Ammonium acetate which is 0.01 M,is hydrolys | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 $pH = pK_a$.

Similar Questions

When a solution of $NH_4Cl$ is diluted,its $pH$ .....

An aqueous solution of $CH_3COONa$ will be

The degree of hydrolysis of which salt is independent of the concentration of the salt solution?

Which among the following salts turns blue litmus red in its aqueous solution?

The aqueous solution of $AlCl_3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module