Question

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

Answer 1

Correct Answer: 76

The dissociation of HX is described as follows:
\[\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-\]
The initial concentration of HX is 0.03 M.
At equilibrium, the concentrations are:
\[[\text{HX}] = 0.03 - x, \quad [\text{H}^+] = x, \quad [\text{X}^-] = x\]
Using the dissociation constant $K_a$:
\[K_a = \frac{x^2}{0.03 - x}\]
Given $K_a = 1.2 \times 10^{-5}$, and assuming $0.03 - x \approx 0.03$ due to the small $K_a$ value:
\[1.2 \times 10^{-5} = \frac{x^2}{0.03}\]
Solving for $x^2$:
\[x^2 = 1.2 \times 10^{-5} \times 0.03 = 3.6 \times 10^{-7}\]
Calculating $x$:
\[x = \sqrt{3.6 \times 10^{-7}} = 6 \times 10^{-4}\]
The total solute concentration is calculated as:
\[C_{\text{total}} = [\text{HX}] + [\text{H}^+] + [\text{X}^-] = 0.03 - x + x + x = 0.03 + x\]
Substituting the value of $x$:
\[C_{\text{total}} = 0.03 + 6 \times 10^{-4} = 0.0306 \, \text{M}\]
The osmotic pressure $\Pi$ is determined using the formula:
\[\Pi = C_{\text{total}}RT\]
Substituting the values:
\[\Pi = (0.0306) \times (0.083) \times (300)\]
\[\Pi = 76.19 \, \text{bar}\]
The nearest integer value for $\Pi$ is:
\[\Pi = 76 \times 10^{-2} \, \text{bar}\]