Gaokao Exam Gaokao Chemistry: Chemical Equilibrium and Reactions 4

Question 1

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

Accepted Answer

4.74

Answer

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

Question 2

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

Accepted Answer

Sodium metal (Na)

Answer

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?

Accepted Answer

Decreasing pressure

Answer

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:

Accepted Answer

1.34×10⁻⁵ M

Answer

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:

Accepted Answer

Buffer solutions

Answer

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

Gaokao Exam Practice Test

Gaokao Exam Practice Test

Gaokao Exam Gaokao Chemistry: Chemical Equilibrium and Reactions 4