AIIMS 2019 Chemistry Conductivity and Molar Conductivity MCQ Question

To calculate the acid dissociation constant (Kₐ) for the weak monobasic acid, we first determine the degree of dissociation (α) using the formula:

$$\text{Conductivity} = c \cdot \lambda^0 + \alpha c \cdot \lambda^+ + \alpha c \cdot \lambda^-$$

Here, $c$ is the concentration (0.05 M), and $\lambda^0$ is the molar conductivity at infinite dilution (500 S cm² mol⁻¹). The conductivity of the solution is given as $10^{-3}$ S cm⁻¹. Solving for α gives us the degree of dissociation, which we can then use to find Kₐ using the relationship:

$$K_a = \frac{c \cdot \alpha^2}{1 - \alpha}$$

Substituting the values leads to Kₐ = $8 \times 10^{-5}$, confirming option A as correct. The other options are incorrect because they do not match the calculated Kₐ value based on the provided data and relationships.