Question

Calculate the degree of hydrolysis and pH of $0.1M$ sodium acetate solution. Hydrolysis constant of sodium acetate is $5.6\times {10}^{-10}$.

Answer 1

Equilibrium reaction is

${CH}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons {CH}_{3}COOH+{OH}^{-}0.1\times (1-h)0.1\times h0.1\times h$

$K_h=\frac{\left[{CH}_{3}COOH\right]\left[{OH}^{-}\right]}{\left[{CH}_{3}CO{O}^{-}\right]}=\frac{(0.1\times h)(0.1\times h)}{0.1(1-h)}$
$h$ is small $(1-h\to 1)$
$5.6\times {10}^{-10}=0.1\times h^2$
or $h^2=\frac{5.6\times {10}^{-10}}{0.1}=56\times {10}^{-10}$
Degree of hydrolysis is $h=7.48\times {10}^{-5}$
$\left[{OH}^{-}\right]=Ch=0.1\times 7.48\times {10}^{-5}=7.48\times {10}^{-6}M$
$\left[H^{+}\right]=\frac{K_w}{\left[{OH}^{-}\right]}=\frac{{10}^{-14}}{7.48\times {10}^{-6}}=1.33\times {10}^{-9}M$
$pH=-\log \left[H^{+}\right]=-\log (1.33\times {10}^{-9})=8.88$