If at `25^(@)` the ionization constant of acetic acid is `2 xx 10^(5)` the hydrolysis constant of sodium acetate will be

Options

A. `5 xx 10^(-8)`
B. `5 xx 10^(-9)`
C. `5 xx 10^(-10)`
D. `4 xx 10^(-10)`

Answer

To find the hydrolysis constant (K_h) of sodium acetate (CH₃COONa) given the ionization constant (K_a) of acetic acid (CH₃COOH) at 25°C, we can use the following relationship:

1. **Understanding the Relationship**:
The hydrolysis of sodium acetate can be represented as:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]
The hydrolysis constant (K_h) for this reaction can be expressed in terms of the ionization constant (K_a) of acetic acid and the ion product of water (K_w).

2. **Using the Formula**:
The relationship between K_a, K_h, and K_w is given by:
\[ K_w = K_a \times K_h \]
Where:
- K_w is the ion product of water at 25°C, which is \(1.0 \times 10^{-14}\).
- K_a is the ionization constant of acetic acid, given as \(2.0 \times 10^{-5}\).

3. **Rearranging the Formula**:
To find K_h, we can rearrange the formula:
\[ K_h = \frac{K_w}{K_a} \]

4. **Substituting the Values**:
Now, substituting the known values into the equation:
\[ K_h = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} \]

5. **Calculating K_h**:
Performing the calculation:
\[ K_h = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10} \]

Thus, the hydrolysis constant of sodium acetate at 25°C is \(5.0 \times 10^{-10}\).