If the dissociation constant of HCN is `1.3 xx 10^(-9)` the value of hydrolysis constant of HCN will be

Options

A. `1.3 xx 10^(-9)`
B. `10^(-14)`
C. `7.7 xx 10^(-5)`
D. `0.77 xx 10^(-5)`

Answer

To find the hydrolysis constant (K_h) of HCN given its dissociation constant (K_a), we can use the following formula:

\[ K_h = \frac{K_w}{K_a} \]

Where:
- \( K_h \) is the hydrolysis constant.
- \( K_w \) is the ionization constant of water, which is \( 1.0 \times 10^{-14} \) at 25°C.
- \( K_a \) is the dissociation constant of the acid, which is given as \( 1.3 \times 10^{-9} \).

### Step-by-step Solution:

1. **Identify the values**:
- \( K_a = 1.3 \times 10^{-9} \)
- \( K_w = 1.0 \times 10^{-14} \)

2. **Substitute the values into the formula**:
\[ K_h = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-9}} \]

3. **Perform the division**:
- First, calculate the numerator and denominator:
\[ K_h = \frac{1.0}{1.3} \times \frac{10^{-14}}{10^{-9}} = \frac{1.0}{1.3} \times 10^{-5} \]

4. **Calculate \( \frac{1.0}{1.3} \)**:
\[ \frac{1.0}{1.3} \approx 0.7692 \]

5. **Combine the results**:
\[ K_h \approx 0.7692 \times 10^{-5} \approx 7.7 \times 10^{-6} \]

6. **Final answer**:
The hydrolysis constant \( K_h \) of HCN is approximately \( 7.7 \times 10^{-6} \).