1. Chapter 17 Acid-Base Equilibria

2. Water molecules undergo a process called autoprotolysis (a.k.a. self−ionization) in which hydronium and hydroxide ions are formed. 2H 2 O (l) H 3 O + (aq) + OH − (aq) This equilibrium is the foundation of acid−base chemistry in aqueous solution. The equilibrium expression for this reaction is [H 3 O + ][OH − ] K c = —————— [H 2 O] 2 Noting, however, that the concentration of water in an aqueous solution is essentially constant we can define a new constant K w = K c [H 2 O] 2 This reduces our expression to K w = [H 3 O + ][OH − ] = (1.0 × 10 −7 )(1.0 × 10 −7 ) = 1.0 × 10 −14

3. When a strong acid or base is added to water it dominates the equilibrium. For example, if 0.010 moles of hydrochloric acid (HCl) is added to 1.0 L of water it produces a concentration of 0.01 M hydrochloric acid. Hydrochloric acid dissociates as follows HCl + H 2 O → Cl − + H 3 O + The one−to−one stoichiometric ratio between hydrochloric acid and hydronium means that hydronium also has a concentration of 0.010 M. K w = [H 3 O + ][OH − ] 1.0 × 10 −14 = (0.010)[OH − ] [OH − ] = 1.0 × 10 −12 M

4. The 7 Strong Acids

5. Determine the pH of a 0.10 M solution of acetic acid. (K a = 1.8 × 10 −5 ) HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq) [H 3 O + ][C 2 H 3 O 2 − ] K a = ———————— [HC 2 H 3 O 2 ]

6. Substituting into the equilibrium expression we get (x)(x) 1.8 × 10 −5 = ———— (0.10 − x) (1.8 × 10 −5 )(0.10 – x) = x 2 1.8 × 10 −6 – (1.8 × 10 −5 )x = x 2 x 2 + (1.8 × 10 −5 )x − 1.8 × 10 −6 = 0 This is a quadratic equation that can easily be solved. The two roots are 1.3 × 10 −3 and −1.4 × 10 −3

7. Our equilibrium concentrations are [H 3 O + ] = [C 2 H 3 O 2 − ] = 1.3 × 10 −3 M [HC 2 H 3 O 2 ] = 0.10 − 1.3 × 10 −3 = 0.099 M Finally, the pH is pH = −log(1.3 × 10 −3 ) = 2.89

8. Let’s revisit our equilibrium expression (x)(x) 1.8 × 10 −5 = ———— (0.10 − x) Since we have just said the concentration of the acid does not change appreciably we can state that 0.10 – x ≈ 0.10 Which makes the expression become (x)(x) 1.8 × 10 −5 = ——— (0.10) This is much easier to solve than the original x 2 = 1.8 × 10 −5 x = 1.3 × 10 −3 M and… pH = −log(1.3 × 10 −3 ) = 2.89

9. ROT: when x is added to or subtracted from any number that is at least 100 times larger than K a, that x can be discarded without introducing any significant error.

10. Determine the pH of a 0.0500 M solution of cyanic acid (HCNO) if the K a of the acid is 3.5 × 10 -4 HCNO (aq) + H 2 O (l) H 3 O + (aq) + CNO − (aq) [H 3 O + ][CNO − ] K a = —————— [HCNO]

11. (x)(x) 3.5 × 10 −4 = ————— (0.0500 − x) Since 0.0500 >> 3.5 × 10 −4, 0.0500 − x ≈ 0.0500 (x)(x) 3.5 × 10 −4 = ———— (0.0500) x 2 = 1.8 × 10 -5 x = 4.2 × 10 -3 pH = −log(4.2 × 10 -3 ) = 2.38

12. Determine the pH of a 2.5 × 10 -2 M solution of benzoic acid (HC 7 H 5 O 2 ) if the K a of the acid is 6.5 × 10 -5 HC 7 H 5 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 7 H 5 O 2 − (aq) [H 3 O + ][C 7 H 5 O 2 − ] K a = ——————— [HC 7 H 5 O 2 ]

13. (x)(x) 6.5 × 10 −5 = ——————— ((2.5 × 10 -2 ) − x) Since 2.5 × 10 -2 >> 6.5 × 10 −5, (2.5 × 10 -2 ) − x ≈ 2.5 × 10 -2 (x)(x) 6.5 × 10 −5 = ————— (2.5 × 10 -2 ) x 2 = 1.6 × 10 -6 x = 1.3 × 10 -3 pH = −log(1.3 × 10 -3 ) = 2.89

14. Determine the pH of a 0.051 M solution of ammonia if the K b is 1.8 × 10 -5 NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH − (aq) [NH 4 + ][OH − ] K b = —————— [NH 3 ]

15. (x)(x) 1.8 × 10 −5 = ———— 0.051 − x Since 0.051 >> 1.8 × 10 −5, 0.051 − x ≈ 0.051 (x)(x) 1.8 × 10 −5 = ——— 0.051 x 2 = 9.2 × 10 -7 x = 9.6 × 10 -4 pOH = −log(9.6 × 10 -4 ) = 3.02 pH = 14 – 3.02 = 10.98

16. Determine the pH of a 0.156 M solution of NaF. (K a of HF = 6.8 × 10 -4 ) F − (aq) + H 2 O (l) HF (aq) + OH − (aq) [HF][OH − ] K b = ————— [F − ]

17. 1 × 10 −14 K b = ————— = 1.5 × 10 −11 6.8 × 10 −4 (x)(x) 1.5 × 10 −11 = ———— 0.156 − x Since 0.156 >> 1.5 × 10 −11, 0.156 − x ≈ 0.156 (x)(x) 1.5 × 10 −11 = ——— 0.156 x 2 = 2.3 × 10 -12 x = 1.5 × 10 -6 pOH = −log(1.5 × 10 -6 ) = 5.82 pH = 14 – 5.82 = 8.18

18. Determine the pH of a 0.30 M solution of sodium hydrogen selenite (NaHSeO 3 ). (H 2 SeO 3 : K a 1 = 1.7 × 10 −2, K a 2 = 6.4 × 10 −8 ) There are two possible reactions: 1) HSeO 3 − (aq) + H 2 O (l) SeO 3 2− (aq) + H 3 O + (aq) or 2) HSeO 3 − (aq) + H 2 O (l) H 2 SeO 3(aq) + OH − (aq) 1) K a 2 = 6.4 × 10 −8 2) 1 × 10 −14 K b = ────── = 5.9 × 10 -13 1.7 × 10 −2

19. HSeO 3 − (aq) + H 2 O (l) SeO 3 − (aq) + H 3 O + (aq) [SeO 3 − ][H 3 O + ] K a = —————— [HSeO 3 − ] (x)(x) 6.4 × 10 -8 = ———— (0.30 - x) Since 0.30 >> 6.4 × 10 −8, 0.30 − x ≈ 0.30

20. x 2 = 1.9 × 10 −8 x = 1.4 × 10 −4 pH = -log(1.4 × 10 −4 ) = 3.86

21. Determine the pH of a solution that is 0.10 M ammonia and 0.25 M ammonium nitrate. For NH 3 K b = 1.8 × 10 −5 1 × 10 −14 For NH 4 + K a = ────── = 5.6 × 10 −10 1.8 × 10 −5 The pH of the solutions separately would be 11.13 and 4.93 respectively Since 1.8 × 10 −5 >> 5.6 × 10 −10 the solution will be basic

22. The reaction will be NH 3 + H 2 O → NH 4 + + OH − [NH 4 + ][OH − ] K b = ──────── [NH 3 ] NH 3 +H2OH2O→NH 4 + +OH − Initial0.10-0.250 Change-x-x-+x+x+x+x Equilibrium0.10 – x-0.25 + xx

23. (0.25 + x)(x) 1.8 × 10 −5 = ──────── (0.10 – x) Since 0.10 and 0.25 >> 1.8 × 10 −5, we can say 0.10 – x ≈ 0.10 0.25 + x ≈ 0.25 (0.25)(x) 1.8 × 10 −5 = ───── 0.10

24. x = 7.2 × 10 −6 pOH = –log(7.2 × 10 −6 ) = 5.14 pH = 14 – 5.14 = 8.86

25. Determine the pH of a 0.200 M solution of ammonia. NH 3 + H 2 O → NH 4 + + OH − [NH 4 + ][OH − ] K b = ──────── [NH 3 ] NH 3 +H2OH2O→NH 4 + +OH − Initial0.200-00 Change-x-x-+x+x+x+x Equilibrium0.200 – x-xx

26. (x)(x) 1.8 × 10 −5 = ─────── (0.200 – x) Since 0.20 >> 1.8 × 10 −5, we can say 0.200 – x ≈ 0.200 x 2 1.8 × 10 −5 = ──── 0.200

27. x 2 = 3.6 × 10 −6 x = 1.9 × 10 −3 pOH = –log(1.9 × 10 −3 ) = 2.72 pH = 14 – 2.72 = 11.28

28. Now determine the pH of 75 mL of this solution when 10. mL of 0.10 M HCl is added. First we must dilute the ammonia because we are adding 10. mL of solution M 1 V 1 = M 2 V 2 (0.200 M)(75 mL) = M 2 (85 mL) M 2 = 0.18 M

29. Now we have to do the same thing for the HCl M 1 V 1 = M 2 V 2 (0.10 M)(10 mL) = M 2 (85 mL) M 2 = 0.012 M Strong acids (or bases) react COMPLETELY!!

30. NH 3 +H2OH2O→NH 4 + +OH − Initial0.200–00 Dilution0.18––– Change ( HCl)–0.012–+0.012– After Reaction0.168–0.012– Change–x–x–+x+x+x+x Equilibrium0.168 – x–0.012 + xx

31. (0.012 + x)(x) 1.8 × 10 −5 = ───────── (0.168 – x) Since 0.012 and 0.168 >> 1.8 × 10 −5, we can say 0.012 – x ≈ 0.012 0.168 – x ≈ 0.168 (0.012)x 1.8 × 10 −5 = ───── 0.168

32. x = 2.5 × 10 −4 pOH = –log(2.5 × 10 −4 ) = 3.60 pH = 14 – 3.60 = 10.40

33. Determine the pH of a buffer solution made to be 0.19 M acetic acid and 0.11 M sodium acetate. (K a = 1.8 × 10 −5 ) 1 × 10 −14 K b = ────── = 5.6 × 10 −10 1.8 × 10 −5 HC 2 H 3 O 2 + H 2 O → C 2 H 3 O 2 − + H 3 O + [C 2 H 3 O 2 − ][H 3 O + ] K b = ────────── [HC 2 H 3 O 2 ]

34. (0.11 + x)(x) 1.8 × 10 −5 = ─────── 0.19 – x since 0.11 and 0.19 >> 1.8 × 10 −5 (0.11)(x) 1.8 × 10 −5 = ───── 0.19 HC 2 H 3 O 2 +H2OH2O→C2H3O2−C2H3O2− +H3O+H3O+ Initial0.19-0.110 Change-x-x-+x+x+x+x Equilibrium0.19 – x-0.11 + xx

35. x = 3.1 × 10 −5 pH = –log(3.1 × 10 −5 ) = 4.51 Determine the pH when 20. mL of 0.20 M NaOH is added to 125 mL of the buffer.

36. HC 2 H 3 O 2 + H 2 O → C 2 H 3 O 2 − + H 3 O + Initial 0.19 –0.110 Dilution 0.164–0.095– Change (OH − ) –0.028–+0.028– After Reaction 0.136–0.123– Change –x–+x+x Equilibrium 0.136 – x –0.123 + xx

37. (0.123 + x)(x) 1.8 × 10 −5 = ──────── 0.136 – x since 0.123 and 0.136 >> 1.8 × 10 −5 (0.123)(x) 1.8 × 10 −5 = ────── 0.136 x = 2.0 × 10 −5 pH = –log(2.0 × 10 −5 ) = 4.70

38. Titration Curves If a strong acid is titrated with a strong base and the pH of the mixture is plotted against the mL of base added, what will the curve look like?

39. Titration Curves

40. Strong acid titrated with a strong base

41. Strong base titrated with a strong acid

42. What will the titration curve look like if a weak base is titrated with a strong acid?

43. Weak base titrated with a strong acid

44. The pH falls rapidly at first, but after a buffer forms the decline in pH moderates. Note that the equivalence point is in the acidic region What will the curve look like when a weak acid is titrated with a strong base?

45. Titration of a weak acid with a strong base

46. Once again, initial rapid decline, then it slows considerably. Note that the equivalence point is in the basic region. Lastly, what will it look like if we titrate a weak acid with a weak base?

48. The equivalence will be near 7, but will only actually be 7 if the K a and K b for the acid and base are equal. If K a > K b the equivalence point will be below 7. If K b > K a the equivalence point will be below 7. What will the curve look like when a weak diprotic acid is titrated with a strong base?

49. Titration of a weak diprotic acid with a strong base

50. Na 2 CO 3 titrated with HCl

51. Indicators

52. pK a & pK b pK a = –log(K a ) pK b = –log(K b ) When a buffer is created with equal concentrations of an acid and its conjugate base, the pH of the buffer is equal to the pK a of the acid (or pOH = pK b ).

53. When a weak acid is titrated with a strong base, the pH half way to the equivalence point will be the pK a of the acid. When a weak base is titrated with a strong acid, the pH half way to the equivalence point will be the pK b of the base.

54. Determine the pH of a 0.10 M solution of cinnamic acid (HC 9 H 7 O 2 ). (K a = 3.6 × 10 −5 ) pH = 2.72 30.0 mL of this solution is now titrated with 0.060 M NaOH. Determine the volume of NaOH need to reach the equivalence point.

55. M A V A E A = M B V B E B (0.10 M)(30.0 mL)(1) = (0.060 M)V B (1) V B = 50. mL Determine the pH after the addition of a) 10.0 mL

56. HC 9 H 7 O 2 + H 2 O → C 9 H 7 O 2 – + H 3 O + [C 9 H 7 O 2 – ][H 3 O + ] K a = ─────────── [HC 9 H 7 O 2 ] HC 9 H 7 O 2 + H 2 O → C 9 H 7 O 2 – + H 3 O + Initial 0.10 – 0 0 Dilution 0.075 – 0 0 Add NaOH –0.015 – +0.015 0 Total 0.060 – 0.015 0 Change –x – +x +x Equilibrium 0.060 – x – 0.015 + x x

57. (0.015 + x)(x) 3.6 × 10 −5 = ──────── (0.060 – x) Since 0.015 & 0.060 >> 3.6 × 10 −5 0.015 + x ≈ 0.015 0.060 – x ≈ 0.060 (0.015)(x) 3.6 × 10 −5 = ────── (0.060) x = 1.4 × 10 −4 pH = 3.84

58. b) 25.0 mL HC 9 H 7 O 2 + H 2 O → C 9 H 7 O 2 – + H 3 O + Initial 0.10 – 0 0 Dilution 0.0545 – 0 0 Add NaOH –0.0273 – +0.0273 0 Total 0.0272 – 0.0273 0 Change –x – +x +x Equilibrium 0.0272 – x – 0.0273 + x x

59. (0.0273 + x)(x) 3.6 × 10 −5 = ───────── (0.0272 – x) (0.0273)(x) 3.6 × 10 −5 = ─────── (0.0272) x = 3.6 × 10 −5 pH = 4.44 NOTE: This is pK a for the acid!

60. c) 50.0 mL C 9 H 7 O 2 – + H 2 O → HC 9 H 7 O 2 + OH – Initial 0.10 – 0 0 Dilution 0.0375 – 0 0 Change –x – +x +x Equilibrium 0.0375 – x – x x 1 × 10 −14 K b = ────── = 2.8 × 10 −10 3.6 × 10 −5

61. (x)(x) 2.8 × 10 −10 = ──────── (0.0375 – x) x 2 2.8 × 10 −10 = ───── 0.0375 x = 1.1 × 10 −11 pOH = 5.49 pH = 8.51

62. d) 60.0 mL This is 10 mL beyond the equivalence point. Dilute the 10 mL of NaOH (10 mL)(0.060 M) = (90 mL)M 2 M 2 = 6.7 × 10 −3 pOH = 2.18 pH = 11.82

63. The Henderson−Hasselbalch Equation The Henderson−Hasselbalch equation is a useful equation for handling buffer problems.