1. Chapter 16 Acids and Bases Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php

2. 2 Arrhenius theory of acids and bases The Arrhenius concept: acids dissociate in water (aqueous solution) to produce hydrogen ions H +, and bases dissociate in water to give hydroxide ions, OH -. Arrhenius acid: HA (aq)  H + (aq) + A - (aq) Arrhenius base: MOH (aq)  M + (aq) + OH - (aq)

3. 3 Neutralization reactions Arrhenius acids and bases react with each other to form water and aqueous salts in neutralization reactions. H + (aq) + A - (aq) + M + (aq) + OH - (aq)  H 2 O (l) + M + (aq) + A - (aq) The net ionic equation is H + (aq) + OH - (aq)  H 2 O (l)

4. 4 Aqueous salts The aqueous salt in the reaction comes from the spectator ions in the reaction. These ions are present to balance the proton positive charge or the hydroxide ion negative charge in the acid or base. If we evaporate all the water, we are left with an ionic solid called a salt e.g. NaCl

5. 5 Arrhenius theory of acids and bases Many substances that do not contain OH - act like bases! The key to the Arrhenius description is that we need water to act as a solvent to promote the dissociation of the acid or base.

6. 6 Brønsted-Lowry theory of acids and bases Many substances (like NH 3 ) that do not contain OH - act like bases in water! The Brønsted-Lowry Theory: an acid is any substance that donates protons (H + ) while a base is any substance that can accept protons. This means that Brønsted-Lowry acid- base reactions are proton transfer reactions.

7. 7 Proton transfer reactions Pairs of compounds are related to each other through Brønsted-Lowry acid-base reactions. These are conjugate acid-base pairs. Generally, an acid HA has a conjugate base A - (a proton has transferred away from the acid ). Conversely, a base B has a conjugate acid BH + (a proton has transferred toward the base ).

8. 8 Water as an acid in BL reactions When a Brønsted-Lowry base is placed in water, it reacts with the water (which acts as a Brønsted-Lowry acid) and establishes an acid-base equilibrium.

9. 9 BL base strength The strength of a Brønsted-Lowry base can be quantified by the equilibrium constant as it relates to completeness of reaction with water. For the reaction of ammonia with water

10. 10 BL base strength We give the equilibrium constant a special name for reactions like these – the base dissociation (or ionization) constant K b. Since the value of the constant is less than one, the ammonia does not dissociate to a great extent – it is a weak base!

11. 11 Water as a base in BL reactions When a Brønsted-Lowry acid is placed in water, it reacts with the water (which acts as a Brønsted-Lowry base) and establishes an acid-base equilibrium.

12. 12 BL acid strength The strength of a Brønsted-Lowry acid can be quantified by the equilibrium constant as it relates to completeness of reaction with water. For the reaction of acetic acid with water

13. 13 BL acid strength We give the equilibrium constant a special name for reactions like these – the acid dissociation (or ionization) constant K a. Since the value of the constant is less than one, the acetic acid does not dissociate to a great extent – it is a weak acid!

14. 14 Strong BL acids and bases Strong BL acids and bases have the exact same reaction with water as do weak acids and bases, they just are much more complete. This is reflected in the acid or base dissociation constant – it is much larger than one! Hydrochloric acid is a strong acid.

15. 15 Hydrated protons and hydronium ions What is the strongest Brønsted-Lowry acid there is? The strongest Brønsted- Lowry acid is H +. The ultimate proton-donor is a proton itself! In water there is no such thing as H +.

16. 16 Hydrated protons and hydronium ions Often more than one water molecule will crowd around the hydronium ion (H 3 O + ) to give hydrates with the formula [H 3 O(H 2 O) n ] + where n is 1 to 4.

17. 17 Requirements of Brønsted-Lowry bases For a molecule or ion to accept a proton (to act as a base) requires it to have an unshared pair of electrons which can then be used to create a bond to the H +. All Brønsted-Lowry bases have at least one lone pair of electrons. In the previous reactions we’ve seen NH 3 has a lone pair and can act as a base. Also, water has two lone pairs, and can act as a base.

18. 18 Amphiprotic substances Some substances, like water, have protons that can be donated (BL acid), and lone pairs of electrons that can accept protons (BL base). This is why it can act like an acid AND a base. Such substances are said to be amphiprotic.

19. 19 Problem Write a balanced equation for the dissociation of each of the following Brønsted-Lowry acids in water: a) H 2 SO 4 b) HSO 4 - c) H 3 O + d) NH 4 +

20. 20 Problem What is the conjugate acid of each of the following Brønsted- Lowry bases? a) HCO 3 - b) CO 3 2- c) OH - d) H 2 PO 4 -

21. 21 Problem Of the following species, one is acidic, one is basic, and one is amphiprotic in their reactions with water: HNO 2, PO 4 3-, HCO 3 -. Write the four equations needed to represent these facts.

22. 22 Problem For each of the following reactions, identify the acids and bases in both the forward and reverse directions: Acid Base Base Acid Base Acid Acid Base

23. 23 A 2 nd look at acid and base strength Acid-base equilibria are competitions! The equilibrium is a tug-of war between the two bases in the system as they fight for protons given away by the two acids. The acid that is “better at giving away protons” (or the base that is “better at taking protons”) will be found in lesser amounts at equilibrium than the other acid (or base).

24. 24 Strong acids in water A strong acid (HA) is one that almost completely dissociates in water (which acts as a weak base). The conjugate base of the strong acid (A - ) will be a very weak base. At equilibrium, there will be very little to no HA present in the system, and the concentration of A - will essentially be the same as the initial concentration of HA.

25. 25 Strong bases in water A strong base (A - ) is one that almost completely dissociates in water (which acts as an acid). The conjugate acid of the strong base (HA) will be a very weak acid. At equilibrium, there will be very little to no A - present in the system, and the concentration of HA will essentially be the same as the initial concentration of A -.

26. 26 Relationship of acid/base strengths The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. We’ll look at this a bit more in depth later.

27. 27

28. 28 Differentiating strong acids Very strong acids like HClO 4 and HCl dissociate in water so completely it is almost impossible to find accurate K a values, and therefore determine which is the stronger acid. We must place the acids in a solvent that is a much weaker base than water (poorer at taking protons). Whichever reaction in the new solvent has a higher K value tells us which is really the stronger acid.

29. 29 Differentiating strong acids HClO 4 must be a stronger acid than HCl because it forces the very weak base diethyl ether to accept protons much more readily (reaction is essentially complete – large K) than does HCl, which establishes an equilibrium – smaller K.

30. 30 Self-ionization of water Water can act as an acid or a base because the molecule has both protons and lone pairs available – its amphiprotic! It is possible for one water molecule to act as an acid while another water molecule acts as a base at the same time. This leads to the auto-dissociation (or self- ionization) of water equilibrium reaction: H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) The equilibrium constant for this reaction is called the ion-product constant for water, K w. K w = [H 3 O + ] [OH - ]

31. 31 Autoionization of water

32. 32 At 25 °C, K w = 1.0 x 10 -14 so [H 3 O + ] = [OH - ] = 1.0 x 10 -7 mol/L Relatively few water molecules are dissociated at equilibrium at room temperature! We will always assume that [H 3 O + ] [OH - ] = 1.0 x 10 -14 at 25 °C. [H 3 O + ] > 1.0 x 10 -7 is acidic ([OH - ] < 1.0 x 10 -7 ) [H 3 O + ] 1.0 x 10 -7 ) [H 3 O + ] = 1.0 x 10 -7 is neutral ([OH - ] = 1.0 x 10 -7 )

33. 33 We also find, since [H 3 O + ] [OH - ] = 1.0 x 10 -14 = K w then [H 3 O + ] = 1.0 x 10 -14 / [OH - ] and [OH - ] = 1.0 x 10 -14 / [H 3 O + ] at 25 °C

34. 34

35. 35 Problem The concentration of OH - in a sample of seawater is 5.0 x 10 -6 mol  L -1. Calculate the concentration of H 3 O + ions, and classify the solution as acidic, neutral, or basic. Answer: [H 3 O + ] = 2.0 x 10 -9. Solution is basic.

36. 36 Problem At 50 °C the value of K w is 5.5 x 10 -14. What are the [H 3 O + ] and [OH - ] in a neutral solution at 50 °C? Answer: Both are 2.3 x 10 -7 mol  L -1

37. 37 The pH scale Because [H 3 O + ] in water solutions can range from very small (strongly basic) to very large (strongly acidic) it is sometimes easier to use a logarithmic (power of 10) scale to express [H 3 O + ]. Additionally, since using a negative number is sometime awkward, we actually use a negative logarithmic scale to express the hydronium ion concentration with a term we call the pH of a solution. pH = - log [H 3 O + ]

38. 38 pH and acidity For [H 3 O + ] = 1.0 x 10 -7 mol  L -1 (neutral), pH = 7.00 For [H 3 O + ] = 1.0 x 10 -4 mol  L -1 (acidic), pH = 4.00 For [H 3 O + ] = 1.0 x 10 -11 mol  L -1 (basic), pH = 11.00 In general, pH > 7 is basic pH < 7 is acidic pH = 7 is neutral

39. 39 pOH and acidity We occasionally see the pOH mentioned, which is pOH = - log [OH - ] or [OH - ] = 10 -pOH pOH < 7 is basic pOH > 7 is acidic pOH = 7 is neutral

40. 40 K w = 1.0 x 10 -14 = [H 3 O + ] [OH - ] 14.00 = pH + pOH at 25  C

41. 41 Problem Calculate the pH of each of the following solutions: a) A sample of seawater that has an OH - concentration of 1.58 x 10 -6 mol  L -1 b) A sample of acid rain that has an H 3 O + concentration of 6.0 x 10 -5 mol  L -1 Answers: a) pH = 8.19 b) pH = 4.22

42. 42 Problem Calculate [H 3 O + ] and [OH - ] in each of the following solutions: a) Human blood (pH 7.40) b) A cola beverage (pH 2.8) Answers: a) [H 3 O + ] = 4.0 x 10 -8 M and [OH - ] = 2.5 x 10 -7 M b) [H 3 O + ] = 1. 6 x 10 -3 M and [OH - ] = 6. 3 x 10 -12 M

43. 43 Problem The pH of a solution of HCl in water is found to be 2.50. What volume of water would you add to 1.00 L of this solution to raise the pH to 3.10? Answer: You would add 3.0 L of water

44. 44 Strong acids and strong bases Most of the strong acids are monoprotic acids, that are capable of donating only one proton. Sulphuric acid (H 2 SO 4 ), which is a diprotic acid capable of donating two protons, is also a strong acid (for the first proton!). Strong monoprotic acids (HA) essentially dissociate 100% in water to give H 3 O + and A -, leaving virtually no HA in solution at equilibrium. If we know the initial concentration of HA, then the equilibrium concentration of H 3 O + will be the same, and we can calculate the pH.

45. 45 pH and strong bases Strong bases, such as the alkali metal (Li, Na, K, etc.) hydroxides MOH completely dissociate in water to give metal ions and hydroxide ions. H 2 O MOH (s)  M + (aq) + OH - (aq) Again, calculating the pH is straight-forward, as the final concentration of OH - will be the same as the initial concentration of MOH. Of course, if we know [OH - ], we can calculate [H 3 O + ], and the pH.

46. 46 pH and strong bases Alkaline earth (Ca, Mg, etc) metal hydroxides M(OH) 2 are strong bases and will completely dissociate in water up to the point of their low solubility.

47. 47 pH and strong bases Alkaline earth (Ca, Mg, etc) metal oxides MO are stronger bases than the equivalent hydroxides because O 2- is a very strong base. In fact, much like bare H + can’t exist in water, neither can bare O 2-. O 2- (aq) + H 2 O (l)  2 OH - (aq), so H 2 O MO (s)  M 2+ (aq) + 2 OH - (aq)

48. 48 Be careful! We are assuming the pH of the solution will be determined solely by the initial concentration of the strong acid or strong base. This is true if the initial concentration is large enough that the autoionization of water contributes insignificant amounts of H 3 O + and OH - ! What is the pH of 1.0 x 10 -8 M HCl?

49. 49 Common strong acids and bases

50. 50 Problem Calculate the pH of: a) 0.050 M HClO 4 b) 6.0 M HCl c) 4.0 M KOH d) 0.010 M Ba(OH) 2 Answers: a) pH = 1.30 b) pH = -0.78 c) pH = 14.60 d) pH = 12.30

51. 51 Problem If 535 mL of gaseous HCl at 26.5  C and 747 mmHg is dissolved in enough water to prepare 625 mL of solution, what is the pH of this solution? The gas constant R = 0.08206 L  atm  K -1  mol -1 and 760 mmHg = 1 atm exactly! Answer: pH = 1.467

52. 52 Problem Milk of magnesia is a saturated solution of Mg(OH) 2. Its solubility is 9.63 mg Mg(OH) 2 per 100.0 mL of solution at 20  C. What is the pH of saturated Mg(OH) 2 at 20  C? The molar mass of Mg(OH) 2 is 58.3197 g  mol -1 Answer: pH = 11.52

53. 53 Problem Calculate the pH of an aqueous solution that is 3.00% KOH by mass and has a density of 1.0242 g  mL -1. The molar mass of KOH is 56.1056 g  mol -1. Answer: pH = 13.74

54. 54 pK a and pK b We can define pK a = - log K a and pK b = - log K b exactly like pH = - log [H 3 O + ] A very small K a or K b value is the same as a large positive pK a or pK b which means the acid or base is weak (partially dissociated in water). As K a  (or pK a  ) or as K b  (or pK b  ) acid strength or base strength increases.

55. 55 pK a and pK b

56. 56 Problem The pH of 0.10 mol  L -1 HOCl is 4.27. Calculate K a for hypochlorous acid HOCl (aq) + H 2 O (l)  H 3 O + (aq) + ClO - (aq) Answer: 2.9 x 10 -8

57. 57 Identifying weak acids Generally there are three categories of weak acids: carboxylic acids oxoacids miscellaneous acids

58. 58 Carboxylic acids Carboxylic acids contain the -COOH (carboxyl) group. The hydrogen of this group is the proton that is donated.

59. 59 Carboxylic acids

60. 60 Oxoacids Oxoacids are generally weak acids with the formula H m XO n where m = 1 to 3 and n = 1 to 4 This formula is often quite misleading because the structure is actually usually (HO) m XO n where (m + n) = 1 to 4

61. 61 Strong oxoacids Nitric acid – HNO 3 Sulfuric acid – H 2 SO 4 Perchloric acid – HClO 4

62. 62 Some weak oxoacids Nitrous acid – HNO 2 Phosphoricic acid – H 3 PO 4 Chlorous acid – HClO 2

63. 63 Miscellaneous weak acids There are other weak acids that are not carboxylic acids or oxoacids. Some of the more common ones are Hydrofluoric acid – HF Hydrocyanic acid – HCN Hydrazoic acid – HN 3

64. 64 Identifying weak bases Many weak bases are amines, which contain nitrogen. The lone pair on the nitrogen allow the amine to be a proton acceptor (Brønsted-Lowry base).

65. 65 Identifying weak bases

66. 66 Amino acids Amino acids have both a carboxyl group and an amine group, meaning different parts of the molecule can act as an acid and a base.

67. 67 Equilibrium in solutions of weak acids and bases We can calculate equilibrium concentrations of reactants and products in acid-base reactions with known values for K a or K b. We need to figure out what is an acid and what is a base in our system. Water will be an acid or base depending on whether we added a base or an acid to it. For example, if we start with 0.10 mol  L -1 HCN, then HCN is an acid, and water is a base. HCN (aq) + H 2 O (l)  H 3 O + (aq) + CN - (aq) K a = 4.9 x 10 -10

68. 68 Equilibrium in solutions of weak acids and bases However, since the acid-base reaction we are looking at takes place in water, we must include the autoionization of water reaction as a source of H 3 O + and OH - ! For our example reaction (HCN is a weak acid) HCN (aq) + H 2 O (l)  H 3 O + (aq) + CN - (aq) K a = 4.9 x 10 -10 H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) K w = 1.0 x 10 -14

69. 69 The strongest acid or base (larger K value) will dominate a system. We call this equilibrium reaction with the larger K value the principal reaction. Any other reactions are subsidiary reactions. Equilibrium in solutions of weak acids and bases

70. 70 Since all reactions take place in one container, then [H 3 O + ] = [H 3 O + ] (principal) + [H 3 O + ] (subsidiary) OR [OH - ] = [OH - ] (principal) + [OH - ] (subsidiary) If the K of the principal reaction is much greater than K for the subsidiary reactions, then we assume [H 3 O + ]  [H 3 O + ] (principal reaction) OR [OH - ]  [OH - ] (principal reaction) Equilibrium in solutions of weak acids and bases

71. 71 We can solve this equation using the quadratic formula and get the right answer, but it might be possible to do it more simply. Let’s assume the initial concentration of the acid and the equilibrium concentration of HCN are essentially the same (that is x << 0.10 in this case). Equilibrium in solutions of weak acids and bases

72. 72 4.9 x 10 -10 = x 2 / 0.10 x 2 = (4.9 x 10 -10 )(0.10) x 2 = 4.9 x 10 -11 x =  4.9 x 10 -11 x =  7.0 x 10 -6 mol  L -1 Based on the assumption we’ve made, at equilibrium [H 3 O + ] = [CN - ] = 7.0 x 10 -6 mol  L -1 and [HCN] = 0.10 mol/L. However, any time we make an assumption, we must check it! (5% rule) Equilibrium in solutions of weak acids and bases

73. 73 We also assumed that [H 3 O + ]  [H 3 O + ] (principal reaction) so we must check this. For the subsidiary reaction H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) K w = 1.0 x 10 -14 If the assumption we’ve made is good, then [H 3 O + ] = 7.0 x 10 -6 mol  L -1 and so [OH - ] subsidiary = K w / [H 3 O + ] principal [OH - ] subsidiary = 1.0 x 10 -14 / 7.0 x 10 -6 [OH - ] subsidiary = 1.4 x 10 -9 mol  L -1 Equilibrium in solutions of weak acids and bases

74. 74 From the subsidiary balanced equation we see if [OH - ] subsidiary = 1.4 x 10 -9 mol  L -1, then [H 3 O + ] subsidiary = 1.4 x 10 -9 mol  L -1. Lets check our assumption The ratio [H 3 O + ] subidiary / [H 3 O + ] principal, is 1.4 x 10 -9 mol  L -1 / 7.0 x 10 -6 mol  L -1 = 0.02%. Our assumption is valid (5% rule)! Equilibrium in solutions of weak acids and bases

75. 75 [H 3 O + ]  [H 3 O + ] (principal reaction) = 7.0 x 10 -6 mol  L -1 pH = - log [H 3 O + ] pH = - log 7.0 x 10 -6 pH = 5.15 Equilibrium in solutions of weak acids and bases

76. 76 Generally it turns out that if our principal acid-base calculation gives a [H 3 O + ] or [OH - ] (depending on reaction type!) less than 10 -6 or so then the auto-dissociation of water reaction will actually contribute a significant amount of [H 3 O + ] or [OH - ] to our system and we must include the subsidiary reaction contribution to pH See page 673 (9 th ed.) or 682 (8 th ed.) of textbook

77. 77 Problem Acetic acid CH 3 COOH is the solute that gives vinegar its characteristic odour and sour taste. Calculate the pH and the concentration of all species present in: a) 1.00 mol  L -1 CH 3 COOH b) 0.0100 mol  L -1 CH 3 COOH

78. 78 Problem a) Let’s assume that x << 1.00 and so 1.8 x 10 -5 = x 2 / 1.00 x 2 = (1.8 x 10 -5 )(1.00) x =  1.8 x 10 -5 x =  4.2 x 10 -3 mol  L -1 (must be +ve value since x = [H 3 O + ] ) CHECK ASSUMPTION

79. 79 We also assumed that [H 3 O + ]  [H 3 O + ] (principal reaction) so we must check this. For the subsidiary reaction H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) K w = 1.0 x 10 -14 If the assumption we’ve made is good, then [H 3 O + ] = 4.2 x 10 -3 mol  L -1 and so [OH - ] subsidiary = K w / [H 3 O + ] principal [OH - ] subsidiary = 1.0 x 10 -14 / 4.2 x 10 -3 [OH - ] subsidiary = 2.4 x 10 -12 mol  L -1 Problem a)

80. 80 From the subsidiary balanced equation we see if [OH - ] subsidiary = 2.4 x 10 -12 mol  L -1, then [H 3 O + ] subsidiary = 2.4 x 10 -12 mol  L -1. Lets check our assumption The ratio [H 3 O + ] subidiary / [H 3 O + ] principal, is 2.4 x 10 -12 mol  L -1 / 4.2 x 10 -3 mol  L -1 = 0.00000006%. Our assumption is valid (5% rule)! Problem a)

81. 81 Problem a) pH = - log [H 3 O + ] pH = - log 4.2 x 10 -3 pH = 2.38

82. 82 Let’s assume that x << 0.00100 and so 1.8 x 10 -5 = x 2 / 0.00100 x 2 = (1.8 x 10 -5 )(0.00100) x =  1.8 x 10 -8 x =  1.3 x 10 -4 mol  L -1 (must be +ve value since x = [H 3 O + ]) CHECK ASSUMPTION Problem b)

83. 83 Problem b) We can not assume that x << 0.00100 and so

84. 84 Problem b) pH = - log [H 3 O + ] pH = - log 1.2 5 x 10 -4 pH = 3.90

85. 85 Problem Piperidine (C 5 H 11 N; molar mass = 85.149 g  mol -1 ) is a base found in small amounts in black pepper. What is the pH of 315 mL of an aqueous solution containing 114 mg of piperidine if K b is 1.6 x 10 -3 ? Answer: pH = 11.29

86. 86 Percent ionization of weak acids The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of the weak acid. Therefore, we can define a second measure of the strength of a weak acid by looking of the percent ionization (or dissociation) of the acid. % ionized = [H 3 O + ] eqm / [HA] initial x 100%

87. 87 Percent ionization We saw on slide 78 that an acetic acid solution with initial concentration of 1.00 mol  L -1 at equilibrium had [H 3 O + ] = 4.2 x 10 -3 mol  L -1 % ionized = [H 3 O + ] eqm / [HA] initial x 100% % ionized = 4.2 x 10 -3 mol  L -1 / 1.00 mol  L -1 x 100% % ionized = 0.42%

88. 88 Percent ionization We saw on slide 84 that an acetic acid solution with initial concentration of 0.0010 mol  L -1 at equilibrium had [H 3 O + ] = 1.2 5 x 10 -4 mol  L -1 % ionized = [H 3 O + ] eqm / [HA] initial x 100% % ionized = 1.2 5 x 10 -4 mol  L -1 / 0.0010 mol  L -1 x 100% % ionized = 12.5%

89. 89

90. 90 Polyprotic acids Acids that contain more than one dissociable protons are polyprotic acids. Each dissociable proton has its own K a value. Carbonic acid (H 2 CO 3 ) regulates blood pH. H 2 CO 3 (aq) + H 2 O (l)  H 3 O + (aq) + HCO 3 - (aq) HCO 3 - (aq) + H 2 O (l)  H 3 O + (aq) + CO 3 2- (aq)

91. 91 Polyprotic acids K a1 > K a2 > K a3 is always true!

92. 92 Problem Calculate the pH and the concentration of all species, including OH - present in 0.10 mol  L -1 H 2 SO 3. Values of K a are given in the table on the previous slide. K a1 = 1.3 x 10 -2 and K a2 = 6.2 x 10 -8

93. 93 Problem Potential reactions that can occur in our system are H 2 SO 3 (aq) + H 2 O (l)  H 3 O + (aq) + HSO 3 - (aq) HSO 3 - (aq) + H 2 O (l)  H 3 O + (aq) + SO 3 2- (aq) H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) We’ll first have to assume [H 3 O + ]  [H 3 O + ] (H 2 SO 3 dissociation)

94. 94 Problem We can’t assume x << 0.10 so

95. 95 Problem x = [H 3 O + ] principal = [HSO 3 - ] = 3.0 x 10 -2 mol  L -1 [H 2 SO 3 ] = (0.10 – 3.0 x 10 -2 ) mol  L -1 = 0.07 mol  L -1 The second dissociation reaction is HSO 3 - (aq) + H 2 O (l)  H 3 O + (aq) + SO 3 2- (aq) With our assumption that almost all H 3 O + comes from the first dissociation, if we substitute our equilibrium concentrations from our first reaction into the equilibrium constant expression for this reaction we should see little change if the assumption is true…

96. 96 Problem x = [SO 3 2- ] = [H 3 O + ] sub = 6.3 x 10 -8 mol/L Our assumption is valid! Since the second proton dissociation contributes a insignificant amount of H 3 O + then the autoionization of water won’t either, because it has an even smaller K value.

97. 97 Problem pH = - log [H 3 O + ] pH = - log 3.0 x 10 -2 pH = 1.52 [OH - ] = K w / [H 3 O + ] [OH - ] = 1.0 x 10 -14 / 3.0 x 10 -2 [OH - ] = 3.3 x 10 -13 mol  L -1

98. 98 Relation between K a and K b The strength of an acid in water is expressed through K a. while the strength of a base can be expressed through K b But a Brønsted-Lowry acid-base reaction involves conjugate acid-base pairs so there should be a connection between the K a value and the K b value. HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) A - (aq) + H 2 O (l)  OH - (aq) + HA (aq)

99. 99 Let’s add the reactions together The sum of the reactions is the dissociation of water reaction, which has the ion-product constant for water K w = [H 3 O + ] [OH - ] = 1.0 x 10 -14 at 25 °C Closer inspection shows us that

100. 100 As the strength of an acid increases (larger K a ) the strength of the conjugate base must decrease (smaller K b ) because their product must always be the dissociation constant for water K w. Strong acids always have very weak conjugate bases. Strong bases always have very weak conjugate acids. Since K a x K b = K w then K a = K w / K b and K b = K w / K a

101. 101 Problem a) Piperidine (C 5 H 11 N) is an amine found in black pepper. If K b = 1.6 x 10 -3 for piperidine then calculate K a for the C 5 H 11 NH + cation. b) If K a = 2.9 x 10 -8 for HOCl then calculate K b for OCl -. Answers: a) K a = 6.3 x 10 -12 b) K b = 3.4 x 10 -7

102. 102 Ions as acids and bases Consider the strong acid anions such as Cl -, Br -, I -, NO 3 -, and ClO 4 -. They must be very weak bases because their conjugate acids are so strong. They will not react with water to form the strong acid… Cl - (aq) + H 2 O (l)  no reaction!

103. 103 Ions as acids and bases Consider the strong base cations such as Na +, K +, Ca 2+, Mg 2+, etc. They can not be acids because they have no protons. They will not react with water to form the strong base… H 2 O (l) + Na + (aq)  no reaction!

104. 104 Ions as acids and bases Consider some weak acid anions such as OCl -, CH 3 COO -, HCO 3 -, or NO 2 -. They will be weak bases because their conjugate acids are weak. They can react with water to form the weak acid and OH - … OCl - (aq) + H 2 O (l)  OH- (aq) + HOCl (aq)

105. 105 Ions as acids and bases Consider some weak base cations such as NH 4 +, C 6 H 5 NH +, CH 3 NH 3 +, etc. They are weak acids because their conjugate bases are weak. They can react with water to form the weak base and H 3 O + … H 2 O (l) + NH 4 + (aq)  H 3 O + (aq) + NH 3 (aq)

106. 106 Salts that yield neutral solutions Any ionic salt that contains neither an acidic cation OR a basic anion will give a neutral solution because neither ion can react with water. NaCl, KNO 3, Ca(ClO 4 ) 2, MgI 2, etc.

107. 107 Salts that yield acidic or basic solutions Any ionic salt that contains either an acidic cation OR a basic anion while the other ion can not react with water will give an acidic or basic solution because of the ability of the ion to react with water. NH 4 Cl, CH 3 COONa, C 6 H 5 NHNO 3, KOCl etc.

108. 108 If a salt is composed of an acidic cation and a basic anion, the acidity or basicity of the salt solution depends on the relative strengths of the acid and base. Salts that yield acidic or basic solutions

109. 109 Salts that yield acidic or basic solutions If the acid cation is “stronger” than the base anion the solution will be acidic. If the base anion is “stronger” than the acid cation the solution will be basic.

110. 110 Salts that yield acidic or basic solutions K a > K b the acid is better at reacting with water than the base and the solution is acidic. K a < K b the base is better at reacting with water than the acid and the solution is basic. K a  K b the solution is close to neutral.

111. 111 Problem Predict whether the following salt solution is basic, neutral or acidic, and calculate the pH. 0.25 mol  L -1 NH 4 Br K b of NH 3 is 1.8 x 10 -5 Answer: The solution is acidic and pH is 4.93

112. 112 Problem Classify each of the following salts as acidic, basic, or neutral: a) KBr b) NaNO 2 c) NH 4 Br d) NH 4 F neutral basic acidic K a for HF = 6.6 x 10 -4 K b for NH 3 = 1.8 x 10 -5 acidic

113. 113 Factors that affect acid strength There are several factors that can affect acid strength, and the importance of the factors can be variable. However, two trends are notable. 1) Bond strength – The strength of the bond between the acidic proton and the rest of the molecule will have an effect on acidity. The weaker the bond, the stronger the acid will generally be.

114. 114 Factors that affect acid strength 2) Bond polarity – The polarity of a bond is the distribution of the electrons between the two bonded atoms. A highly polar bond between an acidic hydrogen and another atom tends to make it more easy for the proton to leave the molecule than would happen for a non- polar bond.

115. 115 Polarity versus strength In the binary acids the bond strength is the more important factor. Bond strength tends to decrease down a column in the periodic table. HF is the weakest binary acid even though it has the most polar bond because it has the strongest bond.

116. 116 Polarity versus strength For acids of elements in the same row, the bond strengths tend to be similar, and so the polarity of the bond plays the greater role in determining acid strength.

117. 117 Polarity versus strength Combining the decrease of bond strength down a column and increase of bond polarity across a row we find the strongest acids tend to be those of the elements in the bottom right of the periodic table.

118. 118 Oxoacids and acid strength For oxoacids, acid strength tends to increase with the electronegativity of the central atom, and with an increase in the central atom oxidation number (which generally increases with the number of other atoms bonded to the central atom).

119. 119 Oxoacids and acid strength Here we see three oxoacids with different central atoms. Oxoacid strength increases with electronegativity of X

120. 120 Oxoacids and acid strength Oxoacids with the same central atom X will be strongest when many other atoms are bonded to X (the oxidation number of X increases.)

121. 121 Organic compounds and acid strength The acid strength of organic compounds can be rationalized by the bond strength between the proton and the atoms it’s bonded to (weaker bond - stronger acid) but more correctly it is determined by the stability of the conjugate base. Any factor that tends to stabilize the conjugate base increases organic acid strength.

122. 122 Organic compounds and acid strength

123. 123 Organic compounds and acid strength fluoroacetic acid K a = 2.7 x 10 -3 chlororoacetic acid K a = 1.4 x 10 -3

124. 124 Amines and base strength We require a lone pair of electrons to have a BL base, so any factor that tends to reduce the availability of the lone pair will weaken the base while any factor that tends to increase the availability of the lone pair will strengthen the base.

125. 125 Amines and base strength Alkyl groups are slightly electron donating, so secondary and tertiary amines tend to be slightly stronger bases than ammonia and primary amines.

126. 126 Amines and base strength Factors that stablize the structure of amines will decrease the base strength due to reduced electron availability.

127. 127 Amines and base strength Factors that stablize the structure of amines will decrease the base strength due to reduced electron availability.

128. 128 Lewis acids and bases A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. These definitions are more general than the BL definitions because protons aren’t involved which means there exist Lewis acids that are not Brønsted- Lowry acids.

129. 129 Lewis acids and bases We’ve seen that all Brønsted-Lowry bases must all have at least one lone pair of electrons, so any Brønsted-Lowry base must also be a Lewis base, and any Lewis base must also be a Brønsted-Lowry base!

130. 130 In general LA + :LB  LA-LB

131. 131 In general LA + :LB  LA-LB

132. 132 Coordination compounds We’ve already seen that coordination compounds have complex ions that are formed from a central metal ion (a Lewis acid) surrounded by ligands (Lewis bases)

133. 133 Acidic solutions of metal ions Small and/or highly charged metal ions, like Al 3+, Be 2+, and Li + form acidic solutions because they form complex ions with water. The protons of the water ligands see less electron density than in free water, and so the O-H bond is weaker, leading to increased acidity!

134. 134 Acidic solutions of metal ions There is a second benefit as well because the complex ion has a smaller charge in a larger volume…. K a = 1.74 x 10 -5 – about the same as acetic acid!