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 $p H = p K_{a} .$
$l o g 9 = 0.354, l o g 3 = 0.477$

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Solution

The correct option is A -1
$C H_{3} C O O^{-} + N H_{4}^{+} + H_{2} O ⇌ C H_{3} C O O H + N H_{4} O H I n i t i a l 0.1 0.1 a t e q b 0.001 0.001 0.1 - 0.001 0.1 - 0.001$

$K_{h} = \frac{[C H_{3} C O O H] [N H_{4} O H]}{[C H_{3} C O O^{-}] [N H_{4}^{+}]} = \frac{(0.009)^{2}}{(0.001)^{2}} = 9^{2}$
it is a salt of weak acid and weak base so,
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}}$
$∴ K_{b} = \frac{K_{w}}{K_{a} \times K_{h}}$
Also, $[H^{+}] = \sqrt{\frac{K_{a} \times K_{w}}{K_{b}}} = \sqrt{K_{a}^{2} \times 81} = K_{a} \times 9$
$(p H)_{i n i t i a l} = p K_{a}$
$(p H)_{F i n a l} = - l o g [H^{+}] = - l o g (K_{a} \times 9) = p K_{a} - l o g 9$
$Change in pH = (p H)_{F i n a l} - (p H)_{I n i t i a l}$
$= p K_{a} - l o g 9 - p K_{a}$
$= - l o g 9$
$= - 0.954$

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