1. 11111 Chemistry 132 NT Instead of having “answers” on a math test, they should just call them “ impressions”, and if you got a different “impression”, so what, can’t we all be brothers? Jack Handey

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3. 33333 Acid-Base Equilibria Chapter 16 Module 3 Sections 16.5, 16.6, and 16.7 Reaction of zinc metal with hydrochloric acid.

4. 44444 Review Calculating the concentration of a species in a weak base solution using K b Predicting whether a salt solution is acidic, basic, or neutral Obtaining K a from K b or K b from K a Calculating concentrations of species in a salt solution

5. 55555 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. Consider a solution of acetic acid (HC 2 H 3 O 2 ), in which you have the equilibrium above.

6. 66666 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. If we were to add NaC 2 H 3 O 2 to this solution, it would provide C 2 H 3 O 2 - ions which are present on the right side of the equilibrium.

7. 77777 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased.

8. 88888 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.

9. 99999 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. Consider the equilibrium below. Starting 0.025 00.018 Change -x +x Equilibrium 0.025-x x0.018+x

10. 10 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. The equilibrium constant expression is:

11. 11 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. Substituting into this equation gives:

12. 12 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10 -4. Assume that x is small compared with 0.018 and 0.025. Then

13. 13 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. The equilibrium equation becomes

14. 14 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. Hence,

15. 15 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. Note that x was much smaller than 0.018 or 0.025. For comparison, the pH of 0.025 M formic acid alone (without the sodium formate) is 2.69.

16. 16 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. Note that x was much smaller than 0.018 or 0.025. (see Example 16.10 and Problem 16.61)

17. 17 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer contains both an acid species and a base species in equilibrium.

18. 18 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A -. When H 3 O + is added to the buffer it reacts with the base A -.

19. 19 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A -. When OH - is added to the buffer it reacts with the acid HA.

20. 20 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Two important characteristics of a buffer are its buffer capacity and its pH. Buffer capacity depends on the amount of acid and conjugate base present in the solution. The next example illustrates how to calculate the pH of a buffer.

21. 21 The pH of a Buffer The solution described in Example 16.10 in your text is a buffer, because it is composed of a weak acid (0.10 M acetic acid) and its conjugate base (0.20 M acetate ion). The example described how to calculate the pH of such a buffer (we obtained a pH of 5.07) Given the recipe for making a buffer from volumes of solutions, we should then be able to calculate its pH. Let’s try one.

22. 22 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? This buffer contains a base (NH 3 ) and its conjugate acid (NH 4 + ) in equilibrium. To do the equilibrium calculation, we need the starting concentrations in the solution obtained by mixing the NH 3 and NH 4 Cl solutions.

23. 23 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? To do this, we must calculate the moles of NH 3 and NH 4 Cl present and divide by the total volume of the solution. This buffer contains a base (NH 3 ) and its conjugate acid (NH 4 + ) in equilibrium.

24. 24 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Note that the instructions say to add 60.0 mL (0.0600 L) of 0.100 M NH 3, so… In the same way, we find we have added 0.00400 mol NH 4 + (from NH 4 Cl).

25. 25 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Total volume of buffer = 60.0 mL + 40.0 mL = 100.0 mL (0.100 L) Therefore, the concentration of base and conjugate acid are:

26. 26 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Fill in the concentration table for the acid-base equilibrium (base ionization of NH 3 ). Starting 0.060 0.0400 Change -x +x Equilibrium 0.060-x 0.040+xx

27. 27 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? The equilibrium constant expression is:

28. 28 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Substituting into this equation gives:

29. 29 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Assuming that x is very small compared to 0.040 and 0.060, we can simplify

30. 30 A Problem To Consider Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? Therefore, Thus, the hydroxide concentration is 2.7 x 10 -5. The pH of the buffer is: (see Exercise 16.12 and Problem 16.65)

31. 31 Adding Acid or Base to a Buffer Example 16.10 in your text describes a buffer composed of acetic acid and sodium acetate. This buffer controls the equilibrium below. Adding a strong acid to this buffer would convert some of the C 2 H 3 O 2 - (a base) back into HC 2 H 3 O 2. Adding a strong base to this buffer would convert more HC 2 H 3 O 2 (an acid) into C 2 H 3 O 2 - (its conjugate base)

32. 32 Adding Acid or Base to a Buffer Example 16.10 in your text describes a buffer composed of acetic acid and sodium acetate. This buffer controls the equilibrium below. Let’s try an example where we add a strong acid to this buffer.

33. 33 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. When the hydronium ion (from the HCl) is added to the buffer, it reacts completely with the acetate ion. This initial reaction goes to completion until all of the H 3 O + (from HCl) has been used up.

34. 34 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. We must first calculate the amounts of hydrogen ion, acetate ion and acetic acid present in the solution before reaction. since HCl is a strong acid.

35. 35 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. We must first calculate the amounts of hydrogen ion, acetate ion and acetic acid present in the solution before reaction.

36. 36 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. You can now assume that all of the hydronium ion added (0.00095 mol) reacts with acetate ion. Therefore, 0.00095 mol of acetate ion is used up and 0.00095 mol of acetic acid is produced.

37. 37 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. You can now assume that all of the hydronium ion added (0.00095 mol) reacts with acetate ion. Therefore, 0.00095 mol of acetate ion is used up and 0.00095 mol of acetic acid is produced.

38. 38 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. Now that we know the amounts of acetic acid and acetate ion remaining after the addition of the HCl, we can solve the equilibrium calculation. Note that the total volume of solution (the buffer and the HCl) is 75 mL + 9.5 mL, or 85 mL (0.085 L) to two significant figures.

39. 39 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. Hence the concentrations after the acid addition are:

40. 40 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. Now we can construct the following table. Starting 0.10 ~00.16 Change -x +x Equilibrium 0.10-x x0.016+x 

41. 41 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. The equilibrium constant expression is:

42. 42 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. Substituting, you get: If we assume x is small, then: and

43. 43 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. And simplifying, we get or

44. 44 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. The H 3 O + concentration is 1.1 x 10 -5, so the pH is The pH of the original buffer in Example 16.10 was 5.07.

45. 45 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. The H 3 O + concentration is 1.1 x 10 -5, so the pH is The addition of strong acid to the buffer dropped its pH by 0.11 pH units.

46. 46 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. The H 3 O + concentration is 1.1 x 10 -5, so the pH is This illustrates the stability of the buffer.

47. 47 A Problem To Consider Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2 ) to which 9.5 mL of 0.10 M HCl is added. The H 3 O + concentration is 1.1 x 10 -5, so the pH is Adding the same amount of acid to 75 mL of pure water results in a pH of 1.96. (see Exercise 16.13 and Problem 16.67)

48. 48 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? A buffer must be prepared from a conjugate acid-base pair in which the K a of the acid is approximately equal to the desired H 3 O + concentration. To illustrate, consider a buffer of a weak acid HA and its conjugate base A -. The acid ionization equilibrium is:

49. 49 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? The acid ionization constant is: By rearranging, you get an equation for the H 3 O + concentration.

50. 50 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? Taking the negative logarithm of both sides of the equation we obtain: The previous equation can be rewritten

51. 51 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? More generally, you can write This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation.

52. 52 A Problem To Consider A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The K a for HF is 6.8 x 10 -4 ) The Henderson-Hasselbalch equation is ideal for solving buffer pH problems. The negative log of the K a, which was given. The concentration of HF. The concentration of F - (from NaF).

53. 53 A Problem To Consider A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The K a for HF is 6.8 x 10 -4 ) First, let’s calculate the “new” molarities of HF and F -, once the solutions are mixed together. Total volume = 0.035 L + 0.045 L = 0.080 L

54. 54 A Problem To Consider A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The K a for HF is 6.8 x 10 -4 ) First, let’s calculate the “new” molarities of HF and F -, once the solutions are mixed together.

55. 55 A Problem To Consider A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The K a for HF is 6.8 x 10 -4 ) Now, we need the “pK a ”. Now, we’re ready to assemble the Henderson- Hasselbalch equation.

56. 56 A Problem To Consider A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The K a for HF is 6.8 x 10 -4 ) Substituting into the equation we get: Note that the pH of a buffer will always be close to its pK a. The log term simply “fine tunes” the buffer.

57. 57 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid- base pair with a pK a close to the desired pH. The K a for acetic acid is 1.7 x 10 -5, and its pK a is 4.77. You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid].

58. 58 Acid-Ionization Titration Curves An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use titration curves to choose an appropriate indicator that will show when the titration is complete.

59. 59 Titration of a Strong Acid by a Strong Base Figure 16.11 shows a curve for the titration of HCl with NaOH. Note that the pH changes slowly until the titration approaches the equivalence point. The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added.

60. Curve for the titration of a strong acid by a strong base. Figure 16.11 Some unneutralized acid. In this case simple stoichiometry tells you the concentration of remaining H+ from which you calculate pH. Equivalence point. For a strong acid and strong base, this will always be pH 7.00

61. 61 Titration of a Strong Acid by a Strong Base Figure 16.11 shows a curve for the titration of HCl with NaOH. At the equivalence point, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze. However, the pH changes rapidly from a pH of about 3 to a pH of about 11.

62. 62 Titration of a Strong Acid by a Strong Base Figure 16.11 shows a curve for the titration of HCl with NaOH. To detect the equivalence point, you need an acid-base indicator that changes color within the pH range 3-11. Phenolphthalein can be used because it changes color in the pH range 8.2-10. (see Figure 16.8)

63. 63 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. Because the reactants are a strong acid and a strong base, the reaction is essentially complete.

64. 64 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities.

65. 65 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. All of the OH - reacts, leaving an excess of H 3 O +

66. 66 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. You obtain the H 3 O + concentration by dividing the mol H 3 O + by the total volume of solution (0.0250 L + 0.0100 L=0.0350 L)

67. 67 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. Hence, (see Exercise 16.14 and Problem 16.75)

68. 68 Titration of a Weak Acid by a Strong Base The titration of a weak acid by a strong base gives a somewhat different curve. The pH range of these titrations is shorter. Before the equivalence point is reached, we have some un-neutralized acid and it’s conjugate base anion, that is, a buffer. The pH in this “buffer” region can be determined using the Henderson-Hasselbalch equation.

69. 69 Titration of a Weak Acid by a Strong Base The titration of a weak acid by a strong base gives a somewhat different curve. The equivalence point will be on the basic side since the salt produced contains the anion of a weak acid. Figure 16.12 shows the curve for the titration of nicotinic acid with NaOH.

70. Curve for the titration of a weak acid by a strong base. Figure 16.12 Buffer region. Some un- neutralized acid and some conjugate base anion…you must use the Henderson Hasselbaclch equation to determine pH. Equivalence point…only the conjugate base anion remains…must solve hydrolysis problem to determine pH.

71. 71 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. At the equivalence point, equal molar amounts of acetic acid and sodium hydroxide react to give sodium acetate.

72. 72 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. First, calculate the concentration of the acetate ion. In this case, 25.0 mL of 0.10 M NaOH is needed to react with 25.0 mL of 0.10 M acetic acid.

73. 73 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. The molar amount of acetate ion formed equals the initial molar amount of acetic acid.

74. 74 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. The total volume of the solution is 50.0 mL. Hence,

75. 75 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. You find the K b for the acetate ion to be 5.9 x 10 -10 and that the concentration of the hydroxide ion is 5.4 x 10 -6. The pH is 8.73 The hydrolysis of the acetate ion follows the method given in an earlier module of this chapter.

76. 76 Titration of a Weak Base by a Strong Acid The titration of a weak base with a strong acid is a reflection of our previous example. Figure 16.13 shows the titration of NH 3 with HCl. In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3. Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator.

77. Curve for the titration of a weak base by a strong cid. Figure 16.13

78. 78 A Little “Trick” At the “half-way” point in the titration of a weak acid with a weak base, we can make a significant simplification. At the halfway point in such a titration, exactly half of the original acid has been converted to its conjugate base anion. This means that [A - ] = [HA]

79. 79 A Little “Trick” At the “half-way” point in the titration of a weak acid with a weak base, we can make a significant simplification. At the halfway point in such a titration, exactly half of the original acid has been converted to its conjugate base anion. The pH = pK a only at the halfway point. We’ll use this fact in lab to determine the K a of an unknown acid.

80. 80 Operational Skills Calculating the common-ion effect on acid ionization Calculating the pH of a buffer from given volumes of solution Calculating the pH of a solution of a strong acid and a strong base Calculating the pH at the equivalence point in the titration of a weak cid with a strong base

81. 81 Homework Chapter 16 Homework: collected at the first exam. Review Questions: none. Problems: 37, 41, 45, 49.

82. 82