1. Chapter 15 Acids and Bases

2. What are acids and bases?Learned in Chem 1211 Acid: gives H + in aqueous solution HCl(aq)  H + (aq) + Cl − (aq) Base: gives OH − in aqueous solution NaOH(aq)  Na + (aq) + OH − (aq) Arrhenius acids and bases

3. A more general definition Brønsted-Lowrey acids and bases Acid: proton (H + ) donor Base: proton (H + ) acceptor

4. HCl(aq) + H 2 O(l) H+H+  H 3 O + (aq) + Cl − (aq) acidbase H 3 O + — hydronium ion actual form of H + in aqueous solutions Lewis structure?

5. HAc(aq) + H 2 O(l) H+H+ ⇌ H 3 O + (aq) + Ac − (aq) acidbase NH 3 (aq) + H 2 O(l) H+H+ ⇌ NH 4 + (aq) + OH − (aq) acidbase conjugate base conjugate acid conjugate base conjugate acid H+H+ H+H+

6. Conjugate acid-base pair AcidBase − H + + H +

7. HAcAc − − H + + H + NH 4 + NH 3 − H + + H + acidbaseacidbase

8. H2OH2OOH − − H + + H + − H + + H + H3O+H3O+ Amphiprotic or Amphoteric acidbase acidbase

9. Brønsted-Lowrey acids and bases are not limited to aqueous solutions NH 3 (g) + HCl(g) ⇌ NH 4 + + Cl −  NH 4 Cl(s) H+H+

10. Structural requirement on Brønsted-Lowrey acids and bases Base: must have a pair of nonbonding electrons. Acid: must have a removable proton. HCl, Cl −, NH 3, NH 4 +, H 2 O

11. Strong acids: complete dissociation/ionization Weak acids: partial dissociation/ionization

14. 7 common strong acids to remember HCl, HBr, HI, HNO 3, HClO 3, HClO 4, H 2 SO 4 Names Dissociation equations

15. HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl − (aq) HCl(aq)  H + (aq) + Cl − (aq) H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 − (aq) HSO 4 − (aq) + H 2 O(l) ⇌ H 3 O + (aq) + SO 4 2− (aq) HCl: monoprotic acid;H 2 SO 4 : polyprotic acid HClO 4 (aq)  H + (aq) + ClO 4 − (aq)

16. HA(aq) + H 2 O(l) ⇌ H 3 O + (aq) + A − (aq) Weak monoprotic acid

17. HA(aq) ⇌ H + (aq) + A − (aq) Acid dissociation/ionization constant Relative strength of weak acids larger K a ↔ stronger acid (HA) HA(aq) + H 2 O(l) ⇌ H 3 O + (aq) + A − (aq) Strength of an acid: how easy it dissociates to give a H +

19. HA(aq) ⇌ H + (aq) + A − (aq) Reverse reaction is base accepting H + Relative strength larger K a ↔ stronger acid (HA) ↔ weaker base (A − ) acidbase Strength of a base: its ability to obtain a H +

20. Use table 15.5, arrange the following species according to their strength as bases: F −, Cl −, NO 2 −, and CN −. AcidBase − H + + H +

22. Quiz for 2 extra points Write down the chemical formulae for the 7 common strong acids Write down the dissociation reaction of nitric acid

23. 7 common strong acids to remember HCl, HBr, HI, HNO 3, HClO 3, HClO 4, H 2 SO 4 Names Dissociation equations

24. H2OH2OOH − − H + + H + − H + + H + H3O+H3O+ Amphiprotic or Amphoteric

25. H 2 O(l) + H 2 O(l) acidbase conjugate base conjugate acid H+H+ ⇌ H 3 O + (aq) + OH − (aq) Autoionization of Water Dissociation constant or ion-product constant of water

26. H 2 O(l) ⇌ H + (aq) + OH − (aq) At 25 °C in water, [H + ] = [OH − ] = 1.0 x 10 −7 mol/L K w = 1.0 x 10 −14 True for ANY aqueous solution at 25 °C.

27. [H + ] = [OH − ] —— neutral solution or water [H + ] > [OH − ] —— acidic solution [H + ] < [OH − ] —— basic solution [H + ] ≠ 0, [OH − ] ≠ 0 in any aqueous solution Note:

28. Calculate [OH – ] at 25  C for each solution and determine whether it is acidic, basic, or neutral. (a) [H 3 O + ] = 7.5  10 –5 M (b) [H 3 O + ] = 1.5  10 –9 M (c) [H 3 O + ] = 1.0  10 –7 M Example 15.2, page 669

29. At 60 °C, the value of K w is 1.0 x 10 −13. a) Use Le Châtelier’s principle, predict whether the reaction 2H 2 O(l) ⇌ H 3 O + (aq) + OH − (aq) is exothermic or endothermic. b) Calculate [H + ] and [OH − ] in a neutral solution at 60 °C.

30. pH = − log [H + ] [H + ] = 1.0 x 10 −7 mol/L pH = − log (1.0 x 10 −7 ) = 7.00 Number of decimal places of log is equal to the number of sig figs of the original number. [H + ] = 2.35 x 10 −7 mol/L pH = − log (2.35 x 10 −7 ) = 6.629

31. pH = − log [H + ] pK a = − log K a pOH = − log [OH − ] pK w = − log K w

32. pH = − log [H + ] One unit change of pH corresponds to a change of [H + ] by a factor of 10 e.g. [H + ] = 0.1 mol/L [H + ] = 0.01 mol/L → pH = 1.0 → pH = 2.0

33. pH [H + ] pH = − log [H + ] e.g. [H + ] = 0.1 mol/L → pH = 1.0 [H + ] = 0.01 mol/L → pH = 2.0

34. pOH[OH − ] pK a KaKa pH [H + ] pK w KwKw

35. Quiz for 2 extra points Write down the chemical formulae for the 7 common strong acids Write down the dissociation reaction of nitric acid

36. 7 common strong acids to remember HCl, HBr, HI, HNO 3, HClO 3, HClO 4, H 2 SO 4 Names Dissociation equations

37. Calculate the pH of each solution at 25  C and indicate whether it is acidic or basic. (a)[H + ] = 1.8  10 –4 M (b) [OH – ] = 1.3  10 –2 M Example 15.3, page 671 pOH? Same questions at 60 °C, K w = 1.0 x 10 −13.

38. Relationship between pH and pOH pH + pOH = pK w = 14.00 At 25 °C

39. The pH of a sample of human blood was measured to be 7.41 at 25 °C. Calculate pOH, [H + ], and [OH − ] for the sample. Same questions at 60 °C, K w = 1.0 x 10 −13.

40. Acidity in terms of pH at 25 °C pH = 7, neutral pH < 7, acidic pH > 7, basic

42. pH Meters are Used to Measure Acidity

43. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts Default: aqueous solution, 25 °C

44. a)Calculate the pH of 0.10 mol/L HNO 3. a’) Calculate the pH of 0.15 mol/L HCl. b) Calculate the pH of 1.0 x 10 −10 mol/L HCl. pH of strong acids

46. highly solubleslightly soluble

47. Calculate the pH of a 5.2 x 10 −2 mol/L NaOH solution at 25 °C. pH of strong bases pH for strong base: [OH − ] → pOH → pH Calculate the pH of a 5.2 x 10 −2 mol/L NaOH solution at 60 °C. At 60 °C, K w = 1.0 x 10 −13.

48. pH of strong bases Calculate the pH of a 6.2 x 10 −4 mol/L Ba(OH) 2 solution at 25 °C and at 60 °C, respectively.

49. Quiz 8: up to this point

50. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts

51. pH of weak acids HA(aq) ⇌ H + (aq) + A − (aq) Calculate the pH of a 1.00 mol/L HF solution. K a = 7.2 x 10 −4.

52. HF(aq) ⇌ H + (aq) + F − (aq) I:1.0000 C:−x+x E:1.00 − xxx Approx: x = 2.7 x 10 −2 (mol/L) Valid → pH = −log[H + ] = −log(2.7 x 10 −2 ) = 1.57 < 5%

53. pH of weak acids HA(aq) ⇌ H + (aq) + A − (aq) Calculate the pH of a 1.00 mol/L HF solution. K a = 7.2 x 10 −4. Calculate the pH and percent dissociation of a 0.50 mol/L HF solution. K a = 7.2 x 10 −4.

54. HF(aq) ⇌ H + (aq) + F − (aq) I:0.5000 C:−x+x E:0.50 − xxx Approx: x = 1.9 x 10 −2 (mol/L) Valid → pH = −log[H + ] = −log(1.9 x 10 −2 ) = 1.72

55. pH of weak acids HA(aq) ⇌ H + (aq) + A − (aq) Calculate the pH of a 1.00 mol/L HF solution. K a = 7.2 x 10 −4. Calculate the pH and percent dissociation of a 0.50 mol/L HF solution. K a = 7.2 x 10 −4. Calculate the pH and percent dissociation of a 0.10 mol/L HF solution. K a = 7.2 x 10 −4.

56. HF(aq) ⇌ H + (aq) + F − (aq) I:0.1000 C:−x+x E:0.10 − xxx Approx: x = 8.5 x 10 −3 (mol/L) Invalid → Literally solve the original quadratic equation for x

57. HF(aq) ⇌ H + (aq) + F − (aq) I:0.1000 C:−x+x E:0.10 − xxx x = 8.1 x 10 −3 (mol/L) pH = −log[H + ] = −log(8.1 x 10 −3 ) = 2.09

58. Percent dissociation[HF] 0.10 mol/L 0.50 mol/L 1.00 mol/L 8.1 % 3.8 % 2.7 % 1) 2) 3)

59. Dissociation

60. Try Examples 15.5, 15.6 and 15.7 on page 674

61. value of K c or K p equilibrium concentrations or pressures

62. A 0.100 M weak acid (HA) solution has a pH of 4.25. Find K a for the acid. Example 15.8, page 677

63. In a 0.100 mol/L lactic acid (HC 3 H 5 O 3 ) aqueous solution, lactic acid is 3.7 % dissociated. Calculate the value of K a for this acid. K a = 1.4 x 10 −4

64. HA(aq) ⇌ H + (aq) + A − (aq) Monoprotic acids Polyprotic acids H 3 PO 4

65. H 3 PO 4 (aq) ⇌ H + (aq) + H 2 PO 4 − (aq)K a1 H 2 PO 4 − (aq) ⇌ H + (aq) + HPO 4 2− (aq)K a2 HPO 4 2− (aq) ⇌ H + (aq) + PO 4 3− (aq)K a3 K a1 >> K a2 >> K a3

67. Calculate the pH of a 5.0 mol/L H 3 PO 4 solution and the equilibrium concentrations of the species H 3 PO 4, H 2 PO 4 −, HPO 4 2−, and PO 4 3−.

68. H 3 PO 4 (aq) ⇌ H + (aq) + H 2 PO 4 − (aq)K a1 H 2 PO 4 − (aq) ⇌ H + (aq) + HPO 4 2− (aq)K a2 HPO 4 2− (aq) ⇌ H + (aq) + PO 4 3− (aq)K a3 K a1 >> K a2 >> K a3 first step is the major contribution of H +

69. Calculate the pH of a 5.0 mol/L H 3 PO 4 solution and the equilibrium concentrations of the species H 3 PO 4, H 2 PO 4 −, HPO 4 2−, and PO 4 3−. [H + ] = 0.19 M, pH = 0.72

70. H 3 PO 4 (aq) ⇌ H + (aq) + H 2 PO 4 − (aq)K a1 H 2 PO 4 − (aq) ⇌ H + (aq) + HPO 4 2− (aq)K a2 HPO 4 2− (aq) ⇌ H + (aq) + PO 4 3− (aq)K a3 K a1 >> K a2 >> K a3

72. Calculate the pH of a 0.0100 M H 2 SO 4 solution. Example 15.18, page 695

73. Try Examples 15.17, 15.18 and 15.19 on page 695

74. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts

75. H+H+ B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) acidbase conjugate base conjugate acid H+H+ higher K b ↔ stronger base

76. NH 3 (aq) + H 2 O(l) H+H+ ⇌ NH 4 + (aq) + OH − (aq) acidbase conjugate base conjugate acid H+H+ Base: must have a pair of nonbonding electrons.

77. NH 3 CH 3 NH 2 C5H5NC5H5N

78. B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) Learn how to write the equation of a weak base!

79. H+H+ B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) acidbase conjugate base conjugate acid H+H+ pH for weak base: ICE → [OH − ] → pOH → pH

80. Calculate the pH of a 0.100 mol/L NH 3 solution. K b = 1.76 x 10 −5 Example 15.12, page 684

81. Calculate the pH of a 1.0 mol/L methylamine CH 3 NH 2 solution. K b = 4.38 x 10 −4 [OH − ] = 0.021 M, pH = 12.32

82. HA(aq) ⇌ H + (aq) + A − (aq) Reverse reaction is base accepting H + Relative strength larger K a ↔ stronger acid (HA) ↔ weaker base (A − ) acidbase Strength of a base: its ability to obtain a H +

83. H+H+ B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) acidbase conjugate base conjugate acid H+H+ higher K b ↔ stronger base

84. B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) BH + (aq) + H 2 O(l) ⇌ H 3 O + (aq) + B(aq) KbKb KaKa 2H 2 O(l) ⇌ H 3 O + (aq) + OH − (aq) KwKw Sum of reactions  Product of equilibrium constants Relation between K a and K b for a conjugate acid-base pair

85. B(aq) + H 2 O(l) ⇌ BH + (aq) + OH − (aq) BH + (aq) + H 2 O(l) ⇌ H 3 O + (aq) + B(aq) KbKb KaKa 2H 2 O(l) ⇌ H 3 O + (aq) + OH − (aq) KwKw K a K b = K w pK a + pK b = pK w Relation between K a and K b for a conjugate acid-base pair

86. Acid strengthBase strength high low

87. Find K a, pK a, pK b from this table, and write down the acid dissociation reactions.

88. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts: cation + anion

89. Quiz 9, this week after lab K a ↔ pH and percent dissociation of a weak acid pH of polyprotic acid pH of weak base

90. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts: cation + anion

91. 1. Cation is from strong base; anion is from strong acid. Solution is neutral Other examples? NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) Cations from strong bases and anions from strong acids have no effect on pH.

92. 2. Cation is from strong base; anion is from weak acid. Solution is basic Other examples? NaOH(aq) + HF(aq)  NaF(aq) + H 2 O(l) Cation has no effect on pH, Anion acts as a base.

93. Calculate the pH of a 0.30 mol/L NaF solution. K a for HF is 7.2 x 10 −4 pH for weak base: ICE → [OH − ] → pOH → pH K a K b = K w x = [OH − ] = 2.0 x10 −6, pH = 8.31

94. Find the pH of a 0.100 M NaCHO 2 solution. The salt completely dissociates into Na + (aq) and CHO 2 – (aq) and the Na + ion has no acid or base properties. K a for HCHO 2 is 1.8 x 10 −4. Example 15.14, page 689

95. 3. Cation is conjugate acid of weak base; anion is from strong acid.

96. 3. Cation is conjugate acid of weak base; anion is from strong acid. NH 4 Cl, NH 4 NO 3, CH 3 NH 3 Br Solution is acidic Cation acts as an acid, Anion has no effect on pH.

97. Calculate the pH of a 0.10 mol/L NH 4 Cl solution. K b for NH 3 is 1.76 x 10 −5 x = [H + ] = 7.5 x 10 −6 M, pH = 5.13

98. 4. Cation is conjugate acid of weak base; anion is from weak acid. Qualitative Prediction K a > K b, Acidic, pH < 7 K a 7 K a = K b, neutral, pH = 7 NH 4 F, NH 4 Ac, NH 4 CN, CH 3 NH 3 CN

99. NH 3 K b = 1.76 x 10 −5 CH 3 NH 2 K b = 4.4 x 10 −4 HAcK a = 1.76 x 10 −5 HFK a = 7.2 x 10 −4 HCNK a = 6.2 x 10 −10

100. 5. Cation is highly charged metal ions; anion is from strong acid. AlCl 3, Fe(NO 3 ) 3 Al 3+ (aq) + 6H 2 O(l)  Al(H 2 O) 6 3+ (aq)

102. Calculate the pH of a 0.010 mol/L AlCl 3 solution. K a for Al(H 2 O) 6 3+ is 1.4 x 10 −5 Al(H 2 O) 6 3+ (aq) ⇌ Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) HF(aq) ⇌ H + (aq) + F − (aq) Same method as Answer: pH = 3.43

103. Acid-Base Properties of Various Types of Salt Type of SaltExamplespH of SolutionReason Cation is from strong base; anion is from strong acid KCl, KNO 3, NaCl, NaNO 3 Neutral Neither acts as an a acid or a base Cation is from strong base; anion is from weak acid NaF, KCN, NaAc, KNO 2 BasicAnion acts as a base; cation has no effect on pH Cation is conjugate acid of weak base; anion is from strong acid NH 4 Cl, NH 4 NO 3 AcidicCation acts as acid; anion has no effect on pH Cation is conjugate acid of weak base; anion is conjugate base of weak acid NH 4 Ac, NH 4 CN Acidic if K a > K b, Basic if K b > K a, Neutral if K a = K b Cation acts as an acid; anion acts as a base Cation is highly charged metal ion; anion is from strong acid Al(NO 3 ) 3, FeCl 3 AcidicHydrated cation acts as an acid; anion has no effect on pH Try Example 15.16 on page 692

104. Determine whether the solution formed by each salt is acidic, basic, or neutral: (a)SrCl 2 (b)AlBr 3 (c) CH 3 NH 3 NO 3 (d) NaCHO 2 (e) NH 4 F Example 15.16 on page 692

105. Brønsted-Lowrey acids and bases Arrhenius acids and bases Lewis acids and bases Acids: electron pair acceptor Bases: electron pair donor

108. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts: cation + anion

109. Quiz 10, next week K a K b = K w Acid/base strength comparison Predict acidity of a salt solution pH calculation for salt solutions