A 0.1 M solution of HF is 1 % ionized. What | vedclass

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(A) For a weak acid $HF$,the dissociation is $HF \rightleftharpoons H^+ + F^-$. The dissociation constant $K_a$ is given by the formula $K_a = \frac{C

Vedclass · Accepted Answer

$10^{-5}$

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(A) For a weak acid $HF$,the dissociation is $HF ightleftharpoons H^+ + F^-$. The dissociation constant $K_a$ is given by the formula $K_a = \frac{C \alpha^2}{1-\alpha}$. Given concentration $C = 0.1 \ M$ and degree of ionization $\alpha = 1 \% = 0.01$. Substituting the values: $K_a = \frac{0.1 	imes (0.01)^2}{1 - 0.01} = \frac{0.1 	imes 10^{-4}}{0.99} \approx 1.01 	imes 10^{-6}$. However,using the approximation $1 - \alpha \approx 1$ for weak acids: $K_a = C \alpha^2 = 0.1 	imes (0.01)^2 = 0.1 	imes 10^{-4} = 10^{-5}$.

$A$ $0.1 \ M$ solution of $HF$ is $1 \%$ ionized. What is the $K_a$?

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