First slide

Salt hydrolysis :

Question

The hydrolysis constant of ammonium acetate is given by

Moderate

A

$\large \frac{{{K_w}}}{{{K_a}}}$
B

$\frac{{{K_w}}}{{{K_b}}}$
C

$\frac{{{K_w}}}{{{K_a}.\,{K_b}}}$
D

K_a. K_b

Solution

$\large C{H_3}COON{H_4}\xrightarrow{{}}C{H_3}CO{O^ - }\left( {aq} \right) + NH_4^ + \left( {aq} \right)$

Both the acid radical and the basic radical are derivative of weak electrolytes, so both of them will undergo hydrolysis.
$\large C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons C{H_3}COOH + O{H^-}$
$\large NH_4^ + + {H_2}O \rightleftharpoons N{H_4}OH + {H^ + }$
$\large OR$
$\large C{H_3}CO{O^ - } + NH_4^ + + {H_2}O \rightleftharpoons C{H_3}COOH + N{H_4}OH$
For the above hydrolysis reaction, we need to find the hydrolysis cosntant, K_h
Consider the following equilibria,
$\large NH_4^ + + O{H^ - } \rightleftharpoons N{H_4}OH;\frac{1}{{{K_b}}} \to \left( 2 \right)$
$\large C{H_3}CO{O^ - } + {H^ + } \rightleftharpoons C{H_3}COOH;\frac{1}{{Ka}} \to \left( 3 \right)$
$\large {H_2}O \rightleftharpoons {H^ + } + O{H^ - };{K_w} \to \left( 4 \right)$
$\large Adding{\text{ 2, 3 and 4,}}$
$\large C{H_3}CO{O^ - } + NH_4^ + + {H_2}O \rightleftharpoons N{H_4}OH + C{H_3}COOH$
$\large {K_{eq}} = {K_w} \times \frac{1}{{{K_a}}} \times \frac{1}{{{K_b}}}$

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