1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is ___________. (Nearest integer)
[Given : Density of acetic acid is 1.02 g mL⁻¹, Molar mass of acetic acid is 60 g mol⁻¹, K f (H₂O) = 1.85 K kg mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Compare the observed freezing point drop with the no-dissociation drop. The extra part is α.

1. Moles of acid. Mass = 1.2 mL × 1.02 g mL⁻¹ = 1.224 g, so moles = 1.224/60 = 0.0204 mol. In 2.0 L of a dilute water solution, this gives a molality of about 0.0102 mol kg⁻¹.
2. Find i. If the acid did not dissociate, Δ T_f = 1.85 × 0.0102 = 0.01887 K. The observed drop is larger, and the ratio of the two is i.

   i = (Δ T_f(observed))/(K_f m) = 0.0198/0.01887 = 1.049

3. Link i to α. CH₃COOH ⇌ CH₃COO⁻ + H⁺ turns one particle into two, so i = 1 + α. That gives α = 1.049 - 1 = 0.049.
4. So. The acid is 4.9% dissociated, which rounds to 5.

The common slip: 7. Taking 1.2 g as the mass and ignoring the density gives i ≈ 1.07 and an answer of 7, not 5.

The idea to keep. For a weak electrolyte, i = observed/calculated colligative effect, and α = (i - 1)/(n - 1). For a weak acid HA, n = 2, so α = i - 1.

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