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Chapter 17 : Additional Aspects of Aqueous Equilibria PowerPoint Presentation
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Chapter 17 : Additional Aspects of Aqueous Equilibria

Chapter 17 : Additional Aspects of Aqueous Equilibria

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Chapter 17 : Additional Aspects of Aqueous Equilibria

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  1. Section 2: Buffered Solutions Chapter 17: Additional Aspects of Aqueous Equilibria

  2. Introduction • Solutions prepared with common ions have a tendency to resist drastic pH changes even when subjected to the addition of strong acids or bases. • Such solutions are called buffered solutions. • Blood is buffered to a pH of about 7.4. • Surface seawater is buffered to a pH between 8.1 and 8.3.

  3. Composition and Action • A buffer resists pH change because it contains … • an acid to neutralize added OH− ions • a base to neutralize added H+ ions • But we cannot have the acid and base consume one another through neutralization.

  4. Composition and Action • Therefore, we use weak conjugate acids and bases to prepare our buffers. • For example: • HCH3COO and CH3COO− • HF and F− • NH4+ and NH3 • HClO and ClO−

  5. Composition and Action • If we have a solution of HF and NaF, we have HF(aq) and F−(aq) in solution. • If we add a small amount of a strong acid, • we use the F− to neutralize it: F−(aq) + H+(aq) → HF(aq) • If we add a small amount of a strong base, • we use the HF to neutralize it: HF(aq) + OH−(aq) → F−(aq) + H2O(l)

  6. Composition and Action • Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example). • The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H+(aq) + X−(aq)

  7. Composition and Action • Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example). • The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H+(aq) + X−(aq) • The acid dissociation constant is Ka = [H+][X−] [HX]

  8. Composition and Action • Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example). • The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H+(aq) + X−(aq) • Solving for [H+] [H+] = Ka [HX] [X−]

  9. Composition and Action [HX] [H+] = Ka [X−]

  10. Composition and Action [HX] [H+] = Ka • The [H+] and pH depend on two things: [X−]

  11. Composition and Action [HX] [H+] = Ka • The [H+] and pH depend on two things: • Ka • the ratio of the concentrations of conjugate acid-base pair. [X−]

  12. Composition and Action [HX] [H+] = Ka • If OH− ions are added, they will react with the acid component to produce water and X−. [X−]

  13. Composition and Action [HX] [H+] = Ka • If OH− ions are added, they will react with the acid component to produce water and X−. OH−(aq) + HX(aq) → H2O(l) + X−(aq) [X−]

  14. Composition and Action [HX] [H+] = Ka • If OH− ions are added, they will react with the acid component to produce water and X−. OH−(aq) + HX(aq) → H2O(l) + X−(aq) • This causes a decrease in [HX] and an increase in [X−]. [X−]

  15. Composition and Action [HX] [H+] = Ka • If OH− ions are added, they will react with the acid component to produce water and X−. OH−(aq) + HX(aq) → H2O(l) + X−(aq) • This causes a decrease in [HX] and an increase in [X−]. • So long as the amounts of HX and X− are large, the ratio of [HX]/[X−] does not change. [X−]

  16. Composition and Action [HX] [H+] = Ka • If H+ ions are added, they will react with the base component to produce the acid. [X−]

  17. Composition and Action [HX] [H+] = Ka • If H+ ions are added, they will react with the base component to produce the acid. H+(aq) + X−(aq) → HX(aq) [X−]

  18. Composition and Action [HX] [H+] = Ka • If H+ ions are added, they will react with the base component to produce the acid. H+(aq) + X−(aq) → HX(aq) • This causes a decrease in [X−] and an increase in [HX]. [X−]

  19. Composition and Action [HX] [H+] = Ka • If H+ ions are added, they will react with the base component to produce the acid. H+(aq) + X−(aq) → HX(aq) • This causes a decrease in [X−] and an increase in [HX]. • So long as the amounts of HX and X− are large, the ratio of [HX]/[X−] does not change. [X−]

  20. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.

  21. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method.

  22. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We start with the expression for finding [H+] of a buffer.

  23. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We start with the expression for finding [H+] of a buffer. [HX] [H+] = Ka [X−]

  24. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We then find the negative log of both sides. [HX] [H+] = Ka [X−]

  25. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We then find the negative log of both sides. [HX] −log[H+] = −log(Ka) [X−]

  26. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We then find the negative log of both sides. [HX] −log[H+] = −logKa − log [X−]

  27. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We note that −log[H+] = pH and −logKa = pKa [HX] −log[H+] = −logKa − log [X−]

  28. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We note that −log[H+] = pH and −logKa = pKa [HX] pH = pKa − log [X−]

  29. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We note that − log[HX]/[X−] = + log[X−]/[HX] [HX] pH= pKa − log [X−]

  30. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We note that − log[HX]/[X−] = + log[X−]/[HX] [X−] pH= pKa + log [HX]

  31. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We can generalize HX as an acid and X− as a base. [X−] pH= pKa + log [HX]

  32. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • We can generalize HX as an acid and X− as a base. [base] pH= pKa + log [acid]

  33. Calculating the pH of a Buffer • We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer. • But, there is an alternative method. • This is called the Henderson-Hasselbach equation. [base] pH= pKa + log [acid]

  34. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.

  35. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known:

  36. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4

  37. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa

  38. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85

  39. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M

  40. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M

  41. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = pKa + log [base] [acid]

  42. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log 0.10 0.12

  43. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 + log(0.83) 0.10 0.12

  44. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 − 0.08 0.10 0.12

  45. Sample Exercise 17.3 (pg. 725) • What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4. • Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 − 0.08 = 3.77 0.10 0.12

  46. Sample Exercise 17.3 (pg. 725) • Use Sample Exercise 17.3 to help solve homework exercises 17.21 through 17.24 (pages 759 − 760).

  47. Sample Exercise 17.4 (pg. 726) • How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)

  48. Sample Exercise 17.4 (pg. 726) • How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.) • Known:

  49. Sample Exercise 17.4 (pg. 726) • How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.) • Known: Kb = 1.8 × 10−5

  50. Sample Exercise 17.4 (pg. 726) • How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.) • Known: Kb = 1.8 × 10−5 ⇒