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2. Strength of Vinegar by Acid-Base Titration Final Exercise – 105 points CH 3 COOH NaOH Vinegar Lye 2

3. ? QUESTIONS ? How, and with what accuracy and precision can you determine the concentration of a solution of NaOH? Using that NaOH solution, with what accuracy and precision can you determine the concentration of a solution of acetic acid - by titration? 3

4. Concepts: Strong/Weak Acids Acid Dissociation/K a Concentration End point / Equivalence point Titration curves Logarithms Indicator Mole Relationships pH & pK a Titration Techniques: Titration / Back-titration pH Measurement Apparatus: BuretspH Meter Primary Standard Weighing by Difference * Preparing solutions of precise concentration Standardization Volumetric Flask Analytical Balance Pipet From last week This week 4 * Remember the first exercise!

5. KHP potassium hydrogen phthalate (KHP) Last exercise involved acid-base titration in which concentration of NaOH was given. In this exercise, you STANDARDIZE the NaOH. Standardization verified composition and purity Analytical determination of purity or concentration of substance through reaction with substance of verified composition and purity (Primary Standard ) This STANDARDIZATION involves reaction of the base, NaOH, with a PRIMARY STANDARD, the weak acid: H phthalic acid 5 High purity Stable Not hygroscopic Not efflorescent High solubility High molar mass

6. KHP KHP is a monoprotic weak acid. One available proton. [K + ] + HC 8 H 4 O 4 - + OH -  C 8 H 4 O 4 = + H 2 O + [K + ] The stoichiometry of the reaction is: Acetic acid, CH 3 COOH, is also a monoprotic weak acid. 1 mol KHP  1 mol NaOH 1 mol CH 3 COOH  1 mol NaOH STOICHIOMETRY CH 3 COOH + OH -  CH 3 COO - + H 2 O It’s pK a is 4.7 pK a is 5.4 6

7. An aqueous solution, can have [ H + ] = 0.1 and [ OH - ] = 0.1 at the same time A.True B.False 7

8. An aqueous solution, can have [ H + ] = 0.1 and [ OH - ] = 0.1 at the same time B False 8 [H + ] and [OH - ] are dependent on one another in aqueous solutions. [ H + ] [ OH - ] = 1.0 X 10 -14

9. pH at equivalence point is ~8.5 Phenolphthalein is a reasonable indicator Equivalence Point mL of NaOH added Half equivalence point pK a = 4.7 pH ~ 8.5 [ H + ] [ A - ] K a = --------------- [ HA ] [ H + ]= K a ---------- [ A - ] [ HA ] -log [ H + ] = - log K a - log ------- [ A - ] [ HA ] pH = pK a - log ----------- [ A - ] pH = pK a 9 Titration curve for Acetic Acid with NaOH looks like this

10. The pH at the equivalence point in the titration of acetic acid with sodium hydroxide is 10 A.the pK a of acetic acid B.½ the pK a of acetic acid C. 7 D.Larger than 7

11. Equivalence Point mL of NaOH added pH ~ 8.5 11 D. Larger than 7

12. It does not matter in which order you A.) Standardize NaOH, B.) Titrate Unknown C.) Determine pH of unknown PROCEDURE A. STANDARDIZATION Prepare solution of known concentration of primary standard, KHP Weigh sample BY DIFFERENCE* ERLENMEYER Dissolve fully in ERLENMEYER VOLUMETRIC FLASK** Bring to total volume in VOLUMETRIC FLASK** TA’s will assign order to minimize waiting Must transfer ALL the solution *Web supplement **Follow manual procedure 12

13. CALCULATIONS – KHP Solution Weight of vial + KHP15.4371 g Weight of vial + remaining KHP12.3495 g Weight of KHP transferred 3.0876 g Volume of KHP Solution 0.2500 L Weight of KHP 250 mL Weigh by difference!!! 13

14. CALCULATIONS – KHP Solution Weight of vial + KHP15.4371 g Weight of vial + remaining KHP12.3495 g Weight of KHP transferred 3.0876 g Volume of KHP Solution 0.2500 L Molarity of KHP Solution: (3.0876 g / 204.22 g/mol) / 0.2500 L = 0.06047 M Molar Mass of KHP Weight of KHP Volume of Volumetric Flask 4 Sig Figs because of 0.015119 Mols of KHP    If calculated concentration of KHP differs more than 10% from 0.06 M, you have either made an error in weight of KHP or made a calculation error. 14 0.06  0.006 or between 0.054 and 0.066

15. PROCEDURE (cont’d) Determine concentration of stock NaOH Solution Nominally ( Nominally 0.1 M ) with (Delivered from buret 1 buret 1 ) measured volumes of NaOH Solution unknown (unknown concentration) Titrate known measured volumes of standard KHP solution (known concentration) (Delivered from buret 2 buret 2 ) 15

16. In STANDARDIZATION, BOTH REAGENTS are delivered by a BURET BACKTITRATE  Can "BACKTITRATE“ I.e., if end point is overshot, can recover by adding more KHP and continuing titration, e.g., KHP NaOH 1. Add measured volume of KHP(~35mL) 2. Titrate to Phenolphthalein Endpoint 3. If overshoot endpoint, add more KHP 4. Titrate to Phenolphthalein Endpoint again Make sure you label the burets. Do at most 3 titrations for the standardization 16

17. Having determined concentration of KHP solution, calculate volume of ~ 0.1 M NaOH needed to react with 35 mL of KHP. 35 mL of our KHP solution (0.06047 M) contains: What volume of 0.1 M NaOH will contain 2.1 mmol of NaOH? m moles = Molarity (mmol/mL) X Volume (m L) 2.1 mmol = 0.1 (mmol/mL) X V (mL) V = 2.1 / 0.1 = 21 mL  20 mL 35 mL X 0.06047 mmol / mL = 2.1 mmol of KHP Amt = Concentration X Volume Why do we make this calculation? So we will know when to begin to add the NaOH slowly! That will require 2.1 mmol of NaOH 17

18. How many mL of 0.10 M NaOH would 35 mL of 0.10 M KHP require? A.25 mL B.35 mL C.0.006407 M D.Whatever 18

19. B 35 mL Stoichiometry is 1 ↔ 1 35 mL X 0.10 M = 3.5 mmol KHP 3.5 mmol KHP ↔ 3.5 mmol NaOH 0.10 M = 0.10 mmol/mL 3.5 mmol / 0.10 mmol/mL = 35 mL So, the answer is, How many mL of ~0.10 M NaOH would 35 mL of 0.10 M KHP require? 19

20. PROCEDURE - UNKNOWN Unknowns are solutions of acetic acid in water. We titrate aliquots* of (undiluted) unknown. In this exercise, aliquot is a 5 mL pipet-ful (5.00  0.01) of solution’ * aliquots are fixed, repetitive fractions of a solution Dilute unknown with distilled water (~ 40 mL) Add Phenolphthalein as indicator Titrate unknown to phenolphthalein end point. Since aliquots are identical, can track precision of titration by calculating percent error involumes of NaOH used. Remember: you cannot backtitrate in this case amount Adding water doesn’t change the amount of acetic acid. Mandrake Root Extract 20

21. E.g., suppose volumes of NaOH are Avg Vol = (25.37 + 25.84 + 24.93) / 3 = 25.38 Avg Dev= ( + 0.01 + 0.46 + 0.45) / 3 = 0.31 % Dev= 100 X 0.31 / 25.38 = 1.2 % 25.37,25.84, 24.93 Do at most 5 titrations for unknown (incl prelim) Since all other numbers in subsequent calculations are the same for any run, that will be percent deviation in the final result. 21

22. darkest The darkest pink color is the best signal for the equivalence point in the titration of acetic acid with NaOH using phenolphthalein 22 A.True B.False

23. lightest The lightest pink color is the best signal for the equivalence point in the titration of acetic acid with NaOH using phenolphthalein B B = False A few drops after that point is reached, the pH of the solution increases rapidly and the solution gets as deeply colored as it is able. On further addition of base, solution lightens due to dilution. 23

24. Before end point At end point Just past Just past end point KHP & Unknown solutions are colorless. No interference from beverage color. Way past Way past end point 24

25. CALCULATIONS – NaOH STANDARDIZATION Molarity of KHP Solution: 0.06047 M KHP buret reading, final 37.44 mL 38.77 KHP buret reading, initial 3.68 mL 4.73 Volume of KHP titrated 33.76 mL 34.04 NaOH buret reading, final 28.73 mL 31.83 NaOH buret reading, initial 4.52 mL 7.88 Volume of NaOH used 24.21 mL 23.95 mmol of KHP titrated 33.76 mL X 0.06047 M = 2.042 mmol 2.202 mmol of NaOH used 2.042 mmol 2.202 Molarity of NaOH 2.042 mmol / 24.21mL= 0.08435 M 0.08526 M Stoichiometry is 1 to 1 From Part 1 25

26. CALCULATIONS – NaOH STANDARDIZATION Molarity of KHP Solution: 0.06047 M KHP buret reading, final 37.44 mL 38.77 KHP buret reading, initial 3.68 mL 4.73 Volume of KHP titrated33.76 mL 34.04 NaOH buret reading, final28.73 mL 31.83 NaOH buret reading, initial 4.52 mL 7.88 Volume of NaOH used24.21 mL 23.95 mmol of KHP titrated 33.76 mL X 0.06047 M = 2.042 mmol 2.058 mmol of NaOH used 2.042 mmol 2.058 Molarity of NaOH 2.042 mmol / 24.21mL= 0.08435 M 0.08592 M 26

27. We have three values for our NaOH concentration. Are they in reasonable agreement? Average = (0.08435 + 0.08592 + 0.08539) / 3 = 0.08522 M Avg Dev = (0.00087 + 0.00070 + 0.00017) / 3 = 0.00058 M Pct Dev = 100 X 0.00058 / 0.08522 = 0.68% Suppose the 3rd titration produces 0.08539 M 27

28. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL 5.00 5.00 Concentration of NaOH solution 0.08522 M NaOH buret, final22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used19.32 mL 18.7919.50 mmol of NaOH used 19.32 * 0.08522 = 1.646 mmol 1.585 1.645 mmol of Acetic Acid titrated = 1.646 mmol 1.585 1.645 Acetic Acid Concentration 1.646 / 5.00 = = 0.329 M 0.317 0.329 36 From Standardization 3 Sig Figs Stoichiometry is 1 to 1 28

29. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL 5.00 5.00 Concentration of NaOH solution 0.08522 M NaOH buret, final22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used19.32 mL 18.7919.50 mmol of NaOH used 18.79 * 0.08522 = 1.646 mmol 1.601 1.645 mmol of Acetic Acid titrated = 1.646 mmol 1.601 1.645 Acetic Acid Concentration 1.601 / 5.00 = = 0.329 M 0.320 0.329 36 29

30. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL 5.00 5.00 Concentration of NaOH solution 0.08522 M NaOH buret, final22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used19.32 mL 18.79 19.50 mmol of NaOH used 19.50 * 0.08522 = 1.646 mmol 1.601 1.662 mmol of Acetic Acid titrated = 1.646 mmol 1.601 1.662 Acetic Acid Concentration 1.662 / 5.00 = = 0.329 M 0.320 0.332 36 30

31. The concentration of the sodium hydroxide to be used in computations in this exercise is the value given on the stock solution container. A.True B.False 31

32. The concentration of the sodium hydroxide to be used in computations in this exercise is the value given on the stock solution container.. B = False The purpose of the standardization is to determine the precise concentration of the NaOH solution. 32

33. Avg Conc of Acetic Acid= (0.329 M + 0.320 M + 0.332 M) / 3 = 0.327 M Avg Deviation= (0.002 + 0.007 + 0.005) / 3 = 0.005 M Pct Deviation= 100 X 0.005 / 0.327 = 1.5% 33

34. Read & Record Burets to to 2 DECIMAL PLACES Read & Record Weights to 4 DECIMAL to 4 DECIMAL PLACES 24.64 mL 4.6427 g 34 Begin each titration with buret reading between 0.00 and 5.00 mL

35. STANDARDIZATION – Precision 1. WEIGH 3g of solid KHP using the analytical balance (  0.0004 g ) Precision: 100 X 0.0004 / 3.0000 0.01 % 2. PREPARE A SOLUTION of KHP using 250 mL volumetric flask (  0.05 mL ) Precision:100 X 0.05 / 250.000.02 % 3. TITRATE measured volumes of KHP and NaOH using burets (  0.05 mL ) Precision: KHP 100 X 0.05 / 30.000.2 % NaOH 100 X 0.05 / 30.000.2 % Concentration of NaOH determined with PRECISION of approximately 0.01 + 0.02 + 0.2 + 0.2 = 0.43 % ANALYSIS OF ERRORS 35

36. TITRATION OF UNKNOWN – Precision 1. TRANSFER aliquot of unknown using a 5 mL transfer pipet (  0.01 mL ) Precision:100 X 0.01 / 5.00= 0.2 % Note that 1 drop (0.05 mL) ↔ 1% 2. TITRATE aliquot using standardized NaOH, again using a buret (  0.05 mL ) Precision: (Buret)100 x 0.05 / 30.00 = 0.2 % OVERALL PRECISION of determination of concentration of unknown = 0.2 + 0.2 = 0.4 % ANALYSIS OF ERRORS 36

37. 37 Suppose you over-titrate by 10 drops I.e., you go past the end point by 10 drops (0.5 mL). Unknowns range in concentration from 0.5 M to 1.0 M. 5 mL of the unknown contains between 2.5 and 5.0 mmol of acetic acid. That will require between 25 and 50 mL of ~0.1 M NaOH The error represented by 0.5 mL will range from: 100 X 0.5 / 25 = 2 % to 100 X 0.5 / 50 = 1 % Errors in indentifying the end point are not a major contribution to errors in the accuracy of the concentration of the vinegar unknown. How large an error in accuracy results?

38. 38 What about the pipet? The volume of the sample you are using is 5.00 mL, delivered by pipet. The intrinsic precision of the 5 mL pipet is 0.01 mL The volume of a drop is ~ 0.05 mL.  Each drop represents: 100 X 0.05/5.00 = 1% error in the volume delivered! BUT 100 X 0.01 / 5.00 = 0.2%

39. 1.Incomplete transfer of solid KHP into flask (caused by weighing using a watch glass or paper) Weigh by difference! 2. Failure to dissolve KHP completely Make sure no KHP particles remain undissolved 3. Incomplete transfer of KHP solution from Erlenmeyer to volumetric flask Transfer completely and rinse with distilled water from wash bottle 4. Failing to bring KHP solution to correct volume Use dropper for last 0.5 mL ! PITFALLS TO AVOID 39

40. 5. Incomplete mixing of KHP solution in volumetric flask Mix thoroughly both before and after bringing to final volume 6. Improper use of the pipet Make sure you use the pipet as instructed 7. Contamination of unknown by allowing contact with syringe while pipeting Discard any sample for which this happens 8. Over-titrating unknown Stop at earliest detectable pink color AccuracyPrecision GRADE ON THIS EXERCISE WILL DEPEND ON both Accuracy and Precision 40

41. NEXT WEEK SUSB-054 – Part 1 Gasometric Determination of Sodium Bicarbonate in a Mixture DO PRELAB to SUSB-054 - 1 41

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