Hydrofluoric acid is a weak acid. At 25°C, the molar conductivity of 0.002 M HF is 176.2 Ω⁻¹cm²mol⁻¹. If its λ₀ₘ = 405 Ω⁻¹cm²mol⁻¹. Equilibrium constant at the given concentration is:
6.7×10⁻⁸M
3.2×10⁻⁸M
6.7×10⁻⁹M
3.2×10⁻⁹M
Solution:
α = Λc/λ₀ₘ = 176.2/405 ≈ 0.435 K = [H⁺][F⁻]/[HF] = Cα²/1-α = 0.002M × 0.435²/1 - 0.435 = 6.7 × 10⁻⁸ M Hence, option A is correct.