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(B) For a weak base $BOH$, the dissociation equilibrium is $BOH \rightleftharpoons B^+ + OH^-$.Let $C$ be the initial concentration and $\alpha$ be th

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The degree of dissociation of a weak base is inversely proportional to the square root of its concentration.

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(B) For a weak base $BOH$, the dissociation equilibrium is $BOH \rightleftharpoons B^+ + OH^-$.Let $C$ be the initial concentration and $\alpha$ be the degree of dissociation.The equilibrium concentrations are $[BOH] = C(1-\alpha)$, $[B^+] = C\alpha$, and $[OH^-] = C\alpha$.The dissociation constant $K_b$ is given by $K_b = \frac{[B^+][OH^-]}{[BOH]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$.Since the base is weak, $\alpha \ll 1$, so $1-\alpha \approx 1$.Thus, $K_b \approx C\alpha^2$, which gives $\alpha = \sqrt{K_b/C}$.Therefore, $\alpha \propto \frac{1}{\sqrt{C}}$.

What is the relationship between the degree of dissociation of a weak base and its concentration?

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