What is the dissociation constant of 0.002 M methanoic acid if its molar conductivity is 40.42 S cm² mol⁻¹ and Lambda_m⁰

Practice question

Question

What is the dissociation constant of 0.002 M methanoic acid if its molar conductivity is 40.42 S cm² mol⁻¹ and Lambda_m⁰ = 404.2 S cm² mol^{-1 ?

Options

Choose one · Correct answer highlighted

1.11 × 10⁻⁵ mol L⁻¹ 4.44 × 10⁻⁵ mol L⁻¹ 1.67 × 10⁻⁵ mol L⁻¹ 2.22 × 10⁻⁵ mol L⁻¹ Correct

Explanation

Given: What is the dissociation constant of 0.002 M methanoic acid if its molar conductivity is 40.42 S cm² mol⁻¹ and Lambda_m⁰ = 404.2 S cm² mol^{-1 ? Formula: α = 40.42/404.2 = 0.1. Substitution & Calculation: K_a = c α²/1 - α = 0.002 × (0.1)²/0.9 = 0.002 × 0.01/0.9 = 2.22 × 10⁻⁵mol L^{-1 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

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