1. CHE 116 Prof. T. L. Heise 1 CHE 116: General Chemistry u CHAPTER SIXTEEN Copyright © Tyna L. Heise 2001-2002 All Rights Reserved

2. CHE 116 Prof. T. L. Heise 2 Acids and Bases: Review Properties of Acids 3sour taste 3change with litmus Properties of Bases 3 3bitter taste 3 3change with litmus

3. CHE 116 Prof. T. L. Heise 3 Acids and Bases: Review 1830 - scientists have recognized that all acids contain hydrogen, but not all hydrogen bearing compounds are acids Svante Arrhenius - linked acid behavior with the presence of an H + and base behavior with the presence of an OH -

4. CHE 116 Prof. T. L. Heise 4 Bronsted-Lowry Acids and Bases Arrhenius’ definition is useful, but restricts acid base reactions to aqueous conditions Johannes Bronsted and Thomas Lowry proposed a more general definition which involves the transfer of an H + ion from one molecule to another

5. CHE 116 Prof. T. L. Heise 5 Bronsted-Lowry Acids and Bases H + ion is simply a proton with no surrounding valence electron. 3Small particle interacts strongly with the nonbonding pairs of water molecules to form hydrated hydrogen ions 3chemists use H + and H 3 O + interchangeably

6. CHE 116 Prof. T. L. Heise 6 Bronsted-Lowry Acids and Bases Fig 16.1 Demonstrates the interconnections possible between hydrogenated water

7. CHE 116 Prof. T. L. Heise 7 Bronsted-Lowry Acids and Bases Definitions: Acid - any compound which transfers an H+ to another molecule Base - any compound which accepts a transfer of an H+ from another molecule * an acid and base always work together

8. CHE 116 Prof. T. L. Heise 8 Bronsted-Lowry Acids and Bases Definitions: Amphoteric - some substances can be an acid or a base depending on reaction Conjugate Acid Base pairs - two compounds that differ only in the presence of an H +. The molecule with the extra H is the acid.

9. CHE 116 Prof. T. L. Heise 9 Bronsted-Lowry Acids and Bases Fig 16.7, 16.8

10. CHE 116 Prof. T. L. Heise 10 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO 3 - F - PO 4 3- CO

11. CHE 116 Prof. T. L. Heise 11 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO 3 - F - PO 4 3- CO Given a base, bases accept H +, so add an H + to each molecule.

12. CHE 116 Prof. T. L. Heise 12 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO 3 HSO 3 - F - PO 4 3- CO Given a base, bases accept H +, so add an H + to each molecule.

13. CHE 116 Prof. T. L. Heise 13 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO 3 HSO 3 - HF F - PO 4 3- CO Given a base, bases accept H +, so add an H + to each molecule.

14. CHE 116 Prof. T. L. Heise 14 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO 3 HSO 3 - HF F - HPO 4 2- PO 4 3- CO Given a base, bases accept H +, so add an H + to each molecule.

15. CHE 116 Prof. T. L. Heise 15 Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO 3 HSO 3 - HF F - HPO 4 2- PO 4 3- HCO + CO Given a base, bases accept H +, so add an H + to each molecule.

16. CHE 116 Prof. T. L. Heise 16 Bronsted-Lowry Acids and Bases Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.

17. CHE 116 Prof. T. L. Heise 17 Bronsted-Lowry Acids and Bases Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. O 2- + H 2 O  OH - + OH -

18. CHE 116 Prof. T. L. Heise 18 Bronsted-Lowry Acids and Bases Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. O 2- + H 2 O  OH - + OH -

19. CHE 116 Prof. T. L. Heise 19 Bronsted-Lowry Acids and Bases Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. O 2- + H 2 O  OH - + OH - acid base acid base baseacid baseacid

20. CHE 116 Prof. T. L. Heise 20 Bronsted-Lowry Acids and Bases Relative Strengths of Acids and Bases: 3the more readily a substance donates an H +, the less readily it’s conjugate base will accept one 3the more readily a substance accepts an H +, the less readily it’s conjugate acid will donate one 3the stronger one of the substances is, the weaker it’s conjugate

21. CHE 116 Prof. T. L. Heise 21 Bronsted-Lowry Acids and Bases Relative Strengths of Acids and Bases: 3strong acids completely transfer their protons to water, leaving no undissociated molecules 3weak acids are those that only partly dissociate in aqueous solution and therefore exist in the solution as a mixture of acid molecules and component ions

22. CHE 116 Prof. T. L. Heise 22 Bronsted-Lowry Acids and Bases Relative Strengths of Acids and Bases: 3negligible acids are those that have hydrogen but do not donate them at all, their conjugate bases would be extremely strong, reacting with water to complete their octet and leaving OH - behind. 3In every acid base reaction, the position of the equilibrium favors transfer of H + from stronger side to weaker side

23. CHE 116 Prof. T. L. Heise 23 Bronsted-Lowry Acids and Bases Fig. 16.4

24. CHE 116 Prof. T. L. Heise 24 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3- (aq)+H 2 O(l)  HPO 4 2- (aq)+OH - (aq) b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l)

25. CHE 116 Prof. T. L. Heise 25 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3- (aq)+H 2 O(l)  HPO 4 2- (aq)+OH - (aq) 2 acids are: H 2 O and HPO 4 2- 2 bases are: PO 4 3- and OH -

26. CHE 116 Prof. T. L. Heise 26 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3- (aq)+H 2 O(l)  HPO 4 2- (aq)+OH - (aq) 2 acids are: H 2 O and HPO 4 2- 2 bases are: PO 4 3- and OH - red indicates strength

27. CHE 116 Prof. T. L. Heise 27 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3- (aq)+H 2 O(l)  HPO 4 2- (aq)+OH - (aq) 2 acids are: H 2 O and HPO 4 2- 2 bases are: PO 4 3- and OH - reverse reaction favored

28. CHE 116 Prof. T. L. Heise 28 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3- (aq)+H 2 O(l)  HPO 4 2- (aq)+OH - (aq) 2 acids are: H 2 O and HPO 4 2- 2 bases are: PO 4 3- and OH - shifts left

29. CHE 116 Prof. T. L. Heise 29 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l)

30. CHE 116 Prof. T. L. Heise 30 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH -

31. CHE 116 Prof. T. L. Heise 31 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH - red indicates strength

32. CHE 116 Prof. T. L. Heise 32 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH - favors forward reaction

33. CHE 116 Prof. T. L. Heise 33 Bronsted-Lowry Acids and Bases Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH - (aq)  NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH - shifts right

34. CHE 116 Prof. T. L. Heise 34 The Autoionization of Water One of the most important properties of water is its ability to ac as either a Bronsted acid or Bronsted base, depending on circumstances. 3One water molecule can donate a proton to another water molecule Fig 16.10

35. CHE 116 Prof. T. L. Heise 35 The Autoionization of Water The autoionization of water, although rapid and weak, does exist as an equilibrium, and therefore has an equilibrium constant expression: K eq = [H 3 O + ][OH - ] [H 2 O] 2 * because water is a liquid, it can be excluded from the equation...

36. CHE 116 Prof. T. L. Heise 36 The Autoionization of Water K eq [H 2 O] 2 = [H 3 O + ][OH - ] K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 * this equation is not only applicable to water, but to all aqueous solutions, and it is upon this fact that the pH scale was built.

37. CHE 116 Prof. T. L. Heise 37 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 b) [OH - ] = 3 x 10 -9 c) [OH - ] = 1 x 10 -7

38. CHE 116 Prof. T. L. Heise 38 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 then [OH - ] must equal 1.0 x 10 -14 which is 2 x10 -5 [OH - ] = 5.0 x 10 -10

39. CHE 116 Prof. T. L. Heise 39 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 then [OH - ] must equal 1.0 x 10 -14 which is 2 x10 -5 [OH - ] = 5.0 x 10 -10 [H + ] > [OH - ] so acidic

40. CHE 116 Prof. T. L. Heise 40 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH - ] = 3 x 10 -9 then [H + ] must equal 1.0 x 10 -14 which is 3 x10 -9 [H + ] = 3.3 x 10 -6

41. CHE 116 Prof. T. L. Heise 41 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH - ] = 3 x 10 -9 then [H + ] must equal 1.0 x 10 -14 which is 3 x10 -9 [H + ] = 3.3 x 10 -6 [H + ] > [OH - ] so acidic

42. CHE 116 Prof. T. L. Heise 42 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH - ] = 1 x 10 -7 then [H + ] must equal 1.0 x 10 -14 which is 1 x10 -7 [H + ] = 1.0 x 10 -7

43. CHE 116 Prof. T. L. Heise 43 The Autoionization of Water Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH - ] = 1 x 10 -7 then [H + ] must equal 1.0 x 10 -14 which is 1 x10 -7 [H + ] = 1.0 x 10 -7 [H + ] = [OH - ] so neutral

44. CHE 116 Prof. T. L. Heise 44 The pH Scale For convenience, we can use a logarithmic version of concentration to turn the very small concentrations of [H + ] and [OH - ] into whole numbers. p(anything) = - log[anything] p(H) = -log[H + ] p(OH) = - log[OH - ] ** pH + pOH = 14

45. CHE 116 Prof. T. L. Heise 45 The pH Scale

46. CHE 116 Prof. T. L. Heise 46 The pH Scale Common household products and their relative pH’s.

47. CHE 116 Prof. T. L. Heise 47 The pH Scale Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH?

48. CHE 116 Prof. T. L. Heise 48 The pH Scale Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH? pH = -log[H + ]

49. CHE 116 Prof. T. L. Heise 49 The pH Scale Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH? pH = -log[H + ] pH = -log[3.8 x 10 -4 ]

50. CHE 116 Prof. T. L. Heise 50 The pH Scale Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH? pH = -log[H + ] pH = -log[3.8 x 10 -4 ] pH = 3.42

51. CHE 116 Prof. T. L. Heise 51 The pH Scale Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH?

52. CHE 116 Prof. T. L. Heise 52 The pH Scale Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH? pH = -log[H + ]

53. CHE 116 Prof. T. L. Heise 53 The pH Scale Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH? pH = -log[H + ] pH = -log[5.3 x 10 -9 ]

54. CHE 116 Prof. T. L. Heise 54 The pH Scale Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH? pH = -log[H + ] pH = -log[5.3 x 10 -9 ] pH = 8.28

55. CHE 116 Prof. T. L. Heise 55 The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].

56. CHE 116 Prof. T. L. Heise 56 The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. pH = -log[H + ]

57. CHE 116 Prof. T. L. Heise 57 The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. pH = -log[H + ] 9.18 = -log[H + ]

58. CHE 116 Prof. T. L. Heise 58 The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. pH = -log[H + ] 9.18 = -log[H + ] 10 -9.18 = [H + ]

59. CHE 116 Prof. T. L. Heise 59 The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. pH = -log[H + ] 9.18 = -log[H + ] 10 -9.18 = [H + ] 6.61 x 10 -10 = [H + ]

60. CHE 116 Prof. T. L. Heise 60 The pH Scale Measuring pH: a pH can be measured quickly and accurately using a pH meter. 3A pair of electrodes connected to a meter capable of measuring small voltages 3a voltage which varies with pH is generated when the electrodes are placed in a solution 3calibrated to give pH

61. CHE 116 Prof. T. L. Heise 61 The pH Scale Measuring pH: a pH can be measured quickly and accurately using a pH meter. 3Electrodes come in a variety of shapes and sizes 3a set of electrodes exists that can be placed inside a human cell 3acid base indicators can be used, but are much less precise

62. CHE 116 Prof. T. L. Heise 62 The pH Scale Fig 16.7

63. CHE 116 Prof. T. L. Heise 63 Strong Acids and Bases Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Strong Acids HCl HBr HImonoprotic HNO 3 HClO 3 HClO 4 H 2 SO 4 diprotic

64. CHE 116 Prof. T. L. Heise 64 Strong Acids and Bases Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Calculating the pH of a solution made up entirely of ions means the [H+] is proportional to [acid]

65. CHE 116 Prof. T. L. Heise 65 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?

66. CHE 116 Prof. T. L. Heise 66 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H + ]

67. CHE 116 Prof. T. L. Heise 67 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H + ] 2.66 = -log[H + ]

68. CHE 116 Prof. T. L. Heise 68 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66 = [H + ] 2.2 x 10 -3 = [H + ]

69. CHE 116 Prof. T. L. Heise 69 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66 = [H + ] 2.2 x 10 -3 = [H + ] 2.2 x 10 -3 M H + 1 mole HNO 3 1 mole H +

70. CHE 116 Prof. T. L. Heise 70 Strong Acids and Bases An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66 = [H + ] 2.2 x 10 -3 = [H + ] 2.2 x 10 -3 M H + 1 mole HNO 3 1 mole H + 2.2 x 10 -3 HNO 3

71. CHE 116 Prof. T. L. Heise 71 Strong Acids and Bases The most soluble common bases are the ionic hydroxides of the alkali and alkaline earth metals. Due to the complete dissociation of the base into its ion components makes the pH calculation equally straightforward

72. CHE 116 Prof. T. L. Heise 72 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?

73. CHE 116 Prof. T. L. Heise 73 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ]

74. CHE 116 Prof. T. L. Heise 74 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ] 11.89 = -log[H + ]

75. CHE 116 Prof. T. L. Heise 75 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89 = [H + ]

76. CHE 116 Prof. T. L. Heise 76 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89 = [H + ] 1.29 x 10 -12 = [H + ]

77. CHE 116 Prof. T. L. Heise 77 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89 = [H + ] 1.29 x 10 -12 = [H + ] 1.0 x 10 -14 = [OH - ] 1.29 x 10 -12

78. CHE 116 Prof. T. L. Heise 78 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89 = [H + ] 1.29 x 10 -12 = [H + ] 1.0 x 10 -14 = [OH - ] =7.8 x 10 -3 1.29 x 10 -12

79. CHE 116 Prof. T. L. Heise 79 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? 1.0 x 10 -14 = [OH - ] =7.8 x 10 -3 1.29 x 10 -12 7.8 x 10 -3 M OH - 1 mol KOH 1 mol OH -

80. CHE 116 Prof. T. L. Heise 80 Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? 1.0 x 10 -14 = [OH - ] =7.8 x 10 -3 1.29 x 10 -12 7.8 x 10 -3 M OH - 1 mol KOH 1 mol OH - 7.8 x 10 -3 M KOH

81. CHE 116 Prof. T. L. Heise 81 Weak Acids Most acids are weak acids and only partially dissociate in aqueous solution. The extent to which a weak acid dissociates can be expressed using an equilibrium constant for the ionization reaction HX + H 2 O  H 3 O + + X - Ka = [H 3 O + ][X - ] [HX] *the larger the K a the stronger the acid

82. CHE 116 Prof. T. L. Heise 82 Weak Acids Calculating K a from pH: a much more complicated calculation is required for the determinations of weak acids, in many cases, due to the extremely small magnitude of the values, some simpler approximations can be made.

83. CHE 116 Prof. T. L. Heise 83 Weak Acids Sample exercise: Niacin, one of the B vitamins, has the following molecular structure: C O H O N A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? b) What is the acid-dissociation constant?

84. CHE 116 Prof. T. L. Heise 84 Weak Acids Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? pH = -log[H + ] 3.26 = -log[H + ] e -3.26 = [H + ] 5.5 x 10 -4 = [H + ]

85. CHE 116 Prof. T. L. Heise 85 Weak Acids Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? C 6 H 4 NOOH  C 6 H 4 NOO - + H + initial0.020 0 0 change -5.5 x 10 -4 +5.5 x 10 -4 +5.5 x 10 -4 equil 0.01945 5.5 x 10 -4 5.5 x 10 -4

86. CHE 116 Prof. T. L. Heise 86 Weak Acids Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? % = part/total x 100 = 5.5 x 10 -4 /0.01945 x 100 = 2.8%

87. CHE 116 Prof. T. L. Heise 87 Weak Acids Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 b) What is the acid-dissociation constant? C 6 H 4 NOOH  C 6 H 4 NOO - + H + K a = [C 6 H 4 NOO - ][H + ] [C 6 H 4 NOOH] = (5.5 x 10 -4 )(5.5 x 10 -4 ) 0.01945 = 1.55 x10 -5

88. CHE 116 Prof. T. L. Heise 88 Weak Acids Using K a to calculate pH: Using the value of K a and knowing the initial concentration of the weak acid, we can calculate the concentration of H + (aq). Example: Calculate the pH of a 0.30 M solution of acetic acid. HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq)

89. CHE 116 Prof. T. L. Heise 89 Weak Acids Example: Calculate the pH of a 0.30 M solution of acetic acid. HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq) From Table 16.2, K a = 1.8 x 10 -5 Ka = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] set up data table of concentrations involved...

90. CHE 116 Prof. T. L. Heise 90 Weak Acids Ka = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] set up data table of concentrations involved… [HC 2 H 3 O 2 ] [H + ] [C 2 H 3 O 2 - ] Initial 0.30 M 0 0 Change -x +x +x Equilibrium 0.30 - x x x

91. CHE 116 Prof. T. L. Heise 91 Weak Acids Input concentrations in formula Ka = 1.8 x 10 -5 = [x][x] [0.30 -x] This will lead to a quadratic equation, but we can simplify the problem a bit. All weak acids dissociate so little that we can assume the initial concentration of the acid remains essentially the same, that means we can rewrite the formula to read...

92. CHE 116 Prof. T. L. Heise 92 Weak Acids Ka = 1.8 x 10 -5 = [x][x] [0.30] 1.8 x 10 -5 (0.30) = x 2 1.8 x 10 -5 (0.30) = x 0.0023 = x 0.0023 = [H + ]pH = -log [H + ] = -log[0.0023] = 2.64

93. CHE 116 Prof. T. L. Heise 93 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin?

94. CHE 116 Prof. T. L. Heise 94 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin? C 6 H 4 NOOH  C 6 H 4 NOO - + H + K a = 1.5 x10 -5 = [C 6 H 4 NOO - ][H + ] [C 6 H 4 NOOH]

95. CHE 116 Prof. T. L. Heise 95 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin? C 6 H 4 NOOH C 6 H 4 NOO - H + initial 0.010 0 0 change -x +x +x equilibrium 0.010 -x x x

96. CHE 116 Prof. T. L. Heise 96 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin? C 6 H 4 NOOH  C 6 H 4 NOO - + H + K a = 1.5 x10 -5 = [x][x] [0.010]

97. CHE 116 Prof. T. L. Heise 97 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin? 1.5 x10 -5 [0.010] = x 2 1.5 x10 -5 [0.010] = x 3.87 x 10 -4 = x

98. CHE 116 Prof. T. L. Heise 98 Weak Acids Sample Exercise: The K a for niacin is 1.5 x 10 -5. What is the pH of a 0.010 M solution of niacin? 3.87 x 10 -4 = x 3.87 x 10 -4 = [H + ] pH = -log[H + ] pH = -log[3.87 x 10 -4 ] pH = 3.41

99. CHE 116 Prof. T. L. Heise 99 Weak Acids The results seen in the two previous examples are typical of weak acids. The lower number of ions produced during the partial dissociation causes less electrical conductivity and a slower reaction rate with metals. Percent ionization is a good way to discover the actual conductivity, however...

100. CHE 116 Prof. T. L. Heise 100 Weak Acids As the concentration of a weak acid increases, the % ionized decreases.

101. CHE 116 Prof. T. L. Heise 101 Weak Acids As the concentration of a weak acid increases, the % ionized decreases.

102. CHE 116 Prof. T. L. Heise 102 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise b) a 1.0 x 10-3 M solution

103. CHE 116 Prof. T. L. Heise 103 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise Our approximation was good so % = part x 100 total = 3.87 x 10 -4 x 100 = 3.9% 0.010

104. CHE 116 Prof. T. L. Heise 104 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution K a = 1.5 x10 -5 = [x][x] [1.0 x 10 -3 ] 1.5 x10 -5 (1.0 x 10 -3 ) = x 2 1.2 x 10 -4

105. CHE 116 Prof. T. L. Heise 105 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution K a = 1.5 x10 -5 = [x][x] [1.0 x 10 -3 ]but, 1.2 x 10 -3 x 100 1.5 x10 -5 (1.0 x 10 -3 ) = x 2 1.0 x 10 -3 1.2 x 10 -3 is greater than 5% so use quadratic...

106. CHE 116 Prof. T. L. Heise 106 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution K a = 1.5 x10 -5 = [x][x] [1.0 x 10 -3 - x] -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0

107. CHE 116 Prof. T. L. Heise 107 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0 x = -b ± b 2 - 4ac 2a x =

108. CHE 116 Prof. T. L. Heise 108 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0 x = -b ± b 2 - 4ac 2a x = 1.1 x 10 -4 or -1.3 x 10 -4

109. CHE 116 Prof. T. L. Heise 109 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10 -4 1.1 x 10 -4 x 100 1.0 x 10 -3

110. CHE 116 Prof. T. L. Heise 110 Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10 -4 1.1 x 10 -4 x 100= 11% 1.0 x 10 -3

111. CHE 116 Prof. T. L. Heise 111 Weak Acids Polyprotic Acids: many acids have more than one ionizable H atom. H 2 SO 3 (aq)  H + (aq) + HSO 3 - (aq) K a1 HSO 3 - (aq)  H + (aq) + SO 3 -2 (aq) K a2 The Ka are labeled according to which proton is dissociating. -it is always easier to remove the first proton than the second

112. CHE 116 Prof. T. L. Heise 112 Weak Acids

113. CHE 116 Prof. T. L. Heise 113 Weak Acids If K a values differ by 10 3 or more, only use K a1 to determine calculations.

114. CHE 116 Prof. T. L. Heise 114 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid.

115. CHE 116 Prof. T. L. Heise 115 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. H 2 C 2 O 4  HC 2 O 4 - + H + Ka = 5.9 x 10 -2 = [HC 2 O 4 - ][H + ] [H 2 C 2 O 4 ]

116. CHE 116 Prof. T. L. Heise 116 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. H 2 C 2 O 4  HC 2 O 4 - + H + strong acid so 100% dissociation 0.020 M H+

117. CHE 116 Prof. T. L. Heise 117 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. HC 2 O 4 -  C 2 O 4 -2 + H + I 0.020 0 0.020  -x +x +x E 0.020 - x x 0.020 + x

118. CHE 116 Prof. T. L. Heise 118 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. Ka = 6.4 x 10 -5 = [0.020 + x ][x] [0.020 - x] use your assumption

119. CHE 116 Prof. T. L. Heise 119 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. Ka = 6.4 x 10 -5 = [0.020 ][x] [0.020] 6.4 x 10 -5 = x = [C 2 O 4 2- ]

120. CHE 116 Prof. T. L. Heise 120 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. [H + ]= 0.020

121. CHE 116 Prof. T. L. Heise 121 Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2- ] in a 0.020 M solution of oxalic acid. pH = -log[H + ] pH = - log[0.020] pH = 1.70

122. CHE 116 Prof. T. L. Heise 122 Weak Bases Many substances behave as weak bases in water. Such substances react with water, removing protons from water, leaving the OH - ion behind. NH 3 + H 2 O  NH 4 + + OH - K b = [NH 4 + ][OH - ] [NH 3 ] * K b is the base dissociation constant utilizing the [OH - ]

123. CHE 116 Prof. T. L. Heise 123 Weak Bases * K b is the base dissociation constant utilizing the [OH - ] 3bases must contain one or more lone pair to bond with the H + from water. 3as before, the larger the K b the stronger the base 3stronger base have low pOH, but high pH

124. CHE 116 Prof. T. L. Heise 124 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine B) methylamine C) nitrous acid

125. CHE 116 Prof. T. L. Heise 125 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine K b = 1.7 x 10 -9 B) methylamine K b = 4.4 x 10 -4 C) nitrous acid K b = 2.2 x 10 -11

126. CHE 116 Prof. T. L. Heise 126 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine K b = 1.7 x 10 -9 K b = 1.7 x 10 -9 = [x][x] [0.05] x = [OH - ] = 9.2 x 10 -6 pOH = 5.0 so pH = 9.0

127. CHE 116 Prof. T. L. Heise 127 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? B) methylamine K b = 4.4 x 10 -4 K b = 4.4 x 10 -4 = [x][x] [0.05] x = [OH - ] = 4.6 x 10 -3 pOH = 2.32 so pH = 11.68

128. CHE 116 Prof. T. L. Heise 128 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? C) nitrous acid K b = 2.2 x 10 -11 K b = 2.2 x 10 -11 = [x][x] [0.05] x = [OH - ] = 1.0 x 10 -6 pOH = 5.97 so pH = 8.02

129. CHE 116 Prof. T. L. Heise 129 Weak Bases Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine pH = 9.0 B) methylamine pH = 11.68 C) nitrous acid pH = 8.02

130. CHE 116 Prof. T. L. Heise 130 Weak Bases Identifying a Weak Base 3Neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor. 3 Most of these are nitrogen atoms 3Anions of weak acids

131. CHE 116 Prof. T. L. Heise 131 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?

132. CHE 116 Prof. T. L. Heise 132 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution? pH = 10.50 so pOH = 3.50

133. CHE 116 Prof. T. L. Heise 133 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution? pOH = 3.50NH 3 + H 2 0  NH 4 + + OH - [OH - ] = 3.16 x 10 -4

134. CHE 116 Prof. T. L. Heise 134 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution? NH 3 + H 2 0  NH 4 + + OH - x 0 0 -3.16 x 10 -4 +3.16 x 10 -4 +3.16 x 10 -4 x - 3.16 x 10 -4 3.16 x 10 -4 3.16 x 10 -4

135. CHE 116 Prof. T. L. Heise 135 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution? NH 3 + H 2 0  NH 4 + + OH - K b = 1.8 x 10 -5 = [3.16 x 10 -4 ][3.16 x 10 -4 ] [x - 3.16 x 10 -4 ]

136. CHE 116 Prof. T. L. Heise 136 Weak Bases Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution? NH 3 + H 2 0  NH 4 + + OH - K b = 1.8 x 10 -5 = [3.16 x 10 -4 ][3.16 x 10 -4 ] [x - 3.16 x 10 -4 ] x = 0.0058 M

137. CHE 116 Prof. T. L. Heise 137 Relationship Between K a and K b When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is equal to the product of the equilibrium constants for the two added reactants. NH 4 + (aq)  NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) H 2 O(l)  H + (aq) + OH - (aq)

138. CHE 116 Prof. T. L. Heise 138 Relationship Between K a and K b NH 4 + (aq)  NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) H 2 O(l)  H + (aq) + OH - (aq) K a x K b = K w pK a + pK b = pK w = 14

139. CHE 116 Prof. T. L. Heise 139 Relationship Between K a and K b Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO 2 - B) PO 4 3- C) N 3 -

140. CHE 116 Prof. T. L. Heise 140 Relationship Between K a and K b Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO 2 - K a = 4.5 x 10 -4 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 2.2 x 10 -11 4.5 x 10 -4

141. CHE 116 Prof. T. L. Heise 141 Relationship Between K a and K b Sample exercise: Which of the following anions has the largest base-dissociation constant? B) PO 4 3- K a = 4.2 x 10 -13 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 2.4 x 10 -2 4.2 x 10 -13

142. CHE 116 Prof. T. L. Heise 142 Relationship Between K a and K b Sample exercise: Which of the following anions has the largest base-dissociation constant? C) N 3 - K a = 1.9 x 10 -5 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 5.2 x 10 -10 1.9 x 10 -5

143. CHE 116 Prof. T. L. Heise 143 Relationship Between K a and K b Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO 2 - K b = 2.2 x 10 -11 B) PO 4 3- K b = 2.4 x 10 -2 C) N 3 - K b = 5.2 x 10 -10

144. CHE 116 Prof. T. L. Heise 144 Relationship Between K a and K b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline?

145. CHE 116 Prof. T. L. Heise 145 Relationship Between K a and K b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline? pK a + pK b = 14 4.90 + x = 14 x = 9.1

146. CHE 116 Prof. T. L. Heise 146 Relationship Between K a and K b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline? pK b = 9.1 pK b = -log[K b ] 9.1 = -log[K b ] [K b ] = 7.9 x 10 -10

147. CHE 116 Prof. T. L. Heise 147 Acid Base Properties of Salt Soln’s Salt solutions have the potential to be acidic or basic. 3Hydrolysis of a salt 3acid base properties are due to the behavior of their cations and anions 3perform the necessary double replacement reaction and examine the products using your strength rules

148. CHE 116 Prof. T. L. Heise 148 Acid Base Properties of Salt Soln’s If a strong acid and strong base are produced, the resultant solution will be neutral. If a strong acid and weak base are produced, the resultant solution will be acidic. If a strong base and a weak acid are produced, the resultant solution will be basic.

149. CHE 116 Prof. T. L. Heise 149 Acid Base Properties of Salt Soln’s If a weak acid and weak base are produced, the resultant solution will be dependent on the K a values.

150. CHE 116 Prof. T. L. Heise 150 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO 3, Fe(NO 3 ) 3 NaNO 3 + H 2 O  NaOH + HNO 3 SB SA

151. CHE 116 Prof. T. L. Heise 151 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO 3, Fe(NO 3 ) 3 Fe(NO 3 ) 3 + 3H 2 O  Fe(OH) 3 + 3HNO 3 WB SA

152. CHE 116 Prof. T. L. Heise 152 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO 3, Fe(NO 3 ) 3 Fe(NO 3 ) 3 + 3H 2 O  Fe(OH) 3 + 3HNO 3 WB SA

153. CHE 116 Prof. T. L. Heise 153 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO

154. CHE 116 Prof. T. L. Heise 154 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBr + H 2 O  KOH + HBr SB SA

155. CHE 116 Prof. T. L. Heise 155 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBrO + H 2 O  KOH + HBrO SB WA

156. CHE 116 Prof. T. L. Heise 156 Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO

157. CHE 116 Prof. T. L. Heise 157 Acid-Base Behavior & Chem Structure How does the chemical structure determine which of the behaviors will be exhibited? Acidic 3 a molecule containing H will transfer a proton only if the H-X bond is polarized like H -- Xthe stronger the bond the weaker the acid and vice versa

158. CHE 116 Prof. T. L. Heise 158 Acid-Base Behavior & Chem Structure How does the chemical structure determine which of the behaviors will be exhibited? Acidic 3 oxyacids exist when the H is attached to an oxygen bonded to a central atom if the OH’s attached are equal in number to the O’s present, acid strength increases with electronegativity

159. CHE 116 Prof. T. L. Heise 159 Acid-Base Behavior & Chem Structure How does the chemical structure determine which of the behaviors will be exhibited? Acidic 3 oxyacids exist when the H is attached to an oxygen bonded to a central atom the more O’s present compared to the OH’s, the more polarized the OH bond becomes and the stronger the acid is

160. CHE 116 Prof. T. L. Heise 160 Acid-Base Behavior & Chem Structure How does the chemical structure determine which of the behaviors will be exhibited? Acidic 3 carboxylic acids exist when the functional group COOH is present the strength also increases as the number of electronegative atoms in the molecule increase