Question

pH of 0.5 M monobasic acid solution is found to be 2. Thus its osmotic pressure at T K is

• A. 0.01 RT

Answer 1

0.51 RT

Answer 2

The problem asks for the osmotic pressure of a 0.5 M monobasic acid solution with a pH of 2 at temperature T K.

1. Calculate the hydrogen ion concentration ([H⁺]) from pH: Given pH = 2, $[H^+] = 10^{-pH} = 10^{-2} \, M = 0.01 \, M$

2. Determine the degree of dissociation (α): For a monobasic acid (HA), the dissociation equilibrium is: $HA \rightleftharpoons H^+ + A^-$ The initial concentration of the acid is $C = 0.5 \, M$. At equilibrium, the concentration of $H^+$ ions produced is $C\alpha$. Therefore, $[H^+] = C\alpha$ $0.01 \, M = 0.5 \, M \times \alpha$ $\alpha = \frac{0.01}{0.5} = 0.02$

3. Calculate the van't Hoff factor (i): For the dissociation of a weak electrolyte into 'n' ions, the van't Hoff factor is given by $i = 1 + (n-1)\alpha$. For a monobasic acid, $n=2$ (one $H^+$ ion and one $A^-$ ion are produced per molecule). So, $i = 1 + (2-1)\alpha = 1 + \alpha$ $i = 1 + 0.02 = 1.02$

4. Calculate the osmotic pressure (π): The osmotic pressure is given by the formula $\pi = iCRT$, where $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature in Kelvin. $\pi = 1.02 \times 0.5 \, M \times R \times T$ $\pi = 0.51 \, RT$

The calculated osmotic pressure is $0.51 \, RT$. The given option is (A) $0.01 \, RT$. There is a discrepancy between the calculated value and the provided option. Based on the standard principles of colligative properties and weak acid dissociation, the calculated value is $0.51 \, RT$.

The final answer is 0.51 RT However, since the question provides options and expects one of them to be correct, and the calculated answer is not among them, it indicates a potential error in the question or the options provided. If forced to choose from the given options, and assuming a common conceptual error might be tested (where osmotic pressure is mistakenly taken as [H⁺]RT), then 0.01 RT would be the result. But this is chemically incorrect.