The degree of hydrolysis in hydrolytic equili | vedclass

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Question

(A) For the hydrolysis of a salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.Given $K_w = 1 	imes 10

Vedclass · Accepted Answer

$1 	imes 10^{-3}$

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(A) For the hydrolysis of a salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.Given $K_w = 1 	imes 10^{-14}$ and $K_a = 1 	imes 10^{-5}$,we have $K_h = \frac{10^{-14}}{1 	imes 10^{-5}} = 10^{-9}$.The degree of hydrolysis $\alpha$ is related to $K_h$ and concentration $C$ by the formula $\alpha = \sqrt{\frac{K_h}{C}}$ (assuming $\alpha \ll 1$).Given $C = 0.001 \ M = 10^{-3} \ M$,we have $\alpha = \sqrt{\frac{10^{-9}}{10^{-3}}} = \sqrt{10^{-6}} = 1 	imes 10^{-3}$.

The degree of hydrolysis in hydrolytic equilibrium $A^{-} + H_2O \rightleftharpoons HA + OH^{-}$ at salt concentration of $0.001 \ M$ is $(K_a = 1 \times 10^{-5})$

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