If a 0.01 M NaCN solution is 3.37 % hydroly | vedclass

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(C) $NaCN$ is a salt of a strong base and a weak acid. For the hydrolysis of such a salt,the hydrolysis constant $K_h$ is given by the formula $K_h =

Vedclass · Accepted Answer

$1.13 	imes 10^{-5}$

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(C) $NaCN$ is a salt of a strong base and a weak acid. For the hydrolysis of such a salt,the hydrolysis constant $K_h$ is given by the formula $K_h = C h^2$,where $C$ is the concentration and $h$ is the degree of hydrolysis. Given: $C = 0.01 \ M$ and $h = 3.37 \% = 0.0337$. Substituting the values: $K_h = 0.01 	imes (0.0337)^2$. $K_h = 0.01 	imes 0.00113569 = 1.13569 	imes 10^{-5} \approx 1.13 	imes 10^{-5}$.

If a $0.01 \ M \ NaCN$ solution is $3.37 \%$ hydrolyzed,calculate the value of $K_h$.

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