Specific conductance of 0.1 MHA is `3.75xx10^(-4)ohm^(-1)cm^(-1)`. If `lamda^(oo)` of HA is `250 ohm^(-1)cm^2mol^(-1)`, then dissociation constant `K_a` of HA is

Answer

To solve for the dissociation constant \( K_a \) of the weak acid HA, we will follow these steps:

### Step 1: Calculate the Degree of Dissociation (\( \alpha \))

The degree of dissociation (\( \alpha \)) can be calculated using the formula:

\[ \alpha = \frac{\text{Specific conductance at given concentration}}{\text{Specific conductance at infinite dilution}} \]

Given:
- Specific conductance of 0.1 M HA = \( 3.75 \times 10^{-4} \, \Omega^{-1} \, \text{cm}^{-1} \)
- Specific conductance at infinite dilution (\( \lambda^{\infty} \)) = \( 250 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)

Substituting the values:

\[ \alpha = \frac{3.75 \times 10^{-4}}{250} \]

Calculating \( \alpha \):

\[ \alpha = 1.5 \times 10^{-6} \]

### Step 2: Calculate the Dissociation Constant (\( K_a \))

The dissociation constant \( K_a \) can be calculated using the formula:

\[ K_a = C \alpha^2 \]

Where:
- \( C \) is the concentration of the acid (0.1 M)
- \( \alpha \) is the degree of dissociation calculated in Step 1

Substituting the values:

\[ K_a = 0.1 \times (1.5 \times 10^{-6})^2 \]

Calculating \( K_a \):

\[ K_a = 0.1 \times (2.25 \times 10^{-12}) = 2.25 \times 10^{-13} \]

### Final Answer

Thus, the dissociation constant \( K_a \) of HA is:

\[ K_a = 2.25 \times 10^{-13} \]

---