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Strength of Vinegar by Acid-Base Titration

Vinegar. Strength of Vinegar by Acid-Base Titration. Final Exercise – 105 points. CH 3 COOH. NaOH. Lye. ? QUESTIONS ?. How, and with what accuracy and precision can you determine the concentration of a solution of NaOH?.

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Strength of Vinegar by Acid-Base Titration

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  1. Vinegar Strength of Vinegar by Acid-Base Titration Final Exercise – 105 points CH3COOH NaOH Lye

  2. ? 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. This week From last week Concepts: Strong/Weak Acids Acid Dissociation/Ka Concentration End point / Equivalence point Titration curves Logarithms Indicator Mole Relationships pH & pKa Titration Primary Standard Techniques: Titration pH Measurement Weighing by Difference * Standardization Preparing solutions of precise concentration Apparatus: Burets pH Meter Pipet Volumetric Flask Analytical Balance * Remember the first exercise!

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

  5. STOICHIOMETRY KHP is a monoprotic weak acid. One available proton. [K+]+ HC8H4O4- + OH- C8H4O4= + H2O + [K+] pKais 5.4 The stoichiometry of the reaction is: 1 mol KHP  1 mol NaOH Acetic acid, CH3COOH, is also a monoprotic weak acid. CH3COOH + OH- CH3COO- + H2O 1 mol CH3COOH  1 mol NaOH It’s pKa is 4.7

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

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

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

  9. The pH at the equivalence point in the titration of acetic acid with sodium hydroxide is • the pKa of acetic acid • ½ the pKa of acetic acid • 7 • Larger than 7

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

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

  12. CALCULATIONS – KHP Solution Weight of vial + KHP 15.4371 g Weight of vial + remaining KHP 12.3495 g Weight of KHP transferred 3.0876 g Volume of KHP Solution 0.2500 L Weigh by difference!!! Weight of KHP 250 mL

  13. CALCULATIONS – KHP Solution Weight of vial + KHP 15.4371 g Weight of vial + remaining KHP 12.3495 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 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. 0.06  0.006 or between 0.054 and 0.066 Weight of KHP Molar Mass of KHP Volume of Volumetric Flask 4 Sig Figs because of    0.015119 Mols of KHP

  14. PROCEDURE (cont’d) Determineconcentration of stock NaOH Solution ( Nominally0.1 M ) Titrate measured volumes of standard KHP solution (known concentration) • with measured volumes of NaOH Solution (unknown concentration) (Delivered from buret 1 ) (Delivered from buret 2)

  15. In STANDARDIZATION, BOTH REAGENTS are delivered by a BURET KHP • Can "BACKTITRATE“ I.e., if end point is overshot, can recover by adding more KHP and continuing titration, e.g., NaOH Do at most 3 titrations for the standardization 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 youlabel the burets.

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

  17. How many mL of 0.10 M NaOH would 35 mLof 0.10 M KHP require? • 25 mL • 35 mL • 0.006407 M • Whatever

  18. How many mL of ~0.10 M NaOH would 35 mLof 0.10 M KHP require? 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, B 35 mL

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

  20. E.g., suppose volumes of NaOH are 25.37 ,25.84 , 24.93 AvgVol = (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 % Since all other numbers in subsequent calculations are the same for any run, that will be percent deviation in the final result. Do at most 5 titrations for unknown (incl prelim)

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

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

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

  24. CALCULATIONS – NaOH STANDARDIZATION Molarity of KHP Solution: 0.06047 M KHP buret reading, final 37.44 mL 38.77 KHP buret reading, initial 3.68mL 4.73 Volume of KHP titrated 33.76 mL34.04 NaOH buret reading, final 28.73 mL 31.83 NaOH buret reading, initial 4.52mL 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 2.042mmol / 24.21mL = 0.08435M 0.08526 M From Part 1 Stoichiometry is 1 to 1

  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 2.058 mmol of NaOH used 2.042 mmol 2.058 Molarity of NaOH 2.042mmol / 24.21mL = 0.08435M 0.08592 M

  26. Suppose the 3rd titration produces 0.08539 M 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%

  27. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL5.00 5.00 Concentration of NaOH solution 0.08522 M NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.374.22 Volume NaOH used 19.32mL18.79 19.50 mmol of NaOH used 19.32 * 0.08522 = 1.646 mmol 1.585 1.645 mmol of Acetic Acid titrated = 1.646mmol1.585 1.645 Acetic Acid Concentration 1.646/ 5.00 = =0.329 M 0.317 0.329 36 Stoichiometry is 1 to 1 3 Sig Figs From Standardization

  28. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL 5.005.00 Concentration of NaOH solution 0.08522 M NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.374.22 Volume NaOH used 19.32 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 1.6011.645 Acetic Acid Concentration 1.601/ 5.00 = =0.329 M 0.320 0.329 36

  29. CALCULATIONS - UNKNOWN Volume of Unknown 5.00 mL 5.00 5.00 Concentration of NaOH solution 0.08522 M NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.374.22 Volume NaOH used 19.32 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. The concentration of the sodium hydroxide to be used in computations in this exercise is the valuegiven on the stock solutioncontainer. • True • False

  31. 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. 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. Read & Record Burets to 2 DECIMAL PLACES 24.64 mL 4.6427 g Read & Record Weights to 4 DECIMAL PLACES

  34. ANALYSIS OF ERRORS • STANDARDIZATION – Precision • 1. WEIGH 3g of solid KHP using the analytical balance ( 0.0004 g ) • Precision: 100 X0.0004 / 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 / 30.00 0.2 % • NaOH100 X0.05 / 30.00 0.2 % • Concentration of NaOH determined with PRECISION of approximately 0.01 + 0.02 + 0.2 + 0.2 = 0.43 %

  35. ANALYSIS OF ERRORS • TITRATION OF UNKNOWN – Precision • 1. TRANSFERaliquot 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 %

  36. Suppose you over-titrate by 10 drops I.e., you go past the end point by 10 drops (0.5 mL). How large an error in accuracy results? • 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.

  37. 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 100 X 0.01 / 5.00 = 0.2% BUT The volume of a drop is ~ 0.05 mL. Each drop represents: 100 X 0.05/5.00 = 1% error in the volume delivered!

  38. PITFALLS TO AVOID • 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 !

  39. 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 GRADE ON THIS EXERCISE WILL DEPEND ON both Accuracy and Precision

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

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