Question

Selenious acid $\left(H_{2}Se{O}_3\right),$ a diprotic acid has $K_{a1}=3.0\times {10}^{-3}$ and $K_{a2}=5.0\times {10}^{-8}.$ What is the $\left[O{H}^{-}\right]$ of a $0.30M$ solution of a selenious acid?

• A. $2.85\times {10}^{-3}$
• B. $5.0\times {10}^{-6}$
• C. $3.5\times {10}^{-12}$
• D. $3.5\times {10}^{-13}$

Answer 1

The correct option is D $3.5\times {10}^{-13}$

$H_{2}Se{O}_3\rightleftharpoons HSe{O}_3^{-}+H^{+}$
$HSe{O}_3^{-}\rightleftharpoons Se{O}_3^{2-}+H^{+}$
Since $K_{a1}$ has a very high value compared to $K_{a2}$, thus
$[H^{+}{]}_{total}=[H^{+}]$

Atgain, $H_{2}Se{O}_3\rightleftharpoons HSe{O}_3^{-}+H^{+}$
$t=0C00$
$t=t_{eq}C-C\alpha C\alpha C\alpha$
$\therefore K_{a_1}=\frac{C{a}^2}{(1-\alpha )}\alpha \text{is not negligible with respect to 1. So, after solving quadratic equation,}$
$\alpha =0.095$
$\left[H^{+}\right]=C\alpha =0.095\times 0.3$
$\left[O{H}^{-}\right]=\frac{k_w}{\left[H^{+}\right]}$
$\left[O{H}^{-}\right]=\frac{{10}^{-14}}{0.095\times 0.3}$
$=3.5\times {10}^{-13}.$