Question

$H_2{CO}_3$ ionises as
$H_2{CO}_3\rightleftharpoons H^{+}+H{CO}_3^{-};K_1=4.3\times {10}^{-7}$
$H{CO}_3^{-}\rightleftharpoons H^{+}+{CO}_3^{2-};K_2=5.6\times {10}^{-11}$
Calculate the degree of hydrolysis and pH value of $0.12M$ ${Na}_2{CO}_3$ solution.

Answer 1

Salt of a strong base and weak acid: ${Na}_2{CO}_3$

Thus, ${CO}_3^{2-}$ ion is hydrolysed

${CO}_3^{2-}+H_{2}O\rightleftharpoons H{CO}_3^{-}+{OH}^{-}$

$K_h=C{h}^2$

or $h^2=\frac{K_h}{C}=\frac{K_h}{0.12}$

We know that

$K_h=\frac{K_w}{K_a}=\frac{K_w}{K_2}=\frac{{10}^{-14}}{5.6\times {10}^{-11}}=1.7857\times {10}^{-4}$

So, $h^2=\frac{1.7857\times {10}^{-4}}{0.12}=14.88\times {10}^{-4}$

Degree of hydrolysis $=h=3.85\times {10}^{-2}$

$\left[{OH}^{-}\right]=C\times h=0.12\times 3.85\times {10}^{-2}=4.62\times {10}^{-3}$

$\left[H^{+}\right]=\frac{{10}^{-14}}{4.62\times {10}^{-3}}=2.164\times {10}^{-12}M$

$pH=-\log \left[H^{+}\right]=-\log (2.164\times {10}^{-12})=11.665$