Gaokao Exam Gaokao Chemistry: Chemical Equilibrium and Reactions 4 — Questions and Answers
Question 1: The Ka of acetic acid is 1.8×10⁻⁵. The pKa is approximately:
- 4.74 (Correct answer)
- 5.26
- 9.26
- 14.0
Correct answer: 4.74
pKa = −log(Ka) = −log(1.8×10⁻⁵) ≈ 4.74.
Question 2: In the electrolysis of molten NaCl, the product at the cathode is:
- Sodium metal (Na) (Correct answer)
- Chlorine gas (Cl₂)
- Hydrogen gas (H₂)
- Oxygen gas (O₂)
Correct answer: Sodium metal (Na)
At the cathode, reduction occurs: Na⁺ + e⁻ → Na. Chloride is oxidised at the anode to Cl₂.
Question 3: For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which change increases the equilibrium yield of PCl₃?
- Decreasing pressure (Correct answer)
- Adding more PCl₃
- Decreasing temperature (if exothermic)
- Adding an inert gas at constant volume
Correct answer: Decreasing pressure
The forward reaction increases moles of gas (1 → 2); decreasing pressure shifts equilibrium to the side with more moles of gas, increasing PCl₃ yield.
Question 4: The solubility product Ksp of AgCl is 1.8×10⁻¹⁰. The molar solubility of AgCl in pure water is:
- 1.34×10⁻⁵ M (Correct answer)
- 1.8×10⁻¹⁰ M
- 9.0×10⁻¹¹ M
- 3.6×10⁻¹⁰ M
Correct answer: 1.34×10⁻⁵ M
AgCl → Ag⁺ + Cl⁻; if s = molar solubility, then Ksp = s² → s = √(1.8×10⁻¹⁰) ≈ 1.34×10⁻⁵ M.
Question 5: The Henderson-Hasselbalch equation pH = pKa + log([A⁻]/[HA]) is used to calculate the pH of:
- Buffer solutions (Correct answer)
- Strong acid solutions
- Strong base solutions
- Pure water
Correct answer: Buffer solutions
The Henderson-Hasselbalch equation applies to buffer solutions containing a weak acid and its conjugate base.
Question 6: In the reaction 2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + 5O₂ + K₂SO₄ + 8H₂O, the oxidation state of Mn changes from:
- +7 to +2 (Correct answer)
- +7 to +4
- +2 to +7
- +4 to +2
Correct answer: +7 to +2
In KMnO₄, Mn is +7; in MnSO₄, Mn is +2. This is a 5-electron reduction of manganese.
Question 7: For a first-order reaction, the half-life t₁/₂ is related to the rate constant k by:
- t₁/₂ = ln 2 / k (Correct answer)
- t₁/₂ = 1/k
- t₁/₂ = k/ln 2
- t₁/₂ = 2/k
Correct answer: t₁/₂ = ln 2 / k
For a first-order reaction: [A] = [A₀]e^(−kt); at t₁/₂, [A] = [A₀]/2, giving t₁/₂ = ln 2 / k ≈ 0.693/k.
Gaokao Exam Gaokao Chemistry: Chemical Equilibrium and Reactions 4
The Ka of acetic acid is 1.8×10⁻⁵.
The pKa is approximately: