Question

$K_1$ and $K_2$ for oxalic acid $C_2{H}_2{O}_4$ are $6.5\times {10}^{-2}$ and $6.1\times {10}^{-5}$ respectively. What will be $\left[O{H}^{-}\right]$ in a 0.01 M solution of sodium oxalate?

• A. $9.6\times {10}^{-16}$
• B. $1.4\times {10}^{-1}$
• C. $1.26\times {10}^{-6}$
• D. $3.3\times {10}^{-9}$

Answer 1

The correct option is C $1.26\times {10}^{-6}$

Concentration of oxalate ion= 0.01 M
The hydrolysis of $C_2{O}_4^{2-}$ is represented by:
$C_2{O}_4^{2-}+H_{2}O\rightleftharpoons H{C}_2{O}_4^{-}+O{H}^{-}{K}_h=\frac{\left[H{C}_2{O}_4^{-}\right]\left[O{H}^{-}\right]}{[C_2{O}_4{]}^{2-}}Also,H{C}_2{O}_4^{-}\rightleftharpoons H^{+}+C_2{O}_4^{2-}{K}_2=\frac{\left[C_2{O}_4^{2-}\right]\left[H^{+}\right]}{\left[H{C}_2{O}_4^{-}\right]}{K}_w=\left[H^{+}\right]\left[O{H}^{-}\right]\text{We see that,}{K}_w=K_h\times K_2{K}_h=\frac{K_w}{K_2}=\frac{1.0\times {10}^{-14}}{6.1\times {10}^{-5}}=1.6\times {10}^{-10}\text{At equilibrium,}\left[H{C}_2{O}_4^{-}\right]=\left[O{H}^{-}\right]\therefore K_h=\frac{[O{H}^{-}{]}^2}{\left[C_2{O}_4^{2-}\right]}=1.6\times {10}^{-10}\left[C_2{O}_4\right]=0.01M\therefore \left[O{H}^{-}\right]=\sqrt{1.6\times {10}^{-12}}=\sqrt{\frac{16\times {10}^{-12}}{10}}\Rightarrow \frac{4\times {10}^{-6}}{\sqrt{10}}=1.26\times {10}^{-6}$