If 0.1 M `H_(2)SO_(4)`(aq.) solution shows freezing point `-0.3906^(@)C` then what is the `K_(a2)` for `H_(2)SO_(4) `? (Assume m = M and `K_(f(H_(2)O)` = `1.86 K kg ``mol^(-1)`)
(a)0.122
(b)0.0122
(d)1.11x `10^(-3)`
(d)None of these

Options

A. 0.122
B. 0.0122
C. 1.11x `10^(-3)`
D. None of these

Answer

To solve the problem, we need to determine the second dissociation constant \( K_{a2} \) for sulfuric acid \( H_2SO_4 \) given the freezing point depression of a 0.1 M solution.

### Step-by-Step Solution:

1. **Understand Freezing Point Depression**:
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \( K_f \) is the cryoscopic constant (given as \( 1.86 \, \text{K kg mol}^{-1} \)),
- \( m \) is the molality (assumed equal to molarity, so \( m = 0.1 \, \text{mol/kg} \)),
- \( i \) is the van 't Hoff factor.

2. **Calculate the Freezing Point Depression**:
Given that the freezing point is \( -0.3906^\circ C \), the depression is:
\[ \Delta T_f = 0 - (-0.3906) = 0.3906^\circ C \]

3. **Substitute Values into the Freezing Point Depression Formula**:
Rearranging the formula to solve for \( i \):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the known values:
\[ i = \frac{0.3906}{1.86 \cdot 0.1} = \frac{0.3906}{0.186} \approx 2.1 \]

4. **Determine the Ionization of \( H_2SO_4 \)**:
\( H_2SO_4 \) dissociates in two steps:
- First dissociation: \( H_2SO_4 \rightarrow H^+ + HSO_4^- \) (complete ionization, \( i = 2 \))
- Second dissociation: \( HSO_4^- \rightarrow H^+ + SO_4^{2-} \)

From the first dissociation, we have:
- Initial concentration \( C = 0.1 \, \text{M} \)
- After complete dissociation, \( [H^+] = [HSO_4^-] = 0.1 \, \text{M} \)

5. **Consider the Second Dissociation**:
Let \( \alpha \) be the degree of ionization for the second dissociation:
- \( [HSO_4^-] = 0.1(1 - \alpha) \)
- \( [H^+] = 0.1 + 0.1\alpha \)
- \( [SO_4^{2-}] = 0.1\alpha \)

6. **Write the Expression for \( K_{a2} \)**:
The expression for the second dissociation constant \( K_{a2} \) is:
\[ K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} = \frac{(0.1 + 0.1\alpha)(0.1\alpha)}{0.1(1 - \alpha)} \]

7. **Substituting Values**:
Since \( i = 2.1 \), we can express:
\[ i = 2 + \alpha \implies \alpha = 2.1 - 2 = 0.1 \]
Now substituting \( \alpha = 0.1 \):
\[ K_{a2} = \frac{(0.1 + 0.1 \cdot 0.1)(0.1 \cdot 0.1)}{0.1(1 - 0.1)} = \frac{(0.1 + 0.01)(0.01)}{0.1 \cdot 0.9} \]
\[ K_{a2} = \frac{0.11 \cdot 0.01}{0.09} = \frac{0.0011}{0.09} \approx 0.0122 \]

8. **Final Answer**:
Thus, the value of \( K_{a2} \) for \( H_2SO_4 \) is \( 0.0122 \).

### Conclusion:
The correct answer is (b) \( 0.0122 \).