Question

Carbonic acid $\left(H_{2}C{O}_3\right)$, 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 $0.025M$ 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 $1.0\times {10}^{-4}$

Since ${Ka}_2<<<{Ka}_1$, So, second ionisation's slow or not observed

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

at initially $\to \underset{–}{c}$

at eqm $\to$ $\overline{c}$ $\overline{c}$

$\therefore {Ka}_1=7.0\times {10}^{-11}=\frac{\left[{HCO}_3^{-}\right]\left[H^{+}\right]}{\left[H_2{CO}_3\right]}=\frac{{\left[{HCO}_3^{-}\right]}^2}{0.025}=\frac{c^2}{0.025}$

$\therefore \left[{HCO}_3^{-}\right]=c=1.0\times {10}^{-4}M$