The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be :

Options

A. `10^(-9)`
B. `10^(-7)`
C. `10^(-5)`
D. `10^(-3)`

Answer

To find the ionization constant (K) for the dissociation of HCN, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Concentration of HCN (C) = 0.1 M
- Degree of dissociation (α) = 0.01%

2. **Convert the Degree of Dissociation to Decimal:**
- α = 0.01% = 0.01/100 = 0.0001

3. **Write the Expression for Ionization Constant (K):**
- The ionization of HCN can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]
- The equilibrium expression for K is given by:
\[ K = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

4. **Determine the Concentrations at Equilibrium:**
- Initial concentration of HCN = C = 0.1 M
- At equilibrium, the concentration of HCN will be:
\[ [\text{HCN}] = C(1 - \alpha) = 0.1(1 - 0.0001) \approx 0.1 \, \text{M} \, (\text{since } \alpha \text{ is very small}) \]
- The concentrations of the ions at equilibrium will be:
\[ [\text{H}^+] = [\text{CN}^-] = C \cdot \alpha = 0.1 \cdot 0.0001 = 0.00001 \, \text{M} \]

5. **Substitute the Values into the K Expression:**
- Now substituting the values into the K expression:
\[ K = \frac{(0.00001)(0.00001)}{0.1} = \frac{0.0000000001}{0.1} = 0.000000001 = 10^{-9} \]

6. **Final Result:**
- The ionization constant (K) for HCN is:
\[ K = 10^{-9} \]

### Summary:
The ionization constant (K) for the 0.1 M HCN solution with a degree of dissociation of 0.01% is \( K = 10^{-9} \).