1. Acids and Bases Chapter 15

2. Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + Mg (s) MgCl 2 (aq) + H 2 (g) 2HCl (aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) Aqueous acid solutions conduct electricity.

3. Have a bitter taste. Feel slippery. Many soaps contain bases. Bases Cause color changes in plant dyes. Aqueous base solutions conduct electricity.

4. Pg 129

6. Common Household acids and bases.

7. Types of Chemical Reactions The Arrhenius Concept Acid-Base Reactions The Arrhenius concept defines acids as substances that produce hydrogen ions, H +, when dissolved in water. An example is nitric acid, HNO 3, a molecular substance that dissolves in water to give H + and NO 3 -.

8. Types of Chemical Reactions Acid-Base Reactions The Arrhenius concept defines bases as substances that produce hydroxide ions, OH -, when dissolved in water. An example is sodium hydroxide, NaOH, an ionic substance that dissolves in water to give sodium ions and hydroxide ions. The Arrhenius Concept

9. Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

10. Hydronium ion, hydrated proton, H 3 O +

11. Types of Chemical Reactions Acid-Base Reactions The Brønsted-Lowry concept of acids and bases involves the transfer of a proton (H + ) from the acid to the base. In this view, acid-base reactions are proton-transfer reactions. The Brønsted-Lowry Concept

12. A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase A Brønsted acid must contain at least one ionizable proton!

13. Types of Chemical Reactions Acid-Base Reactions The Brønsted-Lowry concept defines an acid as the species (molecule or ion) that donates a proton (H + ) to another species in a proton-transfer reaction. A base is defined as the species (molecule or ion) that accepts the proton (H + ) in a proton-transfer reaction. The Brønsted-Lowry Concept

14. Types of Chemical Reactions Acid-Base Reactions the H 2 O molecule is the acid because it donates a proton. The NH 3 molecule is a base, because it accepts a proton. H+H+ In the reaction of ammonia with water, The Brønsted-Lowry Concept

15. Types of Chemical Reactions Acid-Base Reactions This “mode of transportation” for the H + ion is called the hydronium ion. The H + (aq) ion associates itself with water to form H 3 O + (aq). The Brønsted-Lowry Concept

16. Types of Chemical Reactions Acid-Base Reactions The dissolution of nitric acid, HNO 3, in water is therefore a proton-transfer reaction The Brønsted-Lowry Concept where HNO 3 is an acid (proton donor) and H 2 O is a base (proton acceptor). H+H+

17. Types of Chemical Reactions Acid-Base Reactions –The Arrhenius concept acid: proton (H + ) donor base: hydroxide ion (OH - ) donor In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution.

18. Types of Chemical Reactions Acid-Base Reactions –The Brønsted-Lowry concept acid: proton (H + ) donor base: proton (H + ) acceptor In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution.

19. Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH 3 COO -, (c) H 2 PO 4 - HI (aq) H + (aq) + I - (aq)Brønsted acid CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq)Brønsted base H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) H 2 PO 4 - (aq) + H + (aq) H 3 PO 4 (aq) Brønsted acid Brønsted base

20. Brønsted-Lowry Concept of Acids and Bases Some species can act as an acid or a base. For example, HCO 3 - acts as a proton donor (an acid) in the presence of OH - H+H+ An amphoteric species is a species that can act either as an acid or a base (it can gain or lose a proton).

21. Brønsted-Lowry Concept of Acids and Bases Some species can act as an acid or a base. An amphoteric species is a species that can act either as an acid or a base (it can gain or lose a proton). Alternatively, HCO 3 can act as a proton acceptor (a base) in the presence of HF. H+H+

22. Brønsted-Lowry Concept of Acids and Bases The amphoteric characteristic of water is important in the acid-base properties of aqueous solutions. Water reacts as an acid with the base NH 3. H+H+

23. Brønsted-Lowry Concept of Acids and Bases The amphoteric characteristic of water is important in the acid-base properties of aqueous solutions. Water can also react as a base with the acid HF. H+H+

24. Self-ionization of Water Self-ionization is a reaction in which two like molecules react to give ions. In the case of water, the following equilibrium is established. The equilibrium-constant expression for this system is:

25. O H H+ O H H O H HH O H - + [] + Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid autoionization of water

26. Self-ionization of Water Self-ionization is a reaction in which two like molecules react to give ions. The concentration of ions is extremely small, so the concentration of H 2 O remains essentially constant. This gives: constant

27. Self-ionization of Water Self-ionization is a reaction in which two like molecules react to give ions. We call the equilibrium value for the ion product [H 3 O + ][OH - ] the ion-product constant for water, which is written K w. At 25 o C, the value of K w is 1.0 x 10 -14. Like any equilibrium constant, K w varies with temperature.

28. Self-ionization of Water These ions are produced in equal numbers in pure water, so if we let x = [H + ] = [OH - ] Thus, the concentrations of H + and OH - in pure water are both 1.0 x 10 -7 M. If you add acid or base to water they are no longer equal but the K w expression still holds.

29. Solutions of Strong Acid or Base By dissolving substances in water, you can alter the concentrations of H + (aq) and OH - (aq). In a neutral solution, the concentrations of H + (aq) and OH - (aq) are equal, as they are in pure water. In an acidic solution, the concentration of H + (aq) is greater than that of OH - (aq). In a basic solution, the concentration of OH - (aq) is greater than that of H + (aq).

30. Solutions of Strong Acid or Base At 25 o C, you observe the following conditions. In an acidic solution, [H + ] > 1.0 x 10 -7 M. In a neutral solution, [H + ] = 1.0 x 10 -7 M. In a basic solution, [H + ] < 1.0 x 10 -7 M.

31. The pH of a Solution Although you can quantitatively describe the acidity of a solution by its [H + ], it is often more convenient to give acidity in terms of pH. The pH of a solution is defined as the negative logarithm of the molar hydrogen- ion concentration.

32. The pH of a Solution For a solution in which the hydrogen-ion concentration is 1.0 x 10 -3, the pH is: Note that the number of decimal places in the pH equals the number of significant figures in the hydrogen-ion concentration.

33. The pH of a Solution In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10 -7, the pH = 7.00. For acidic solutions, the hydrogen-ion concentration is greater than 1.0 x 10 -7, so the pH is less than 7.00. Similarly, a basic solution has a pH greater than 7.00.

34. The pH Scale

35. A Problem to Consider A sample of orange juice has a hydrogen-ion concentration of 2.9 x 10 -4 M. What is the pH?

36. A Problem to Consider The pH of human arterial blood is 7.40. What is the hydrogen-ion concentration?

37. The pH of a Solution A measurement of the hydroxide ion concentration, similar to pH, is the pOH. The pOH of a solution is defined as the negative logarithm of the molar hydroxide- ion concentration.

38. The pH of a Solution A measurement of the hydroxide ion concentration, similar to pH, is the pOH. Then because K w = [H + ][OH - ] = 1.0 x 10 -14 at 25 o C, you can show that

39. A Problem to Consider An ammonia solution has a hydroxide-ion concentration of 1.9 x 10 -3 M. What is the pH of the solution? You first calculate the pOH: Then the pH is:

40. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = 10 -4.82 = 1.5 x 10 -5 M The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

41. The pH of a Solution The pH of a solution can accurately be measured using a pH meter

42. Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq)

43. HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) H2OH2O KOH (s) K + (aq) + OH - (aq) H2OH2O Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O

44. F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Conjugate acid-base pairs: The conjugate base of a strong acid has no measurable strength. H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqueous solution.

46. Strong AcidWeak Acid

47. What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

48. Solutions of Strong Acid or Base As an example, calculate the concentration of OH - ion in 0.10 M HCl. Because you started with 0.10 M HCl (a strong acid) the reaction will produce 0.10 M H + (aq). Substituting [H + ]=0.10 into the ion-product expression, we get:

49. Solutions of Strong Acid or Base Similarly, in a solution of a strong base you can normally ignore the self-ionization of water as a source of OH - (aq). The OH - (aq) concentration is usually determined by the strong base concentration. However, the self-ionization still exists and is responsible for a small concentration of H + ion.

50. Solutions of Strong Acid or Base As an example, calculate the concentration of H + ion in 0.010 M NaOH. Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH - (aq). Substituting [OH - ]=0.010 into the ion- product expression, we get:

51. Solutions of Strong Acid or Base As an example, calculate the concentration of H + ion in 0.010 M NaOH. Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH - (aq). Substituting [OH - ]=0.010 into the ion- product expression, we get:

52. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength

54. What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M

55. When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation.

56. 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.

57. 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

58. Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution.

59. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M. We will abbreviate the formula for nicotinic acid as HNic.

60. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. Starting 0.012 00 Change -x +x Equilibrium 0.012-x xx Let x be the moles per liter of product formed.

61. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. The equilibrium-constant expression is:

62. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. Substituting the expressions for the equilibrium concentrations, we get

63. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. We can obtain the value of x from the given pH.

64. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. Substitute this value of x in our equilibrium expression. Note first, however, that the concentration of unionized acid remains virtually unchanged.

65. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. Substitute this value of x in our equilibrium expression.

66. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 o C. Calculate the acid- ionization constant for this acid at 25 o C. To obtain the degree of dissociation: The percent ionization is obtained by multiplying by 100, which gives 3.4%.

67. Calculations With K a How do you know when you can use this simplifying assumption? then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%. It can be shown that if the acid concentration, C a, divided by the K a exceeds 100, that is,

68. Calculations With K a How do you know when you can use this simplifying assumption? If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation. The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC 9 H 7 O 4, a common headache remedy.

69. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. The molar mass of HC 9 H 7 O 4 is 180.2 g. From this we find that the sample contained 0.00180 mol of the acid.

70. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. The molar mass of HC 9 H 7 O 4 is 180.2 g. Hence, the concentration of the acetylsalicylic acid is 0.00180 mol/0.500 L = 0.0036 M (Retain two significant figures, the same number of significant figures in K a ).

71. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. Note that which is less than 100, so we must solve the equilibrium equation exactly.

72. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. We will abbreviate the formula for acetylsalicylic acid as HAcs and let x be the amount of H 3 O + formed per liter. The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is (0.0036-x) mol.

73. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. Starting 0.0036 00 Change -x +x Equilibrium 0.0036-x xx These data are summarized below.

74. The equilibrium constant expression is A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C.

75. If we substitute the equilibrium concentrations and the K a into the equilibrium constant expression, we get A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C.

76. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. Rearranging the preceding equation to put it in the form ax 2 + bx + c = 0, we get You can solve this equation exactly by using the quadratic formula.

77. Now substitute into the quadratic formula. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C.

78. Now substitute into the quadratic formula. A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. The lower sign in ± gives a negative root which we can ignore

79. Taking the upper sign, we get A Problem To Consider What is the pH at 25 o C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25 o C. Now we can calculate the pH.

80. percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration

81. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

83. Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak Acid and Its Conjugate Base Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa

85. Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows. The corresponding equilibrium constant is:

86. Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows. The concentration of water is nearly constant.

87. Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. In general, a weak base B with the base ionization has a base ionization constant equal to

88. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. 1.Write the equation and make a table of concentrations. 2.Set up the equilibrium constant expression. 3.Solve for x = [OH - ]. As before, we will follow the three steps in solving an equilibrium.

89. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. Pyridine ionizes by picking up a proton from water (as ammonia does). Starting 0.20 00 Change -x +x Equilibrium 0.20-x xx

90. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. Note that which is much greater than 100, so we may use the simplifying assumption that (0.20-x)  (0.20).

91. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. The equilibrium expression is

92. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. If we substitute the equilibrium concentrations and the K b into the equilibrium constant expression, we get

93. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. Using our simplifying assumption that the x in the denominator is negligible, we get

94. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. Solving for x we get

95. A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. Solving for pOH Since pH + pOH = 14.00

96. Molecular Structure and Acid Strength H X H + + X - The stronger the bond The weaker the acid HF << HCl < HBr < HI

97. Molecular Structure and Acid Strength Two factors are important in determining the relative acid strengths. One is the polarity of the bond to which the hydrogen atom is attached. The H atom should have a partial positive charge: ++ -- The more polarized the bond, the more easily the proton is removed and the greater the acid strength.

98. Molecular Structure and Acid Strength Two factors are important in determining the relative acid strengths. The second factor is the strength of the bond. Or, in other words, how tightly the proton is held. This depends on the size of atom X. ++ -- The larger atom X, the weaker the bond and the greater the acid strength.

99. Molecular Structure and Acid Strength Consider a series of binary acids from a given column of elements. As you go down the column of elements, the radius increases markedly and the H-X bond strength decreases. You can predict the following order of acidic strength.

100. Molecular Structure and Acid Strength As you go across a row of elements, the polarity of the H-X bond becomes the dominant factor. As electronegativity increases going to the right, the polarity of the H-X bond increases and the acid strength increases. You can predict the following order of acidic strength.

101. Molecular Structure and Acid Strength Z O HZ O-O- + H + -- ++ The O-H bond will be more polar and easier to break if: Z is very electronegative or Z is in a high oxidation state

102. Molecular Structure and Acid Strength 1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. Acid strength increases with increasing electronegativity of Z H O Cl O O H O Br O O Cl is more electronegative than Br HClO 3 > HBrO 3

103. Molecular Structure and Acid Strength 2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO 4 > HClO 3 > HClO 2 > HClO

104. Molecular Structure and Acid Strength Consider the oxoacids. An oxoacid has the structure: The acidic H atom is always attached to an O atom, which in turn is attached to another atom Y. Bond polarity is the dominant factor in the relative strength of oxoacids. This, in turn, depends on the electronegativity of the atom Y.

105. Molecular Structure and Acid Strength Consider the oxoacids. An oxoacid has the structure: If the electronegativity of Y is large, then the O-H bond is relatively polar and the acid strength is greater. You can predict the following order of acidic strength.

106. Molecular Structure and Acid Strength Consider the oxoacids. An oxoacid has the structure: Other groups, such as O atoms or O-H groups, may be attached to Y. With each additional O atom, Y becomes effectively more electronegative.

107. Molecular Structure and Acid Strength Consider the oxoacids. An oxoacid has the structure: As a result, the H atom becomes more acidic. The acid strengths of the oxoacids of chlorine increase in the following order.

108. Molecular Structure and Acid Strength Consider polyprotic acids and their corresponding anions. Each successive H atom becomes more difficult to remove. Therefore the acid strength of a polyprotic acid and its anions decreases with increasing negative charge.

109. 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 COOH (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq)

110. 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+

111. Acid Hydrolysis of Al 3+

112. 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

113. Acid-Base Properties of a Salt Solution One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. A 0.1 M solution has a pH of 11.1 and is therefore fairly basic. Consider a solution of sodium cyanide, NaCN.

114. Acid-Base Properties of a Salt Solution One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Sodium ion, Na +, is unreactive with water, but the cyanide ion, CN -, reacts to produce HCN and OH -.

115. Acid-Base Properties of a Salt Solution One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. From the Brønsted-Lowry point of view, the CN - ion acts as a base, because it accepts a proton from H 2 O.

116. Acid-Base Properties of a Salt Solution One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. You can also see that OH - ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic. The reaction of the CN - ion with water is referred to as the hydrolysis of CN -.

117. Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The CN- ion hydrolyzes to give the conjugate acid and hydroxide.

118. Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The hydrolysis reaction for CN - has the form of a base ionization so you writ the K b expression for it.

119. Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The NH 4 + ion hydrolyzes to the conjugate base (NH 3 ) and hydronium ion.

120. Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. This equation has the form of an acid ionization so you write the K a expression for it.

121. Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral? The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs. Consequently, the anions of weak acids (poor proton donors) are good proton acceptors. Anions of weak acids therefore, are basic.

122. Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral? One the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. For example, the Cl - ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water.

123. Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral? Conversely, the cations of weak bases are acidic. One the other hand, the cations of strong bases have virtually no acidic character, that is, they do not hydrolyze. For example,

124. Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Consider potassium acetate, KC 2 H 3 O 2. The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

125. Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Consider potassium acetate, KC 2 H 3 O 2. The acetate ion, however, is the anion of a weak acid (HC 2 H 3 O 2 ) and is basic. A solution of potassium acetate is predicted to be basic.

126. Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 1.A salt of a strong base and a strong acid. The salt has no hydrolyzable ions and so gives a neutral aqueous solution. An example is NaCl.

127. Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution. An example is NaCN. 2.A salt of a strong base and a weak acid.

128. Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution. An example is NH 4 Cl. 3.A salt of a weak base and a strong acid.

129. Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) Both ions hydrolyze. You must compare the K a of the cation with the K b of the anion. If the K a of the cation is larger the solution is acidic. If the K b of the anion is larger, the solution is basic. 4.A salt of a weak base and a weak acid.

130. The pH of a Salt Solution To calculate the pH of a salt solution would require the K a of the acidic cation or the K b of the basic anion. The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. Thus the K b for CN - is related to the K a for HCN.

131. The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb When these two reactions are added you get the ionization of water. KwKw

132. The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb When two reactions are added, their equilibrium constants are multiplied. KwKw

133. The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb Therefore, KwKw

134. The pH of a Salt Solution For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. The only difference is first obtaining the K a or K b for the ion that hydrolyzes. The next example illustrates the reasoning and calculations involved.

135. A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 o C? The K a for HCN is 4.9 x 10 -10. Sodium cyanide gives Na + ions and CN - ions in solution. Only the CN - ion hydrolyzes.

136. A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 o C? The K a for HCN is 4.9 x 10 -10. The CN - ion is acting as a base, so first, we must calculate the K b for CN -. Now we can proceed with the equilibrium calculation.

137. A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 o C? The K a for HCN is 4.9 x 10 -10. Let x = [OH - ] = [HCN], then substitute into the equilibrium expression.

138. A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 o C? The K a for HCN is 4.9 x 10 -10. This gives

139. A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 o C? The K a for HCN is 4.9 x 10 -10. Solving the equation, you find that Hence, As expected, the solution has a pH greater than 7.0.

140. Oxides of the Representative Elements In Their Highest Oxidation States CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) N 2 O 5 (g) + H 2 O (l) 2HNO 3 (aq)

141. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO 4 -. Sulfuric acid, for example, can lose two protons in aqueous solution.

142. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. For a weak diprotic acid like carbonic acid, H 2 CO 3, two simultaneous equilibria must be considered.

143. Each equilibrium has an associated acid- ionization constant. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. For the loss of the first proton

144. Each equilibrium has an associated acid- ionization constant. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. For the loss of the second proton

145. In the case of a triprotic acid, such as H 3 PO 4, the third ionization constant, K a3, is smaller than the second one, K a2. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. In general, the second ionization constant, K a2, for a polyprotic acid is smaller than the first ionization constant, K a1.

146. However, reasonable assumptions can be made that simplify these calculations as we show in the next example. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.

147. For diprotic acids, K a2 is so much smaller than K a1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

148. The pH can be determined by simply solving the equilibrium problem posed by the first ionization. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

149. If we abbreviate the formula for ascorbic acid as H 2 Asc, then the first ionization is: A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

150. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12. Starting 0.10 00 Change -x +x Equilibrium 0.10-x xx

151. The equilibrium constant expression is A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

152. Substituting into the equilibrium expression A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

153. Assuming that x is much smaller than 0.10, you get A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

154. The hydronium ion concentration is 0.0028 M, so A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

155. The ascorbate ion, Asc 2-, which we will call y, is produced only in the second ionization of H 2 Asc. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

156. Assume the starting concentrations for HAsc - and H 3 O + to be those from the first equilibrium. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

157. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12. Starting 0.0028 0 Change -y +y Equilibrium 0.0028-x 0.0028+yy

158. The equilibrium constant expression is A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

159. Substituting into the equilibrium expression A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

160. Assuming y is much smaller than 0.0028, the equation simplifies to A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

161. Hence, A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12. The concentration of the ascorbate ion equals K a2.

162. Arrhenius acid is a substance that produces H + (H 3 O + ) in water A Brønsted acid is a proton donor A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons Definition of An Acid H+H+ H O H + OH - acidbase N H H H H+H+ + acidbase N H H H H +

163. Lewis Concept of Acids and Bases The reaction of boron trifluoride with ammonia is an example. Boron trifluoride accepts the electron pair, so it is a Lewis acid. Ammonia donates the electron pair, so it is the Lewis base. + N H H H : : : : B F F F :: :: : : : : : B F F F :: :: : : N H H H

164. Chemistry In Action: Antacids and the Stomach pH Balance NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) Mg(OH) 2 (s) + 2HCl (aq) MgCl 2 (aq) + 2H 2 O (l)

165. WORKED EXAMPLES

166. Worked Example 15.1

167. Worked Example 15.2

171. Worked Example 15.6

172. Worked Example 15.7

173. Worked Example 15.8a

175. Worked Example 15.9

176. Worked Example 15.10

178. Worked Example 15.11b

180. Worked Example 15.12

181. Worked Example 15.13