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Answer: \(K_a = \dfrac{\alpha^2 c}{1-\alpha} \approx \alpha^2 c\), so \(\alpha = \sqrt{K_a/c}\).

Consider a weak acid HA of concentration \(c\) mol dm–3 with degree of dissociation \(\alpha\):

HA⇌ H++ A–
Initial\(c\)00
At equilibrium\(c(1-\alpha)\)\(c\alpha\)\(c\alpha\)

\(K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \dfrac{c\alpha \cdot c\alpha}{c(1-\alpha)} = \dfrac{\alpha^2 c}{1-\alpha}\)

For a weak acid \(\alpha\) is very small, so \(1-\alpha \approx 1\): \(K_a = \alpha^2 c\) and \(\alpha = \sqrt{\dfrac{K_a}{c}}\). Thus the degree of dissociation increases with dilution (Ostwald’s dilution law).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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