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(A) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = c \alpha^2$,where $c$ is the concentration and $\alpha$ is the degr

Vedclass · Accepted Answer

$4 	imes 10^{-7}$

Vedclass · Answer

(A) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = c \alpha^2$,where $c$ is the concentration and $\alpha$ is the degree of dissociation.Given: $c = 0.001 \ M = 10^{-3} \ M$ and $\alpha = 2 \% = 0.02 = 2 	imes 10^{-2}$.Substituting the values: $K_b = (10^{-3}) 	imes (2 	imes 10^{-2})^2$.$K_b = 10^{-3} 	imes 4 	imes 10^{-4} = 4 	imes 10^{-7}$.

Calculate the dissociation constant of a $0.001 \ M$ weak monoacidic base undergoing $2 \%$ dissociation.

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