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