1. Chapter 16 Acids and Bases

2. Arrhenius Definition Acids produce hydrogen ions in aqueous solution. Bases produce hydroxide ions when dissolved in water. Definition is limited to aqueous solutions. Only one kind of base for this definition. NH 3, ammonia could not be an Arrhenius base.

3. Bronsted-Lowry Definitions An acid is a proton (H + ) donor and a base is a proton acceptor. Acids and bases always come in pairs. HCl is an acid. When it dissolves in water it gives its proton to water. HCl + H 2 O(l)  H 3 O + + Cl - Water is a base making the hydronium ion

4. A Brønsted-Lowry acid… …must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

5. Types of Acid-Base reactions Recall…. Strong acid – strong base H + (aq) + OH - (aq)  H 2 O Weak acid – strong base (1) HA  H + + A - (weak acid) (2) H + + OH -  H 2 O So, HA + OH -  A - + H 2 O

6. Types of acid base reactions Strong acid – weak base (1) B + H 2 O  BH + + OH - (weak base) (2) H + + OH -  H 2 O So, B + H +  BH + Common weak bases: ammonia (NH 3 ), methylamine (CH 3 NH 2 )

7. Acid Base Pairs General equation HA(aq) + H 2 O(l)  H 3 O + (aq) + A - (aq) Acid + Base  Conjugate acid +Conjugate base This is an equilibrium. Competition for H + between H 2 O and A - ion If H 2 O is a stronger base it will take the H + Equilibrium moves to right.

8. If it can either add or lose a proton it is amphiprotic. Such as…. HCO 3 - HSO 4 - H 2 O

9. What Happens When an Acid Dissolves in Water? Water acts as a Brønsted-Lowry base and attracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

10. Conjugate Acids and Bases Reactions between acids and bases always yield their conjugate bases and acids.

11. Acid and Base Strength Strong acids are completely dissociated in water. –Their conjugate bases are quite weak. Weak acids only dissociate partially in water. –Their conjugate bases are stronger weak bases.

12. Acid and Base Strength Substances with negligible acidity do not dissociate in water. –Their conjugate bases are exceedingly strong.

13. Acid dissociation constant K a HA(aq)  H + (aq) + A - (aq) K a = [H + ][A - ] [HA] We can write the expression for any acid. Strong acids dissociate completely. Equilibrium far to right. Conjugate base must be weak.

14. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) H 2 O is a much stronger base than Cl -, so the equilibrium lies so far to the right that K is not measured (K>>1).

15. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1). CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 - (aq)

16. To review…. Strong acids K a is large [H + ] = [HA] A - is a weaker base than water Weak acids K a is small [H + ] <<< [HA] A - is a stronger base than water

17. Types of Acids Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic). Oxyacids - Proton is attached to the oxygen of an ion. Organic acids contain the Carboxyl group -COOH with the H attached to O, generally very weak.

18. Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq)

19. Ion-Product Constant The equilibrium expression for this process is K c = [H 3 O + ] [OH - ] This special equilibrium constant is referred to as the ion-product constant for water, K w. At 25  C, K w = 1.0  10 -14

20. Aqueous solutions In EVERY aqueous solution three possibilities exist… Neutral solution [H + ] = [OH - ]= 1.0 x10 -7 Acidic solution [H + ] > [OH - ] Basic solution [H + ] < [OH - ]

21. pH pH is defined as the negative base-10 logarithm of the concentration of hydronium ion. pH = -log [H 3 O + ]

22. pH Used because [H + ] is usually very small As pH decreases, [H + ] increases exponentially [H + ] = 1.0 x 10 -8 pH= 8.00 pOH= -log[OH - ] pKa = -log K

23. pH In pure water, K w = [H 3 O + ] [OH - ] = 1.0  10 -14 Since in pure water [H 3 O + ] = [OH - ], [H 3 O + ] = 1.0  10 -14 = 1.0  10 -7

24. pH Therefore, in pure water, pH = -log (1.0  10 -7 ) = 7.00 An acid has a higher [H 3 O + ] than pure water, so its pH is <7. A base has a lower [H 3 O + ] than pure water, so its pH is >7.

25. Relationships K W = [H + ][OH - ]; pH= -log[H + ] [H + ] = 10 -pH -log K W = -log[H + ] + -log[OH - ] pK W = pH + pOH K W = 1.0 x10 -14 14.00 = pH + pOH [H + ],[OH - ],pH and pOH Given any one of these we can find the other three.

26. pH These are the pH values for several common substances.

27. How Do We Measure pH? For less accurate measurements, we can use –Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 Blue to red = “ahsaid” –Or an indicator.

28. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

29. Calculating pH of Solutions Always write down the major ions in solution. Remember these are equilibrium problems. Remember the chemistry. Don’t try to memorize, there is no one way to do this.

30. Recall Strong Acids HBr, HI, HCl, HNO 3, H 2 SO 4, HClO 4 Completely dissociate [H + ] = [HA] [OH - ] is going to be small because of equilibrium 10 -14 = [H + ][OH - ]

31. Recall Weak Acids Ka will be small. It will be an equilibrium problem from the start.

32. Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A - ] [HA] K c = HA (aq) + H 2 O (l) A - (aq) + H 3 O + (aq)

33. Dissociation Constants The greater the value of K a, the stronger is the acid.

34. Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a for acetic acid at 25  C is 1.8  10 -5.

35. Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K a =

36. Calculating pH from K a We next set up a table… [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 - ], M Initially0.3000 Change-x+x At Equilibrium 0.30 - x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

37. Calculating pH from K a Now, (x) 2 (0.30) 1.8  10 -5 = (1.8  10 -5 ) (0.30) = x 2 5.4  10 -6 = x 2 2.3  10 -3 = x

38. Calculating pH from K a pH = -log [H 3 O + ] pH = -log (2.3  10 -3 ) pH = 2.64

39. Example Calculate the pH of 2.0 M acetic acid HC 2 H 3 O 2 with a Ka 1.8 x10 -5 Calculate pOH, [OH - ], [H + ]

40. A mixture of Weak Acids The process is the same. Determine the major species. The stronger will predominate. Bigger Ka if concentrations are similar

41. Calculate the pH of a mixture 1.0 M HF (Ka = 7.2 x 10 -4 ) and 1.0 M HOC 6 H 5 (Ka =1.6x10 -10 )

42. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COO - ] [HCOOH] K a =

43. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO - ], from the pH.

44. Calculating K a from the pH pH = -log [H 3 O + ] 2.38 = -log [H 3 O + ] -2.38 = log [H 3 O + ] 10 -2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 -3 = [H 3 O + ] = [HCOO - ]

45. Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO - ], M Initially0.1000 Change - 4.2  10 -3 + 4.2  10 -3 At Equilibrium 0.10 - 4.2  10 -3 = 0.0958 = 0.10 4.2  10 -3

46. Calculating K a from pH [4.2  10 -3 ] [0.10] K a = = 1.8  10 -4

47. Percent dissociation = amount dissociated x 100 initial concentration For a weak acid percent dissociation increases as acid becomes more dilute. As [HA] 0 decreases [H + ] decreases but % dissociation increases. Le Chatelier’s principle

48. Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 -3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent Ionization =  100 4.2  10 -3 0.10 = 4.2%

49. Calculate the % dissociation of 1.00 M and 1.0 x 10 -3 M Acetic acid (Ka = 1.8 x 10 -5 )

50. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution.

51. Bases The OH - is a strong base. Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. The hydroxides of alkaline earths Ca(OH) 2 etc. are strong dibasic bases, but they don’t dissolve well in water. Used as antacids because [OH - ] can’t build up.

52. Bases without OH - Bases are proton acceptors. NH 3 + H 2 O  NH 4 + + OH - It is the lone pair on nitrogen that accepts the proton. Many weak bases contain nitrogen B(aq) + H 2 O(l)  BH + (aq) + OH - (aq) K b = [BH + ][OH - ] [B]

53. Weak Bases Bases react with water to produce hydroxide ion.

54. Strength of Bases Hydroxides are strong. Others are weak. The smaller the K b, the weaker the base.

55. Weak Bases The equilibrium constant expression for this reaction is [HB] [OH - ] [B - ] K b = where K b is the base-dissociation constant.

56. Weak Bases K b can be used to find [OH - ] and, subsequently, pH.

57. pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH - ] [NH 3 ] K b = = 1.8  10 -5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)

58. pH of Basic Solutions Tabulate the data. [NH 3 ], M[NH 4 + ], M[OH - ], M Initially0.1500 At Equilibrium 0.15 - x  0.15 xx

59. pH of Basic Solutions (1.8  10 -5 ) (0.15) = x 2 2.7  10 -6 = x 2 1.6  10 -3 = x 2 (x) 2 (0.15) 1.8  10 -5 =

60. pH of Basic Solutions Therefore, [OH - ] = 1.6  10 -3 M pOH = -log (1.6  10 -3 ) pOH = 2.80 pH = 14.00 - 2.80 pH = 11.20

61. Calculate the pH of a solution of 0.20 M pyridine (Kb = 1.7 x 10 -9 )

62. K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

63. K a tells us K b The anion of a weak acid is a semi-strong base. Calculate the pH of a solution of 0.050M NaCN. Ka of HCN is 6.2 x 10 -10 The CN - ion competes with OH - for the H +

64. Polyprotic Acids… …have more than one acidic proton If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

65. Polyprotic acids Always dissociate stepwise. The first H + comes of much easier than the second. Ka for the first step is much bigger than Ka for the second. Denoted Ka 1, Ka 2, Ka 3

66. Polyprotic acid H 2 CO 3  H + + HCO 3 - Ka 1 = 4.3 x 10 -7 HCO 3 -  H + + CO 3 -2 Ka 2 = 4.3 x 10 -10 Base in first step is acid in second. In calculations we can normally ignore the second dissociation because it’s so small.

67. Of all the ions in a solution of 0.20 M Arsenic acid H 3 AsO 4 Ka 1 = 5.0 x 10 -3 Ka 2 = 8.0 x 10 -8 Ka 3 = 6.0 x 10 -10

68. Sulfuric acid is special In first step it is a strong acid. Ka 2 = 1.2 x 10 -2 Calculate the concentrations in a 1.0 M solution of H 2 SO 4

69. Salts as acids an bases Salts are ionic compounds. Salts of the cation of strong bases and the anion of strong acids are neutral. for example NaCl, KNO 3 There is no equilibrium for strong acids and bases. We ignore the reverse reaction.

70. Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH - and the conjugate acid: X - (aq) + H 2 O (l) HX (aq) + OH - (aq)

71. Basic Salts If the anion of a salt is the conjugate base of a weak acid, it will create a basic solution. In an aqueous solution of NaF The major species are Na +, F -, and H 2 O F - + H 2 O  HF + OH - K b =[HF][OH - ] [F - ] but Ka = [H 3 O + ][F - ] [HF]

72. Reactions of Cations with Water Cations with acidic protons (like NH 4 + ) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution.

73. Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic.

74. Acidic salts A salt with the cation of a weak base and the anion of a strong acid will be basic. K a x K b = K W Calculate the pH of a solution of 0.40 M NH 4 Cl (the K b of NH 3 1.8 x 10 -5 ).

75. Calculate the pH of a solution of 0.10 M NH 4 Cl (the K b of NH 3 1.8 x 10 -5 ).

76. Anion of weak acid, cation of weak base K a > K b acidic K a < K b basic K a = K b Neutral

77. Effect of Cations and Anions 1.An anion that is the conjugate base of a strong acid will not affect the pH. 2.An anion that is the conjugate base of a weak acid will increase the pH. 3.A cation that is the conjugate acid of a weak base will decrease the pH.

78. Effect of Cations and Anions 4.Cations of the strong Arrhenius bases will not affect the pH. 5.Other metal ions will cause a decrease in pH. 6.When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the K a and K b values.

79. Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group.

80. Factors Affecting Acid Strength In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

81. Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

82. Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

83. Hydrated metals Highly charged metal ions pull the electrons of surrounding water molecules toward them. Make it easier for H + to come off. Al +3 O H H

84. Acid-Base Properties of Oxides Non-metal oxides dissolved in water make acids. SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) Metal oxides dissolved in water produce bases. CaO(s) + H 2 O(l) Ca(OH) 2 (aq)

85. Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

86. Lewis Bases Lewis bases are defined as electron-pair donors. Anything that could be a Brønsted-Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however.