`0.5 M` ammonium benzoate is hydrolysed to `0.25` precent, hence its hydrolysis constant is

Options

A. `2.5xx10^(-5)`
B. `1.5xx10^(-4)`
C. `3.125xx10^(-6)`
D. `6.25xx10^(-4)`

Explanation

To find the hydrolysis constant of ammonium benzoate, we will follow these steps:

### Step 1: Understand the Reaction
Ammonium benzoate (C6H5COONH4) hydrolyzes in water to form benzoate ions (C6H5COO^-) and ammonium ions (NH4^+). The reaction can be represented as:
\[ \text{C}_6\text{H}_5\text{COONH}_4 \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{NH}_4^+ \]

### Step 2: Define Initial Concentration and Degree of Hydrolysis
Given:
- Initial concentration (C) = 0.5 M
- Degree of hydrolysis (α) = 0.25%

Convert the degree of hydrolysis from percentage to a fraction:
\[ \alpha = \frac{0.25}{100} = 0.0025 \]

### Step 3: Set Up the Hydrolysis Constant Expression
The hydrolysis constant (K_h) can be expressed as:
\[ K_h = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{NH}_4^+]}{[\text{C}_6\text{H}_5\text{COONH}_4]} \]

At equilibrium:
- Concentration of C6H5COO^- = Cα = 0.5 × 0.0025 = 0.00125 M
- Concentration of NH4^+ = Cα = 0.5 × 0.0025 = 0.00125 M
- Concentration of C6H5COONH4 = C(1 - α) = 0.5(1 - 0.0025) ≈ 0.5 M (since α is very small)

### Step 4: Substitute Values into the Hydrolysis Constant Expression
Now substitute these values into the hydrolysis constant expression:
\[ K_h = \frac{(0.00125)(0.00125)}{0.5} \]

### Step 5: Calculate the Hydrolysis Constant
Calculate the value:
\[ K_h = \frac{(0.00125)^2}{0.5} = \frac{0.0000015625}{0.5} = 0.000003125 \]

Expressing this in scientific notation:
\[ K_h = 3.125 \times 10^{-6} \]

### Conclusion
The hydrolysis constant of ammonium benzoate is:
\[ K_h = 3.125 \times 10^{-6} \]

---