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 Strong Acid and Strong Base HCl + NaOH  HCl + H 2 O --->H 3 O + + Cl - NaOH(aq) ---> Na + (aq) + OH - (aq) H 3 O + + Cl - + Na + + OH -  2 H 2 O + Cl - + Na + Net ionic Equation doesn’t show spectator ions!

2. 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 H 2 O + H 2 O  H 3 O + + OH - K = 1 x 10- 14 H 3 O + + OH -  2 H 2 O K = 1/ 1 x 10- 14 = 1 x 10 14 NEUTRALIZATION REACTION, pH = 7

3. 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 Strong Acid and Weak Base NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) HCl + NH 3  H 3 O + + OH -  2 H 2 O H 3 O + + NH 3  H 2 O + NH 4 + K net = K b (1/Kw) = 1.8 x 10 9

4. 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 STRONG Acid and Weak Base Mixing equal molar quantities gives an acidic solution. Weak Acid and STRONG Base Mixing equal molar quantities gives a basic solution.

5. 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 Weak Acid and Weak Base pH depends on Kb and Ka of conjugate base and acid.

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Stomach Acidity & Acid-Base Reactions

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Example 18.1—Calculate pH 100 mL of 0.10 M HCl with 50 mL of 0.20 M NH 3 HCl + NH 3  Cl - + NH 4 + 0.100 L x 0.10 mol/L = 0.0100 mol HCl =0.0100 mol Cl - 0.050 L x 0.20 mol/L = 0.0100 mol NH 3 =0.0100 mol NH 4 +

8. 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Example 18.1—Calculate pH 100 mL of 0.10 M HCl with 50 mL of 0.20 M NH 3 HCl + NH 3  Cl - + NH 4 + 0.0100 mol Cl - / 0.150 L solution = 0.0667 M Cl - 0.0100 mol NH 4 + / 0.150 L solution = 0.0667 M NH 4 + H 2 O + NH 4 +  H 3 O + + NH 3

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Example 18.1—Calculate pH Equilibria of NH 4 + contributes to pH EQUIL Ka = x (x) / (0.0667 – x) Ka = 5.6 x 10 -10 and x = 6.1 x 10 -6 M H 2 O + NH 4 +  H 3 O + + NH 3 Ka = [ H 3 O + ] [ NH 3 ] / [ NH 4 + ] pH = 5.21

10. 10 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 -

11. 11 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

12. 12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Strategy — find the conc. of the conjugate base B - 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 B - 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?

13. 13 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?

14. 14 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?

15. 15 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?

16. 16 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?

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 Acid-Base Reactions HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10 -5 At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = 1 (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 At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = 1 (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?

19. 19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation Take the negative log of both sides of this equation

20. 20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation or This is called the Henderson-Hasselbalch equation.

21. 21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation This shows that the pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base.