The percentage degree of hydrolysis of a salt | vedclass

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Question

(B) For a salt of a weak acid $(HA)$ and a weak base $(BOH)$,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w

Vedclass · Accepted Answer

$10$

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(B) For a salt of a weak acid $(HA)$ and a weak base $(BOH)$,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a 	imes K_b}}$.Since the expression for the degree of hydrolysis $(h)$ does not contain the concentration term $(C)$,it is independent of the concentration of the salt solution.Therefore,if the degree of hydrolysis is $10 \%$ at $0.1 \ M$,it will remain $10 \%$ at $0.05 \ M$.

The percentage degree of hydrolysis of a salt of weak acid $(HA)$ and weak base $(BOH)$ in its $0.1 \ M$ solution is found to be $10 \%$. If the molarity of the solution is $0.05 \ M$,the percentage hydrolysis of the salt should be ....... $\%$.

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