Volumetric AnalysisIntroduction The Equipment The Process Calculations Dr.S.Kumar Author: J R Reid Volumetric analysis A titration is a lab procedure where a measured volume of one solution (burette) is added to a known volume (flask) of another solution until the reaction is complete Dr.S.Kumar Author: J R Reid Introduction Often in chemistry we need to work out the concentration of a solution. There are a number of methods we could use, but they all involve working out the amount of the substance in a certain volume. Volumetric analysis involves using volumes of liquids to analyse a concentration. To do this we need the following things: A chemical of a known concentration that will react with our ‘unknown’ concentration chemical An indicator that will tell us when all the chemical has been reacted A number of pieces of equipment that we can use to measure volume accurately .Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration that is consumed by the analyte. Titrimetry is a term which includes a group of analytical methods based on determining the quantity of a reagent of known concentration that is required to react completely with the analyte. but the mass is measured instead of the volume. The benefits of these methods are that they are rapid. convenient. gravimetric titrimetry. accurate. . the time required to complete the electrochemical reaction is measured. Volumetric titrimetry is used to measure the volume of a solution of known concentration that is needed to react completely with the analyte. Coulometric titrimetry is where the reagent is a constant direct electrical current of known magnitude that consumes the analyte.There are three main types of titrimetry: volumetric titrimetry. and coulometrtic titrimetry. Gravimetric titrimetry is like volumetric titrimetry. and readily available. Quantitative Classical Chemical Analysis Gravimetry Titrations Acid-base Precipitation Complexometric Redox 08/06/11 Lecture Notes: Dr.Santosh Kumar . Quantitative Classical Chemical Analysis Gravimetry Titrations Acid-base Precipitation Complexometric Redox Permanganimetric Dichromatometric Titrations involving iodine (I2) Iodimetry Iodometry Iodometric titration of copper 08/06/11 Lecture Notes: Dr.Santosh Kumar . Santosh Kumar .Titrations Acid-base Examples Quantification of acetic acid in vinegar Complexometric Quantification of chloride (Cl-) in water Water Hardness (Calcium Precipitation and magnesium) Quantification of hydrogen Redox peroxide (H2O2) 08/06/11 Lecture Notes: Dr. Eriochrome black T Murexide Precipitation AgNO3 (silver nitrate)Mohr.Santosh Kumar . Fajans Redox Quantification of Hydrogen peroxide hydrogen peroxide (H2O2) (H2O2) KMnO4 (potassium permanganate) No indicator 08/06/11 Lecture Notes: Dr.Titration example Acid-base Complexometric Quantification of acetic acid in avinegar Water Hardness (Calcium and magnesium) Quantification of chloride (Cl-) in water Analyte Acetic acid (CH3COOH) Calcium and magnesium (Ca Mg 2+ ) Chlordie Titrant NaOH (sodium hydroxide) EDTA 2 + Indicator Phenolphthalein . Volhard. More Defining Terms Titration: This is performed by adding a standard solution from a buret or other liquid. and the amount of the excess is determined by back titration with a second standard titrant.Titration: This is a process that is sometimes necessary in which an excess of the standard titrant is added. In this instance the equivalence point corresponds with the amount of initial titrant is chemically equivalent to the amount of analyte plus the amount of back.titrant. . Back. Equivalence Point: Occurs in a titration at the point in which the amount of added titrant is chemically equivalent to the amount of analyte in a sample.dispensing device to a solution of the analyte until the point at which the reaction is believed to be complete. To determine the titration error: Et= Vep . End point: The point in titration when a physical change occurs that is associated with the condition of chemical equivalence. The difference between the end point and equivalence point should be very small and this difference is referred to as titration error. Indicators are used to give an observable physical change (end point) at or near the equivalence point by adding them to the analyte.Veq Et is the titration error Vep is the actual volume used to get to the end point Veq is the theoretical value of reagent required to reach the end point .Equivalence Points and End Points One can only estimate the equivalence point by observing a physical change associated with the condition of equivalence. Reasonably large molar mass Compounds that meet or even approach these criteria are few. . and only a few primary standards are available. Absence of hydrate water 4. High purity 2. The accuracy depends on the properties of a compound and the important properties are: 1. Readily available at a modest cost 5.Primary Standards A primary standard is a highly purified compound that serves as a reference material in all volumetric and mass titrimetric properties. Reasonable solution in the titration medium 6. Atmospheric stability 3. Standard solutions A standard is a solution of precisely known concentration It must be available in a highly pure state It must be stable in air It must dissolve easily in water It should have a fairly high relative molecular wt It should under go a complete and rapid reaction Endure a selective reaction with analyte . Non standard solutions Sodium hydroxide absorbs carbon dioxide from atmosphere HCl can produce chlorine gas in reactions and liberate hydrogen when exposed to air Nitric acid can act as an oxidising agent interfering with reactions Sulphuric acid absorbs water form the air . fsu.sb.Example of titration and set up http://wine1.htm .edu/chm1045/notes/Aqueous/Stoich/Aqua02. Fact File 1: Introduction to iodometric and iodimetric titrations Titrations: Direct Titrations Indirect Titrations Back Titrations Iodometry . Santosh Kumar .Fact File 1: Introduction to iodometric and iodimetric titrations Titrations Acid-base Example Quantification of acetic acid in vinegar Type of reaction ■ Direct Titration □ Indirect Titration □ Back Titration Complexometric Water Hardness (Calcium and magnesium) ■ Direct Titration □ Indirect Titration □ Back Titration Precipitation Quantification of Cl inMohr Method Water ■ Direct Titration □ Indirect Titration □ Back Titration Fajans Method ■ Direct Titration □ Indirect Titration □ Back Titration Volhard Method □ Direct Titration □ Indirect Titration ■ Back Titration Redox Quantification of hydrogen peroxide (H2O2) ■ Direct Titration □ Indirect Titration □ Back Titration 08/06/11 Lecture Notes: Dr. EQUIPMENT . The Equipment Volumetric analysis involves a few pieces of equipment that you may not have seen before: Pipette – for measuring accurate and precise volumes of solutions Burette – for pouring measured volumes of solutions Conical flask – for mixing two solutions Wash bottles – these contain distilled water for cleaning equipment Funnel – for transfer of liquids without spilling Volumetric flasks – a flask used to make up accurate volumes for solutions of known concentration . Apparatus used Burette Volumetric flask Beaker Pipette Funnel Indicator White tile . Burette titration procedures . The Process - Preparation Two solutions are used: The solution of unknown concentration; The solution of known concentration – this is also known as the standard solution Write a balanced equation for the reaction between your two chemicals Clean all glassware to be used with distilled water. The pipettes and burettes will be rinsed with the solutions you are adding to them Process – The Setup The burette is attached to a clamp stand above a conical flask The burette is filled with one of the solutions (in this case a yellow standard solution) A pipette is used to measure an aliquot of the other solution (in this case a purple solution of unknown concentration) into the conical flask Prepare a number of flasks for repeat tests Last, an indicator is added to the conical flask Process – The Titration Read the initial level of liquid in the burette Turn the tap to start pouring out liquid of the burette into the flask. Swirl the flask continuously. When the indicator begins to change colour slow the flow. When the colour changes permanently, stop the flow and read the final volume. The volume change needs to be calculated (and written down). This volume is called a titre Repeat the titration with a new flask now that you know the ‘rough’ volume required. Repeat until you get precise results Do not blow out remainder of liquid into flask and keep tip of pipette in contact with flask .Precautions when using equipment Burette must be vertical. use and then remove funnel. Check meniscus. check meniscus. rinse with de-ionised water and then given solution. In using a Pipette rinse with de-ionised water first and then with given solution. Precautions Conical flask should not be rinsed with solution it is to contain and swirl In using a Volumetric flask the last few cm³ must be added so that the meniscus rest on calibration mark Invert stoppered flask to ensure solution is homogeneous/uniform . rather than a beaker.Why is a conical flask. used in the experiment? To allow easy mixing of the contents. by swirling. . Why is the funnel removed from the burette after adding the acid solution? So that drops of solution from the funnel will not fall into the burette. . In using a burette. (b) to clamp it vertically. (c) to have the part below the tap full? Author: J R Reid . why is it important (a) to rinse it with a little of the solution it is going to contain. Solution (a) Rinsing To remove any residual water. and so avoid dilution of the acid solution when it is poured into the burette. . Solution (b) clamp vertically To enable the liquid level to be read correctly . .Solution (c) Full tap To ensure that the actual volume of liquid delivered into the conical flask is read accurately. The following procedures were carried out during the titration: The sides of the conical flask were washed down with deionised water. Give one reason for carrying out each of these procedures. Author: J R Reid . The conical flask was frequently swirled or shaken. To ensure that all of the acid added from the burette can react with the base. To ensure complete mixing of the reactants . This information enables the subsequent titrations to be carried out more quickly.Why is a rough titration carried out? To find the approximate end-point. . Why is more than one accurate titration carried out? To minimise error by getting accurate readings within 0.1 cm3 of each other. . Cb is the concentration of the base na is the mol of full balanced per litre .Calculations Volume of acid Va (cm3) is the titration figure from burette The concentration of acid is Ca (mol) na is the mol of full balanced equation per litre Volume of base is Vb (cm3). Usually placed in the conical flask. Calculations USE FORMULA Va. Cb na nb Va = 37cm3 Ca is unknown na = 2 Vb = 25 cm3 Cb = 0. Ca = Vb.1 mol nb = 1 . Santosh Kumar 25cm3 × 0. Ca = Vb.1 37 08/06/11 Lecture Notes: Dr. Cb na nb 37 cm3× Ca = 2 Ca = 25 × 2 × 0.Va.13 mol/L .1mol 1 = 0. Cb na nb .g vinegar Choice of indicator Type of vol flask given 1L OR 250 Cm³ as you have to adjust in your calculations Take titre reading from burette and given vol of solution is taken from conical flask USE FORMULA Va.EXAM QUESTIONS Look out for dilution factors e. Ca = Vb. Volumetric Analysis Calculations Author: J R Reid . 0ml 13.6ml 13.9ml = Sum of the titres / number of titres = (13.1 + 12. The first thing we do is to calculate the mean (average) titre: Titres Mean = 12.1ml 12.Calculations – Mean Titre We will have a number of titres for each solution we analysed.6ml reading? .9) / 3 = 13.0 + 13.0ml Why did we discard the 12. c. 2. b. H2SO4 + 2NaOH → Na2SO4 + 2H2O Write down everything else we know. Write down the balanced equation e. d.Calculations – The Unknown Concentration – Preparation 1.g. This will be: a. Volume of liquid in the pipette Mean titre (from burette) The concentration of the standard solution Was the standard solution in the pipette or in the burette? . Remember: the millilitres must always be converted into litres for these formulae Now that you know how many moles of the standard you used. Use the n = cv formula. look at the balanced equation. one third? We can calculate the amount of the unknown: We multiply if we need more i. …etc . then how much less – half as much.e.Calculations – The ‘Unknown’ Amount 3. 3x. …etc We divide if we need less i.e. Would you need more or less of the ‘unknown’ substance in a balanced reaction? If more. 2x. three times? If less. then how much more – two times. Now calculate the amount in the standard solution you used. ½ = divide by 2. 4. We can now rearrange our n = cv formula to say c = n/v Remember: All the calculations must be in litres (not millilitres) The final value must have units (molL-1 ) written after it . Now we have the volume and amount of the ‘unknown’ substance.Calculations – The ‘Unknown’ Concentration 5. 0013/2 = 0.0+12.0ml 2.6ml. 4. 13.325molL-1 5. 13.1+13.9) / 3 = 13.0.1ml. H2SO4 + 2NaOH → Na2SO4 + 2H2O Standard solution = NaOH (in burette) = 0.0013mol Amount of H2SO4 = half of NaOH = 0. 12. 3.1 x (13/1000) = 0. .Example: 1. Amount of NaOH = cv = 0.00065mol Concentration H2SO4 = n/v = 0.1molL-1 Unknown concentration = H2SO4 (from 20ml pipette) Titres = 12.00065/(20/1000) = 0.9ml Average titre = (13. 12.Titration examples A HCl + NaOH → NaCl + H2O Titres: 12.0mL Known solution details: HCl in the burette.1mL.3mL. Concentration = 0. 12. 12.522 molL-1 Unknown solution details: NaOH 15mL aliquots Calculations: .1mL. Concentration = 0.Titration examples B CH3COOH + NaOH → NaCH3COO + H2O Titres: 17. 18.4mL. 17. 17.5mL.103 molL-1 Unknown solution details: CH3COOH 15mL aliquots Calculations: .6mL.5mL Known solution details: NaOH in the burette. 9mL Known solution details: HCl in the burette.8mL.8mL. Concentration = 0.8mL.Titration examples C 2HCl + Na2CO3 → 2NaCl + H2O + CO2 Titres: 12.555 molL-1 Unknown solution details: Na2CO3 25mL aliquots Calculations: . 12. 12. 12. 12. Concentration = 1.3mL. 12.0mL Known solution details: NaOH in the burette.04 molL-1 Unknown solution details: H2SO4 10mL aliquots Calculations: .Titration examples D H2SO4 + 2NaOH → Na2SO4 + 2H2O Titres: 12. 12.1mL.1mL. + 2H+ → Cl− + I2 + H2O The iodine produced is then titrated with standardised sodium thiosulfate solution. It shows an example of a chemical reaction that could be used in a titration: Hypochlorite ions react with iodide ions according to the equation. The overall equation for both reactions is: OCl− + 2H+ + 2S2O32− → Cl− + S4O62− + H2O . It reacts according to the equation below. I2 + 2S2O32− → 2I− + S4O62– Since starch turns blue in the presence of iodine.Level 3 – Spot the difference… Here is an extract from a level 3 titration assessment. OCl− + 2I. it is used as an indicator for this final reaction. 0.7mL Balanced equation: → NaOH c= n= v= HCl c= n= v= . 22. 22.8mL.112molL-1 . She decides to use a standard solution of Sodium hydroxide (NaOH).3mL.Example One: Jo has a solution of Hydrochloric acid (HCl) that she does not know the concentration of. 22. She then titrates these with the HCl from the burette until the indicator turns from purple to colourless These are her results: Titres of HCl: 23. for the titration She uses a 15mL pipette to measure aliquots of NaOH into her flasks.9mL. but the label is missing. To work out the concentration he decides to use a standard solution of Hydrochloric acid (HCl) with a concentration of 0. 17.3mL Balanced equation: → HCl c= n= v= Na2CO3 c= n= v= .2mL.9mL.2mL. 18. He uses a 20mL pipette to measure aliquots of HCl into his flasks.322molL-1 . 18.Example Two: Jim has a solution of Sodium carbonate (Na2CO3). He then titrates these with the Na2CO3 from the burette until the indicator turns from colourless to purple These are his results: Titres of Na2CO3 : 18. 1gL-1 of ethanoic acid in it. To work out the real concentration she decides to use a standard solution of Sodium hydroxide (NaOH) with a concentration of 1.Example Three: Jill has a bottle of vinegar (ethanoic acid solution – CH3COOH). She then titrates these with the NaOH from the burette until the indicator turns from colourless to purple These are her results: Titres of NaOH : 22.5mL. The label says that it should have 3.6mL.4mL Balanced equation: → NaOH c= n= v= CH3COOH c= molL-1 gL-1 n= v= . 22. 22. She uses a 25mL pipette to measure aliquots of vinegar into her flasks. 22.22 x 10-2 molL-1 .1mL. 1 mol Na2CO3 = 0.5 mol Na2CO3 x 105. Amount Na2CO3 = n Na2CO3 (mol) = Volume solution x c Na2CO3 (mol/ L) = 5 L x 0. .0 L of 0.99 g Na2CO3 =53 g Na2CO3 mol Na2CO3 The solution is prepared by dissolving 53 g of Na2CO3 in water and diluting to 5 L.99 g/mol) from the primary standard solution.5 mol Na2CO3 L Mass Na2CO3 = mNa2CO3=0.10 M Na2CO3 (105.Example: Calculating the Molarity of Standard Solutions Describe the preparation of a 5. Example: Calculating the Molarity using different algebraic relationships How would you prepare 50mL portions of standard solutions that are 0. and 0.01 M Na+? To solve this the relationship Vconcd x cconcd = Vdil x cdil Vconcd = Vdil x cdil = 50mL x 0.005 M.001 M in a standard 0. 0. . 25mL of the concentration solution should be diluted to 50mL.01mmol Na+ /mL To produce 50mL of 0.005 mmol Na+ /mL = 25mL cconcd 0.005 M Na+.002 M. The second example involves calculating the amount of analyte in a sample from titration data. .How to deal with titration data… The following two examples show the two types of volumetric calculations. The first involves computing the molarity of solutions that have been standardized against either a primary standard or another standard solution. 01963 x 2) mmol HCL 50mL solution = 0.01963 mmol Ba(OH)2 mL Ba(OH)2 Amount HCl = (29.0233M mL solution .71mL of 0.01963) mmol Ba(OH)2 x 2 mmol HCl 1 mmol Ba(OH)2 C HCl = (29.71 x 0.71 x 0.01963 M Ba(OH)2 to reach an end point with bromocresol green indicator.71 mL Ba(OH)2 x 0.023328 mmol HCl = 0. Stoichiometric ratio= 2 mmol HCl/ 1 mmol Ba(OH)2 Amount Ba(OH)2 = 29.Example: Molarity of solutions that have been standardized A 50mL volume of HCl solution required 29. Calculate the molarityof the HCl. 134 5 = 0.2121 g Na2C2O4 x 1 mmol Na2C2O4 0.2121 mmol Na2C2O4 x 2 mmol KMnO4 0. What is the molarity of the KMnO4 solution? Stoichiometric ratio = 2 mmol KmnO4/ 5 mmol Na2C2O4 Amount Na2C2O4= 0.31 mL KMnO4 .134 5 mmol Na2C2O4 C KMnO4 = ( 0.31 mL of KMnO4.134 g na2C2O4 Amount KMnO4 = 0.Example: Amount of analyte in sample from titration Titration of 0.2121 x 2) mmol KMnO4 0.01462M 43.2121 g of pure Na2C2O4 (134 g/mol) required 43. 77% 0.055847 g Fe 2+ mmol Fe2+ % Fe2+ = (47.02242) mmol KMnO4 x 5 mmol Fe2+ 1 mmol KMnO4 Mass Fe2+ = (47.847 g/mol). The iron is then reduced to Fe2+ and titrated with 47.22 x 0.02242 mmol KMnO4 mL KMnO4 Amount Fe2+ = (47.8040g sample of an iron ore is dissolved in acid. Calculate the results of this analysis in terms of percent Fe (55. Stoichiometric ratio = 5 mmol Fe2+/ 1 mmol KMnO4 Amount KMnO4 = 47.02242 x 5 x 0.02242 x 5) mmol Fe2+ x 0.055947) g Fe 2+ x 100% = 36.8040 g sample .22mL of 0.02242 M KMnO4 solution.22 x 0.22mL KMnO4 x 0.Example: Computing analyte concentrations from titration data A 0.22 x 0. Titration Curves Example of a sigmoidal titration curve once calculations of data have been computed.htm .ac.psigate.uk/newsite/ reference/plambeck/chem1/p01173. www. J.psigate.ac.sb. Inc: United States of America.edu/chm1045/notes/Aq ueous/Stoich/Aqua02..htm http://www2. 7th ed.htm www. & Crouch. (2000).uk/newsite/ reference/plambeck/chem1/p01173.html . S.hmc. West. Holler. Thomson Learning.fsu. D.References Skoog.. F.. Analytical Chemistry: An Introduction. D. http://wine1.edu/~karukstis/chem21f 2001/tutorials/tutorialStoichiFrame.