoma. She has won the Presidential Award for Excellence in Science Teaching. and Question Leader at the AP Chemistry Exam Reading. National Board Certification in Chemistry. Smith has contributed to numerous 161 . Valerie Ferguson teaches AP Chemistry at Moore High School in Moore. She has served as chair of the Oklahoma ACS. and is a National Leader for the College Board. She has been a Reader and Table Leader for the AP Chemistry Examination. Table Leader. and Presidential Scholar Teacher. She is currently a Question Leader for the AP Chemistry Examination and has also served as a Reader and Table Leader. and was co-organizer of the Teaching AP Chemistry symposium at the 14th Biennial Conference on Chemistry Education. John’s School in San Juan. She has also served as a consultant for the College Board. and has been a member of the Woodrow Wilson Foundation Teacher Outreach Program and a Dreyfus Master Teacher. he has served as a Reader. About the Authors Lew Acampora currently teaches AP Chemistry at the St. Puerto Rico. Since 1995. Adele Mouakad is an award-winning teacher at St. British Colombia in Canada. Cheri Smith is an award-winning teacher of AP Chemistry at Yale Secondary School in Abbotsford. Kentucky. She has a long list of accomplishments including Teacher of the Year. For the last 16 years he has taught AP Chemistry at independent schools across the country.About the Editor Marian DeWane teaches AP Chemistry at Centennial High School in Boise. He is also involved in the development and scoring of the IB Chemistry Program. Hattori was a College Board consultant in the mid-90s. Hattori received her bachelor’s degree in chemistry and English from Stephen F. pre-AP chemistry. and has presented both day.D. Table Leader. and instrumental analysis. New York. She has been teaching AP Chemistry and/or AP Environmental Science since 2000 and has been an AP Chemistry Reader since 2005. He has been a Reader. Miller is a senior lecturer emeritus in the Chemistry department at the University of California Irvine. and is chair of the National Chemistry Olympiad Examination Task Force. radiochemistry. Arden Zipp is a distinguished faculty member at State University of New York College at Cortland in Cortland.publications and served as a reviewer for the AP Vertical Teams® Guide for Science and the Pre-AP®: Topics for Vertical Teams in Science workshop. George E. and has taught freshman chemistry. He was member and chair of the AP Chemistry Development Committee and a member of the College Board Subject Test in Chemistry Committee. and Development Committee member and chair. where he is also principal scientist and supervisor of the nuclear reactor facility and responsible for K–12 science teacher professional development. . analytical chemistry. He has served in every capacity for the College Board in the development and scoring of the AP Examination: He has been a Reader. Heather Hattori teaches AP Chemistry. Austin State University in 1991 and is currently attending the University of North Texas in pursuit of a Ph. AP Environmental Science. Table Leader. but has been taking a break to raise Arabian horses and most recently attend graduate school in her spare time. in chemistry. He has been on the chemistry faculty at UCI since 1965. Questions Leader. and Question Leader for AP Chemistry over a period of 20 years. Chief Reader. Guyer High School in Denton. Texas.and weeklong workshops for AP and general chemistry teachers. and pre-AP physics at John H. He has more than 60 publications in analytical chemistry and in science education. . F08CMSF131 .