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

Options

A. `2.5 xx 10^(-5)`
B. `1.5 xx 10^(-4)`
C. `3.125 xx 10^(-6)`
D. `6.25 xx 10^(-4)`

Answer

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

### Step 1: Write the hydrolysis reaction
The hydrolysis of ammonium benzoate (C₆H₅COONH₄) 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 concentrations
Let the initial concentration of ammonium benzoate (C) be 0.5 M. At the start, we have:
- Concentration of C₆H₅COONH₄ = C = 0.5 M
- Concentration of C₆H₅COO⁻ = 0
- Concentration of NH₄⁺ = 0

### Step 3: Define changes in concentration
Let α be the degree of hydrolysis. After hydrolysis, the concentrations will change as follows:
- Concentration of C₆H₅COONH₄ = C - Cα = 0.5 - 0.5α
- Concentration of C₆H₅COO⁻ = Cα = 0.5α
- Concentration of NH₄⁺ = Cα = 0.5α

### Step 4: Find the degree of hydrolysis (α)
Given that the hydrolysis is 0.25%, we can express this as:
\[ \alpha = \frac{0.25}{100} = 0.0025 \]

### Step 5: Substitute values into the hydrolysis constant expression
The hydrolysis constant (K_h) can be expressed as:
\[ K_h = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{NH}_4^+]}{[\text{C}_6\text{H}_5\text{COONH}_4]} \]
Substituting the concentrations we found:
\[ K_h = \frac{(0.5α)(0.5α)}{0.5 - 0.5α} \]

### Step 6: Simplify the expression
Since α is very small (0.0025), we can approximate:
\[ K_h \approx \frac{(0.5 \times 0.0025)(0.5 \times 0.0025)}{0.5} \]
\[ K_h \approx \frac{(0.5^2)(0.0025^2)}{0.5} \]
\[ K_h = 0.5 \times (0.5^2) \times (0.0025^2) \]
\[ K_h = 0.5 \times 0.25 \times 0.00000625 \]
\[ K_h = 0.5 \times 0.25 \times 6.25 \times 10^{-6} \]
\[ K_h = 3.125 \times 10^{-6} \]

### Final Answer
Thus, the hydrolysis constant for ammonium benzoate is:
\[ K_h = 3.125 \times 10^{-6} \]

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