JEE Chemistry 2025 Question with Solution

Given:

• Freezing point depression = $0.20 \, ^\circ \text{C}$
• $K_f = 1.8 \, \text{K kg mol}^{-1}$
• Molality = $0.1 \, \text{m}$
• Dissociation: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

Find: $K_a$ for the weak acid HA.

For a monobasic weak acid, the van 't Hoff factor is

$$i = 1 + \alpha$$

Using freezing point depression,

$$\Delta T_f = iK_f m$$

Substitute the given values:

$$0.20 = (1+\alpha)(1.8)(0.1)$$ $$0.20 = 0.18 + 0.18\alpha$$ $$0.18\alpha = 0.02$$ $$\alpha = \frac{0.02}{0.18} = \frac{1}{9}$$

Now use the weak acid relation

$$K_a = \frac{c\alpha^2}{1-\alpha}$$

where $c = 0.1 \, \text{mol L}^{-1}$.

Substitute:

$$K_a = \frac{0.1\left(\frac{1}{9}\right)^2}{1-\frac{1}{9}}$$ $$K_a = \frac{0.1 \cdot \frac{1}{81}}{\frac{8}{9}}$$ $$K_a = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 \times 10^{-3}$$

Therefore, the dissociation constant is $1.38 \times 10^{-3}$ and the correct option is B.

Given: $\Delta T_f = 0.20 \, ^\circ \text{C}$, $K_f = 1.8 \, \text{K kg mol}^{-1}$, $m = 0.1$.

Find: $K_a$.

First calculate the van't Hoff factor:

$$i = \frac{\Delta T_f}{K_f m}$$ $$i = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.11$$

For the dissociation

$$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$$

we have

$$i = 1 + \alpha$$

So,

$$\alpha = 1.11 - 1 = 0.11$$

Now,

$$K_a = \frac{C\alpha^2}{1-\alpha}$$ $$K_a = \frac{0.1 \times (0.11)^2}{1-0.11}$$ $$K_a = \frac{0.1 \times 0.0121}{0.89} = 1.36 \times 10^{-3} \approx 1.38 \times 10^{-3}$$

Therefore, the correct option is B.