For a weak acid HA of concentration `C(mol l^-1)` and degree of dissociation `(alpha)`, Ostwald's dilution law is represented by the equation

Options

A. `K_a=(C^2alpha)/(1-alpha)`
B. `K_a=(alpha^2C)/(1-alpha)`
C. `K_a= Calpha`
D. `K_a= (Calpha^2)/(1-alpha^2)`

Answer

To derive the Ostwald's dilution law for a weak acid HA with concentration \( C \) and degree of dissociation \( \alpha \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the dissociation equation**:
The weak acid HA dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

2. **Define initial concentrations**:
- Initial concentration of HA = \( C \) (mol/L)
- At the start, the concentrations of \( H^+ \) and \( A^- \) are both 0.

3. **Establish changes at equilibrium**:
- When the acid dissociates, let \( \alpha \) be the degree of dissociation. Thus, at equilibrium:
- Concentration of \( HA \) = \( C - C\alpha = C(1 - \alpha) \)
- Concentration of \( H^+ \) = \( C\alpha \)
- Concentration of \( A^- \) = \( C\alpha \)

4. **Write the expression for the equilibrium constant \( K_a \)**:
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

5. **Substitute the equilibrium concentrations into the expression**:
- Substitute the equilibrium concentrations into the \( K_a \) expression:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

6. **Simplify the equation**:
- This simplifies to:
\[ K_a = \frac{C^2\alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

7. **Final expression**:
Thus, the Ostwald's dilution law for the weak acid HA is:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Conclusion:
The final expression derived from Ostwald's dilution law for the weak acid HA is:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]