The degree of dissociation of 0.1 , N , CH_3C | vedclass

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Question

(D) For a weak acid,the degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the concentration in $M$.

Vedclass · Accepted Answer

$10^{-2}$

Vedclass · Answer

(D) For a weak acid,the degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the concentration in $M$. Since $CH_3COOH$ is a monoprotic acid,its normality $(N)$ is equal to its molarity $(M)$. Thus,$C = 0.1 \, M = 10^{-1} \, M$. Given $K_a = 1 	imes 10^{-5}$. Substituting the values: $\alpha = \sqrt{\frac{1 	imes 10^{-5}}{10^{-1}}} = \sqrt{10^{-4}} = 10^{-2}$.

The degree of dissociation of $0.1 \, N \, CH_3COOH$ is (Dissociation constant $K_a = 1 \times 10^{-5}$)

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