2. Acid and Base Equilibria

3. Electrolytes Strong Conduct electricity Weak Poor conductors of electricity Nonelectrolytes Do not conduct electricity

4. Strong Electrolytes Completely ionize or dissociate in dilute aqueous solutions. Strong acids Strong soluble bases Most soluble salts HCl + H 2 O  H 3 O + + Cl - KOH  K + + OH -

5. Concentration of Ions Calculated directly from molarity of strong electrolyte What is the concentration of hydrogen ion in a 0.25 M HCl solution? HCl  H+ H+ + Cl - initial 0.25 M change -0.25 M 0.25M final 0 M0.25M

6. Concentration of Ions Calculated directly from molarity of strong electrolyte What is the concentration of hydroxide ion in a 0.25 M Ba(OH) 2 solution? Ba(OH) 2  Ba 2+ + 2OH - initial 0.25 M change -0.25 M 0.25M 2(0.25M) final 0 M0.25M0.50M

7. Weak Electrolyte Slightly ionized in dilute aqueous solutions. Weak acids Weak bases A few covalent salts

8. Auto-Ionization of Water H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) Kw = [ H 3 O + ][OH - ]

9. Auto-Ionization of Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M H 2 O + H 2 O ¾ H 3 O + + OH -

10. K w = [H 3 O + ] [OH - ] Kw H 2 O + H 2 O ¾ H 3 O + + OH -

11. The Auto-Ionization of Water Calculate the concentrations of H 3 O + and OH - in 0.050 M HCl.

12. The Auto-Ionization of Water Calculate the concentrations of H 3 O + and OH - in 0.050 M HCl.

13. [H 3 O + ], [OH - ] and pH A common way to express acidity and basicity is with pH pH = - log [H 3 O + ] In a neutral solution, [H 3 O + ] = [OH - ] = 1.00 x 10 -7 at 25 o C pH = -log (1.00 x 10 -7 ) = - (-7) pH = 7.00

14. [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00 General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solutionpH < 7

15. If the pH of Coke is 3.12, What is [H 3 O + ]? Because pH = - log [H 3 O + ] then log [H 3 O + ] = - pH Take antilog and get [H 3 O + ] = 10 -pH [H 3 O + ] = 10 -3.12 [H 3 O + ] = 7.6 x 10 -4 M [H 3 O + ], [OH - ] and pH

16. Other pX Scales In generalpX = -log X and so pOH = - log [OH - ] Kw = [H 3 O + ] [OH - ] = 1.00 x 10 -14 Take the -log of both sides -log (10 -14 ) = - log [H 3 O + ] + (-log [OH - ]) 14 = pH + pOH

17. pKa Values The Ka of acetic acid is 1.8 x 10 -5, what is the pKa? pKa = -log Ka = - log (1.8 x 10 -5 ) = 4.74

18. pH [H + ] [OH - ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

19. The pH and pOH scales Develop familiarity with pH scale by looking at a series of solutions in which [H 3 O + ] varies between 1.0 M and 1.0 x 10 -14 M.

20. The pH and pOH scales Calculate [H 3 O + ], pH, [OH - ], and pOH for 0.020 M HNO 3 solution.

21. The pH and pOH scales Calculate [H 3 O + ], pH, [OH - ], and pOH for 0.020 M HNO 3 solution.

22. pH Examples

23. Ionization Constants for Weak Monoprotic Acids Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is:

24. Ionization Constants for Weak Monoprotic Acids and Bases Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant. HCN  H + + CN -

25. Equilibria Involving Weak Acids and Bases Consider acetic acid HOAc + H 2 O ¾ H 3 O + + OAc - Acid Conj. base

26. Equilibria Involving Weak Acids and Bases Consider acetic acid HOAc + H 2 O ¾ H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

27. Ka From Equilibrium Concentrations A solution of a weak acid has the following concentrations at equilibrium. What is the Ka? [HA] = 0.049 M; [H + ] = [A - ] = 0.00084 M HA + H 2 O ¾ H 3 O + + A - Ka = [H 3 O + ][A - ]/[HA] Ka = (0.00084) 2 /0.049 Ka = 1.4 x 10 -5

28. Ka From Percent Ionization In a 0.0100 M solution, acetic acid is 4.2% ionized. What is the Ka? HA + H 2 O ¾ H 3 O + + A - initial 0.0100 change -0.00042 0.00042 0.00042 equil 0.00958 0.00042 0.00042 Ka = [H 3 O + ][A - ]/[HA] Ka = (0.00042) 2 /0.00958 Ka = 1.8 x 10 -5

29. Ka From pH The pH of a 0.115 M weak acid is 1.92. What is the Ka? HA + H 2 O ¾ H 3 O + + A - [H 3 O + ] = 10 -1.92 = 0.012 M initial 0.115 change -0.012 0.012 0.012 equil 0.103 0.012 0.012 Ka = [H 3 O + ][A - ]/[HA] Ka = (0.012) 2 /0.103 Ka = 1.4 x 10 -3

30. Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. [HOAc][H 3 O + ][OAc - ] Initial Change Equilib

31. Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. [HOAc][H 3 O + ][OAc - ] Initial1.0000 Change-x+x+x Equilib1.00-xxx

32. Weak Acid Step 2. Write Ka expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

33. Weak Acid Step 2. Write Ka expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

34. Weak Acid Step 2. Write Ka expression This is a quadratic. Solve using quadratic formula or method of approximations You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

35. Weak Acid Step 3. Solve Ka expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

36. Weak Acid Step 3. Solve Ka expression First assume x is very small because Ka is so small. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

37. Weak Acid Step 3. Solve Ka expression First assume x is very small because Ka is so small. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

38. Weak Acid Step 3. Solve Ka expression First assume x is very small because Ka is so small. And so x = [H 3 O + ] = [OAc-] = [Ka 1.00] 1/2 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

39. Weak Acid Step 3. Solve Ka approximate expression x = [H 3 O + ] = [OAc-] = [Ka 1.00] 1/2 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

40. Weak Acid Step 3. Solve Ka approximate expression x = [H 3 O + ] = [OAc-] = [Ka 1.00] 1/2 x = [H 3 O + ] = [OAc-] = 4.2 x 10 -3 M You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

41. Weak Acid Step 3. Solve Ka approximate expression x = [H 3 O + ] = [OAc-] = [Ka 1.00] 1/2 x = [H 3 O + ] = [OAc-] = 4.2 x 10 -3 M pH = - log [H 3 O + ] = -log (4.2 x 10 -3 ) pH = 2.37 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

42. Percent Ionization Calculate the percent ionization of a 0.10 M solution of acetic acid. HA + H 2 O ¾ H 3 O + + A - initial 0.10 change -x xx equil 0.10 - xxx Ka = [H 3 O + ][A - ]/[HA] = x 2 /0.10 - x 1.8 x 10 -5 = x 2 /0.10 - x = x 2 /0.10 x = 1.3 x 10 -3

43. Calculations Based on Ionization Constants Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Ka = 4.0 x 10 -10 for HCN You do it!

44. Calculations Based on Ionization Constants

45. The percent ionization of 0.15 M HCN solution is calculated as in the previous example.

46. Calculations Based on Ionization Constants Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

47. Calculations Based on Ionization Constants 7.7 x 10 -6 is the VALID chemical solution!

48. Calculations Based on Ionization Constants Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Note that the [H + ] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

49. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. [NH 3 ][NH 4 + ][OH - ] initial0.01000 change-x+x+x equilib0.010 - x x x

50. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. [NH 3 ][NH 4 + ][OH - ] initial0.01000 change-x+x+x equilib0.010 - x x x

51. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - Kb = 1.8 x 10 -5 Step 2. Solve the equilibrium expression

52. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - Kb = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small (100Kb < Co), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M

53. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - Kb = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small (100Kb < Co), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 = 0.010 M

54. Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ¾ NH 4 + + OH - Kb = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

55. Example 18-15 The pH of household ammonia is 11.50. What is its molarity? NH 3 + H 2 O ¾ NH 4 + + OH - pOH = 14 - 11.50 = 2.50 [OH - ] = 10 -2.50 = 0.0032 M initial x change - 0.0032 0.0032 0.0032 equil x - 0.0032 0.0032 0.0032 Kb = [NH 4 + ][OH - ]/[NH 3 ] = (0.0032 ) 2 /x - 0.0032 1.8 x 10 -5 = (0.0032 ) 2 /x x = 0.57 M

56. Polyprotic Acids Many weak acids contain two or more acidic hydrogens. polyprotic acids ionize stepwise ionization constant for each step Consider arsenic acid, H 3 AsO 4, which has three ionization constants 1 K 1 =2.5 x 10 -4 2 K 2 =5.6 x 10 -8 3 K 3 =3.0 x 10 -13

57. Polyprotic Acids The first ionization step is

58. Polyprotic Acids The second ionization step is

59. Polyprotic Acids The third ionization step is

60. Polyprotic Acids Notice that the ionization constants vary in the following fashion: This is a general relationship.

61. Polyprotic Acids Calculate the concentration of all species in 0.100 M arsenic acid, H 3 AsO 4, solution. 1 Write the first ionization ionization step and represent the concentrations.

62. Polyprotic Acids 2 Substitute into the expression for K 1.

63. Polyprotic Acids 3 Use the quadratic equation to solve for x, and obtain two values

64. Polyprotic Acids 4 Now we write the equation for the second step ionization and represent the concentrations.

65. Polyprotic Acids 5 Substitute into the second step ionization expression.

66. Polyprotic Acids

67. 6 Now we repeat the procedure for the third ionization step.

68. Polyprotic Acids 7 Substitute into the third ionization expression.

69. Polyprotic Acids

70. Use K w to calculate the [OH - ] in the 0.100 M H 3 AsO 4 solution.

71. Polyprotic Acids A comparison of the various species in 0.100 M H 3 AsO 4 solution follows.

72. Solvolysis Solvolysis is the reaction of a substance with the solvent in which it is dissolved. Hydrolysis refers to the reaction of a substance with water or its ions.

73. Hydrolysis …the reaction of a substance with water or its ions A - + H 2 O ¾ HA + OH - Ah ha, BASIC! BH + + H 2 O ¾ B + H 3 O +

74. Hydrolysis Hydrolysis refers to the reaction of a substance with water or its ions. Combination of the anion of a weak acid with H 3 O + ions from water to form nonionized weak acid molecules. A - + H 2 O ¾ HA + OH - H 2 O ¾ H + + OH -

75. NaCN Na + + H 2 O  NR CN - + H 2 O  HCN + OH - Basic HCN + NaOH  NaCN + H2OH2O Hydrolysis

76. NH 4 Cl Cl - + H 2 O  NR NH 4 + + H 2 O  NH 3 + H3O+H3O+ Acidic HCl + NH 3  NH 4 Cl

77. NH 4 CN NH 4 + + H 2 O  NH 3 + H3O+H3O+ CN - + H 2 O  HCN + OH - Acidic or Basic HCN + NH 4 OH  NH 4 CN + H2OH2O Hydrolysis

78. Salts of Weak Bases and Weak Acids Acidic Solution Ka > Kb NH 4 F  NH 4 + + F - NH 4 + + H 2 O  NH 3 + H 3 O + F - + H 2 O  HF + OH - Ka = 7.2 x 10 -4 and Kb = 1.8 x 10 -5

79. Salts of Weak Bases and Weak Acids Neutral Solution Ka = Kb NH 4 CH 3 COO  NH 4 + + CH 3 COO - NH 4 + + H 2 O  NH 3 + H 3 O + CH 3 COO - + H 2 O  CH 3 COOH + OH - Ka = 1.8 x 10 -5 and Kb = 1.8 x 10 -5

80. Salts of Weak Bases and Weak Acids Basic Solution Ka < Kb NH 4 CN  NH 4 + + CN - NH 4 + + H 2 O  NH 3 + H 3 O + CN - + H 2 O  HCN + OH - Ka = 4.0 x 10 -10 and Kb = 1.8 x 10 -5

81. Salts of Weak Bases and Weak Acids Acidic Solution Ka > Kb Basic solution Kb > Ka Neutral solution Ka = Kb

82. Hydrolysis The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base. Hydrochloric acid, a typical strong acid, is essentially completely ionized in dilute aqueous solutions.

83. Hydrolysis The conjugate base of HCl, the Cl - ion, is a very weak base. True of all strong acids and their anions.

84. Hydrolysis HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the F - ion, is a much stronger base than the Cl - ion. F - ions combine with H 3 O + ions to form nonionized HF.

85. Salts of Strong Soluble Bases and Strong Acids Salts made from strong acids and strong soluble bases form neutral aqueous solutions. An example is potassium nitrate, KNO 3, made from nitric acid and potassium hydroxide.

86. Salts of Strong Soluble Bases and Weak Acids Salts made from strong soluble bases and weak acids hydrolyze to form basic solutions. Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.

87. Salts of Strong Soluble Bases and Weak Acids Combine these equations into one single equation that represents the reaction:

88. Acid-Base Properties of Salts NH 4 Cl(aq)  NH 4 + (aq) + Cl - (aq) Reaction of NH 4 + with H 2 O NH 4 + + H 2 O  NH 3 + H 3 O + acid base base acid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH 4 + is acidic solution

89. Salts of Weak Bases and Strong Acids Salts made from weak bases and strong acids form acidic aqueous solutions. An example is ammonium bromide, NH 4 Br, made from ammonia and hydrobromic acid.

90. Hydrolysis Constant Relationship of conjugate acid- base pairs K w = K a K b KaKb

91. Example The Ka for acetic acid, CH 3 COOH, is 1.8 x 10 -5, calculate the Kb for the acetate ion, CH 3 COO -.

92. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NH 4 Cl. Cl - + H 2 O is neutral NH 4 + + H 2 O  NH 3 +H 3 O + acidbasebaseacid Ka = 5.6 x 10 -10 Step 1.Set up concentration table [NH 4 + ][NH 3 ][H 3 O + ] initial change equilib 0.1000 -x+x+x 0.10-xxx

93. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NH 4 Cl. Cl - + H 2 O is neutral NH 4 + + H 2 O  NH 3 +H 3 O + acidbasebaseacid Ka = 5.6 x 10 -10

94. Calculate the pH of a 0.10 M solution of NH 4 Cl. Cl - + H 2 O is neutral NH 4 + + H 2 O  NH 3 +H 3 O + acidbasebaseacid Ka = 5.6 x 10 -10 Assume 0.10-x = 0.10 Acid-Base Properties of Salts

95. Calculate the pH of a 0.10 M solution of NH 4 Cl. Cl - + H 2 O is neutral NH 4 + + H 2 O  NH 3 +H 3 O + acidbasebaseacid Ka = 5.6 x 10 -10 Assume 0.10-x = 0.10 x = [NH 3 ] = [H 3 O + ] = 7.5 x 10 -6 M

96. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NH 4 Cl. Cl - + H 2 O Ž neutral NH 4 + +H 2 O ¾ NH 3 +H 3 O + acidbasebaseacid Ka = 5.6 x 10 -10 Step 3.Calculate the pH [H 3 O + ] = 7.5 x 10 -6 M pH = -log [H 3 O + ] = 5.13

97. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 1.Set up concentration table [CH 3 COO - ] [CH 3 COOH][OH - ] initial change equilib

98. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 1.Set up concentration table [CH 3 COO - ] [CH 3 COOH][OH - ] initial0.1000 change-x+x+x equilib0.10-xxx

99. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 2.Solve the equilibrium expression

100. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 2.Solve the equilibrium expression

101. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 2.Solve the equilibrium expression Assume 0.10-x = 0.10

102. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 2.Solve the equilibrium expression Assume 0.10-x = 0.10 x = [CH 3 COO] = [OH - ] = 7.5 x 10 -6 M

103. Acid-Base Properties of Salts Calculate the pH of a 0.10 M solution of NaCH 3 COO. Na + + H 2 O Ž neutral CH 3 COO - +H 2 O ¾ CH 3 COOH+ OH - base acid acid base Kb = 5.6 x 10 -10 Step 3.Calculate the pH [OH - ] = = 7.5 x 10 -6 M pOH = -log [OH - ] = 5.13 pH + pOH = 14.00 pH = 8.87