Gaokao Chemistry: Chemical Equilibrium and Reactions Flashcards

The Ka of acetic acid is 1.8×10⁻⁵. The pKa is approximately:

Answer: 4.74

pKa = −log(Ka) = −log(1.8×10⁻⁵) ≈ 4.74.

In the electrolysis of molten NaCl, the product at the cathode is:

Answer: Sodium metal (Na)

At the cathode, reduction occurs: Na⁺ + e⁻ → Na. Chloride is oxidised at the anode to Cl₂.

For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which change increases the equilibrium yield of PCl₃?

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.

The solubility product Ksp of AgCl is 1.8×10⁻¹⁰. The molar solubility of AgCl in pure water is:

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.

The Henderson-Hasselbalch equation pH = pKa + log([A⁻]/[HA]) is used to calculate the pH of:

Answer: Buffer solutions

The Henderson-Hasselbalch equation applies to buffer solutions containing a weak acid and its conjugate base.

In the reaction 2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + 5O₂ + K₂SO₄ + 8H₂O, the oxidation state of Mn changes from:

Answer: +7 to +2

In KMnO₄, Mn is +7; in MnSO₄, Mn is +2. This is a 5-electron reduction of manganese.

For a first-order reaction, the half-life t₁/₂ is related to the rate constant k by:

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.