0.5 M ammonium benzoate is hydrolysed to 0.25 percent, hence, its hydrolysis constant is ?

Options

A. `1.25×10^(−4)`
B. `3.125×10^(−6)`
C. `2.5×10^(−4)`
D. `6.25×10^(−4)`

Explanation

To find the hydrolysis constant (K_h) for the hydrolysis of ammonium benzoate, we can follow these steps:

### Step 1: Understand the given information
We are given:
- Concentration (C) of ammonium benzoate = 0.5 M
- Degree of hydrolysis (α) = 0.25%

### Step 2: Convert the degree of hydrolysis to a decimal
To use α in our calculations, we need to convert it from a percentage to a decimal:
\[ \alpha = \frac{0.25}{100} = 0.0025 \]

### Step 3: Use the formula for hydrolysis constant
The hydrolysis constant (K_h) is given by the formula:
\[ K_h = C \cdot \alpha^2 \]
Where:
- C is the concentration of the salt (0.5 M)
- α is the degree of hydrolysis (0.0025)

### Step 4: Substitute the values into the formula
Now we can substitute the values into the formula:
\[ K_h = 0.5 \cdot (0.0025)^2 \]

### Step 5: Calculate α squared
First, calculate \( (0.0025)^2 \):
\[ (0.0025)^2 = 0.00000625 \]

### Step 6: Calculate K_h
Now substitute this value back into the equation:
\[ K_h = 0.5 \cdot 0.00000625 = 0.000003125 \]

### Step 7: Convert to scientific notation
To express this in scientific notation:
\[ K_h = 3.125 \times 10^{-6} \]

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

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