Question

When 9.45 g of $ClCH_2COOH$ is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of $ClCH_2COOH$ is $x \times 10^{-3}$. The value of $x$ is __________. (Rounded off to the nearest integer) [$K_f(H_2O)=1.86$ K kg mol$^{-1}$]

Hint:

The van't Hoff factor ($i$) accounts for dissociation. For a weak acid $HA$, $i = 1 + \alpha$.

Answer 1

Correct Answer: 36

Step 1: Moles of solute $n = \frac{9.45}{94.5} = 0.1$ mol. Molality $m = \frac{0.1}{0.5 \text{ kg}} = 0.2$ mol/kg.
Step 2: $\Delta T_f = i K_f m \Rightarrow 0.5 = i \times 1.86 \times 0.2 \Rightarrow i = \frac{0.5}{0.372} \approx 1.344$.
Step 3: For $ClCH_2COOH \to ClCH_2COO^- + H^+$, $i = 1 + \alpha$. So, $\alpha = 0.344$.
Step 4: $K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.2 \times (0.344)^2}{1 - 0.344} = \frac{0.02367}{0.656} \approx 0.0361$.
Step 5: $0.0361 = 36.1 \times 10^{-3} \Rightarrow x = 36$.