The K_a of a weak acid HA is 1.00 times 10^{- | vedclass

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(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\a

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(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\alpha^2$ (assuming $\alpha Given $K_a = 1.00 	imes 10^{-5}$ and $C = 0.100 \ M$.$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.00 	imes 10^{-5}}{0.100}} = \sqrt{1.00 	imes 10^{-4}} = 1.00 	imes 10^{-2}$.Percentage dissociation = $\alpha 	imes 100 = (1.00 	imes 10^{-2}) 	imes 100 = 1\%$.

The $K_a$ of a weak acid $HA$ is $1.00 \times 10^{-5}$. If $0.100 \ mol$ of this acid is dissolved in $1 \ L$ of water,what is the percentage dissociation of the acid at equilibrium?

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