1. TitrationsTitrations pHpH Titrant volume, mL Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2. 2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.

3. 3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Titrations Additional NaOH is added. pH rises as equivalence point is approached.

4. 4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Titrations Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

5. 5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Equivalence point pH of solution of benzoic acid, a weak acid

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: How do we calculate pH at the equivalence point? Equivalence point pH of solution of benzoic acid, a weak acid QUESTION: How do we calculate pH before titration?

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions HBz reacts completely to produce Bz- Part 1 calculation: stoichiometry gives [Bz - ] produced is 0.020 M. Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] equilib0.020 - xxx equilib0.020 - xxx x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25 HBz reacts completely to produce Bz- Part 1 calculation: stoichiometry gives [Bz - ] produced is 0.020 M. Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] equilib0.020 - xxx equilib0.020 - xxx x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?

8. 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Half-way point [HBz] = [Bz - ] At midpoint of titration…

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10 -5 pH = pKa + log ( [Bz - ]/[HBz] ) pH = 4.20 + log (1) pH = 4.20 HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10 -5 pH = pKa + log ( [Bz - ]/[HBz] ) pH = 4.20 + log (1) pH = 4.20 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point?

10. 10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Half-way point …can calculate pH if know [Bz - ]/[HBz] At any point of the titration…

11. 11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved pH = pKa …can calculate pH if know [Bz - ]/[HBz] At any point of the titration… If pH > pKa, [Bz - ]>[HBz] If pH < pKa, [HBz]>[Bz - ]