The hydrolysis constant of `0.1M` aqueous solution of sodium acetate if `K_(a)` of `CH_(3)COOH = 1.8 xx 10^(-5)` is

Options

A. `5.5 xx 10^(-10)`
B. `4.5 xx 10^(-8)`
C. `5.5 xx 10^(-12)`
D. None of these

Explanation

To find the hydrolysis constant (K_h) of a 0.1 M aqueous solution of sodium acetate (CH₃COONa), we can use the relationship between the hydrolysis constant, the dissociation constant of the weak acid (K_a), and the ion-product constant of water (K_w).

### Step-by-Step Solution:

1. **Identify the given values**:
- Concentration of sodium acetate (CH₃COONa) = 0.1 M
- Dissociation constant of acetic acid (K_a) = 1.8 × 10^(-5)
- Ion-product constant of water (K_w) = 1.0 × 10^(-14) (This is a known constant at 25°C)

2. **Write the formula for hydrolysis constant**:
The hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]

3. **Substitute the known values into the formula**:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]

4. **Perform the calculation**:
- First, calculate the division:
\[ K_h = \frac{1.0}{1.8} \times 10^{-14 + 5} = \frac{1.0}{1.8} \times 10^{-9} \]
- Now calculate \(\frac{1.0}{1.8}\):
\[ \frac{1.0}{1.8} \approx 0.5556 \]
- Therefore:
\[ K_h \approx 0.5556 \times 10^{-9} \approx 5.56 \times 10^{-10} \]

5. **Final answer**:
The hydrolysis constant (K_h) of the 0.1 M aqueous solution of sodium acetate is approximately:
\[ K_h \approx 5.56 \times 10^{-10} \]