When 9.45 g of CICH₂COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH₂COOH is x × 10⁻³.
The value of x is ________. (Rounded off to the nearest integer)
[K f(H₂0) = 1.86 K kg mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Use the freezing point drop to find i, turn i into α, then put α into Ostwald's expression for Kₐ.

1. Molality first. Chloroacetic acid, ClCH₂COOH, has molar mass 94.5 g mol⁻¹, so 9.45 g is 0.1 mol. Take 500 mL of water as 0.5 kg, so the molality is m = 0.2 mol kg⁻¹.
2. Find i. From Δ T_f = i K_f m we get 0.5 = i × 1.86 × 0.2 = 0.372 i. So i = 1.344, which is more than 1 because the acid partly ionises.
3. Degree of dissociation. ClCH₂COOH ⇌ ClCH₂COO⁻ + H⁺ gives 2 particles per molecule that ionises, so i = 1 + α. That makes α = 0.344.

   Kₐ = cα²/(1-α) = (0.2 × (0.344)²)/0.656 = 0.02368/0.656 = 3.61 × 10⁻²

4. So. Here the dilute solution's molality, 0.2, is used as c. Then Kₐ = 36.1 × 10⁻³, so x rounds to 36.

The common slip: 24. Dropping the 1-α and using Kₐ = cα² gives about 24, not 36. With α near 0.34, the approximation is far too rough.

The idea to keep. A colligative property gives you i. For a weak electrolyte that splits into n ions, α = (i-1)/(n-1). Then use Kₐ = cα²/(1-α), and keep the 1-α when α is large.

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