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Chapter 17: Additional Aspects of Aqueous Equilibria

Explore how the common-ion effect impacts equilibrium in aqueous solutions, focusing on weak and strong electrolytes. Learn how adding common ions affects ionization and calculate equilibrium concentrations effectively.

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Chapter 17: Additional Aspects of Aqueous Equilibria

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  1. Chapter 17:Additional Aspects of Aqueous Equilibria Section 1: The Common Ion Effect

  2. Objectives • When you complete this presentation, you will be able to • describe the common ion effect.

  3. Introduction • Water is the most common and most important solvent on Earth. • We will be looking in some detail at the application of equilibrium theory and practice to aqueous solutions. • Additional acid-base equilibria • Buffers and acid-base titrations • Solubility of compounds • Formation of complex ions

  4. The Common-Ion Effect • We know that sodium salts are strong electrolytes and dissociate completely in aqueous solution. NaA(aq) → Na+(aq) + A−(aq) • We also know that certain acids are weak electrolytes and dissociate partially in solution. HA(aq) ⇄ H+(aq) + A−(aq)

  5. The Common-Ion Effect • If we start with a solution of acetic acid, we will set up the equilibrium for the weak acid. HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq) • If we add sodium acetate to the solution, the additional acetate will drive the equilibrium to the left, decreasing the equilibrium [H+]. • The presence of the added acetate ion causes the acetic acid to ionize less than it normally would.

  6. The Common-Ion Effect • Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were alone in the system. • This is called the common-ion effect. • We can calculate equilibrium concentrations of systems with common ions.

  7. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • Plan: • Identify the major species in solution. • Identify the major sources that affect the concentration of H+. • Build “i-c-e” table. • Use K expression to calculate [H+].

  8. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • Major species: • HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

  9. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • Major sources that affect the concentration of H+: • HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

  10. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • “i-c-e” table: • HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

  11. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = [H+][CH3COO−] [HCH3COO]

  12. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = x(0.30 + x) 0.30 − x

  13. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = x(0.30 + x) 0.30 − x assume x<<0.30

  14. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = x(0.30) 0.30 assume x<<0.30

  15. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = x(0.30) 0.30

  16. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = = x x(0.30) 0.30

  17. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • K expression: Ka = 1.8 × 10−5 = = x = [H+] x(0.30) 0.30

  18. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • calculate pH: pH = −log[H+]

  19. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • calculate pH: pH = −log[H+] = −log(1.8 × 10−5)

  20. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • calculate pH: pH = −log[H+] = −log(1.8 × 10−5) = 4.74

  21. The Common-Ion Effect • Sample Exercise 17.1 (pg. 720) • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? • calculate pH: pH = −log[H+] = −log(1.8 × 10−5) = 4.74

  22. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Plan: • Identify the major species in solution. • Identify the major sources that affect the concentration of H+ & F−. • Build “i-c-e” table. • Use K expression to calculate [H+] & [F−].

  23. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Major species: HF(aq) ⇄ H+(aq) + F−(aq)

  24. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Major sources that affect [H+]: HF(aq) ⇄ H+(aq) + F−(aq)

  25. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • “i-c-e” table: HF(aq) ⇄ H+(aq) + F−(aq)

  26. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8 × 10−4= [H+][F−] [HF]

  27. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4= (0.10 +x)x 0.20 − x

  28. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4= (0.10 +x)x 0.20 − x assume x<<0.30

  29. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4= (0.10)x 0.20 assume x<<0.30

  30. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) (0.10)x 0.20 0.20 0.10

  31. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] (0.10)x 0.20 0.20 0.10

  32. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M (0.10)x 0.20 0.20 0.10

  33. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M (0.10)x 0.20 0.20 0.10

  34. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3

  35. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3≈ 0.10 M

  36. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3≈ 0.10 M pH = −log[H+]

  37. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3≈ 0.10 M pH = −log[H+] = −log(0.10)

  38. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3≈ 0.10 M pH = −log[H+] = −log(0.10) = 1.00

  39. The Common-Ion Effect • Sample Exercise 17.2 (pg. 722) • Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. • Find pH: [H+] = 0.10 M − 1.4 × 10−3≈ 0.10 M pH = −log[H+] = −log(0.10) = 1.00

  40. The Common-Ion Effect • Sample Exercises 17.1 and 17.2 both involve weak acids. • We can also use the same techniques with weak bases. • For example, adding NH4Cl to an aqueous solution of NH3 will cause the NH3 to dissociate less and lower the pH.

  41. The Common-Ion Effect • The techniques of Sample Exercises 17.1 and 17.2 may be used to solve homework problems 17.15 and 17.17.

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