Topic 25: Equilibria

Question 1

Definitions I:
(a) Conjugate acid
(b) Buffer solution
(c) Solubility product

Answer 1

Definitions I:
(a) Species formed by base accepting H⁺, e.g., H₃O⁺ from H₂O.
(b) Solution resisting pH change, e.g., weak acid + conjugate base.
(c) Equilibrium constant for a solid dissolving, e.g., Ksp = [Ca²⁺][F⁻]².

Question 2

Definitions II:
(a) pH
(b) Ka
(c) Partition coefficient

Answer 2

Definitions II:
(a) Measure of [H⁺], pH = -log[H⁺].
(b) Acid dissociation constant, Ka = [H⁺][A⁻]/[HA].
(c) Ratio of solute concentrations in two solvents, Kpc = [X]₁/[X]₂.

Question 3

Identify the conjugate acid-base pair in:
(a) HCl + H₂O ⇌ H₃O⁺ + Cl⁻ [1 Mark]
(b) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ [1 Mark]

Answer 3

(a) HCl (acid), Cl⁻ (base); H₂O (base), H₃O⁺ (acid) (1 mark)
(b) NH₃ (base), NH₄⁺ (acid); H₂O (acid), OH⁻ (base) (1 mark)

Question 4

Calculate pH of:
(a) 0.1 M HCl [1 Mark]
(b) 0.01 M CH₃COOH (Ka = 1.8×10⁻⁵) [1 Mark]

Answer 4

(a) [H⁺] = 0.1, pH = -log(0.1) = 1 (1 mark)
(b) Ka = x²/(0.01 - x), x ≈ √(1.8×10⁻⁷) = 4.24×10⁻⁴, pH = 3.37 (1 mark)

Question 5

In a lab, a buffer is made with 0.1 M CH₃COOH and 0.2 M CH₃COONa (Ka = 1.8×10⁻⁵). Calculate pH. [2 Marks]

Answer 5

pH = pKa + log([A⁻]/[HA]), pKa = -log(1.8×10⁻⁵) = 4.74 (1 mark)
pH = 4.74 + log(0.2/0.1) = 5.04 (1 mark)

Question 5

Explain how a CH₃COOH/CH₃COO⁻ buffer resists pH change with added HCl. [3 Marks]

Answer 5

CH₃COO⁻ + H⁺ → CH₃COOH, neutralises added H⁺ (1 mark)
Weak acid reforms, pH stable (1 mark)
Buffer capacity from equilibrium (1 mark)

Question 6

Write the Ksp expression for CaF₂ and calculate [Ca²⁺] if Ksp = 3.9×10⁻¹¹, [F⁻] = 0.01 M. [2 Marks]

Answer 6

Ksp = [Ca²⁺][F⁻]² (1 mark)
3.9×10⁻¹¹ = ², [Ca²⁺] = 3.9×10⁻⁷ M (1 mark)

Question 7

In blood, how does HCO₃⁻ act as a buffer? Include an equation. [2 Marks]

Answer 7

HCO₃⁻ + H⁺ ⇌ H₂CO₃, removes excess H⁺ (1 mark)
Maintains blood pH ~7.4 (1 mark)

Question 8

Iodine partitions between water (0.02 M) and hexane (0.4 M). Calculate Kpc. [2 Marks]

Answer 8

Kpc = [I₂]hexane/[I₂]water = 0.4/0.02 (1 mark)
Kpc = 20 (1 mark)

Question 9

Predict how NaCl affects BaSO₄ solubility (Ksp = 1.1×10⁻¹⁰) with the common ion effect. [2 Marks]

Answer 9

NaCl adds Cl⁻, no common ion with BaSO₄ (1 mark)
Solubility unchanged, Ksp constant (1 mark)

Question 10

Calculate pH of 0.05 M NaOH and explain why Kw = 10⁻¹⁴ is used. [2 Marks]

Answer 10

[OH⁻] = 0.05, [H⁺] = 10⁻¹⁴/0.05 = 2×10⁻¹³, pH = 12.7 (1 mark)
Kw = [H⁺][OH⁻] defines water equilibrium (1 mark)

Question 11

For PbCl₂, Ksp = 1.6×10⁻⁵. Calculate solubility in 0.1 M NaCl and explain the common ion effect. [3 Marks]

Answer 11

Ksp = [Pb²⁺][Cl⁻]², [Cl⁻] = 0.1, 1.6×10⁻⁵ = ² (1 mark)
[Pb²⁺] = 1.6×10⁻³ M, solubility reduced (1 mark)
Common ion Cl⁻ shifts equilibrium left (1 mark)