A weak monoacidic base dissociates to 1.5 % i | vedclass

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Question

(A) For a weak monoacidic base,the dissociation constant $K_b$ is given by the formula $K_b = C \alpha^2 / (1 - \alpha)$.Given concentration $C = 0.00

Vedclass · Accepted Answer

$2.25 	imes 10^{-7}$

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(A) For a weak monoacidic base,the dissociation constant $K_b$ is given by the formula $K_b = C \alpha^2 / (1 - \alpha)$.Given concentration $C = 0.001 \ M = 10^{-3} \ M$.Degree of dissociation $\alpha = 1.5 \% = 0.015 = 1.5 	imes 10^{-2}$.Since $\alpha$ is very small,$(1 - \alpha) \approx 1$.Therefore,$K_b \approx C \alpha^2$.$K_b = (10^{-3}) 	imes (1.5 	imes 10^{-2})^2$.$K_b = 10^{-3} 	imes 2.25 	imes 10^{-4} = 2.25 	imes 10^{-7}$.

$A$ weak monoacidic base dissociates to $1.5 \%$ in $0.001 \ M$ solution at $298 \ K$. Calculate the dissociation constant of the weak base.

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