A monoprotic acid in 1.00 M solution is 0.01 | vedclass

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(A) For a weak monoprotic acid,the dissociation constant $K_a$ is given by the formula $K_a = \frac{C \alpha^2}{1 - \alpha}$.Given concentration $C =

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$1 	imes 10^{-8}$

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(A) For a weak monoprotic acid,the dissociation constant $K_a$ is given by the formula $K_a = \frac{C \alpha^2}{1 - \alpha}$.Given concentration $C = 1.00 \ M$ and degree of ionization $\alpha = \frac{0.01}{100} = 10^{-4}$.Since $\alpha$ is very small $(10^{-4} \ll 1)$,we can approximate $1 - \alpha \approx 1$.Thus,$K_a \approx C \alpha^2$.Substituting the values: $K_a = 1.00 	imes (10^{-4})^2 = 1 	imes 10^{-8}$.

$A$ monoprotic acid in $1.00 \ M$ solution is $0.01\%$ ionized. The dissociation constant of this acid is

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