Question

Carbonic acid, $H_2{CO}_3$, is a diprotic acid for which $K_1=4.2\times {10}^{-7}$ and $K_2=4.7\times {10}^{-11}$. Which solution will produce a $pH$ closest to $9$?

• A. $0.1M$ $H_2{CO}_3$
• B. $0.1M$ ${Na}_2{CO}_3$
• C. $0.1M$ $NaH{CO}_3$
• D. $0.1M$ $NaH{CO}_3$ and $0.1M$ ${Na}_2{CO}_3$

Answer 1

The correct option is C $0.1M$ $NaH{CO}_3$

A) $0.1M$ $H_2{CO}_3$

$H_2{CO}_3\rightleftharpoons H^{+}+{HCO}_3^{-}$

Initial : $0.1$ $0$ $0$

At eqm : $0.1-x$ $x$ $x$

$\therefore$ $K_{a_1}=\frac{x^2}{0.1-x}=\frac{x^2}{0.1}$

$\therefore$ $\frac{x^2}{0.1}=4.2\times {10}^{-7}\Rightarrow x=2.04\times {10}^{-4}$

$\therefore$ $\left[H^{+}\right]=2.04\times {10}^{-4}$ $\therefore$ $pH=3.69$

B) ${CO}_3^{2-}+H_{2}O\rightleftharpoons {HCO}_3^{-}+{OH}^{(-)}$ [hydrolysis of ${Na}_2{CO}_3$]

$\therefore$ $pH=7+\frac{1}{2}log{pK}_2+\frac{1}{2}logC$

$=7+\frac{10.32}{2}-\frac{1}{2}=11.66$

C) $pH=\frac{1}{2}({pK}_1+{pK}_2)=8.35$

D) $pH={pK}_2+log\frac{\left[{CO}_3^{2-}\right]}{\left[{HCO}_3^{-}\right]}=10.32+0=10.32$

$\therefore$ The solution $0.1M$ ${NaHCO}_3$ (C) will produce a $pH$ closer to $9$.