Question

Carbonic acid $\left(H_2{CO}_3\right)$, a diprotic acid has $K_{a_1}=4.0\times {10}^{-7}$ and $K_{a_2}=7.0\times {10}^{-11}$. What is the $\left[HC{O}_3^{-}\right]$ of a 0.025 M solution of carbonic acid?

• A. $7.8\times {10}^{-3}$
• B. $6.6\times {10}^{-4}$
• C. ${10}^{-10}$
• D. $1.0\times {10}^{-4}$

Answer 1

The correct option is D $1.0\times {10}^{-4}$

$H_{2}C{O}_3\rightleftharpoons HC{O}_3^{-}+H+k{a}_1$

$HC{O}_3^{-}\rightleftharpoons C{O}_3^{2-}+H^{+}K{a}_2$

$c=0.0025H$

Most will be from $Ka,$

$\alpha =\sqrt{\frac{K{a}_1}{C}}$

$\sqrt{\frac{4\times {10}^{-7}}{25\times {10}^{-3}}}$

$\sqrt{\frac{4\times {10}^{-4}}{\alpha 5}}$

$=\frac{2}{5}\times {10}^{-2}$

$=\frac{0.02}{5}=0.004<0.1$

$\therefore \left[H^{+}\right]=c\alpha$

$=0.0025\times 0.004$

$=25\times 4\times {10}^{-6}$

$={10}^{-4}$

$\frac{\left[h^{+}\right]\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=Ka,$

$=\frac{{10}^{-4}\times \left[HC{O}_3^{-}\right]}{0.025}=4\times {10}^{-7}$

$\Rightarrow \left[HC{O}_3^{-}\right]=4\times {10}^{-3}\times 25\times {10}^{-3}$

$={10}^{-4}$