A monobasic acid is 5 % dissociated in its 0. | vedclass

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(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{	ext{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 	imes 10^{-2}$.Given

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$5 	imes 10^{-5}$

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(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{	ext{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 	imes 10^{-2}$.Given concentration $c = 0.02 \ M = 2 	imes 10^{-2} \ M$.For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$ (since $\alpha$ is very small,$1 - \alpha \approx 1$).Substituting the values: $K_a = (5 	imes 10^{-2})^2 	imes (2 	imes 10^{-2})$.$K_a = (25 	imes 10^{-4}) 	imes (2 	imes 10^{-2}) = 50 	imes 10^{-6} = 5 	imes 10^{-5}$.

$A$ monobasic acid is $5 \%$ dissociated in its $0.02 \ M$ solution. Calculate the dissociation constant of the acid.

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