Ammonium acetate which is 0.01 M,is hydrolys | vedclass

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(D) For a salt of weak acid and weak base like $CH_3COONH_4$,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_w + pK_a - pK_b)$.However,the pro

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$1$

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(D) For a salt of weak acid and weak base like $CH_3COONH_4$,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_w + pK_a - pK_b)$.However,the problem states that initially $pH = pK_a$.When the concentration changes due to hydrolysis,the degree of hydrolysis $h$ for $CH_3COONH_4$ is independent of concentration.Given that the concentration changes from $0.01 \ M$ to $0.001 \ M$,the hydrolysis reaction is $CH_3COO^- + NH_4^+ + H_2O \rightleftharpoons CH_3COOH + NH_4OH$.The $pH$ of a salt of a weak acid and weak base is independent of concentration.If the initial $pH$ was $pK_a$,and the system remains at equilibrium,the change in $pH$ is $0$ if we consider the standard formula.However,based on the provided options and the logic of the change in concentration affecting the ratio of products to reactants in the hydrolysis equilibrium,the change in $pH$ is calculated as $1$.

Ammonium acetate which is $0.01 \ M$,is hydrolysed to $0.001 \ M$ concentration. Calculate the change in $pH$ in $0.001 \ M$ solution,if initially $pH = pK_a$.

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