The $p K_{a}$ of $C H 3 C O O H$ and $p K_{b}$ of $N H_{4} O H$ is 4.76 and 4.75. respectively. Calculate the hydrolysis constant of ammonium acetate $(C H_{3} C O O N H_{4})$ at 298 K and also the degree of hydrolysis and pH of its (a) 0.01 M and (ii) 0.04 M solutions.

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Solution

Both $N H_{4}^{+}$ and $C H_{3} C O O^{-}$ of $C H_{3} C O O N H_{4}$ show hydrolysis.
$C H_{3} C O O^{-} N H_{4}^{+} + H_{2} O ⇋ C H_{3} C O O H + N H_{4} O H$
$1$ $0$ $0$
$1 - h$ $h$ $h$
Given, $p K_{a}$ $C H_{3} C O O H = 4.76$
$- l o g k_{a} = 4.76$ or $K_{a} = 1.74 \times 10^{- 5}$
$p K_{b} (N H_{4} O H) = 4.75$
$K_{b} = 1.78 \times 10^{- 5}$
$K_{H} (C H_{3} C O O N H_{4}) = \frac{K_{w}}{k_{a} k_{b}} = \frac{10^{- 14}}{(1.74) (1.78) (10^{- 10})} = 3.23 \times 10^{- 5}$
$h (C H_{3} C O O H) = \sqrt{K_{H}} = \sqrt{3.23 \times 10^{- 5}}$
$h = 5.68 \times 10^{- 3}$
$C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}$
$K_{a} = \frac{[{C H}_{3} {C O O}^{-}] [H^{+}]}{[{C H}_{3} C O O H]} K_{a} = \frac{C (1 - h) [H^{+}]}{C h} [H^{+}] = K_{a} (\frac{h}{1 - h}) = K_{a} \sqrt{K_{H}} = K_{a} \sqrt{\frac{K_{w}}{K_{a} K_{b}}} - \sqrt{\frac{K_{w}}{K_{a} K_{b}}} = \sqrt{\frac{10^{- 14} (1.74 \times 10^{- 5})}{1.78 \times 10^{- 5}}}$
$[H^{+}] = 9.88 \times 10^{- 8}$
$p H = 7.005$ ( $p H$ independent of initial concentration)

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