1. Acid-Base Reactions 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 Stomach Acidity & Acid-Base Reactions

3. 3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Section 18.4 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? HBz + NaOH ---> Na + + Bz - + H 2 O 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? HBz + NaOH ---> Na + + Bz - + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

4. 4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Section 18.4 The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid. The final solution is basic. Bz - + H 2 O HBz + OH - Bz - + H 2 O HBz + OH - The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid. The final solution is basic. Bz - + H 2 O HBz + OH - Bz - + H 2 O HBz + OH - + + K b = 1.6 x 10 -10

5. 5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Strategy — find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation Strategy — find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation 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?

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions STOICHIOMETRY PORTION 1. Calc. moles of NaOH req’d (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2.Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d STOICHIOMETRY PORTION 1. Calc. moles of NaOH req’d (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2.Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d 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?

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions STOICHIOMETRY PORTION 25 mL of NaOH req’d 3. Moles of Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M STOICHIOMETRY PORTION 25 mL of NaOH req’d 3. Moles of Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M 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 Acid-Base Reactions EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] initial initial change change equilib equilib EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] initial initial change change equilib equilib 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?

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] initial0.02000 initial0.02000 change-x+x+x change-x+x+x equilib0.020 - xxx equilib0.020 - xxx EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] initial0.02000 initial0.02000 change-x+x+x change-x+x+x equilib0.020 - xxx equilib0.020 - xxx 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?

10. 10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions EQUILIBRIUM PORTION 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 Solving in the usual way, we find x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25 EQUILIBRIUM PORTION 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 Solving in the usual way, we find 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?

11. 11 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 [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x 10 -5 pH = 4.20 HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10 - 5 [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x 10 -5 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?

12. 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

13. 13 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.

14. 14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Titrations Additional NaOH is added. pH rises as equivalence point is approached.

15. 15 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.

16. 16 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

17. 17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions EQUILIBRIUM PORTION 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 Solving in the usual way, we find x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25 EQUILIBRIUM PORTION 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 Solving in the usual way, we find 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?

18. 18 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? Half-way point

19. 19 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 [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x 10 -5 pH = 4.20 HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10 -5 [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x 10 -5 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?