Question

The $p{K}_a$ of $C{H}_{3}COOH$ and $p{K}_b$ of $N{H}_{4}OH$ is $4.76$ and $4.75$, respectively. Calculate the hydrolysis constant of ammonium acetate $\left(C{H}_{3}COON{H}_4\right)$ at $298$K and also the degree of hydrolysis and pH of its (a) $0.01$ M and (b) $0.04$ M solutions.

Answer 1

Both $N{H}_4^{+}$ and $C{H}_{3}CO{O}^{-}$ ion of $C{H}_{3}COON{H}_4$

$C{H}_{3}COON{H}_4+H_{2}O\leftrightharpoons C{H}_{3}CO{O}^{-}+N{H}_4^{+}$

$1$ $0$ $0$

$1-h$ $h$ $h$

$p{K}_a(CH-3COOH)=4.76⟹K_a=1.74\times {10}^{-5}$

$p{K}_b\left(N{H}_{4}OH\right)=4.75⟹K_b=1.78\times {10}^{-5}$

$K_H$ for $C{H}_{3}COON{H}_4=\frac{K_w}{K_a{K}_b}=\frac{{10}^{-14}}{(1.74\times {10}^{-5})(1.78\times {10}^{-5})}=3.23\times {10}^{-5}$

also, $h$ for $C{H}_{3}COON{H}_4=\sqrt{K_H}=\sqrt{3.23\times {10}^{-5}}=5.68\times {10}^{-3}$

$C{H}_{3}COOH\leftrightharpoons C{H}_{C}O{O}^{-}+H^{+}$

$K_a=\frac{\left[{CH}_3{COO}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}=\frac{C(1-h)\left[H^{+}\right]}{Ch}$

$\left[H^{+}\right]=K_a\frac{1}{1-h}=K_a\sqrt{K_H}=K_a\sqrt{\frac{K_w}{K_a\times K_b}}-\sqrt{\frac{K_w{K}_a}{K_b}}=\sqrt{\frac{{10}^{-14}\times 1.74\times {10}^{-5}}{1.78\times {10}^5}}$

$\left[H^{+}\right]=9.88\times {10}^{-8}$

$pH=7.005$

[$pH$ and $h$ are independent of initial concentration of Salt]