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Calculate dissociation constant of $0.001 \mathrm{M}$ weak monoacidic base undergoing $2 \%$ dissociation.
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The correct answer is:
$4 \times 10^{-7}$
$\begin{aligned} & \alpha=\frac{\text { Percent dissociation }}{100}=\frac{2}{100}=0.02 \\ & K_a=\alpha^2 c=(0.02)^2 \times 0.001=4 \times 10^{-7}\end{aligned}$
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