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AIIMS 2019 Chemistry Conductivity and Molar Conductivity MCQ Question

Type: MCQ-numerical-Hard-Class 12

The conductivity of a 0.05 M solution of a weak monobasic acid is 10⁻³ S cm⁻¹. If λᵐ∞ for weak acid is 500 S cm² mol⁻¹, calculate Kₐ of weak monobasic acid:

A

8×10⁻⁵

B

4×10⁻⁶

C

16×10⁻⁷

D

14×10⁻⁸

Correct Answer

Option A

Detailed Explanation

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

Conductivity=c⋅λ0+αc⋅λ++αc⋅λ−\text{Conductivity} = c \cdot \lambda^0 + \alpha c \cdot \lambda^+ + \alpha c \cdot \lambda^-

Here, cc is the concentration (0.05 M), and λ0\lambda^0 is the molar conductivity at infinite dilution (500 S cm² mol⁻¹). The conductivity of the solution is given as 10−310^{-3} S cm⁻¹. Solving for α gives us the degree of dissociation, which we can then use to find Kₐ using the relationship:

Ka=c⋅α21−αK_a = \frac{c \cdot \alpha^2}{1 - \alpha}

Substituting the values leads to Kₐ = 8×10−58 \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.

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