1. Titrations of acids and bases

2. HA + H 2 O H 3 O + + A -

3. Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB +

4. Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB + H 3 O + + OH - 2 H 2 O

5. At equivalence point [H 3 O + ] = [OH - ] For strong acid – strong base

6. H 3 O + + OH - 2 H 2 O At equivalence point [H 3 O + ] = [OH - ] K w = [ H 3 O + ] = [OH - ] = 10 -14

7. H 3 O + + OH - 2 H 2 O At equivalence point [H 3 O + ] = [OH - ] K w = [ H 3 O + ] = [OH - ] = 10 -14 pH = 7

8. Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB + H 3 O + + OH - 2 H 2 O

9. Volume of base added Start with acid solution pH 7

11. 4 10

14. 100 ml 0.1 M HCl

15. Titrate with 0.1 M NaOH

16. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O

17. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O Every part of a mole of NaOH added reduces the moles of H 3 O + by an equal amount.

18. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O As NaOH solution is added, the overall volume increases. This further decreases [H 3 O + ].

19. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Neutralization and dilution; both increase the pH of the solution.

20. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution.

21. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. Start with 0.1 L x 0.1 M H 3 O +

22. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. Start with 0.1 L x 0.1 M H 3 O + 0.01 moles H 3 O +

23. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + Add 0.05 L x 0.1 M OH -

24. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + Add 0.05 L x 0.1 M OH - 0.005 moles OH -

25. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - H 3 O + is neutralized by OH - 1:1 ratio

26. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - Remaining H 3 O + = 0.01 moles – 0.005 moles

27. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - Remaining H 3 O + = 0.005 moles Total volume = 150 ml = 0.15 L

28. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - [H 3 O + ]= 0.005 moles 0.15 L

29. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - [H 3 O + ]= 0.005 moles 0.15 L = 0.033 M

30. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. 0.01 moles H 3 O + 0.005 moles OH - [H 3 O + ]= 0.005 moles 0.15 L = 0.033 M pH = 1.48

32. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution.

33. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O +

34. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH -

35. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH - Moles H 3 O + = 0.01 – 0.00999 = 1 x 10 -5

36. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH - Moles H 3 O + = 0.01 – 0.00999 = 1 x 10 -5 Total volume = 0.1 L + 0.0999 L = 0.1999 L

37. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH - [H 3 O + ]= 1 x 10 -5 moles 0.1999 L

38. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH - [H 3 O + ]= 1 x 10 -5 moles 0.1999 L = 5 x 10 -5

39. 100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution. 0.01 moles H 3 O + 0.0999 L x 0.1 M = 0.00999 moles OH - [H 3 O + ]= 1 x 10 -5 moles 0.1999 L = 5 x 10 -5 pH = 4.3

40. 4.3

41. + 0.1 ml will give pH = 7

42. If more 0.1 M NaOH solution is added after the equivalence point, there is no H 3 O + to neutralize it.

43. At pH = 7 the volume is 0.2 L

44. Add 0.1 mL 0.1 M NaOH

45. At pH = 7 the volume is 0.2 L Add 0.1 mL 0.1 M NaOH Add 0.0001 L x 0.1 M = 10 -5 moles OH - [OH - ] = 1 x 10 -5 moles 0.201 L = 5 x 10 -5 M

46. At pH = 7 the volume is 0.2 L Add 0.1 mL 0.1 M NaOH Add 0.0001 L x 0.1 M = 10 -5 moles OH - [OH - ] = 1 x 10 -5 moles 0.2001 L = 5 x 10 -5 M pOH = 4.3 pH = 9.7

47. 4.3 pH = 7 V = 199.9 ml V = 200 ml pH = 9.7 V = 200.1 ml

48. Titrating a weak acid with a strong base

49. HA + H 2 O H 3 O + + A - K a < 1

50. Titrating a weak acid with a strong base HA + H 2 O H 3 O + + A - K a < 1 Much less than 100% dissociation.

51. Titrating a weak acid with a strong base HA + H 2 O H 3 O + + A - K a < 1 Much less than 100% dissociation. Every OH - added neutralizes an H 3 O + and shifts the equilibrium to the right.

52. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH

53. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5

54. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 0 ml NaOH

55. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 0 ml NaOH pH = 2.87

57. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = 0.005 moles NaOH

58. CH 3 COOH + H 2 O H 3 O + + CH 3 COO - + OH - 2 H 2 O

59. CH 3 COOH + H 2 O H 3 O + + CH 3 COO - + OH - 2 H 2 O Net result: for every OH - added, there is one less CH 3 COOH and one more CH 3 COO -

60. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = 0.005 moles NaOH 0.01 - 0.005 moles CH 3 COOH = 0.005 moles

61. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = 0.005 moles NaOH 0.01 - 0.005 moles CH 3 COOH = 0.005 moles 0.005 moles CH 3 COO -

62. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH 0.01 - 0.005 moles CH 3 COOH = 0.005 moles 0.005 moles CH 3 COO - 0.15 L solution

63. [CH 3 COOH] = 0.005 moles 0.005 moles CH 3 COO - 0.15 L = 0.033 M

64. [CH 3 COOH] = 0.005 moles 0.005 moles CH 3 COO - 0.15 L = 0.033 M K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] Buffer solution: pK a = pH = 4.74

66. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 100 ml NaOH 0.01 moles CH 3 COO - 0.20 L solution

67. Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 100 ml NaOH 0.01 moles CH 3 COO - 0.20 L solution [CH 3 COO - ] = 0.05 M

68. Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [CH 3 COO - ] = 0.05 M K b = KwKw KaKa = 10 -14 1.8 x 10 -5 = 5.6 x 10 -10 K b = [CH 3 COOH][OH - ] [CH 3 COO - ]

69. Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [CH 3 COO - ] = 0.05 M K b = KwKw KaKa = 10 -14 1.8 x 10 -5 = 5.6 x 10 -10 K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y2 0.05

70. Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = 10 -14 1.8 x 10 -5 = 5.6 x 10 -10 K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y2 0.05 y 2 = (5.6 x 10 -10 )(0.05) = 2.8 x 10 -11

71. Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = 10 -14 1.8 x 10 -5 = 5.6 x 10 -10 K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y2 0.05 y 2 = (5.6 x 10 -10 )(0.05) = 2.8 x 10 -11 y = 5.3 x 10 -6

72. Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x 10 -5 = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = 10 -14 1.8 x 10 -5 = 5.6 x 10 -10 K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y2 0.05 y 2 = (5.6 x 10 -10 )(0.05) = 2.8 x 10 -11 y = 5.3 x 10 -6 pOH =5.3; pH = 8.7

73. After the equivalence point, OH - is being added to a saturated buffer system.

74. After the equivalence point, OH - is being added to a saturated buffer system. pH increases rapidly