Consider the dissociation of the weak acid HX as given below $$HX(aq) \rightleftharpoons H^+(aq) + X^-(aq)$$, $$K_a = 1.2 \times 10^{-5}$$ [$$K_a$$ : dissociation constant]. The osmotic pressure of $$0.03$$ M aqueous solution of HX at $$300 \text{ K}$$ is _______ $$\times 10^{-2}$$ bar (nearest integer). [Given : $$R = 0.083 \text{ L bar mol}^{-1} \text{K}^{-1}$$]

We need to find the osmotic pressure of a 0.03 M aqueous solution of weak acid HX. The acid dissociates as $$HX \rightleftharpoons H^+ + X^-$$, and we are given $$K_a = 1.2 \times 10^{-5}$$, $$C = 0.03$$ M, $$T = 300$$ K, and $$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$.

We begin by finding the degree of dissociation, $$\alpha$$. At equilibrium, $$[HX] = C(1-\alpha)$$, $$[H^+] = C\alpha$$, and $$[X^-] = C\alpha$$, so

$$K_a = \frac{C\alpha \times C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$

For small $$\alpha$$ ($$\alpha \ll 1$$), we approximate $$K_a \approx C\alpha^2$$, which gives

$$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2 \times 10^{-5}}{0.03}} = \sqrt{4 \times 10^{-4}} = 0.02$$

Since $$\alpha = 0.02$$ (2%), the approximation $$\alpha \ll 1$$ is valid.

The van’t Hoff factor, $$i$$, for a substance that dissociates into 2 ions with degree of dissociation $$\alpha$$ is

$$i = 1 + \alpha(n - 1) = 1 + \alpha(2 - 1) = 1 + \alpha = 1 + 0.02 = 1.02$$

Finally, the osmotic pressure is given by

$$\pi = iCRT$$

Substituting the values,

$$\pi = 1.02 \times 0.03 \times 0.083 \times 300$$

First compute $$CRT$$:

$$CRT = 0.03 \times 0.083 \times 300 = 0.747$$ bar

Thus,

$$\pi = 1.02 \times 0.747 = 0.76194$$ bar

$$= 76.194 \times 10^{-2}$$ bar $$\approx 76 \times 10^{-2}$$ bar

The answer is 76.