A 100 L solution of KCN undergoes 3.7% hydro | vedclass

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(B) $KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.The degree of hydrolysis $h$ is given as $3.7\% = 0.037$.The formula for the hydr

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$0.986 	imes 10^{-14}$

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(B) $KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.The degree of hydrolysis $h$ is given as $3.7\% = 0.037$.The formula for the hydrolysis constant $K_h$ for a salt of a weak acid and strong base is $K_h = \frac{K_w}{K_a}$.Also,$K_h = c 	imes h^2$ (where $c$ is concentration,$c = \frac{1}{V} = \frac{1}{100} \ M$).Equating the two: $\frac{K_w}{K_a} = \frac{h^2}{V}$.$K_w = \frac{h^2 	imes K_a}{V} = \frac{(0.037)^2 	imes 7.2 	imes 10^{-10}}{100}$.$K_w = \frac{0.001369 	imes 7.2 	imes 10^{-10}}{100} = 0.000098568 	imes 10^{-10} = 0.98568 	imes 10^{-14} \approx 0.986 	imes 10^{-14}$.

$A$ $100 \ L$ solution of $KCN$ undergoes $3.7\%$ hydrolysis. If the dissociation constant of $HCN$ is $7.2 \times 10^{-10}$,find the value of $K_w$.

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