1. 1 Acids and Bases Chapter 15

2. Acids & Bases: Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent (used in organic chemistry, wider range of substances)

3. Arrhenius (or Classical) Acid-Base Definition An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion : H + (H 3 O + ) A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH - Neutralization is the reaction of an H + (H 3 O + ) ion from the acid and the OH - ion from the base to form water, H 2 O. The neutralization reaction is exothermic and releases approximately 56 kJ per mole of acid and base. H + (aq) + OH - (aq) H 2 O(l) H 0 rxn = -55.9 kJ

4. The Acid-Dissociation Constant (K a ) Strong acids dissociate completely (100%) into ions in water: HA (g or l) + H 2 O (l) H 3 O + (aq) + A - (aq) In a dilute solution of a strong acid, almost no HA molecules exist: [H 3 O + ] = [HA] initial or [HA] eq = 0 Q c = [H 3 O + ][A - ] [HA][H 2 O] at equilibrium, Q c = K c >> 1 Nitric acid is an example: HNO 3 (l) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq)

5. The Acid-Dissociation Constant (K a ) Weak acids dissociate very slightly (5% – 10%) into ions in water: HA (aq) + H 2 O (aq) H 3 O + (aq) + A - (aq) In a dilute solution of a weak acid, the great majority of HA molecules are undissociated: [H 3 O + ] << [HA] initial or [HA] eq = [HA] initial Q c = [H 3 O + ][A - ] [HA][H 2 O] at equilibrium, Q c = K c << 1

6. The Meaning of K a, the Acid Dissociation Constant For the ionization of an acid, HA: HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) K c = [H 3 O + ] [A - ] [HA] [H 2 O] Since the concentration of water is high, and does not change significantly during the reaction, it’s value is absorbed into the constant. Therefore: K c = [H 3 O + ] [A - ] [HA] The stronger the acid, the higher the [H 3 O + ] at equilibrium, and the larger the K a : Stronger acid higher [H 3 O + ] larger K a For a weak acid with a relative high K a (~10 -2 ), a 1 M solution has ~10% of the HA molecules dissociated. For a weak acid with a moderate K a (~10 -5 ), a 1 M solution has ~ 0.3% of the HA molecules dissociated. For a weak acid with a relatively low K a (~10 -10 ), a 1 M solution has ~ 0.001% of the HA molecules dissociated.

8. The behavior of acids of different strengths in aqueous solution. A Strong Acid A Weak Acid

9. The Six Strong Acids Hydrogen Halides HCl Hydrochloric Acid HBr Hydrobromic Acid HI HydroIodioic Acid Oxyacids H 2 SO 4 Sulfuric Acid HNO 3 Nitric Acid HClO 4 Perchloric Acid

10. STRONG ACID: HNO 3 (aq) + H 2 O(liq)  H 3 O + (aq) + NO 3 - (aq) (100% dissociated in water.) Strong & Weak Acids/Bases

11. Example 15-1: Calculate the concentrations of ions in 0.050 M nitric acid, HNO 3.

12. Weak acids (much less than 100% ionized in water.) Example: acetic acid = CH 3 CO 2 H Strong & Weak Acids/Bases

13. Weak base:Weak base: (much less than 100% ionized in water.) Example: ammonia NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) Strong & Weak Acids/Bases

14. Acids & Bases: Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent (used in organic chemistry, wider range of substances)

15. Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species that donates an H + ion. An acid must contain H in its formula; HNO 3 and H 2 PO 4 - are two examples, all Arrhenius acids are Brønsted-Lowry acids. A base is a proton acceptor, any species that accepts an H + ion. A base must contain a lone pair of electrons to bind the H + ion; a few examples are NH 3, CO 3 2-, F -, as well as OH -. Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain the Brønsted-Lowry base OH -. Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process. Acids donate a proton to water Bases accept a proton from water

17. Molecular model: The reaction of an acid HA with water to form H 3 O+ and a conjugate base. Acid Base Conjugate Conjugate acid base HA H 2 O H 3 O + A -

18. NH 3 BASEwaterACID Using the Brønsted-Lowry definition, NH 3 is a BASE in water and water is itself an ACID Proton acceptor …NH 3 NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) Proton donor…H 2 O Proton donor…H 2 O The Brønsted–Lowry Concept of Acids & Bases Extended

19. Acids such as HF, HCl, HNO 3, and CH 3 CO 2 H (acetic acid) are all capable of donating one proton and so are called monoprotic acids. Other acids, called polyprotic acids are capable of donating two or more protons. The Brønsted–Lowry Concept of Acids & Bases Extended

20. The Stepwise Dissociation of Phosphoric Acid Phosphoric acid is a weak acid, and normally only looses one proton in solution, but it will lose all three when reacted with a strong base with heat. The ionization constants are given for comparison. H 3 PO 4 (aq) + H 2 O (l) H 2 PO 4 - (aq) + H 3 O + (aq) H 2 PO 4 - (aq) + H 2 O (l) HPO 4 2- (aq) + H 3 O + (aq) HPO 4 2- (aq) + H 2 O (l) PO 4 3- (aq) + H 3 O + (aq) H 3 PO 4 (aq) + 3 H 2 O (l) PO 4 3- (aq) + 3 H 3 O + (aq)

23. A conjugate acid–base pair consists of two species that differ from each other by the presence of one hydrogen ion. Every reaction between a Brønsted acid and a Brønsted base involves two conjugate acid–base pairs Conjugate Acid–Base Pairs

24. The Conjugate Pairs in Some Acid-Base Reactions Acid + Base Base + Acid Conjugate Pair Reaction 1 HF + H 2 O F – + H 3 O + Reaction 2 HCOOH + CN – HCOO – + HCN Reaction 3 NH 4 + + CO 3 2– NH 3 + HCO 3 – Reaction 4 H 2 PO 4 – + OH – HPO 4 2– + H 2 O Reaction 5 H 2 SO 4 + N 2 H 5 + HSO 4 – + N 2 H 6 2+ Reaction 6 HPO 4 2– + SO 3 2– PO 4 3– + HSO 3 –

25. Identifying Conjugate Acid-Base Pairs Problem: The following chemical reactions are important for industrial processes. Identify the conjugate acid-base pairs. (a) HSO 4 - (aq) + CN - (aq) SO 4 2- (aq) + HCN (aq) (b) ClO - (aq) + H 2 O (l) HClO (aq) + OH - (aq) (c) S 2- (aq) + H 2 O (aq) HS - (aq) + OH - (aq) Plan: To find the conjugate acid-base pairs, we find the species that donate H + and those that accept it. The acid (or base) on the left becomes its conjugate base (or acid) on the right. Solution: (a) The proton is transferred from the sulfate to the cyanide so: HSO 4 - (aq) /SO 4 2- (aq) and CN - (aq) /HCN (aq ) are the two acid-base pairs. (b) The water gives up one proton to the hypochlorite anion so: ClO - (aq) /HClO (aq) and H 2 O (l) / OH - (aq ) are the two acid-base pairs. (c) One of water’s protons is transferred to the sulfide ion so: S 2- (aq) /HS - (aq) and H 2 O (l) /OH - (aq) are the two acid-base pairs.

26. Conjugate Acid–Base Pairs

27. Relative Strengths of Acids and Bases The stronger an acid, the weaker its conjugate base and vice versa.

32. The Autoionization of Water H 2 O(l) + H 2 O(l) ⇌ H 3 O + (aq) + OH - (aq) H 2 O(l) ⇌ H + (aq) + OH - (aq)  K w = [H 3 O + ][OH - ] = [H + ][OH - ] = 1.0 x 10 -14 (@ 25°C)  K w – ion-product constant (or dissociation constant)  Pure water is neutral. Thus, [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M @ 25°C

33. Autoionization of Water H 2 O (l) + H 2 O (l) H 3 O + + OH - K c = [H 3 O + ][OH - ] [H 2 O] 2 The ion-product for water, K w : K c [H 2 O] 2 = K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 (at 25°C) For pure water the concentration of hydroxyl and hydronium ions must be equal: [H 3 O + ] = [OH - ] = 1.0 x 10 -14 = 1.0 x 10 -7 M (at 25°C) The molarity of pure water is: = 55.5 M 1000g/L 18.02 g/mol

34.  For an aqueous solution: The Autoionization of Water

35. 16.4 The pH Scale  pH represents a solution’s acidity (@ 25°C). 0 ← 7 → 14 acidic neutral basic  pH = −log[H 3 O + ] = −log[H + ] [H 3 O + ] = 10 -pH pOH = −log[OH - ] pH + pOH = 14 [OH - ] = 10 -pOH

36. In a neutral solution, [H 3 O + ] = [OH  ] Both are equal to 1.00  10  7 M In an acidic solution, [H 3 O + ] > [OH  ] [H 3 O + ] > 1.00  10  7 M and [OH  ] < 1.00  10  7 M In a basic solution, [H 3 O + ] < [OH  ] [H 3 O + ] 1.00  10  7 M Water & the pH Scale

37. The pH Scale

38. The pH and pOH scales If either the [H 3 O + ] or [OH - ] is known, the pH and pOH can be calculated. Example 15-2: Calculate the pH of a solution in which the [H 3 O + ] =0.030 M.

39. The pH and pOH scales Example 15-3: The pH of a solution is 4.597. What is the concentration of H 3 O + ?

40. The pH and pOH scales Example 15-4: Calculate the [H 3 O + ], pH, [OH - ], and pOH for a 0.020 M HNO 3 solution. –Is HNO 3 a weak or strong acid? –What is the [H 3 O + ] ?

41. The pH and pOH scales Example 15-5: Calculate the [H 3 O + ], pH, [OH - ], and pOH for a 0.020 M HNO 3 solution.

42. The pH Values of Some Familiar Aqueous Solutions [H 3 O + ] [OH - ] [OH - ] = KWKW [H 3 O + ] neutral solution acidic solution basic solution [H 3 O + ]> [OH - ] [H 3 O + ]< [OH - ] [H 3 O + ] = [OH - ]

43. The Relationship Between K a and pK a (-logK a ) Acid Name (Formula) K a at 25 o C pK a Hydrogen sulfate ion (HSO 4 - ) 1.02 x 10 -2 1.991 Nitrous acid (HNO 2 ) 7.1 x 10 -4 3.15 Acetic acid (CH 3 COOH) 1.8 x 10 -5 4.74 Hypobromous acid (HBrO) 2.3 x 10 -9 8.64 Phenol (C 6 H 5 OH) 1.0 x 10 -10 10.00

45. Figure: (a) Measuring the pH of vinegar. (b) Measuring the pH of aqueous ammonia.

46. Methods for Measuring the pH of an Aqueous Solution (a) pH paper (b) Electrodes of a pH meter

47. p. 663

48. More about pH  pH does not necessarily indicate strength.  Measuring pH pH meters – measures exact pH based on electrochemistry Acid-base indicators – estimates pH based on the appearance of color

49. p. 664

50. Indicator Colors.

51. We will discuss the quantitative aspects of dissociation of weak acids and bases. The relative strengths of weak acids and bases can be ranked based on the magnitude of individual equilibrium constants. Equilibrium Constants for Acids & Bases

52. Strong acids and bases almost completely ionize in water (~100%): K strong >> 1 (product favored) Weak acids and bases almost completely ionize in water (<<100%): K weak << 1 (Reactant favored) Equilibrium Constants for Acids & Bases

53. HA(aq) ⇌ H + (aq) + A - (aq) K a ↑ acid strength ↑ For polyprotic acids: K a1 >> K a2 >> K a3 pK a = −log[K a ] pK a ↑ acid strength↓ Weak Acids & Acid-dissociation Constant

54. Weak Bases & Base-dissociation Constant Weak bases establish an equilibrium in aqueous solution. B(aq) + H 2 O(l) ⇌ BH + (aq) + OH - (aq) K b ↑ base strength ↑ pK b = −log[K b ] pK b ↑ base strength↓

55. % Dissociation (or ionization)  % dissociation increases as acid/base strength increases.  % dissociation decreases as concentration increases.

56. AcidHCO 3  HClOHF KaKa 4.8  10  11 3.5  10  8 7.2  10  4 BaseCO 3 2  ClO  FF KbKb 2.1  10  4 2.9  10  7 1.4  10  11 Equilibrium Constants for Acids & Bases

57. Problem: Calculate the Percent dissociation of a 0.0100M Hydrocyanic acid solution, K a = 6.20 x 10 -10. HCN (aq) + H 2 O (l) H 3 O + (aq) + CN - (aq) HCN H 3 O + CN - Initial 0.0100M 0 0 K a = Change -x +x +x Eq. 0.0100 –x x x K a = = 6.20 x 10 -10 Assume 0.0100-x = 0.0100 K a = = 6.2 x 10 -10 x = 2.49 x 10 -6 % dissociation = x 100% = 0.0249 % [H 3 O + ][CN - ] [HCN] (x)(x) (0.0100-x) x 2 0.0100 2.49 x 10 -6 0.0100

58. Finding the K a of a Weak Acid from the pH of its Solution–I Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the K a of this weak acid. Plan: We are given [HClO] initial and the pH which will allow us to find [H 3 O + ] and, hence, the hypochlorite anion concentration, so we can write the reaction and expression for K a and solve directly. Solution: Calculating [H 3 O + ] :[H 3 O + ] = 10 -pH = 10 -4.19 = 6.46 x 10 -5 M Concentration (M) HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) Initial 0.12 ---- ------- ------- Change -x ---- +x +x Equilibrium 0.12 -x ---- +x +x Assumptions: [H 3 O + ] = [H 3 O + ] HClO since HClO is a weak acid, we assume 0.12 M - x = 0.12 M

59. Finding the K a of a Weak Acid from the pH of its Solution–II x = [H 3 O + ] = [ClO - ] = 6.46 x 10 -5 M K a = = = 348 x 10 -10 HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) [H 3 O + ] [ClO - ] [HClO] (6.46 x 10 -5 M) 0.12 M K a = 3.48 x 10 -8 In text books it is found to be: 3.5 x 10 -8 Checking: 1. For [H 3 O + ] from water : x 100 = 0.155% assumption is OK 1 x 10 -7 M 6.46 x 10 -5 M 2. For [HClO] dissoc : x 100 = 0.0538 % 6.46 x 10 -5 M 0.12 M

60. Determining Concentrations from K a and Initial [HA] Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H 3 O + ] of a 0.125 M HClO solution? K a = 3.5 x 10 -8 Plan: We need to find [H 3 O + ]. First we write the balanced equation and the expression for K a and solve for the hydronium ion concentration. Solution: HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) K a = = 3.5 x 10 -8 [H 3 O + ] [ClO - ] [HClO] Concentration (M) HClO H 2 O H 3 O + ClO - Initial 0.125 ---- 0 0 Change -x ---- +x +x Equilibrium 0.125 - x ---- x x K a = = 3.5 x 10 -8 (x)(x) 0.125-x assume 0.125 - x = 0.125 x 2 = 0.4375 x 10 -8 x = 0.661 x 10 -4

61. Acid–Base Properties of Salts: Practice

63. According to the Brønsted–Lowry theory, all acid– base reactions can be written as equilibria involving the acid and base and their conjugates. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. Predicting the Direction of Acid– Base Reactions

64. When a weak acid is in solution, the products are a stronger conjugate acid and base. Therefore equilibrium lies to the left. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. Predicting the Direction of Acid– Base Reactions

65. Will the following acid/base reaction occur spontaneously? Predicting the Direction of Acid– Base Reactions

66. Will the following acid/base reaction occur spontaneously? K a = 7.5  10  5 K a = 1.8  10  5 K b = 5.6  10  10 K b = 1.3  10  12 Predicting the Direction of Acid– Base Reactions

67. Will the following acid/base reaction occur spontaneously? Equilibrium lies to the right since all proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. K a = 7.5  10  5 K a = 1.8  10  5 K b = 5.6  10  10 K b = 1.3  10  12 Stronger Acid + Stronger BaseWeaker Base + Weaker Acid Predicting the Direction of Acid– Base Reactions

68. Determining pH from K b and Initial [B]–I Problem: Ammonia is commonly used cleaning agent in households and is a weak base, with a K b of 1.8 x 10 -5. What is the pH of a 1.5 M NH 3 solution? Plan: Ammonia reacts with water to form [OH - ] and then calculate [H 3 O + ] and the pH. The balanced equation and K b expression are: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K b = [NH 4 + ] [OH - ] [NH 3 ] Concentration (M) NH 3 H 2 O NH 4 + OH - Initial 1.5 ---- 0 0 Change -x ---- +x +x Equilibrium 1.5 - x ---- x x making the assumption: since K b is small: 1.5 M - x = 1.5 M

69. Determining pH from K b and Initial [B]–II Substituting into the K b expression and solving for x: K b = = = 1.8 x 10 -5 [NH 4 + ] [OH - ] [NH 3 ] (x)(x) 1.5 x 2 = 2.7 x 10 -5 = 27 x 10 -6 x = 5.20 x 10 -3 = [OH - ] = [NH 4 + ] Calculating pH: [H 3 O + ] = = = 1.92 x 10 -12 KwKw [OH - ] 1.0 x 10 -14 5.20 x 10 -3 pH = -log[H 3 O + ] = - log (1.92 x 10 -12 ) = 12.000 - 0.283 pH = 11.717

70. The Relation Between K a and K b of a Conjugate Acid-Base Pair Acid HA + H 2 O H 3 O + + A - Base A - + H 2 O HA + OH - 2 H 2 O H 3 O + + OH - [H 3 O + ] [OH - ] = x [H 3 O + ] [A - ] [HA] [HA] [OH - ] [A - ] K w = K a x K b For HNO 2 K a = 4.5 x 10 -4 K b = 2.2 x 10 -11 K a x K b = (4.5 x 10 -4 )(2.2 x 10 -11 ) = 9.9 x 10 -15 or ~ 10 x 10 -15 = 1 x 10 -14 = K w

72. Table 7.6 (P 259) - I Acid – Base Properties of Aqueous Solutions of Various Types of Salts Type of Salt Examples Comment pH of Solution Cation is from Neither acts as strong base; KCl, KNO 3 an acid or a Neutral anion is from NaCl, NaNO 3 base strong acid Cation is from Anion acts as strong base; NaC 2 H 3 O 2 a base; cation Basic anion is from KCN, NaF has no effect weak acid on pH Cation is conjugate Cation acts as acid of weak base; NH 4 Cl, an acid; anion Acidic anion is from NH 4 NO 3 has no effect strong acid on pH

73. Table 7.6 (P 259) - II Acid – Base Properties of Aqueous Solutions of Various Types of Salts Type of Salt Examples Comment pH of Solution Cation is conjugate Cation acts as Acidic if : acid of weak base NH 4 C 2 H 3 O 2 an acid; anion K a > K b anion is conjugate NH 4 CN acts as a base Basic if : base of weak acid K b > K a Neutral if : K a = K b Cation is highly Hydrated cation charged metal ion; Al(NO 3 ) 3, acts as an acid; Acidic anion is from FeCl 3 anion has no strong acid effect on pH

74. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.10 Change Equilibrium Calculations with Equilibrium Constants

75. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.1000 Change0.10 - x+ x Equilibrium0.10 - xxx Calculations with Equilibrium Constants

76. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.1000 Change0.10 - x+ x Equilibrium0.10 - xxx Calculations with Equilibrium Constants

77. Determining K from Initial Concentrations and pH [H 3 O + ] = [NO 2  ] = 6.76  10  3 Calculations with Equilibrium Constants

78. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial1.00 Change Equilibrium Calculations with Equilibrium Constants

79. [H 2 S][H 3 O + ][HS  ] Initial1.0000 Change- x+ x Equilibrium1.00 - xxx Calculations with Equilibrium Constants Determining K from Initial Concentrations and pH

80. x = 3.2  10  4 pH = 3.50 Calculations with Equilibrium Constants

81. Determining K from Initial Concentrations and pH In general, the approximation that [HA] equilibrium = [HA] initial  x  [HA] initial is valid whenever [HA] initial is greater than or equal to 100  K a. If this is not the case, the quadratic equation must by used. Calculations with Equilibrium Constants

82. Determining pH after an acid/base reaction: Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Calculations with Equilibrium Constants

83. Determining pH after an acid/base reaction: Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Solution: Solution: From the volume and concentration of each solution, the moles of acid and base can be calculated. Knowing the moles after the reaction and the equilibrium constants, the concentration of H 3 O + and pH can be calculated.

84. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. All of the acetic acid is converted to acetate ion.

85. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. All of the acetic acid is converted to acetate ion.

86. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change Equilibrium

87. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

88. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

89. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Since K b  100 > [CH 3 CO 2 - ] initial, the quadratic equation is not needed. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

90. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH.

91. Acids & Bases: Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent (used in organic chemistry, wider range of substances)

92. coordinate covalent bondThe product is often called an acid–base adduct. This type of chemical bond was called a coordinate covalent bond. The Lewis Concept of Acids & Bases

93. Lewis acid a substance that accepts an electron pair Lewis base a substance that donates an electron pair The Lewis Concept of Acids & Bases

94. New bond formed using electron pair from the Lewis base. Coordinate covalent bondCoordinate covalent bond Notice geometry change on reaction. Reaction of a Lewis Acid & Lewis Base

95. The formation of a hydronium ion is an example of a Lewis acid / base reaction H H H BASE O—H O—H H + ACID The H + is an electron pair acceptor. Water with it’s lone pairs is a Lewis acid donor. The Lewis Concept of Acids & Bases

96. Lewis Acid  Base Reactions

97. Metal cations often act as Lewis acids because of open d-orbitals. Lewis Acids & Bases

98. The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 leads to Coordinate Complex ions. Lewis Acids & Bases

99. What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 0.122 – x  0.122 K a << 1 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok.

100. K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09

101. 101 Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be 2+ ) and the conjugate base of a strong acid (e.g. Cl -, Br -, and NO 3 - ). NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Basic Solutions: Salts derived from a strong base and a weak acid. NaCH 3 COO (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq)

102. 102 Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) H2OH2O NH 4 + (aq) NH 3 (aq) + H + (aq) Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid. Al(H 2 O) 6 (aq) Al(OH)(H 2 O) 5 (aq) + H + (aq) 3+ 2+

103. 103 Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: K b for the anion > K a for the cation, solution will be basic K b for the anion < K a for the cation, solution will be acidic K b for the anion  K a for the cation, solution will be neutral

104. 104 What is the molarity of an NH 4 NO 3 (aq) solution that has a pH = 4.80? Strategy Ammonium nitrate is the salt of a strong acid (HNO 3 ) and a weak base (NH 3 ). In NH 4 NO 3 (aq), NH 4 + hydrolyzes and NO 3 – does not. The ICE format must be based on the hydrolysis equilibrium for NH 4 + (aq). In that format [H 3 O + ], derived from the pH, will be a known quantity, and the initial concentration of NH 4 + will be the unknown. Solution We begin by writing the equation for the hydrolysis equilibrium and the equation for K a in terms of K w and K b. As usual, we can calculate [H 3 O + ] from the pH of the solution. log[H 3 O + ] = –pH = –4.80 [H 3 O + ] = 10 –4.80 = 1.6 x 10 –5 M

105. 105 Example 15.15 continued Solution continued If we assume that all the hydronium ion comes from the hydrolysis reaction, we can set up an ICE format in which x represents the unknown initial concentration of NH 4 +. We now substitute equilibrium concentrations into the ionization constant expression for the hydrolysis reaction. We can assume that the ammonium ion is mostly nonhydrolyzed and that the change in [NH 4 + ] is much smaller than the initial [NH 4 + ], so that 1.6 x 10 – 5 << x and we can replace (x – 1.6 x 10 –5 ) by x. Then we can solve for x. The solution is 0.46 M NH 4 NO 3.

106. 106 Example Calculations.

107. 107 Ionization Constants for Weak Monoprotic Acids and Bases Example 15-8: Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant.

108. 108 Ionization Constants for Weak Monoprotic Acids and Bases Example 15-9: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid.

109. 109 Ionization Constants for Weak Monoprotic Acids and Bases Since the weak acid is 5.0% ionized, it is also 95% unionized. Calculate the concentration of all species in solution.

110. 110 Ionization Constants for Weak Monoprotic Acids and Bases Use the concentrations that were just determined in the ionization constant expression to get the value of K a.

111. 111 Ionization Constants for Weak Monoprotic Acids and Bases Example 15-11: Calculate the concentrations of the various species in 0.15 M acetic acid, CH 3 COOH, solution. It is always a good idea to write down the ionization reaction and the ionization constant expression.

112. 112 Ionization Constants for Weak Monoprotic Acids and Bases Next, combine the basic chemical concepts with some algebra to solve the problem.

113. 113 Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with some algebra to solve the problem

114. 114 Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with some algebra to solve the problem

115. 115 Substitute these algebraic quantities into the ionization expression. Ionization Constants for Weak Monoprotic Acids and Bases

116. 116 Ionization Constants for Weak Monoprotic Acids and Bases Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.

117. 117 Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations. Ionization Constants for Weak Monoprotic Acids and Bases

118. 118 Complete the algebra and solve for the concentrations of the species. Ionization Constants for Weak Monoprotic Acids and Bases

119. 119 Ionization Constants for Weak Monoprotic Acids and Bases Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

120. 120 Ionization Constants for Weak Monoprotic Acids and Bases

121. 121 Ionization Constants for Weak Monoprotic Acids and Bases Let us now calculate the percent ionization for the 0.15 M acetic acid. From Example 18-11, we know the concentration of CH 3 COOH that ionizes in this solution. The percent ionization of acetic acid is

122. 122 Ionization Constants for Weak Monoprotic Acids and Bases Example 15-12: Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. K a = 4.0 x 10 -10 for HCN

123. 123 Ionization Constants for Weak Monoprotic Acids and Bases

124. 124 Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization of 0.15 M HCN solution is calculated as in the previous example.

125. 125 Ionization Constants for Weak Monoprotic Acids and Bases Example 15-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

126. 126 Ionization Constants for Weak Monoprotic Acids and Bases

127. 127 Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization for weak bases is calculated exactly as for weak acids.

128. 128 Salts of Strong Bases and Weak Acids Example 15-16: Calculate the hydrolysis constants for the following anions of weak acids. 1.The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, K a =7.2 x 10 -4.

129. 129 Salts of Strong Bases and Weak Acids The cyanide ion, CN-, the anion of hydrocyanic acid, HCN. For HCN, K a = 4.0 x 10 -10.

130. 130 Salts of Strong Bases and Weak Acids Example 15-17: Calculate [OH - ], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

131. 131 Salts of Strong Bases and Weak Acids Set up the equation for the hydrolysis and the algebraic representations of the equilibrium concentrations.

132. 132 Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the hydrolysis constant expression.

133. 133 Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the hydrolysis constant expression.

134. 134 Salts of Strong Bases and Weak Acids The percent hydrolysis for the hypochlorite ion may be represented as:

135. 135 Salts of Weak Bases and Strong Acids Example 15-18: Calculate [H + ], pH, and percent hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH 4 Br, solution. 1.Write down the hydrolysis reaction and set up the table as we have done before:

136. 136 Salts of Weak Bases and Strong Acids 2.Substitute the algebraic expressions into the hydrolysis constant.

137. 137 Salts of Weak Bases and Strong Acids 3.Complete the algebra and determine the concentrations and pH.

138. 138 Salts of Weak Bases and Strong Acids 4.The percent hydrolysis of the ammonium ion in 0.10 M NH 4 Br solution is: