1 / 102

Chapter 17 : Additional Aspects of Aqueous Equilibria

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

kimi
Download Presentation

Chapter 17 : Additional Aspects of Aqueous Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 ⇒

More Related