A weak electrolyte has Lambdam° = 400 S cm² mol⁻¹ and Lambdam = 40 S cm² mol⁻¹ at 0.1 M. What is the dissociation consta

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A weak electrolyte has Lambdam° = 400 S cm² mol⁻¹ and Lambdam = 40 S cm² mol⁻¹ at 0.1 M. What is the dissociation constant Ka ?

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1.0 × 10⁻² 2.5 × 10⁻⁴ 5.0 × 10⁻³ 1.111 × 10⁻³ Correct

α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .

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