Practice question
Question
A weak acid has Lambdam° = 380 S cm² mol⁻¹ and Lambdam = 19 S cm² mol⁻¹ at 0.02 M. What is the dissociation constant Ka ?
Explanation
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
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