HA(a q) ⇌ H⁺(a q)+A⁻(a q)
The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is
Given: K_f(H₂O) = 1.8 K kg mol⁻¹, molality ≡ molarity

Solution

Chemistry ma’am · 3 steps

Freezing point drop gives i, then i gives α, and α gives Kₐ.

1. Van't Hoff factor. Dissociation makes extra particles, so Δ T_f = i K_f m. This gives i = 0.20/(1.8 × 0.1) = 0.20/0.18 = 1.111.
2. Degree of dissociation. For HA ⇌ H⁺ + A⁻, each molecule that dissociates gives 2 particles, so i = 1 + α. That makes α = 0.111, which is exactly 1/9.

   Kₐ = Cα²/(1-α) = (0.1 × (1/9)²)/(1-1/9) = 0.1/72

3. So. Kₐ = 0.1/72 ≈ 1.39 × 10⁻³, which matches 1.38 × 10⁻³, option (b).

The idea to keep. For a weak electrolyte, a colligative property gives i, and i gives α. For a 1:1 acid, i = 1 + α and Kₐ = Cα²/(1-α).

Answer from the NTA final answer key. Solution by PrepWiser.