QUESTIONS0. What do you know about acid by Arrhenius, Brondted Lowry and Lewis 1. Write the chemical equation for the autoionization of water and the equilibrium law for Kw? 2. How are acidic, basic, and neutral solutions in water defined a. in terms of [H+] and [OH-] and b. in terms of pH ? 3. At the temperature of the human body, 37oC, the value of Kw is 2.4 x 10-14. Calculate the [H+], [OH-], pH and pOH of pure water at this temperature. What is the relation between pH, pOH, and Kw at this temperature? Is water neutral at this temperature? 4. Deuterium oxide, D2O, ionizes like water. At 20°C its Kw, or ion product constant analogous to that of water, is 8.9 x 10-16. Calculate [D+] and [OD-] in deuterium oxide at 20°C. Calculate also the pD and the pDO. 5. Calculate the H+ concentration in each of the following solutions in which the hydroxide ion concentrations are : a. 0.0024 M a. 1.4 x 10-5 M a. 5.6 x 10-9 M a. 4.2 x 10-13 M 6. Calculate the OH- concentration in each of following solutions in which the hydrogen ion concentrations are a. 3.5 x 10 -8 M a. 0.0065 M a. 2.5 x 10 -13 M a. 7.5 x 10 -5 M 7. A certain brand of beer had a hydrogen ion concentration equal to 1.9 x 10-5 mol L-1.What is the pH of the beer? 8. A soft drink was put on the market with [ ] = 1,4 x mol 9. Calculate the pH of each of the solutions in Exercises 5 and 6. . What it's pH ? 10. Calculate the molar concentrations of H+ and OH- in solution that have the following pH values. a. 3.14 b. 2.78 c. 9.25 d. 13.24 e. 5.70 11. Calculate the molar concentration of H+ and OH- in solution that have the following pOH values . a. 8.26 b. 10.25 c. 4.65 d. 6.18 e. 9.70 12. What is the pH of 0.010 M HCl ? 13. What is the pH of 0.0050 M solution of HNO3 ? 14. A sodium hydroxide solution is prepared by dissolving 6.0 g NaOH in 1.00 L of solution. What is the pOH and the pH of the solution? 15. A solution was made by dissolving 0.837 g Ba(OH)2 in 100 mL final volume. What is the pOH and the pH of the solution? 16. A solution of Ca(OH)2 has a measured pH of 11.60. What is the molar concentration of Ca(OH)2 in the solution? 17. A solution of HCl has a pH of 2.50. How many grams of HCl are there in 250 mL of this solution. 18. Write the chemical equation for the ionization of each of the following weak acids in water (For any polyprotic acids , write only the equation for the first step in the ionization). a. HNO2 b. H3PO4 c. HAsO42d. (CH3)3NH+ 19. For each of the acids in exercise 18, write the appropriate Ka expression 20. Write the chemical equation for the ionization of each of following weak bases in water. a. (CH3)3N b. AsO43c. NO2d. (CH3)2N2H2 21. For each of the bases in Exercise 20, write the appopriate Kb expression. 22. Benzoic acid, C6H5CO2H, is an organic acid whose sodium salt, C6H5CO2Na, has long been used as a safe foods additive to protect beverages and many foods againts harmful yeasts and bacteria. The acid is monoprotic. Write the equation for it's Ka ! 23. Write the equation for the equilibrium that the benzoate ion, C6H5CO2- (review exercise 22), would produce in water as functions as a Bronsted base. Then write the expression for the Kb of the conjugate base of benzoic acid. 24. The pKa of HCN is 9.21 and that of HF is 3.17. Which is the strong Bronsted base CN− or F−? 25. The Ka for HF is 6.8 x 10x. what is the Kb for F-? 26. The barbiturate ion C4HO has Kb = 1,0 x 10 -10 . What is Ka for Barbituric acid ? 27. Hydrogen peroxide, H2O2 is a week acid with Ka = 1.8 x 10-12. What the value of Kb for the HO2 ion? 28. Methylamine, CH3NH2 resambles ammonia in odor and basicity. Its Kb is 4.4 x 10-4. Calculate the Ka of its conjugate acid! 29. Lactic acid, HC3H5O3, is responsible for the sour taste of sour milk. At 25oC its Ka = 1.4 x 10-4. What is the Kb of its conjugate base, tha lactate ion, C3H5O3- ? 30. Iodic acid, HIO3 has a pKa of 0.77 a. What is the formula an the Kb of its conjugate base? b. Its is conjugate base a stronger or a weaker base than the acetate ion? 31. Periodic acid,HIO4,is an important oxidizing agent and a moderately strong acid. In a 0.10 M solution , [H+] = 3.8 x 10-2 mol L-1. Calculate the Ka and pKa for periodic acid! 32. Choloacetic acid, HC2H2ClO2, is a stronger monoprotic acid than acetic acid. In a 0,10 M solution, this acid is 11 % ionized. Calculate the Ka and pKa for Choloacetic acid. Refer to data in the preceding question to calculate the percentage ionization of the base in 0. a monoprotic acid whose Ka value is 3. What are Kb and pKb for Hidroxylamine? 35. basic.15 M HONH2. Codeine.33. like ammonia. Deuteroammonia. also known as carbolic acid. Like ammonia. does a solution of the sodium salt of aspirin in water test acidic. What will be the pH of a 0.10 M solution has a pH of 11. Hidroxylamine. is a weak base with a pKb of 5. A 0. What is the pH of a 0.020 M solution of codeine? (Use Cod as a symbol for codeine) 41.15 M HN3 ? for HN3.9x10-10.96 at 25oC. Calculate the Kb and pKb for ethylamine. it is a Bronsted base. has a strong.125 M pyruvic acid ? It's Ka is 3. CH3CH2NH2. What is the pH of a 1.86. Phenol. is 1. pungent odor similar to that ammonia.54. Is a solution of K2C2O4 acidic.0 M solution of hydrogen peroxide. 36. A 0. a cough suppressant extracted from crude opium. or neutral? Explain. . 34. H2O2? For this solute. Ka = 1. What is pH of 0.15 M solution has a pH of 10. is a weak base with a pKb of 4. What is the pH of 0. ND3. Ka = 1. is a Bronsted base. or neutral ? Explain 44.2 x 10-3 37. Ethylamine. C2O42-.8 x 10-2 39. HONH2.12. basic. The Kb value of the oxalate ion. What are the concentrations of all of the substance in a 0. Aspirin is acetylsalicyclic acid. Ka= 1.79.050 M solution of phenol.8 x 10-5 38.27 x 10-4. is sometimes used as a disinfectant. A solution of acetic acid has a pH of 2. HC6H50? What percentage of the phenol is ionized? For this acid.3 x 10-10 40.20 M solution of this compound? 42. What is the concentration of acetic acid in this solution ? 43. Basic. is very slightly soluble in water. b. BHY.28. will this compound have on the acidity of the moisture in the ground? Explain. (NH4)2SO4. Calculate the pH of 0. 49.00 or less than 5. Write the formulas of those that have solutions that are a. composed of the ions BH+ and Cl-.4 x 10-4 52. A weak base B forms the salt BHCl. HY.00 L of water at 25oC to have a solution with a pH of 5. have a pKa greater than 5. BH+. We may represent morphine . A 0. The conjugate acid of a molecular base has a hypohetical formula. if any. Calculate the number of grams of NH4Br that have to be dissolved in 1.15 M solution of the salt has a pH of 4. Many drugs that are natural Bronsted bases are put into aqueous solution as their much more soluble salt with strong acids. Calculate the pH of 0. For CH3NH2.15 M CH3NH3Cl. The powerful painkiller morphine. Consider the following compounds and suppose that 0. KC2H3O2. Ammonium nitrate is commonly used in fertilizer mixtures as a source of nitrogen for plant growth. KF. Will an aqueous solution of ALCl3 turn litmus red or blue ? explain? 47. Explain why the beryllium ion is a more acidic cation than the calcium ion.04 M KNO2 ? 51.20 M NaCN. 50. A solution of salt of this cation. and has pKa of 5. 48. CsNO3.45. What effect. Acidic. KCN.00. tests slightly basic. Calculate the pH of 0. 46. Will the conjugate acid of Y.5M solutions are prepared of each : NaI. Kb = 4. and KBr. but morphine nitrate is quite soluble.00? explain 55. Neutral. and c. for example.16 ! 54. What is the value of Kb for the base B? 53. To make it more soluble in water... a. under what conditions are we unable to use the initial concentration of an acid or base as though it were the equilibrium concentration in the mass action expression? 58. 57. Para-Aminobenzoic acid (PABA) is a powerful sunscreening agent whose salt were once used widely in suntanning..010 M HSO4-.. is a moderately strong Bronsted acid with a Ka of 1.(furnished by the salt.oC).010 M HSO4. What is the percentage ionization in a 0. What is the pH of a 1. What is the [H+] and pH of 0.20 M solution of H-Mor+? 56.92 (.0x10-2.by the symbol Mor and its conjugate acid as H-Mor+. it is put into a solution as its conjugate acid. What is the value of [ H+] in 0.0010 M acetic acid ? What is the pH of the solution? 60. What is the calculate pHof a 0... How much error is produced by incorrectly using the simplifying assumption? 62.. solve the quadratic equation. is a weak acid with a pKa of 4.0 x 10-7 M solution of HCl ? 61.. The hydrogen sulfate ion HSO4-. which we may symbolize as H-Paba. What is the calculate of [H+] in 0. The parent acid.15 M solution of H- ? Its pKa is 8. What is the percentage ionization in 0.15 M solution of HF ? What is the pH of the solution ? 59. What is the calculated pH of a 0.52 at 25 0C. The pKb of morphine is 6. obtained by using the usual simplifying assumption? a. a. a. which we may represent as H- . Generally.. is a weak Bronsted base that we may represent as Qu. Quinine. Write the chemical equation for the ionization of the acid and give the appropriate Ka expression.13. NaHSO4) ? Do NOT make simplifying assumptions.030 M solution of this acid? . an important drug in treating malaria. 14 M NH4+ a. calculate the pH of the buffer using the Kb for NH3 a. Buffer 1 is a solution containing 0.25 M C2H3O2-? Use Ka = 1. Rework the preceding problem using the Kb for the acetate ion. calculate the pH of the buffer using the Ka for NH4+ . By how much will the pH change if 0. NaH2PO4 and Na2HPO4 (the "phosphate" buffer in side body cells) c. the barbituretes.8 x 10-5 for HC2H3O2 67.050 M solution of H-Bar? 64. 66. It is the parent compound of widely sleeping drugs. a. Barbara.63. ( be sure to write the poper chemical equation and equilibrium law ) 68. 70. what is the [H+] and pH of a 0.15 M HC2H3O2 and 0.050 mol of HCl is added to 1.10 M NH3.10 M NH4Cl and 1 M NH3.25 M NH3 and 0. Write ionic equation that illustrate how each pair of compounds can serve as a buffer pair. Buffer 2 is a solution containing 1 M NH4Cl and 0. By how much will the pH change if 50. Which buffer would be better able to hold a steady pH on the addition of strong acid. A buffer is prepared containg 0. 69. buffer 1 or buffer 2? Explain. NH4Cl and NH3 65. was discovered by the Nobel Prize-winning organic chemist Adolph von Baeyer and named after his friend. What is the pH of a solution that contains 0. Barbituric acid.01. HC4H3N2O3 (which we will abbreviate H-Bar). Its pKa is 4.00 L off the buffer in Exercise 66. H2CO3 and NaHCO3 (the "carbonate" buffer in blood) b.0 mL of 0.10 M NaOH is added to 500mL of the buffer in Exercise 66. 05 pH unit? 80.100 mol of acetic acidand 0.05 pH unit? How many milliliters of the same HCl solution would. Suppose 25.00 ? Ho w does this possibility affect the choice of an indicator ? 81.20 M NH3 to prepare a solution buffered at pH 10. What can make the titrated solution at the equivalence point in an acid-base titration have a pH not equal to 7.0 L of 0. .15 M acetic acid (pKa 4.80 ? 75.74) to make the solution a buffer for pH 3.110 of sodium acetate in 500 mL of solution.12 M formic acid (Pka 3.25? 76.0 L of 0.00? 74.15 M HCl would have to be added to 100 mL of the buffer described in exercise 78 to make the pH decrease by 0. NaC2H3O2. would have to be added to 1. NaCHO2.15 ? 78. What is a good indicator for titrating potassium hydroxide with hydrobromic acid? Explain. How many grams of sodium acetat. if added to 100 mL of pure water. What mole ratio of NH4Cl to NH3 would buffer a solution at pH 9. would have to be added to 1.00 L of the buffer in Exercise 70? 72. How many grams of ammonium choride would have to be dissolved in 500 mL of 0. What are the initial and final pH value? what would be the pH if the same amount of HCl solution were added to 125 mL of pure water? 79.020 mL of HCl is added to 1.00 mL of 0.71.00? 77.10 M NH3 to make it a buffer with a pH of 9.100 M HCl is added to an acetate buffer prepared by dissolving 0.10 M KOH is added to 200 ml of the buffer in exercize 70? 73. How many milliliters of 0. How many grams of sodium formate. By how much will the pH change if 0. How many grams of ammonium chloride have to be dissolved into 125 mL of 0. By how much will the pH change if 75 ml of 0. make the pH decrease by 0. Explain why ethyl red is a better indicator than phenolphtalein in the titration of dilute ammonia by dilute hydrochloric acid? 82.74) to make the solution a buffer for pH 5. 1000 M acetic acid with 0.00 mL of the base has been added. calculate the pH of the resulting solution after each of the following quantities of base has been added to the original solution (you must take into account the change in total volume). What is a good indicator for this titration? 85.99 mL e.10 M hydrobromic acid. 25. 24.00 mL f.10 M aqueous ammonia is titrated with 0.1000 M HCl with 0.00 mL of 0. calculate the pH: a. what is the pH at the equivalence point? What is a good indicator? 86. In the titration of an acid with base. Construct a graph showing the titration curve for this experiment. When 50 mL of 0. 26.1000 M NaOH.1000 M NaOH.10 M formic acid is titrated with 0.10 mL h.00 mL i. When 25 mL of 0. 25. a. 50. .00 mL c.01 mL g.00 mL 87.10 M sodium hydroxide. For the titratin of 25.00 mL of 0. b. For the titration of 25. 0 mL b. what is the pH at the equivalence point? (Be sure to take into account the change in volume during the titration).what condition concerning the quantities of reactans ought to be true at the equivalence point? 84. Before the addition of any NaOH solution. 10. After 10.90 mL d. 24.83. 25. after 10. acidic is the substance that has pH<7. According to Bronsted Lowry opinion. and d. at the equivalence point ANSWER 0. acid is a species that releases H+ when dissolved in the water.00 mL of the acid has been added. 1. before the addition of any HCl solution. Chemical equation for the autoionization of water H2O (l) →H+(aq) + OH-(aq) K[H2O] = [H+][ OH-] Kw = [H+][ OH-] 2. c. acid a substance that accepts a pair of electrons.1000 M ammonia with 0. basic is the substance that has pH>7 and neutral is the substance that has pH=7 . At the equivalence point. For the titration of 25. acid is a species that gives proton (proton donor) According to Lewis opinion. basic is the substance that [H+]<[OH-] neutral is the substance that [H+]=[OH-] in terms of pH. calculate the pH a. 88. After half of the HC2H302 has been neutralized.c. after half of the NH3 has been neutralized. acidic is the substance where the [H+] > [OH-]. b. and d. In terms of [H+] and [OH-].00 mL of 0.1000 M HCl. According to Arrhenius opinion. 3.Is water neutral at this temperature? Answer : Kw = [H+][ OH-] . pOH.Relation between pH. pH and pOH . Given :T = 37oC Kw = 2.4x10-14 Question : . [OH-]. and Kw at this temperature .[H+]. pH=pOH.98 x 10-16 pD = 8 ..4 x 10-3 M pOH = .? Answer : [D+] = [OD-] = = = 2. a. .98 = 7. both are half of pKw At this temperature. Given : [OH-] = 2. pD.log 2.4 x 10-3 M Questions : [H+] .log [OH-] . the water is almost neutral because pH is 4. Given : Kw = 8.log 2.. [OD-].53 pOD = 8 . pOD = .At this temperature.53 5. .98 = 7.9 x 10-16 T = 20°C D2O → D+ + ODQuestions :[D+]. ? Answer : [OH-] = 2. log 2.4 = 9. . .6 x 10-9 M Questions : [H+] .4 = 11.4 ) = 9 + log 1. Given : [OH-] = 5.4 x 10-5 M pOH = . . Given Questions : [H+] .145 x 10-10 M c.4 x 10-5 M b.4 pH = 14. .146 [H+] = 10-9.38 [H+] = 10-11.146 = 7.log 2.(3.4 ) = 11 + log 2.4 x 10-5 = 5 . ? .= .4 x 10-3 = 3 .log 2.log 1.log 1.log [OH-] = .(5. ? Answer : [OH-] = 1.4 pH = 14.168 x 10-12 M : [OH-] = 1.log 1.38 = 4. log 5.log 5.62 .2 x 10-13 = 13 .2 ) = 1 + log 4.log [OH-] = .log 5.748 = 1.(13.6 = 5.6 x 10-9 M pOH = .log 4.6 pH = 14.Answer : [OH-] = 5.(9. Given Question : [H+] .log 4.748 [H+] = 10-5.2 x 10-13 M d.2 = 1. .2 pH = 14.log [OH-] = .786 x 10-6 M : [OH-] = 4. ? Answer : [OH-] = 4.2 x 10-13 M pOH = .log 4. .6 x 10-9 = 9 .6 ) = 5 + log 5. log 6.5 pOH = 14 .0065 M = 6.log 3.5 x 10-3 Questions : [OH-] ? Answer : pH = 3.]= 10 -6.62 = 2. Given : [H+]=0.5 b.log 3.5 = 11.5) = 11 + log 6.8 [OH. Given = 10 -11.5 [OH.8 : [H+]=2.pH = 14 .5 x 10 -13 M .pH = 14 .5 pOH = 14 .(8 .] c.(3 ..log 6.5 x 10 -8 M 6.5 = 6.[H+] = 10-1.5) = 6 + log 3. a.? Answer : pH = 8.4 x 10-2M : [ H+ ] = 3.]. Given Questions : [OH. 8 [OH.log 2.5 pOH = 14 .8 : [H+] = 1.Questions :[OH-] ? Answer : pH = 13.log 7.log 7.(5 .(13 .39 : [H+]=7.5 = 9.9 x10-5] . Given = 10 -1.? Answer : pH = -log [H+] = -log [1.log 2.] d.5) = 1 + log 2.5 pOH = 14 .9 x 10-5mol L-1 Questions : pH…….5 x 10-5 M Questions : [OH.5) = 9 + log 7.pH = 14 .] ? Answer : pH = 5.] 7. Given = 10 -9.39 [OH.5 = 1.pH = 14 . 4 = 3 – 0.72 8..4 x 10-5 M Asked : pH .28 pH = 4.4 x 10-3 M pOH = .4 x 10-5mol L-1 pH = ..4 x 10-3 M Asked : pH ..146 pH 9..854 Exercise 5 a.log 1.4 x 10-3 M = 3 – log 2.4 x 10-5mol L-1 = 5 – log 1.log [OH-] .. ? Answer : [OH-] = 1.9 pH = 5 . = 4.4 = 5 – 0..= 5-log 1.log [H+] = . Given : [OH-] = 0. Given = 11.log 2. ? Answer : [H+] = 1.38 pOH = 2.4 x 10-5mol L-1 Questions : pH .62 pH b.log [OH-] = .38 : [OH-] = 1. Given : [H+] = 1.4 x 10-5 M pOH = .62 pH = 14 – pOH = 14 – 2.0.0024 M = 2. ? Answer : [OH-] = 2. 854 pH = 14 – pOH = 14 – 4.6 x 10-9 M = 9 – log 5. Given = 9. Given = 5.2 = 13 – 0.= .377 pH = 1. ? Answer : [OH-] = 5.748 : [OH-] = 4.146 pOH = 4.854 pH c.252 pH d.6 x 10-9 M Asked : pH .748 pOH = 8...4 x 10-5 M = 5 – log 1..6 = 9 – 0.252 pH = 14 – pOH = 14 – 8.6 x 10-9 M pOH = .log [OH-] = .4 = 5 – 0.146 : [OH-] = 5.log 1.2 x 10-13 M = 13 – log 4.623 pOH = 12. ? Answer : [OH-] = 4.log 5.log [OH-] = .377 pH = 14 – pOH = 14 – 12.2 x 10-13 M Asked : pH ..623 .log 4.2 x 10-13 M pOH = . 5 x 10-13 M = 13 – log 2..log [H+] = .log 3..0065 M = 6.log 6.602 : [H+] = 7.5 x 10-8 M pH = . Given : [H+] = 3.5 x 10-8 M = 8 – log 3. ? Answer : [H+] = 3.5 x 10-13 M pH = .... ? Answer : [H+] = 2.log 2.log [H+] = . ? Answer : . Given = 12.456 : [H+] = 0.398 pH d.5 = 13 – 0.5 = 3 – 0..5 x 10-3 M Asked : pH .5 x 10-3 M = 3 – log 6.813 pH c.5 x 10-8 M Asked : pH .5 x 10-3 M pH = .log [H+] = .187 : [H+] = 2..5 x 10-5 M Asked : pH . Given = 7. Given = 2..544 pH b.5 x 10-13 M Asked : pH .Exercise 6 a. ? Answer : [H+] = 6.5 = 8 – 0. 86 [H+] = 4 – log 7. Given = 4.5 = 5 – 0.22 pOH = 12 – 0.24 [H+] = 7.5 x 10-5 M pH = . a.66 [H+] = 1..78 pOH = 11.125 : pH = 3.14 [OH-] = 11 – log 1.log 7.875 pH 10.38 x 10-11 M b.78 Asked : [H+] and [OH-] ...86 pOH = 11 – 0.026 .14 Asked : [H+] and [OH-] .22 [H+] = 3 – log 1. Given : pH = 2.14 pH = 4 – 0.14 pOH = 10.log [H+] = .24 x 10-4 M pOH = 14 – pH = 14 – 3.38 [OH-] = 1. ? Answer : pH = 3.78 pH = 3 – 0.5 x 10-5 M = 5 – log 7..66 x 10-3 M pOH = 14 – pH = 14 – 2.[H+] = 7. ? Answer : pH = 2.78 [OH-] = 12 – log 6. [OH-] = 6.70 Asked : [H+] and [OH-] ... Given : pH = 13...24 pH = 14 – 0.026 x 10-12 M c.25 Asked : [H+] and [OH-] .62 [H+] = 5. Given : pH = 9. ? Answer : pH = 5. ? Answer : pH = 13.76 [H+] = 14 – log 5.174 M e.74 [OH-] = 1.24 [OH-] = 1 – log 1.76 pOH = 1 – 0.25 [OH-] = 5 – log 1.75 [H+] = 5.75 pOH = 5 – 0.62 x 10-10 M pOH = 14 – pH = 14 – 9.24 pOH = 0.25 pOH = 4. Given : pH = 5.75 [H+] = 10 – log 5. ? Answer : pH = 9.78 [OH-] = 1.74 x 10-1 M = 0.70 ..75 x 10-14 M pOH = 14 – pH = 14 – 13.24 Asked : [H+] and [OH-] .25 pH = 10 – 0..78 x 10-5 M d. 70 [OH-] = 9 – log 5.62 x 10-11 M pH = 14 – pOH = 14 – 10.495 x 10-9 M = 14 – pOH pH = 14 – 8.26 = 9 – 0. Given [H ] = 6 – log 1.995 [H+] = 1.30 pOH = 9 – 0.26 pH + b.25 Asked : [H+] and [OH-] .495 [OH-] = 5.01 x 10-9 M 11.26 Asked : [H+] and [OH-] ..75 [OH-] = 11 – log 5.995 x 10-6 M pOH = 14 – pH = 14 – 5. a. ? Answer : pOH = 10.74 = 6 – 0.26 pH = 5..82 x 10-6 M : pOH = 10. ? Answer : pOH = 8.74 pOH - [OH ] = 9 – log 5.82 [H+] = 1..= 6 – 0.25 .30 pH [H+] = 6 – log 1..01 [OH-] = 5.62 [OH-] = 5.25 pOH = 11 – 0.70 pOH = 8. Given : pOH = 8. Given : pOH = 4...82 pH = 8 – 0.78 x 10-4 M [H ] c.24 x 10-5 M = 14 – pOH pH = 14 – 4..65 Asked : [H+] and [OH-] .35 pOH - [OH ] = 5 – log 2.47 [H+] = 4.82 [OH-] = 7 – log 6.. Given [H ] = 10 – log 4.47 x 10-10 M : pOH = 6.18 pOH = 7 – 0.18 [H+] = 8 – log 1.51 x 10-8 M : pOH = 9.65 pH = 9.24 [OH-] = 2.18 Asked : [H+] and [OH-] .61 [OH-] = 6.25 [H+] = 4 – log 1.65 pH + d.70 . ? Answer : pOH = 4.65 = 5 – 0. ? Answer : pOH = 6.75 pH = 4 – 0.61 x 10-7 M pH = 14 – pOH = 14 – 6.35 = 10 – 0.18 e. Given pH = 7.78 + = 1.51 [H+] = 1.pH = 3. log 5 x 10-3 .30 [OH-] = 10 – log 1. 1 x 10-2 mol/liter = 1 x 10-2mol/liter pH = . Ma = 1 .70 [H+] = 5 – log 5.log [ H+] = .Asked : [H+] and [OH-] . Given : pH = 4..30 pH = 5 – 0.995 [OH-] = 1.005M Questions : pH ? Answer : [H+] = x .01 [H+] = 5. Given : M = 0.log [1 x 10-2] =2 13.995 x 10-10 M pH = 14 – pOH = 14 – 9.70 pOH = 10 – 0. Ma = 1 x 5 x 10-3 pH = .01 x 10-5 M Ma = 1 x 10-2mol/liter Questions : pH ? Answer : [H+] = x . ? Answer : pOH = 9..70 12. . Given : m NaOH V larutan MrNaOH = 6.5 x 10-1 M pOH = .log 5 = 2.15 M = 1.176 .15 M [OH-] = 0..? Answer : [NaOH] = x = x = 0.3 14.824 = 13.= 5 x 10-3 = 3 ..log 1.0.0 gram = 1.log [OH-] = .5 = 0.00 L = 40 Question : pOH and pH = .log 1.5 x 10-1 = 1 .824 pH = 14 .. 60 . Given : pH Ca(OH)2 = 11.15.837 g Mr Ba(OH)2 = 171 V = 100 mL Question: pOH and pH of solution = …? Answer : 16. Given : Mass of Ba(OH)2 = 0.60 Questions :Molaritas ? Answer : pOH = 14 – 11. 99 x 10-3 M 17.4 = 3 – 11. 2 M = 1.5 [H+] = 3.16 x 10-3 .98 .50 V = 250 ml = 0. Given : pH HCl = 2. 10-3 [OH-] =M x 3.60 [OH-] =3.= 2.98 .25 L Questions : grams of HCl? Answer : [H+] = 3 – 0. 10-3 = M . V n = 3. HNO2 ↔ H+ + NO2- b. 10-3 = M . 10-3 18. H3PO4 ↔ H+ + H2PO4- c.↔ H+ AsO43- . HAsO42.25 n = 0. 0.16 .[H+] = M .16 . 10-3 .16 . HNO2 ↔ H+ + NO2- b.79 . Answer : a. H3PO4 ↔ H+ + H2PO4- c. HAsO42. 1 3. (CH3)3NH+ ↔ H+ (CH3)3N 19.x 3. 10-3 = M M = n/V n = M. Answer : a.↔ H+ AsO43d. + OHc. (CH3)2N2H2 + H2O ↔ H — N — NH2 + OHCH3 21.+ H2O ↔HAsO42. AsO43c. (CH3)3NH+ ↔ H+ (CH3)3N CH3 20. (CH3)3N + H2O (CH3)3NH+ + OH- . AsO43.+ H2O ↔ HNO2 + OHCH3 d. (CH3)3N b. (CH3)2N2H2 Asked : Kb expression? Answer : a. NO2. (CH3)3N + H2O ↔ CH3 — N — H+ + OHCH3 b. Given: a.d. NO2d. a. (CH3)3NH3+ + OH- C6H5COONa + H2O ↔ C6H5COO. AsO43. the strong Bronsted base is CN− .+ H+ C6H5CO2. NO2.+H2O H3AsO4 + H2O c.17 → Ka HF = 6.76×10-4 Kb CN− = Kb F− = = 0.b.16×10-4 = 0.+ H2O ↔ C6H5OH + OH- Kb = 24.+ H2O As(OH)5 C6H5COOH + NaOH C6H5COOH 23. H3AsO4 + 3OH- HNO2 + OH- d. (CH3)2N2H2 + H2O 22. pKa HCN = 9.15×10-10 So.21 → Ka HCN = 6.17×10-10 pKa HF = 3. 25.0 x 10 -10 x Ka Ka = Ka = 10 -4 28.4 x 10-4 T =25oC Question : Kb . .? Answer : 29. Given : Ka = 1.. Given : Kb of CH3NH2 = 4. Given : Ka for HF = 6.? . .4 x 10-4 Questions: Ka= .47x10-11 Kw = Kb x Ka 10-14 = 1.8x10-4 x Kb Kb = 1. 26..8x10-4 Questions : What is Kb for F-? Answer : HF H+ + F- Kw = Ka x Kb 10-14 = 6. Its is conjugate base a stronger or a weaker base than the acetate ion? Answer : a. HIO3= is stronger conjugate base then acetate ion. formula Kb ? b.88 x 10 -1 Ka x Kb = 10-14 Kb = 1.8 x 10-5 Kb CH3COO.1 M HIO4 [H+] = 3.77 Ka = 5. Given : the formula : HIO3 →H+ + IO3pKa = 0.= 5. Ka CH3COO.4 x 10-4 x Kb Kb = 7.Answer : Kw = Ka xKb 10-14 = 1. Given : 0.77 Questions : a.7 x 10-14 b.8 x 10-2mol L-1 .14 x 10-11 Kb 30.= 1.5 x 10-10 SO. 31. pKa = 0. . 10-3 Ka= 3 .10 M danα =11 % Asked : KadanpKa.1 1.44 x 10-3= Ka x 0.21 .1 Ka = Ka = 1.44 x 10-2) = 2 .log 1.44 x 10-2 pKa= -log Ka = -log (1.8 x 10-2 )2 = Ka x 0.21 pKa = .11 Ka = (0.? Answer : ( 3.Question : Ka and pKa…….log 1. Given : [HC2H2ClO2] = 0..log Ka .? Answer :α 0.1 Ka = 1.11)2 x 0..44 32. 86 so pOH = 14 .88 → [OH-] = 10-3.184 x 10-4 34.21.76 = Kb Kb = 1.88 = 10 -7.86 = 2.88 Asked : Kb ?andpKb ? Answer : [OH-] = 10-3.184 x 10-5 : 0. so pOH = 3.86 Asked : Kb and pKb ? Answer : pH = 11.11. 10-3 33.1 mol/L pH = 11.1 = Kb Kb = 5.12. Given : M HONH2 = 0.14 [OH-] = 7.2 x 10-3 7.15 M pH = 10.log 1.2 x 10-3 5. Given : Ethylamine M = 0.pKa= .158 x 10 -7 . = 0.000878 Given : Ma = 0.15 mol/lt pH = 10.125 mol/liter Ka = 3.Log 1.158 = 6.Log Kb = . Given : HONH2 M = 0.15x10-7 (from exercise 34) Asked :α? Answer : = 36.12 Kb = 1.936 35.2 x 10-3 Asked : pH…? Answer : [H+] = = = .pKb = .158 x 10 -7 = 7 – Log 1. = 2 x 10-2 pH = - log [2 x 10-2] = 2 - log 2 = 2 - 0.301 = 1.699 37. Given : Ma HN3 : 0,15 mol/liter Ka :1,8 x 10-5 Asked : pH…? Answer : [H+] = = = = 1.64 x 10-3 pH = - log [ H+] = -log 1.64 x 10-3 = 3 - log 1.64 38. Given : [H2O2] = 1.0 M Ka = 1.8 x 10-2 Asked : pH = ....? Answer : H2O2 [H+] O2 + 2H+ + 2e= = = 0.134 M pH = - log [H+] = - log 0.134 = 0.87 39. Given : M HC6H5O = 0.050 mol/L = 1.3 x 10-10 Ka Asked : [H+] = …? % HC6H5O ioniozed= …? Answer: HC6H5O C6H5O 0 + H+ M 0.05 0 R 2.55 x 10-6 2.55 x 10-6 2.55 x 10-6 S 0.05 - 2.55 x 10-6 2.55 x 10-6 2.55 x 10-6 40. Given : pKb of Cod = 5,79 M of Cod = 0,020 M Asked : pH of Cod = …? Answer : pKb = - log Kb Kb = 10 -5,79 = 10 -6 pOH = - log [OH-] = - log (1,41.10-4) = 4 – log 1,41 = 4 – 0,15 = 3,85 pH = pKw – pOH = 14 – 3.85 = 10,15 41. Given : pKb of ND3= 4,96 Kb = 1,096.10-5 M of ND3 = 0,20 M T = 25°C Asked : pH of ND3 = … ? Answer : pOH = - log [OH-] = - log (1,48.10-3) = 3 – log 1,48 = 3 – 0,17 = 2,83 pH = pKw – pOH = 14 – 2,83 = 11,17 42. Given : pH of CH3COOH = 2,54 Asked : M of CH3COOH = ...? Answer : pH = - log [H+] 2,54 = - log [H+] [H+] = 10-2,54 [H+] = 2,88 x 10-3 a. K2C2O4 is base because K2C2O4 a salt formed from a strong base and a weak acid to be base salt. KCN. Acidic is (NH4)2SO4 and CsNO3 b. Basic is KF. In the periodic system of elements. The sodium salt of aspirin is basic. Thus. Neutral is NaI and KBr 46. 47. Be is located above Ca so it is more likely to be acidic because of the periodic system of elements in one group the greater the atomic number or from top to the down it will be more likely to be basic. reaction: H2C2O4 + 2 KOH → K2C2O4 + 2 H2O base 45. AlCl3 is acidic compounds because it cannot yet reached the central atom an octet configuration or can be said to still have vacant orbitals (empty). 48. beryllium ion is a more acidic cation than the calcium ion. a solution ofAlCl3 can change the color of litmus paper blue to red due to the acid characteristic. because it from acetylsalyclic acid and sodium hydroxide. So. K2C2O4 can be obtained from the reaction of a strong base and a weak acid that is KOH H2C2O4. Weak acid with strong base want to produce basic salt or if sodium hydroxide dissolved in water to produce acetylsalicyclic acid and OHHAcetylsalicyclic + OH- NaAcetylsalicyclic + H2O 44. KC2H3O2 c.43. Based onthe Lewis acid-base theory. Ammonium nitrate (NH4NO3) NH4NO3 → NH4+ + NO3− . the acidity of the moisture in the ground will increase.Log 1.796 x10-3 pOH = 3 .796 pH = 14 – (3 . NaCN CN CN- HCN + OH- [NaCN] = 0.Log 1.796) = 11 + Log 1. There is H3O+ as a product from the chemical equations above.20 M Ka = 6.3 x 10-4 [OH-] = 50. → Na+ + Na+ + H2O → - + H2O ↔ 49.25 → K+ + K+ + H2O → NO2- + H2O ↔ [KNO2] = 0.NH4+ + H2O ↔ NO3− + H2O → NH3 + H3O+ If any this compound in the ground.04 M Ka = 4.2 x 10-10 [OH-] = = = = 1. KNO2 NO2HNO2 + OH- .796 = 11. pH = 4.846 = 5.4 x 10-4 [H+] = = = = 1.15 H+ .64 = 7.15 Questions :Kb for the base B = ..64 x10-7 pOH = 7 .? BHCl 0.64) = 7 + Log 9.= = = 9.Log 9.73 52.15 M + Cl- 0.846 x10-6 pH = 6 . Given : 0. CH3NH3Cl → ClCH3NH3 + CH3NH3+ + Cl- + H2O → ↔ CH3NH2 + [CH3NH3Cl] = 0..15 M BHCl.64 pH = 14 – (7 .Log 9.15 M Kb = 4.Log 1.28 BH+ 0.98 51. 16 Mr NH4Br = 107 T 25oC Kb = 1.067 x 10-20 = 2.Answer : pOH = 14 .61 x 10-20 = Kb x 0.8 x 10-5 Question : massa of NH4Br .? Answer : pH = 5.00 L pH = 5.9 x 10-10 = 3.72 [OH-] = 10-9.log . .15 Kb = 24. Given : V H2O = 1.28 = 9.16 = .4.4067 x 10-19 53.27 = 1. .9 x 10-10 [OH-] = 1. Conjugate acid is BH+ that pKa = 5 BHOH↔ BH+ + OHBHOH + HY ↔BHY + H2O BHY ↔ BH+ + Y- [ OH-] = = = = = 10-5 x [BHY] x 10 pOH = 5 .log = = = = = 9. Ka = 10-5 .= .218 gram 54.log = .log [BHY] pKa = 5. Given : 0.13 Question : pH…….6.pOH = pKa HY So. pH = 14 .2 M [OH-] = = 7.41 x 10-7 pOH = 7 . Given : Reaction : Qu + H2O [H- H- ] = 0.52 + H3O+ .2 M H-Mor+ pKb = 6.13 Kb = 7. pKa less than 5 55.15 M. pKa = 8.? Answer : pKb = 6.41 =6.2 M H-Mor+ + OH- 0.13 Kb = 10-pKb = 10-6.log 7.13 So.41 x 10-7 Mor → + H2O 0.87 56.13 = 7. log Ka Ka = 3.33 57..? Answer :pKa = 8.52 = . Given : M HF= 0.log Ka 8.15 M Ka HF = 6.5 x 10-4 Asked Solution : : α and pH ? α= = = 6. Initial concentration is unable to calculate equilibrium concentration when mole of both of components which react is same.Questions :pH. 58.02 x 10-9 [ H+ ] = = = 2..13 = 0.87 x 10-3 .52 pKa = .58 x 10-2 [H+] = = = 9.13 x 10-5 pH = 5 – log 2.. 87 .13 = 3.Log 9.= .Log 9.Log 1.01 59.Log 1.0010 M Ka = 1. Given : CH3COOH M = 0.34 x 10-1 [H+] = = = 1.Log [H+] = .87 = 2.Log [H+] pH = .34 x 10-4 pH = .8 x 10-5 Ask Solution : α and pH ? :α= = = 1.87 x 10-3 = 3 .34 x 10-4 = 4 .34 = 4 – 0. The error is produced by incorrectly using the simplifying assumption is 0% because the result of point B and C is same.2 = 1..92 Ka = 5-log 1.2 x 10-5 . 62. pKa = 4.log 1 = 7 d. 66 64.77 x 10-5 pH = 3 – log 2. H2CO3(aq) + NaOH(aq) NaHCO3(aq) + H2O(l) Ionic equation: 2H+(aq) + CO32-(aq) + Na+(aq) + OH-(aq) Na+(aq) + HCO3-(aq) + H2O (l) Weak acid : H2CO3 Conjugation base : HCO3- . a. pKa = 4.21 = 3 – 0.01 Ka = 5 – log 9.22 63.34 = 2.pH = 4 – log 6 = 4 – 0.78 = 3.77 = 9. NH3(aq) + HCl(aq) NH4Cl (aq) Ionic equation: NH3(aq) + H+(aq) + Cl-(aq) NH4+(aq) + Cl-(aq) Weak base : NH3 Conjugation acid : NH4+ 65. 10-4 .10 M NH4Cl and 1 M NH3 b. = 1.8 . buffer1 M NH4Cl and 0.10-5 = 1.(aq) 2Na+(aq) + HPO42-(aq) + H2O (l) Weak acid : H2PO4Conjugation base : HPO42c. buffer 0.b.8. H3PO4(aq)+ NaOH(aq) NaH2PO4(aq) + H2O Ionic equation: 3H+(aq) + PO43-(aq) + Na+(aq) + OH-(aq) Na+(aq) + H2PO4-(aq) + H2O(l) NaH2PO4(aq) + NaOH(aq) Na2HPO4(aq) + H2O (l) Ionic equation: Na+(aq) + H2PO4-(aq) + Na+(aq) + OH. Given : a.10 M NH3 Questions : buffer would be better able to hold a steady pH=? Answer : pH solution before added strong base [OH-] = Kb NH3. 745 pH = 10. 10-6 = 6.74 pH = 8. = 1.= 4.255 Ph of b solution before added strong acid [OH-] = Kb NH3.8.8 .log 1.78 .8 .10-5 = 6.log 6 pOH = 5.05 M [OH-] = Kb NH3.10-5 =1.26 The pH of a solution after the adding of strong acid For example the strong acid that added HCl 0. 10-6 = 6.8 pOH = 5.22 pH = 8.10-5 = 1.8 pOH = 3.log 1. = 1.8 . 6.10-5 = 8.25 M Question Answer : pH ….8. Given : C2H2H3O2 H+ +C2H3O2M HC2H3O2: 0.10-5 =1.log 8.475 The pH of b solution before the adding of strong acid and pH of b solution after the adding of strong base = (8.325 The conclution is buffer solution that the pH is stabile is b solution because the pH changes only 0.8 .6 pOH = 6.: 0.065 pH = 7.? : .935 The pH of a solution changes before the adding of strong acid and pH of a solution after the adding of strong acid = (10.78) = 1. 10-7 = 7. = 1.325 66.8 .15 M M C2H3O2.255 .935) = 0.26 .7.the pH of b solution after the adding of strong base [OH-] = Kb NH3. 08 = 5-0.97 67.1.25 M Ka = 1.8 x 10-5 = 1.08 x 10-5 = 5-log 1.8 x 10-5 x 0.08 x 10-5 pH = [H+] = -log 1.25 x [H+] = [H+] 1.15 = 0.8 x 10 -5 Question : pH use Kb from acetate ion ? Answer : .033 = 4.15 M M C2H3O2- = 0.967 = 4. Given : M HC2H3O2 = 0. 075 mol C2H3O2.05 mol [H+] = Ka .15 M x 0.05 mol s 0 0.(aq) HC2H3O2 (aq) + HCl(aq) Cl.125 mol + Cl.005 mol + HC2H3O2(aq) 0.(aq) m 0.20 mol 0.8 x 10-5 Asked : the pH change after 0.98 mol of HCl = 0.125 mol NaOH(aq) m 0.(aq) 0 .(aq) 0.25 mol Ka HC2H3O2 = 1. Given : pH buffer = 4.8 x 10-5 Asked : the pH change after 0.05 mol 0.8 x 10-5 x = log 1.= 0.20 mol 0. Given : pH buffer = 4.= 0.005 mol of NaOH is added to the buffer Answer : mol of HC2H3O2 = 0.05 mol -0.05 mol +0.15 mol mol of C2H3O2.8 x 10-5 = 4.25 mol 0.050 mol of HCl is added to the buffer Answer : + C2H3O2.05 mol +0.8 x 10-5 = 1.05 mol volume = 1.5 L = 0.98 – 4. pH = log [H+] = 1.005 mol volume = 0.5 = 0.68.00 L mol of HC2H3O2 = 0.15 mol 0 r -0.075 mol mol of C2H3O2.25 M x 0.5 L = 0.5 Change of pH = 4.48 69.50 L Ka HC2H3O2 = 1.98 mol of NaOH = 0. 69 x 10-6 = 9.25 mol mol of NH4+ : 0.21428574 pOH = 4.7857143 = 3.r -0.02 70.log 2. 1. Given : Kb NH3 = 1.005 mol +0.21428574 x 10-5 pOH = 5 . = 1.8 x 10-5 We assumed the volume of solution : 1L M NH3 : 0. = Kb .14 M Question : pH ? Answer : mol of NH3 : 0.005 mol +0.130 mol 0.005 mol [H+] = Ka .070 mol 0.69 x 10-6 =5 Change of pH = 5 – 4.005 mol s 0 0.8 x 10-5 x = log 9.14 M x 1 = 0.8 x 10-5 . NH3(aq) + H2O(l) NH4+(aq) + OH[OH-] =Kb .25 M MNH4+ : 0.14 mol a. = Kb .492915518 pH = 9.25 M x 1 = 0. pH = log [H+] = 1.005 mol -0.98 = 0.5 . b.14 mol *If using Kb of NH3 If added 0.16mol [OH-] = Kb . = 1.020 mol HCl NH3 (aq) + H+ (aq) NH4+ B 0.00 L = 0.11 x 10-10 pH = 10 .56 = 3.14 M x 1.020 mol HCl is added to 1. Given : Buffer = 0.00 L.00 L pH Buffer : 9.56 x 10 -10 Ka = [H+] = Ka .5 71.25mol 0.56 x 10 -10 x 0.5 Asked : the change of pH if 0.56 x 10 -10 x = 5.11 = 9.23mol - 0.25 M NH3 and 0.14 M NH4+ 1.25 mol The mol of NH4+ = 0.8 x 10-5 x .02mol A 0.00 L = 0. = 5.02mol +0.14 mol C -0.log 3.020mol 0.02mol -0.25 M x 1.? Answered : the mol of NH3= 0. = 5. 4.log 3.41 the change of pH = 9.892 x 10 -10 pH = 10 .25 mol The mol of conjugate acid = 0.25 M x 1.25 M NH3 1.5 .41 = 0.00 L 1.log 2.020mol HCl [H+] = Ka .892 pH = 9.09 72.6 x 10-5 pOH = 5 .00 L NH4+ 0.56 x 10 -10 x = 5.9.14 mol .0.= 2.00 L = 0.14 M x 1.59 = 9.56 x 10 -10 x 0.02 ml HCl 0.14M Question: the change of pH?? Answer: The mol of base = 0.59 pH = 14 .5 .09 *If using Ka of NH4+ If added 0.9.41 = 4. = 5.7 = 3.6 = 5 .1 M 0.00 L = 0. Given : 0.41 = 0.41 Change of pH = 9. = 3.41 = 4. = 1.00 M HCl + NH3 NH4Cl M 0.2 x10-5 pOH = -log 3.023 mmol 0.58 = 5 – 0.5= 9.8 x10-5 .02 ml x 1. = 2.8 x10-5 .5 Mol HCl 0.50 = 4.4.020 mmol 0.58 x10-5 pOH = -log 2.020 mmol - [OH-] = Kb .2 x 10-5 = 5 – log 3.58 x 10-5 = 5 – log 2. = 1.020 mmol +0.020 mmol 0.25 mmol 0.160 mmol S -0.59 .14 mmol R -0.2 = 5 – 0.With NH3 [OH-] = Kb .5 pH = 14 . Given: 75 ml 0.41= 0.41 the change of pH = 9.5 mmol Mol of NH3 200 ml x 0.10 M KOH added to 200 ml buffer of 0.25 M NH3 and 0.7 x10-5 pOH = -log 0.pH = 14 .5 – 9.15 = 4.4.75 = 5. = 10-5 .7 = 5 – 0. = 0.56 x10-5 pH = -log 0. = 10-5 .56 x 10-6 = 6 – log 5.75 [H+] = Ka .14 M NH4+ Question : The Change of pH?? Answer : Mol of KOH 75 mlx 0.59 = 9.84 73. = 0.25 M = 50 mmol .7 x 10-5 = 5 – log 0.1 M = 7.09 With NH4+ [H+] = Ka .6 = 6 – 0. 8 x10-5 .78 x10-5 pOH = -log 1. = 1.5 mmol 7.78 = 5 – 0.05 = 5 – 0.05 x 10-5 = 5 – log 5.4.25 = 4.70-9.25 NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l) m 28 mmol 7.4.5 = 9. = 5.25 = 0.5 mmol 7.30 pH = 14 .8 x10-5 .05 x10-5 pOH = -log 5.45 .30= 9.70 = 4.5 mmol − 57.70 the change of pH is 9.78 x 10-5 = 5 – log 1. = 1. = 1.Mol of NH4+ 200 ml x 0.75 pH = 14 .14 M = 28 mmol [OH-] = Kb .5 mmol s 20.5 mmol [OH-] = Kb .5 mmol 50 mmol r 7. Given massa = 0. 10-5 x = 0.80 = 0.15 M acetic acid (pKa 4.2 x 10-5 = 1.27 x 10-5 x = 0.27mol x 82 massa = 22.74 = 1.2 x 10-4 Question : how many grm of sodium formate?? Answer : Mol of formic acid = 1. 0.00 = 10-5 Question : how many grams od sodium asetate?? Answer : Mol of asetic acid 1.8 x 10-4) a buffer for pH 3.27 mol Mass of sodium asetate = mol x mr of sodium asetate 75. = 1.12 mol 10-5 x [H+] = Ka .0 L x 0.0 L x 0.12 M formic acid (Pka 3.8 x 10-5) A buffer pH is 5.12 M = 0.14 gram : 1.0 L of 0. = 1.8 x10-5 . Given : 1.15 M= 0.0 L of 0.15 mol [H+] 10-5 = Ka .74.74 = 1.08 x 10-4 .8 x10-4 . 108mol x 68 massa = 7.x = 0.20 = 100 mmole pH buffer = 10.00 Kb NH3 = 1..344 gram 75.6 x 10-10= 1. Known : 500 ml of NH3 0. Given : buffer a solution have pH 9.? . 5.108 mol Mass of sodium asetate = mol x mr of sodium formate massa = 0. = = 76.8 x 10-5 Mr NH4Cl = 53.20 M Mole of NH3 = V x M = 500 ml x 0..25 Question : what the ratio of NH4Cl to NH3?? Answer: [H+] = Kb.8 x10-5 .5 Question : Answer : mass of NH4Cl = . Known : 125 ml of NH3 0..5 = 963 mgram = 0.15 mass of NH4Cl = .4 x 10-5 .8 x 10-5 pH = 9..85 [OH-] = 1.5mmol Question : Kb NH3 = 1.pH = 10 pOH = 4 [OH-] = 10-4 = [OH-] = 10-4 10-5 x Mole of NH4Cl = Mole of NH4Cl = Mole of NH4Cl = 18 mmol Mass of NH4Cl = mole x Mr NH4Cl = 18 x 53.10 M Mole of NH3 = V x M = 125 x 0.15 pOH = 4.963 gram 77.? Answer : pH = 9.1 = 12. 5mmol = 0..07mmol Mass of NH4Cl = mole x Mr NH4Cl = 16.[OH-] = 1.1 mol Mole of HCl = V x M = 25 x 0.75mgram = 0. Known : 25 ml of HCl 0.1mol Mole of CH3COONa = 0.0025mol Mole of CH3COOH = 0.? The final of pH =.1 = 2.859 gram 78..4 x 10-5 = 10-5 x Mole of NH4Cl = Mole of NH4Cl = Mole of NH4Cl = 16.5 = 859..82 x 10-5 ..? = 1.11mol Ka CH3COOH Question : a.07 x 53. The initial of pH =. 735 x 10-5 pH CH3COOH = 5 – log 1. the pH if the same amount of HCl solution were added to 125 ml of pure water =. [H+] = = 10-5 x = 16.0025 S 0 0. the initial of pH = 4.54 x 10-6 = 1.21 = 4.11 0.65 x 10-5 = 5 – log 1.? Answer : a.b.0025 0.0025 [H+] = = 10-5 x = 10-5 x = 1.65 pH = 5 – 0.0025 0.79 So.0025 0.0025 0.1 R 0.1075 0..79 HCl + CH3COONa → + NaCl M 0..735 0 .1025 0. pH buffer from no.1 + a a .0. Addition of water in the buffer solution. the final pH = 4. if added to 100 ml of pure water.78 = 4.78 = 4.05 pH unit = .= 5 – 0. Known : HCl = 0.239 = 4. volume of the same HCl solution would.79 Question : a. the pH will be constant bacause the determinant of pH is the number of mol not the concentration of buffer.? Answer : HCl + CH3COONa → M a R a S 0 0...761 b. The pH if the same amount of HCl solution were added to 125 ml of pure water is same 4.79 .1 a a.a = 4. make the pH decrease by 0.11 a 0. Volume of HCl to make the pH decrease by 0.761.05 = 4.11 .15 M Volume of buffer = 100 ml pH buffer from no..? b.761 So.82 x 10-5 + NaCl 0. 79.05 pH unit = ..74 [H+] = 1.79 pH after CH3COOH 0 a 0. 82 x 10-5 a) = (1. In acid base titration. it’s mean that one of solution is same or it’s weak base.005 mol Mole of HCl = 0.82 x 10-5 a) (2 x 10-6) . with the Ph range that estimated until the indicator choosing is not false. 80.(1. 81.82 x 10-5 a) 1.[H+] = 1.11 – a) = 10-5 x (0.(1. The buffer solution is added with water the pH is constant not change.005 mol Volume of HCl = = = 0.82 x 10-6) + (1. equivalence point can reach at pH is not 7.82 x 10-5 = 10-5 x 1. Amount of HCl which added was same is 33 ml.82 x 10-5 a) + (1.1 + a) (2 x 10-6) .033 litre = 33 ml b. The indicator choosing is very important to effort the titration process because to know the end point of titration with the change color of the solution.8 x 10-7 = 3. Methyl red is a better indicator than phenolphtalein in the titration of dilute ammonia by dilute hydrochloric acid because the result of ammonia and hydrochloric acid is a solution .82 x 10-5 x (0.64 x 10-5 a a = = 5x 10-3 a = 0.82 x 10-6) = (1. 3 10. 50 = 0. The indicators are metil red indicator. The quantities of reactans ought to be true at the equivalence point when the equivalence mole of acid is as same as the equivalence mole of base in the titration. Therefore the change of color in the phenolphthalein indicator is sharper (more noticeable). bromtimol blue indicator. V = 0.that has pH < 7 ( influenced by hydrochloric acid as a strong acid and ammonia is a weak base ). 50 = 5 mmol = 5 mmol .0 82. V1= M2 .10 . Titrating potassium hydroxide with hydrobromic acid is example of titration strong base and strong acid.10 M V formic acid : 50 ml M NaOH : 0. V2 0. The color of indicators can be change around the equivalen point.4 -6. 83.10 . so the phenolphthalein indicator is more frequently used. For titrating potassium hydroxide with hydrobromic acid we can use few indicators.10 M Asked : what is the pH at the equivalence point? Answer : Mol HCOOH = MolNaOH M1 .10 . and phenolphthalein indicator. 84. 50 = 0.2 and pH range of phenolphtalein is 8. V Mol HCOOH = M. The equivalen point is occur in value pH 7 (neutral). Which is pH range of methyl red is 4.10 V2 V2 = 50 ml MolNaOH = M. Given : M formic acid : 0. 10 mol/L Asked : pH at equivalence point and good indicator Answer: At equivalence point means that the number of acid moles equal to the moles of base.8 x 10-5 85. .HCOOH + NaOH → NaCOOH + H2O Ka m 5 5 0 0 r -5 -5 +5 +5 s 0 0 5 5 = 1.8 x 10-5 M HBr = 0. Given: V NH3 = 25 mL M NH3 = 0.10 mol/L Kb NH3= 1. 5 mmol = 0.5 mmol A 86. + Given - - : [HCl] 2.5 mmol 2.5 mmol 2.1000 M .5 mmol - R 2.5 mmol 2.NH3 HBR NH4Br B 2. 1000 M V HCl = 25 mL MolHCl = 25 ml x 0.log [H+] = .1000 M = 2..00 ml MolNaOH = 10.[NaOH] = 0.5 mol 0 R 0 0 0 0 0 0 S 0 2. V NaOH = 10.5 mol Asked : pH and the titration curve Solution : = . V NaOH = 0 mL MolNaOH = 0 x 0.? a.log 0.00 ml x 0..1 M [H+] = valence x [HCl] =1 x 0.1000M = 0 HCl + NaOH NaCl + H2O M 2.1 M = 0.1 M pH = .1 =1 b.5 mol [HCl] 0 = = 0.1000 M 0 . 49 mol 2.49mol 2.5mol 1mol 1mol [HCl] 0 1mol = = 0.1000 M = 2.5mol 2.49mol 2.043 M [H+] = valence x [HCl = 1 x 0.log 0.90 mL MolNaOH = 24.01mol 2. V NaOH = 24.49mol [HCl] = 0 .log [H+] pH = .043 = 1.49mol S 0.90 mL x 0.49 mol HCl + NaOH NaCl + H2O M 2.37 c.49mol 2.49mol R 2.5 mol NaCl + H2O 1mol 0 0 R 1mol 1mol 1mol S 1.043 M = .043 M = 0.= 1 mol HCl + NaOH M 2. 0 x 10-4 = 3.log [H+] .499mol NaCl + 0 H2O 0 R 2.0 x 10-4 M [H+] = valence x [HCl] = 1 x (2. V NaOH = 24.0 x 10-4) = 2.499mol S 1x10-3mol 2.0 x 10-5 M [H+] = valence x [HC] =1 x (2.5mol 2.499 mol HCl + NaOH M 2.log [H+] pH = .0 x 10-5 M pH = .0x10-5) = 2.499mol [HCl] 0 = = 2.499mol 2.698 d.= 2.1000 M = 2.499mol 2.499mol 2.log 2.0 x 10-4 M = .99 mL MolNaOH = 24.99 ml x 0.499mol 2. = .1000 M = 2.5mol 0 0 Titration in equivalent point pH = 7 (neutral) f.0 x 10-5 = 4.5mol 2. V NaOH = 25.5mol 2.5mol 2.log 2.5mol S 2.1000M = 2.7 e.5mol 2.00 mL MolNaOH = 25 ml x 0.01 mL MolNaOH = 25.5mol 2.5mol 1x10-3 mol S 0 [NaOH] = 0 2.501mol 0 R 2.5 mol HCl + NaOH NaCl + H2O M 2.5mol 0 0 R 2.5mol 2. V NaOH = 25.5mol 2.5mol 2.501 mol HCl + NaOH NaCl + H2O M 2.01 ml x 0.5mol .5mol 2.5mol 2. 699 = 9.51mol R 2.= 1.9996 x 10-5 M [OH-] = valence x [NaOH] = 1 x (1.3 g.log [OH-] = .5 mol S 0 0.01mol 2.log 1.1000 M = 25.5mol 2.1 mol HCl + NaOH NaCl + H2O M 2.10 ml x 0.5mol 2.996 x 10-4 M [OH-] = valence x [NaOH] = 1 x (1.5mol 2.5mol 2.10 mL MolNaOH = 25.9996 x 10-5 M pOH = .5 mol 2.9996 x 10-5) = 1.996 x 10-4) 0 0 .5 mol [NaOH] = = 1.699 pH = 14 .9996 x 10-5 = 4. V NaOH = 25.4. 96 x 10-3 0 0 .7 pH = 14 . V NaOH = 26.= 1.7 = 10.5mol 2.5 mol 2.00 mL MolNaOH= 26 ml x 0.5mol 2.996 x 10-4 M pOH = .3 h.log [OH-] = .1000 M = 2.5mol S - 0.3.6mol R 2.5mol [NaOH] = = 1.96 x 10-3) = 1.996 x 10-4 = 3.log [OH-] = .5mol 2.96 x 10-3 M pOH = .log 1.log 1.96 x 10-3 M [OH-] = valence x [NaOH] = 1 x (1.5mol 2.1 mol 2.6 mol HCl + NaOH NaCl + H2O M 2. 5mol 5mol R 2.477 pH = 14 .5mol 2.477 = 12.3 i.1.log 0.5mol 2.1000 M = 5 mol HCl + NaOH NaCl + H2O M 2.7 = 11.log [OH-] = .5mol 0 [NaOH] = = 0.5mol 2.= 2.033 M pOH = .5mol S 2.7 pH = 14 .00 mL MolNaOH = 50 ml x 0.523 0 0 .5mol 2. V NaOH = 50.5mol 2.033 = 1.2.033) = 0.033 M [OH-] = valence x [NaOH] = 1 x (0. b. pKa = 1. After 10. Given : V = 25 mL M = 0.1000 mol/L = 10-1 mol/L Ka = 1. Before the addition of any NaOH solution. After half of the HC2H302 has been neutralized.87. . Before the addition of any NaOH solution. c. At the equivalence point.00 mL of the base has been added.8 x 10-5 Question : Calculate pH= a. it means calculate pH of a weak acid. Answer : a. and d. 25 mmol 1. CH3COOH + NaOH → CH3COONa + H2O B 2.25 mmol - 1. After 10. At a half of the HC2H3O2 means that the number of acid moles equal to a half the moles of base.25 mmol 1.5 mmol R 1 mmol A 1. CH3COOH + NaOH B 2.00 mL of the base has been added means that we calculate pH of acid buffer.25 mmol .25 mmol 1.25 mmol 1.b.5 mmol 1.25 mmol A 1.5 mmol 1 mmol 1 mmol - →CH3COONa + H2O 1 mmol 1 mmol 1 mmol 1 mmol c.25 mmol - - R 1. 5 mmol 2.5 mmol 2.5 mmol 2. At equivalence point means that the number of acid moles equal to the moles of base.5 mmol R 2.744 d. CH3COOH + NaOH → B 2.5 mmol A - - CH3COONa + H2O 2.5 mmol .log1.5 mmol 2.8 = 4.5 mmol 2. Known : 25.00 mL of 0.88.pOH .1000 M NH3 Asked : pH….log 10-3 =3 pH = pKw .log [OH-] = .? Answer : [OH-] = = = = 10 -3 pOH = . ? Answer: NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2.0.3 = 11 So.5 x 10-5 = 5 .5 1 - React: 1 1 1 After : 1. pH of NH3 before the addition of any HCl solution are11 a.5 mmol Moles of HCl = n x M = 10 x 0. Known : Moles of NH3 = n x M = 25 x 0.5mmol - [OH-] = Kb x = 10-5 x = 1.176 = 4.= 14 .log 1.log 1.1 = 1 mmol Asked : pH after mixed….5 x 10-5 pOH = .5 = 5 .log [OH-] = .1 = 2.824 1 mmol . pOH = 14 .pOH = 14 .5 1.25 mmol [OH-] = Kb x = 10-5 x = 10-5 pOH = . pH after 10. Asked : pH at the equivalence point….log 10-5 =5 pH = pKw .25 mmol - 1.5 =9 So.pH = pKw .4.25 1.00 mL of HCl has been added were 9.log [OH-] = .? Answer : NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2.25 After : 1.25 - React: 1. pH after half of the NH3 has been neutralized were 9 c.176 So.25 1.824 = 9. Asked : pH after half the NH3 has been neutralized….? .176 b. 7 x 10 -5 = 5 . pH at the equivalence point are 5.5 2.log 0.5 2.log 0.5 2.15) = 5.7 x 10 -5 pH = .(-0.1 V = 25 mL [H+] = = = = 0.15 So.5 After : - - 2.5 mmol Looking for volume total: Moles of NH3 = moles of HCl 25 x 0.1 = V x 0.5 - React: 2.15 .log [H+] = .Answer : NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2.7 = 5 .
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