Question

Given an aqueous solution of $0.034M$ carbonic acid, the ionization constants for carbonic acid are $K_1=4.2\times {10}^{-7}$ and $K_2=4.8\times {10}^{-11}$
Select the correct statement from the following:

• A. The concentration of $C{O}_3^{2-}$ is $0.034M$.
• B. The concentration of $C{O}_3^{2-}$ is greater than that of $HC{O}_3^{-}$.
• C. The concentration of $H^{+}$ and $HC{O}_3^{-}$ are approximately equal.
• D. The concentration of $H^{+}$ is double that of $C{O}_3^{2-}$.

Answer 1

The correct option is C The concentration of $H^{+}$ and $HC{O}_3^{-}$ are approximately equal.

we can also express dissociation of $H_{2}C{O}_3$ simpler way as:
Step 1 :
$H_{2}C{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+HC{O}_3^{-}\left(aq\right)$
$att=0C_{}00$
$att=t_{eq}C(1-x)x+yx-y$
Step 2 :
$HC{O}_3^{-}\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$att=t_{eq}(x-y)x+yy$$x=\left[H^{+}\right]and\left[HC{O}_3^{-}\right]$ and $y=\left[C{O}_3^{2-}\right]$
In many cases with $K_{a1}>>K_{a2}$,
$x>>y$
$\therefore x-y\approx xandx+y\approx x$
Note:- If $C>>x$ so $C-x\approx C$ then $C(1-{\alpha }_1)\approx C$
$K_{a1}\approx \frac{\left(x\right)\left(x\right)}{(C-x)}$ where $C-x=\left[H_{2}C{O}_3\right]$ and $x=\left[H^{+}\right]and\left[HC{O}_3^{-}\right]$
$K_{a2}\approx \frac{\left(y\right)\left(x\right)}{\left(x\right)}\approx y$

Therefore as , $\left[H^{+}\right]=x+y\approx x=\left[HC{O}_3^{-}\right]$. So , Option c is correct.