Question

For a 0.01 M solution of carbonic acid, the concentration of $C{O}_3^{2-}$ would be:
Given: pH of the solution is 4.18.
$K_{a1}for{H}_{2}C{O}_3=4.45\times {10}^{-7}$
$K_{a2}for{H}_{2}C{O}_3=4.69\times {10}^{-11}$
Take ${10}^{-4.18}=6.6\times {10}^{-5}$

• A. $4.8\times {10}^{-11}M$
• B. $6.73\times {10}^{-5}M$
• C. $4.5\times {10}^{-7}M$
• D. $8.8\times {10}^{-11}M$

Answer 1

The correct option is A $4.8\times {10}^{-11}M$

Given, pH=4.18
$\Rightarrow -log\left[H^{+}\right]=4.18$
$\left[H^{+}\right]=6.6\times {10}^{-5}$
Now,
$H_{2}C{O}_3\rightleftharpoons H^{+}+HC{O}_3^{-}$
$K_{a1}=\frac{\left[H^{+}\right]\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}$
$\Rightarrow 4.45\times {10}^{-7}=\frac{(6.6\times {10}^{-5})\left[HC{O}_3^{-}\right]}{0.01}$
$\Rightarrow \left[HC{O}_3^{-}\right]=6.74\times {10}^{-5}M$
Again, $HC{O}_3^{-}\leftrightharpoons H^{+}+C{O}_3^{2-}$
$\Rightarrow K_{a2}=\frac{\left[H^{+}\right]\left[C{O}_3^{2-}\right]}{\left[HC{O}_3^{-}\right]}$
$\Rightarrow \left[C{O}_3^{2-}\right]=\frac{K_{a2}\left[HC{O}_3^{-}\right]}{\left[H^{+}\right]}$
$\Rightarrow \left[C{O}_3^{2-}\right]=4.79\times {10}^{-11}$