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 initaially $p H = p K_{a}$ .

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Solution

The correct option is D 1
$C H_{3} C O O N H_{4} 0.01 M + H_{2} O \to C H_{3} C O O H 0.001 M + N H_{4} O H$
$p H = p K_{a} + l o g \frac{[C H_{3} C O O N H_{4}]}{[C H_{3} C O O H]}$
$= p K_{a} + l o g [\frac{0.01}{0.001}]$
$= p K_{a} + l o g 10$
$= p K_{a} + 1$
$∴$ Change in pH = 1

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