1. QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. The pH will go _____________. After all, NH 4 + is an acid! QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. The pH will go _____________. After all, NH 4 + is an acid! The Common Ion Effect Section 18.2 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 Let us first calculate the pH of a 0.25 M NH 3 solution. [NH 3 ][NH 4 + ][OH - ] initial0.2500 change-x+x+x equilib0.25 - xx x Let us first calculate the pH of a 0.25 M NH 3 solution. [NH 3 ][NH 4 + ][OH - ] initial0.2500 change-x+x+x equilib0.25 - xx x QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) The Common Ion Effect Section 18.2

3. 3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) The Common Ion Effect Section 18.2

4. 4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M The Common Ion Effect Section 18.2

5. 5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 The Common Ion Effect Section 18.2

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH 3 QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH 3 The Common Ion Effect Section 18.2

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initialchangeequilib Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initialchangeequilib The Common Ion Effect Section 18.2

8. 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initial0.250.100 change-x+x+x equilib0.25 - x0.10 + x x Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initial0.250.100 change-x+x+x equilib0.25 - x0.10 + x x The Common Ion Effect Section 18.2

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH 4 Cl. Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH 4 Cl. The Common Ion Effect Section 18.2

10. 10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) [OH - ] = x = (0.25 / 0.10)K b = 4.5 x 10 -5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion. Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) [OH - ] = x = (0.25 / 0.10)K b = 4.5 x 10 -5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion. The Common Ion Effect Section 18.2

11. 11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO 4 2-.

12. 12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid+Conj. Base HOAc+OAc - H 2 PO 4 - +HPO 4 2- Weak Base+Conj. Acid NH 3 +NH 4 + Buffer Solutions

13. 13 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider HOAc/OAc - to see how buffers work ACID USES UP ADDED OH - We know that OAc - + H 2 O HOAc + OH - OAc - + H 2 O HOAc + OH - has K b = 5.6 x 10 -10 Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9 K reverse is VERY LARGE, so HOAc completely snarfs up OH - !!!! Buffer Solutions

14. 14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider HOAc/OAc - to see how buffers work CONJ. BASE USES UP ADDED H + HOAc + H 2 O OAc - + H 3 O + has K a = 1.8 x 10 -5 Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4 K reverse is VERY LARGE, so OAc - completely snarfs up H + ! Buffer Solutions

15. 15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 [HOAc][OAc - ][H 3 O + ] [HOAc][OAc - ][H 3 O + ]initialchangeequilib Buffer Solutions

16. 16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [HOAc][OAc - ][H 3 O + ] [HOAc][OAc - ][H 3 O + ] initial0.7000.600x change-x+x+x equilib0.700 - x0.600 + xx Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions

17. 17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [HOAc][OAc - ][H 3 O + ] equilib0.700 - x0.600 + xx Assuming that x << 0.700 and 0.600, we have Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions

18. 18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [HOAc][OAc - ][H 3 O + ] equilib0.700 - x0.600 + xx Assuming that x << 0.700 and 0.600, we have Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions

19. 19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [HOAc][OAc - ][H 3 O + ] equilib0.700 - x0.600 + xx Assuming that x << 0.700 and 0.600, we have [H 3 O + ] = 2.1 x 10 -5 and pH = 4.68 Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions

20. 20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Notice that the expression for calculating the H + conc. of the buffer is and this leads to a general equation for finding the H + or OH - concs. of a buffer. Buffer Solutions

21. 21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Notice that the expression for calculating the H + conc. of the buffer is and this leads to a general equation for finding the H + or OH - concs. of a buffer. Buffer Solutions

22. 22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Notice that the expression for calculating the H + conc. of the buffer is and this leads to a general equation for finding the H + or OH - concs. of a buffer. Notice that the H + or OH - concs. depend on K and the ratio of acid and base concs. Buffer Solutions

23. 23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation

24. 24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation Take the negative log of both sides of this equation

25. 25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Henderson-Hasselbalch Equation Take the negative log of both sides of this equation or This is called the Henderson-Hasselbalch equation.

26. 26 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.

27. 27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (before HCl, pH = 7.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 M 2 = 1.00 x 10 -3 M M 2 = 1.00 x 10 -3 M pH = 3.00 pH = 3.00

28. 28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b) Step 1 — do the stoichiometry H 3 O + (from HCl) + OAc - (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large. What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

29. 29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H 3 O + ][OAc - ][HOAc] [H 3 O + ][OAc - ][HOAc] Before rxn Change After rxn What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

30. 30 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H 3 O + ][OAc - ][HOAc] [H 3 O + ][OAc - ][HOAc] Before rxn0.001000.6000.700 Change-0.00100-0.00100+0.00100 After rxn00.5990.701 What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

31. 31 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O H 3 O + + OAc - HOAc + H 2 O H 3 O + + OAc - [HOAc] [H 3 O + ][OAc - ] [HOAc] [H 3 O + ][OAc - ] Before rxn0.7010.5990 Change-x+x+x After rxn0.710-x0.599+xx What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

32. 32 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O H 3 O + + OAc - HOAc + H 2 O H 3 O + + OAc - [HOAc] [H 3 O + ][OAc - ] [HOAc] [H 3 O + ][OAc - ] After rxn0.710-x0.599+xx Because [H 3 O + ] = 2.1 x 10 -5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

33. 33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O H 3 O + + OAc - HOAc + H 2 O H 3 O + + OAc - [H 3 O + ] = 2.1 x 10 -5 M ------> pH = 4.68 The pH has not changed on adding HCl to the buffer! What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

34. 34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H 3 O + ] = 10 -pH = 5.0 x 10 -5 M It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ). —then you get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.

35. 35 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Preparing a Buffer Solution Buffer prepared from HCO 3 - weak acid CO 3 2- conjugate base HCO 3 - + H 2 O ¸ H 3 O + + CO 3 2-

36. 36 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M POSSIBLE ACIDSK a POSSIBLE ACIDSK a HSO 4 - / SO 4 2- 1.2 x 10 -2 HSO 4 - / SO 4 2- 1.2 x 10 -2 HOAc / OAc - 1.8 x 10 -5 HOAc / OAc - 1.8 x 10 -5 HCN / CN - 4.0 x 10 -10 HCN / CN - 4.0 x 10 -10 Best choice is acetic acid / acetate.

37. 37 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Preparing a Buffer

38. 38 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Preparing a Buffer

39. 39 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Solve for [HOAc]/[OAc - ] ratio Solve for [HOAc]/[OAc - ] ratio = 2.78/ 1 = 2.78/ 1 Preparing a Buffer

40. 40 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Solve for [HOAc]/[OAc - ] ratio Solve for [HOAc]/[OAc - ] ratio = 2.78/ 1 = 2.78/ 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. Preparing a Buffer

41. 41 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Preparing a Buffer