Ionisation constant of HA (weak acid) and BOH (weak base) are `3.0xx10^(-7)` each at 298 K. the percentage extent of hydrolysis of BA at the dilution of 10 L is :

Options

A. 25
B. 50
C. 75
D. 40

Answer

To solve the problem, we need to find the percentage extent of hydrolysis of the salt BA formed from a weak acid (HA) and a weak base (BOH), given that both have the same ionization constant \( K_a = K_b = 3.0 \times 10^{-7} \) at 298 K.

### Step-by-Step Solution:

1. **Understanding Hydrolysis of Salt**:
The salt BA dissociates in water to give ions:
\[ BA \rightleftharpoons B^+ + A^- \]
where \( B^+ \) is the cation from the weak base and \( A^- \) is the anion from the weak acid.

2. **Setting Up the Equilibrium Expression**:
At equilibrium, the concentrations of the ions will change due to hydrolysis. The hydrolysis reaction can be represented as:
\[ B^+ + H_2O \rightleftharpoons BOH + H^+ \]
\[ A^- + H_2O \rightleftharpoons HA + OH^- \]
The equilibrium constant for hydrolysis \( K_h \) can be expressed as:
\[ K_h = \frac{K_w}{K_a \cdot K_b} \]
where \( K_w \) is the ion product of water, \( K_w = 1.0 \times 10^{-14} \) at 298 K.

3. **Calculating \( K_h \)**:
Substituting the values:
\[ K_h = \frac{1.0 \times 10^{-14}}{(3.0 \times 10^{-7}) \cdot (3.0 \times 10^{-7})} \]
\[ K_h = \frac{1.0 \times 10^{-14}}{9.0 \times 10^{-14}} = \frac{1}{9} \approx 0.111 \]

4. **Using the Hydrolysis Constant**:
Let \( C \) be the initial concentration of the salt BA. At equilibrium, if \( \alpha \) is the degree of hydrolysis, the concentrations will be:
- \( [B^+] = C(1 - \alpha) \)
- \( [A^-] = C(1 - \alpha) \)
- \( [H^+] = C\alpha \)
- \( [OH^-] = C\alpha \)

The expression for \( K_h \) becomes:
\[ K_h = \frac{[H^+][OH^-]}{[B^+][A^-]} = \frac{(C\alpha)(C\alpha)}{(C(1 - \alpha))(C(1 - \alpha))} = \frac{C^2\alpha^2}{C^2(1 - \alpha)^2} = \frac{\alpha^2}{(1 - \alpha)^2} \]

5. **Setting Up the Equation**:
Thus, we have:
\[ K_h = \frac{\alpha^2}{(1 - \alpha)^2} \]
Plugging in the value of \( K_h \):
\[ 0.111 = \frac{\alpha^2}{(1 - \alpha)^2} \]

6. **Solving for \( \alpha \)**:
Taking the square root of both sides:
\[ \sqrt{0.111} = \frac{\alpha}{1 - \alpha} \]
Let \( \sqrt{0.111} \approx 0.333 \):
\[ 0.333 = \frac{\alpha}{1 - \alpha} \]
Rearranging gives:
\[ 0.333(1 - \alpha) = \alpha \]
\[ 0.333 - 0.333\alpha = \alpha \]
\[ 0.333 = \alpha + 0.333\alpha \]
\[ 0.333 = 1.333\alpha \]
\[ \alpha = \frac{0.333}{1.333} \approx 0.25 \]

7. **Calculating Percentage Hydrolysis**:
The percentage extent of hydrolysis is given by:
\[ \text{Percentage Hydrolysis} = \alpha \times 100 = 0.25 \times 100 = 25\% \]

### Final Answer:
The percentage extent of hydrolysis of BA at the dilution of 10 L is **25%**.