Question

Specific conductance of 0.1M HA is 3.75 × 10⁻4Ω⁻1cm⁻1. If Lambdam^∘(HA) = 250Ω^-1cm²mol⁻1, the dissociation constant Kₐ of HA is

• A. 1.0 × 10⁻5
• B. 2.25 × 10⁻4
• C. 2.25 × 10⁻5
• D. 2.25 × 10⁻13

Hint:

For weak electrolytes: Kₐ = (Cα²)/(1-α)

Answer 1

The Correct Option is C

Step 1: Molar conductivity: Lambdam = (κ × 1000)/(C) = frac3.75×10⁻4×10000.1 = 3.75 

Step 2: Degree of dissociation: α = (Lambdam)/(Lambdam^∘) = (3.75)/(250) = 0.015 

Step 3: Dissociation constant: Kₐ = (Cα²)/(1-α) ≈ 0.1 × (0.015)² ≈ 2.25×10⁻5