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(B) $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For the salt of a strong base and a weak acid,the hydrolysis constant $K_h$ i

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$0.01$

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(B) $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For the salt of a strong base and a weak acid,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a}$.Given $K_b$ for $OCN^- = 10^{-10}$,we know $K_a 	imes K_b = K_w = 10^{-14}$.Thus,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.$K_h = \frac{10^{-14}}{10^{-4}} = 10^{-10}$.The degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.01 \ M = 10^{-2} \ M$.$h = \sqrt{\frac{10^{-10}}{10^{-2}}} = \sqrt{10^{-8}} = 10^{-4}$.Percentage of hydrolysis $= h 	imes 100 = 10^{-4} 	imes 100 = 0.01 \%$.

Calculate the percentage of hydrolysis in $0.01 \ M$ aqueous solution of $NaOCN$ ($K_b$ for $OCN^{-} = 10^{-10}$)

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