1. CHE1102, Chapter 15 Learn, 1 Chapter 16 Acid-Base Equilibria in Aqueous Solutions

2. CHE1102, Chapter 16 Learn, 2 Autoionization of Water In pure water, a few molecules act as bases and a few act as acids. Self-ionization – when a substance reacts with itself to form ions H 2 O ( ) H + (aq) + OH - (aq) H 2 O ( ) + H 2 O ( ) H 3 O + (aq) + OH - (aq)

3. CHE1102, Chapter 16 Learn, 3 H 2 O ( ) H + (aq) + OH - (aq) K w = [H + ] [OH - ] K w is called the ion-product constant for water K w is just another equilibrium constant K w = [H + ] [OH - ] = 1.0 x 10 - 14 (at 25  C) Ion-Product Constant for Water

4. CHE1102, Chapter 16 Learn, 4 In pure DI water at 25  C [OH - ] = 1.0 x 10 - 7 M [H + ] = 1.0 x 10 - 7 M and H 2 O ( ) H + (aq) + OH - (aq)

5. CHE1102, Chapter 16 Learn, 5 when [H + ] = [OH - ] solution is neutral when [H + ] > [OH - ] when [H + ] < [OH - ] solution is acidic solution is basic K w = [H + ] [OH - ] = 1.0 x 10 - 14 (at 25  C) K w expression is valid for ALL aqueous solutions (acidic, basic or neutral) at 25  C

6. CHE1102, Chapter 16 Learn, 6 Learning Check In a sample of blood at 25 °C, [H + ] = 4.6  10 –8 M. Find [OH – ] and determine if the solution is acidic, basic or neutral. So2.2 × 10 –7 M > 4.6 × 10 –8 M [OH – ] > [H 3 O + ] so the solution is basic

7. CHE1102, Chapter 16 Learn, 7 The pH Scale Søren Sørensen 1868 – 1939 pH = - log [H + ]

8. CHE1102, Chapter 16 Learn, 8 For Any Aqueous Solution at 25  C When pH = 7 the solution is neutral When pH < 7 the solution is acidic When pH > 7 the solution is basic pH scale range is from 0 - 14

9. CHE1102, Chapter 16 Learn, 9 Dr. Pepper Cherry (it’s amazingly smooth) has [H + ] = 1.2 x 10 - 3 M. What is the pH of this pop?

10. CHE1102, Chapter 16 Learn, 10 Lemon juice has a pH = 2.41 Stomach juice has a pH = 2.09 pH = 2.41 vs. pH = 2.09 Which one is more acidic? Calculate, specifically by how much more acidic one is (vs. the other).

11. CHE1102, Chapter 16 Learn, 11 Learning Check What are [H + ] and [OH – ] of pH = 3.00 solution? [H + ] = 10 –3.00 = 1.0  10 –3 M [OH – ] = = 1.0  10 –11 M What are [H + ] and [OH – ] of pH = 4.00 solution? pH = 4.00[H + ] = 1.0  10 –4 M [OH – ] = = 1.0  10 –10 M pH 4.00 solution has 10 times less H + than pH 3.00 solution

12. CHE1102, Chapter 16 Learn, 12

13. CHE1102, Chapter 16 Learn, 13 pH

14. CHE1102, Chapter 16 Learn, 14 Blood has a [OH - ] = 2.69 x 10 - 7 M at 25  C. What is the pH of blood ?

15. CHE1102, Chapter 16 Learn, 15 There is another scale analogous to pH pOH = - log [OH - ] pH + pOH = 14.00 (at 25  C) Blood has a [OH - ] = 2.69 x 10 - 7 M at 25  C. What is the pH of blood ? Above expression is valid for ALL aqueous solutions (acidic, basic or neutral) at 25  C

16. CHE1102, Chapter 16 Learn, 16 Learning Check K w increases with increasing temperature. At 50 °C, K w = 5.476 ×10 –14. What is the pH of a neutral solution at 50 °C ? A. 7.00 B. 6.63 C. 7.37 D. 15.3 [H + ] = [OH - ] K w =[H + ]x[OH - ] [H + ]=K w 1/2 [H + ]=(5.476x10 -14 ) 1/2 [H + ]=2.34x10 -7 pH=-log(2.34x10 -7 )

17. CHE1102, Chapter 16 Learn, 17 Calculations for strong acids/bases Determine the pH of a 0.15 M HBr (aq) soln

18. CHE1102, Chapter 16 Learn, 18 Learning Check What is the pH of 0.15 M HBr? HBr is a strong acid and dissociates completely, HBr(aq) + H 2 O  H + (aq) + Br – (aq) Int. 0.15 N/A 0 0 Final 0 N/A 0.15 0.15 pH = –log(0.15) = 0.82

19. CHE1102, Chapter 16 Learn, 19 14.2 g Ba(OH) 2 (s) is dissolved in enough H 2 O to form 500.0 mL soln. What’s the resulting pH of the solution ?

20. CHE1102, Chapter 16 Learn, 20 Learning Check What is the pH of 0.083 M Ba(OH) 2 ? A strong base and dissociates completely Ba(OH) 2 (aq)  Ba 2+ (aq) + 2 OH – (aq) Int. 0.0830 0 Final 00.083 0.166 pOH = –log(0.166) = 0.780 pH = 14.00 – pOH = 14.00 – 0.780 = 13.22

21. CHE1102, Chapter 16 Learn, 21 Calculations of Strong Acids and Bases The auto-ionization of H 2 O will always add to [H + ] and [OH – ] of an acid or base. Does this have an effect on the last answer? –The previous problem had 0.166 M [OH – ] from the Ba(OH) 2 but the [H + ] must have come from water. If it came from water an equal amount of [OH – ] comes from water and the total [OH – ] is –[OH – ] total = [OH – ] from Ba(OH) 2 + [OH – ] from H 2 O –[OH – ] total = 0.166 M + 1.47 × 10 –14 M = 0.166 M (when properly rounded) The autoionization of water is insignificant

22. CHE1102, Chapter 16 Learn, 22 CH 3 COOH (aq) CH 3 COO − (aq) + H + (aq) H2OH2O CH 3 COOH (aq) H + (aq) + CH 3 COO − (aq) H2OH2O or [ H + ] [CH 3 COO − ] [CH 3 COOH] K a = K a is called the acid-dissociation constant K a is just an equilibrium constant that is specific for a weak acid Weak Acids

23. CHE1102, Chapter 16 Learn, 23 A 0.050 M aqueous acetic acid solution has a pH = 3.03. What is the K a for acetic acid ?

24. CHE1102, Chapter 16 Learn, 24 Calculating K a from the pH The pH of a 0.050 M solution of acetic acid, CH 3 COOH, at 25  C is 3.03. Calculate K a for acetic acid at this temperature. We know that [H 3 O + ] [HCOO  ] [HCOOH] K a =

25. CHE1102, Chapter 16 Learn, 25 Calculating K a from the pH The pH of a 0.050 M solution of acetic acid, CH 3 COOH, at 25  C is 3.03. Calculate K a for acetic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three components. We can find [H 3 O + ], which is the same as [HCOO  ], from the pH.

26. CHE1102, Chapter 16 Learn, 26 Calculating K a from the pH pH=  log [H 3 O + ] 3.03=  log [H 3 O + ]  3.03= log [H 3 O + ] 10  3.03 = 10 log [H 3 O + ] = [H 3 O + ] 9.3  10  4 = [H 3 O + ] = [HCOO  ]

27. CHE1102, Chapter 16 Learn, 27 Calculating K a from pH Now we can set up an ICE table… [HCOOH], M[H 3 O + ], M[HCOO  ], M Initially0.05000 Change At Equilibrium

28. CHE1102, Chapter 16 Learn, 28 Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO  ], M Initially0.05000 Change  9.3  10  4 +9.3  10  4 At Equilibrium

29. CHE1102, Chapter 16 Learn, 29 Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO  ], M Initially0.05000 Change  9.3  10  4 +9.3  10  4 At Equilibrium 0.050  9.3  10  4 = 0.0491 = 0.049 9.3  10  4

30. CHE1102, Chapter 16 Learn, 30 Calculating K a from pH [9.3  10  4 ] [0.049] K a = = 1.8  10  5

31. CHE1102, Chapter 16 Learn, 31 pK a = - log K a What is the pK a of acetic acid ? What is the percentage of ionization of acetic acid ? pK a = - log 1.8 x 10 -5 = 4.7

32. CHE1102, Chapter 16 Learn, 32 Calculating Percent Ionization Percent ionization =  100 In this example, [H 3 O + ] eq = 9.3  10  4 M [CH 3 COOH] initial = 0.050 M [H 3 O + ] eq [HA] initial Percent ionization =  100 9.3  10  4 0.050 = 1.9%

33. CHE1102, Chapter 16 Learn, 33 The greater the value of K a, the stronger is the acid.

34. CHE1102, Chapter 16 Learn, 34 What is the pH of a 0.60 M HNO 2 (aq) ?

35. CHE1102, Chapter 16 Learn, 35 Calculating pH from K a Calculate the pH of a 0.60 M solution of nitrous acid, HNO 2, at 25  C. HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2  (aq) K a for nitrous acid at 25  C is 4.5  10  4.

36. CHE1102, Chapter 16 Learn, 36 The equilibrium constant expression is [H 3 O + ] [NO 2  ] [HNO 2 ] K a = Calculating pH from K a

37. CHE1102, Chapter 16 Learn, 37 We next set up a table… [HNO 2 ], M[H 3 O + ], M[NO 2  ], M Initially0.6000 Change At equilibrium Calculating pH from K a

38. CHE1102, Chapter 16 Learn, 38 [HNO 2 ], M[H 3 O + ], M[NO 2  ], M Initially0.6000 Change xx +x+x+x+x At equilibrium Calculating pH from K a

39. CHE1102, Chapter 16 Learn, 39 [HNO 2 ], M[H 3 O + ], M[NO 2  ], M Initially0.6000 Change xx +x+x+x+x At equilibrium 0.60  x  0.60 xx Calculating pH from K a We will assume that x is very small compared to 0.60 and can, therefore, be ignored.

40. CHE1102, Chapter 16 Learn, 40 Now, (x) 2 (0.60) 4.5  10  4 = (4.5  10  4 ) (0.60) = x 2 2.7  10  4 = x 2 1.6  10  2 = x Calculating pH from K a

41. CHE1102, Chapter 16 Learn, 41 pH =  log [H 3 O + ] pH =  log (1.6  10  2 ) pH = 1.80 Calculating pH from K a

42. CHE1102, Chapter 16 Learn, 42 When can you legitimately disregard “X” in the denominator to avoid quadratic equation ? Perform 5% check 1. disregard the “X” in the denominator 2. solve the equality with that approximation If value < 5%, approximation OK If value > 5%, approximation is NOT OK and must go back and use quadratic equation  3. “value obtained for X” [original] x 100

43. CHE1102, Chapter 16 Learn, 43 Weak Bases K b is called the base-dissociation constant K b is just an equilibrium constant that is specific for a weak base Hydrolysis – a substance reacts with H 2 O leaving either H + or OH − Weak bases undergo hydrolysis

44. CHE1102, Chapter 16 Learn, 44 Weak Bases The equilibrium constant expression for this reaction is [HB] [OH  ] [B  ] K b = where K b is the base-dissociation constant. B - (aq) + H 2 O (l) HB (aq) + OH  (aq)

45. CHE1102, Chapter 16 Learn, 45 NH 3 (aq) + H 2 O ( ) NH 4 + (aq) + OH − (aq) [ NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8 x 10 −5 Equilibrium favors the reactants (left)

46. CHE1102, Chapter 16 Learn, 46 Weak Bases K b can be used to find [OH  ] and, through it, pH.

47. CHE1102, Chapter 16 Learn, 47 Trimethylamine, (CH 3 ) 3 N has a lovely, putrefying smell (like rotting  dead fish  ). What is the pH of a 0.145 M aqueous solution of trimethylamine ? (CH 3 ) 3 N (aq) + H 2 O ( ) (CH 3 ) 3 NH + (aq) + OH − (aq) hydrolysis

48. CHE1102, Chapter 16 Learn, 48 pH of Basic Solutions Tabulate the data. [(CH 3 ) 3 N], M[(CH 3 ) 3 NH + ], M[OH  ], M Initially0.14500 At equilibrium

49. CHE1102, Chapter 16 Learn, 49 pH of Basic Solutions Tabulate the data. [(CH 3 ) 3 N], M[(CH 3 ) 3 NH + ], M[OH  ], M Initially0.14500 At equilibrium 0.145  x  0.145 xx

50. CHE1102, Chapter 16 Learn, 50 pH of Basic Solutions (7.4  10  5 ) (0.145) = x 2 1.07  10  6 = x 2 1.04  10  3 = x 2 (x) 2 (0.145) 7.4  10  5 =

51. CHE1102, Chapter 16 Learn, 51 pH of Basic Solutions Therefore, [OH  ] = 1.04  10  3 M pOH =  log (1.04  10  3 ) pOH = 2.98 pH = 14.00  2.98 pH = 11.02

52. CHE1102, Chapter 16 Learn, 52 pK b = - log K b The pKb of methylamine is 3.38. What is the Kb ?

53. CHE1102, Chapter 16 Learn, 53 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.

54. CHE1102, Chapter 16 Learn, 54 NO 2 − + H 2 O HNO 2 + OH − [HNO 2 ][OH − ] [NO 2 − ] K b = HNO 2 H + + NO 2 − [ H + ] [NO 2 − ] [HNO 2 ] K = a H2OH2O

55. CHE1102, Chapter 16 Learn, 55 [HNO 2 ] [OH − ] [NO 2 − ] K b = [H + ] [NO 2 − ] [HNO 2 ] K a = (K a )(K b ) = [H + ] [NO 2 − ] [HNO 2 ] [HNO 2 ] [OH − ] [NO 2 − ] (K a )(K b ) = [H + ] [OH − ] = K w = 1.0 x 10 −14 (at 25  C) for any conjugate acid/base pair at 25  C (K a )(K b ) = 1.0 x 10 −14

56. CHE1102, Chapter 16 Learn, 56 What is the K b for NO 2 − ? K a (HNO 2 ) = 4.5 x 10-4

57. CHE1102, Chapter 16 Learn, 57 Household bleach is a 0.65 M NaOCl (aq) solution. What is the pH of bleach?

58. CHE1102, Chapter 16 Learn, 58 Equilibrium Calculations when Simplifications Fail Weak acid, HA, ionizes in water to give ions HA(aq) + H 2 O A – (aq) + H 3 O + (aq) [HA] equilibrium = [HA] initial – x where x = amount dissociated to ions = [A – ] = [H + ] For the solutions we have looked at so far –[HA] equilibrium  [HA] initial

59. CHE1102, Chapter 16 Learn, 59 Simplifications work when Simplifications fail when In latter case, must solve using quadratic equation Exact Mathematically more complex

60. CHE1102, Chapter 16 Learn, 60 The Quadratic Equation For a quadratic equation in the form The values of x that satisfy this equation are Take only the positive root for the answer –Negative [x] is meaningless Always gives you the correct answer

61. CHE1102, Chapter 16 Learn, 61 What is the pH of a 0.15 M solution of dichloroacetic acid, CHCl 2 CO 2 H, in water? For dichloroacetic acid, K a = 5.0 × 10 –2. CHCl 2 CO 2 H (aq) + H 2 O CHCl 2 CO 2 – (aq) + H 3 O + (aq) The equilibrium law for this reaction is [ CHCl 2 CO 2 H ] (M)[ CHCl 2 CO 2 – ] (M)[H 3 O + ] (M) I0.1500 C E – x+ x 0.15 – xxx

62. CHE1102, Chapter 16 Learn, 62 Using Quadratic Equation Or in terms of general quadratic equation –Where a = 1, b = 0.050, and c = –7.5 ×10 –3 Now put into quadratic formula The two roots are: x = 0.065 M and x = – 0.115 M

63. CHE1102, Chapter 16 Learn, 63 Since only positive root has physical meaning, we use this answer [H + ] = x = 0.065 M [C 2 HCO 2 Cl 2 – ] = x = 0.065 M [HC 2 HCO 2 Cl 2 ] = 0.15 – 0.065 = 0.085 M pH = –log(0.065) = 1.19 Using Quadratic Equation

64. CHE1102, Chapter 16 Learn, 64 Buffers Buffer Capacity – the amount of acid or base that can be added before the buffer is overwhelmed and pH dramatically changes – solutions which resist changes in pH 1. A mixture of a weak acid + soluble salt of the conjugate base of that weak acid 2. A mixture of a weak base + soluble salt of the conjugate acid of that weak base or

65. CHE1102, Chapter 16 Learn, 65 Buffers If a small amount of hydroxide is added to an equimolar solution of HF and NaF, for example, the HF reacts with the OH  to make F  and water. Similarly, if acid is added, the F  reacts with it to form HF and water.

66. CHE1102, Chapter 16 Learn, 66 Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A  ] [HA] K a = HA + H 2 OH 3 O + + A 

67. CHE1102, Chapter 16 Learn, 67 Rearranging slightly, this becomes [A  ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A  ] [HA]  log K a =  log [H 3 O + ] +  log pKapKa pH acid base Buffer Calculations

68. CHE1102, Chapter 16 Learn, 68 So pK a = pH  log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch Equation. Buffer Calculations

69. CHE1102, Chapter 16 Learn, 69 Henderson-Hasselbalch Equation Lawrence Henderson 1878 – 1942 Karl Hasselbalch 1874 – 1962

70. CHE1102, Chapter 16 Learn, 70 Henderson-Hasselbalch Equation [base] [acid] pH = pK a + log pH = pH of the buffered solution pK a = pK a of the weak acid [base] and [acid] are initial [ ]’s of the conjugate acid/base pair

71. CHE1102, Chapter 16 Learn, 71 Suppose 3.398 g NaCHO 2 (s) is dissolved into 305 mL of 0.45 M HCHO 2 (aq). What is the pH of the resulting solution ? 1. Determine how all species exist in solution 2. Write the equilibrium reaction describing the species 3. Leave out spectator ions such as Na + and Cl - because they don’t affect the pH Steps to Working Any Buffer Problem

72. CHE1102, Chapter 16 Learn, 72 Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10  4.

73. CHE1102, Chapter 16 Learn, 73 pH = 3.85 + (  0.08) pH = 3.77 pH = pK a + log [base] [acid] pH =  log (1.4  10  4 ) + log (0.10) (0.12) Henderson–Hasselbalch Equation

74. CHE1102, Chapter 16 Learn, 74 Henderson-Hasselbalch equation for base [acid] [base] pOH = pK b + log pOH = pOH of the buffered solution pK b = pK b of the weak base [acid] and [base] are initial [ ]’s of the conjugate acid/base pair

75. CHE1102, Chapter 16 Learn, 75 Determine the pH of a solution of 0.30 M NH 3 (aq) and 0.18 M NH 4 Cl (aq). (Use data on Slide 53) 1. Work this problem using the “base” form of Henderson-Hasselbalch 2. Work this problem using the “acid” form of Henderson-Hasselbalch

76. CHE1102, Chapter 16 Learn, 76 pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH.

77. CHE1102, Chapter 16 Learn, 77 A solution buffered at pH 3.90 is needed for a reaction. Would formic acid and its salt, sodium formate, make a good choice for this buffer? If so, what ratio of moles of the conjugate base anion (HCOO – ) to the acid (HCOOH) is needed? –From table pK a of formic acid is 3.74. –This is within pH = pK a  1 range 2.90 to 4.90 –So we can use this buffer system

78. CHE1102, Chapter 16 Learn, 78 Titration In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).

79. CHE1102, Chapter 16 Learn, 79 Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

80. CHE1102, Chapter 16 Learn, 80 Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

81. CHE1102, Chapter 16 Learn, 81 Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly.

82. CHE1102, Chapter 16 Learn, 82 Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

83. CHE1102, Chapter 16 Learn, 83 Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

84. CHE1102, Chapter 16 Learn, 84 Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations.

85. CHE1102, Chapter 16 Learn, 85 Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle. © 2012 Pearson Education, Inc.

86. CHE1102, Chapter 16 Learn, 86

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88. CHE1102, Chapter 16 Learn, 88