A weak acid HA has a K_a of 1.00 times 10^{-5 | vedclass

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(C) For a weak acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.The dissociation constant $K_a$ is given by the formula $K_

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(C) For a weak acid $HA$,the dissociation equilibrium is $HA ightleftharpoons H^+ + A^-$.The dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2 / (1 - \alpha)$.Since the acid is weak,$\alpha \ll 1$,so we approximate $K_a \approx C \alpha^2$.Given $K_a = 1.00 	imes 10^{-5}$ and $C = 0.100 \ M$.$1.00 	imes 10^{-5} = 0.100 	imes \alpha^2$.$\alpha^2 = 1.00 	imes 10^{-4}$.$\alpha = 0.01$.The percentage of dissociation is $\alpha 	imes 100 = 0.01 	imes 100 = 1 \%$.

$A$ weak acid $HA$ has a $K_a$ of $1.00 \times 10^{-5}$. If $0.100 \ mol$ of this acid is dissolved in one litre of water,the percentage of acid dissociated at equilibrium is closest to.....$\%$

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