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(D) For a weak monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$Given degree of dissociation $\alpha =

Vedclass · Accepted Answer

$5.0 	imes 10^{-9}$

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(D) For a weak monobasic acid $HA$,the dissociation is given by: $HA ightleftharpoons H^+_{(aq)} + A^-_{(aq)}$Given degree of dissociation $\alpha = 0.05 \% = 0.05 	imes 10^{-2} = 5 	imes 10^{-4}$.Concentration $C = 0.02 \ M = 2 	imes 10^{-2} \ M$.For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 C$ (since $\alpha \ll 1$).Substituting the values: $K_a = (5 	imes 10^{-4})^2 	imes (2 	imes 10^{-2})$.$K_a = (25 	imes 10^{-8}) 	imes (2 	imes 10^{-2}) = 50 	imes 10^{-10} = 5.0 	imes 10^{-9}$.

Calculate the dissociation constant of a weak monobasic acid if it is $0.05 \%$ dissociated in a $0.02 \ M$ solution.

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