Ionisation constant of a weak acid (HA) in terms of `A_(m)^(oo)` and `A_(m)` is:

Options

A. `K_(a)=(CA_(m)^(oo))/((A_(m)-A^(oo)))`
B. `k_(a)=(CA_(m)^(2))/(A_(m)^(oo)(A_(m)^(00)-A_(m)))`
C. `K_(a)=(C(A_(m)^(oo))^(2))/(A_(m)^(oo)(A_(m)^(oo)-A_(m)))`
D. None of these

Answer

To derive the ionization constant (Ka) of a weak acid (HA) in terms of molar conductivity at infinite dilution (A_m^∞) and molar conductivity at some specific concentration (A_m), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of a weak acid (HA) can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

2. **Define Initial and Equilibrium Concentrations**:
Let the initial concentration of the acid be \( C \). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- \([H^+] = C \alpha\)
- \([A^-] = C \alpha\)
- \([HA] = C(1 - \alpha)\)

3. **Write the Expression for the Ionization Constant (Ka)**:
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C \alpha \cdot C \alpha}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]

4. **Relate Degree of Dissociation to Molar Conductivity**:
The degree of dissociation \( \alpha \) can be expressed in terms of molar conductivities:
\[ \alpha = \frac{A_m}{A_m^{\infty}} \]
where \( A_m \) is the molar conductivity at concentration \( C \) and \( A_m^{\infty} \) is the molar conductivity at infinite dilution.

5. **Substitute \( \alpha \) into the Ka Expression**:
Substitute the expression for \( \alpha \) into the equation for \( K_a \):
\[ K_a = \frac{C \left(\frac{A_m}{A_m^{\infty}}\right)^2}{1 - \frac{A_m}{A_m^{\infty}}} \]

6. **Simplify the Expression**:
To simplify, let’s rewrite the denominator:
\[ K_a = \frac{C \frac{A_m^2}{A_m^{\infty 2}}}{\frac{A_m^{\infty} - A_m}{A_m^{\infty}}} = \frac{C A_m^2}{A_m^{\infty 2} \cdot \left( \frac{A_m^{\infty} - A_m}{A_m^{\infty}} \right)} \]
This can be further simplified to:
\[ K_a = \frac{C A_m^2}{A_m^{\infty} (A_m^{\infty} - A_m)} \]

### Final Expression:
Thus, the ionization constant \( K_a \) of the weak acid (HA) in terms of molar conductivities is:
\[ K_a = \frac{C A_m^2}{A_m^{\infty} (A_m^{\infty} - A_m)} \]