Given 0.1 M oxalic acid $\left(K_{\mathrm{a}1}=5.9\times {10}^{-2}\mathrm{M}\text{ and }{K}_{\mathrm{a}2}=6.4\times {10}^{-5}\mathrm{M}\right)$ solution. The concentration ofoxalate anion in the solution will be

In diprotic acid ${\mathrm{H}}_2\mathrm{A},\left[{\mathrm{A}}^{2-}\right]=K_{\mathrm{a}2}$

$\begin{array}{l}k{a}_1=\frac{c\alpha \times c\alpha }{c(1-\alpha )}\\ 5.9\times {10}^{-2}=c{\alpha }^2\\ 5.9\times {10}^{-2}=0.1\times {\alpha }^2\\ \alpha =0.768\end{array}$

${\mathrm{H}}^{+}$already in solution = 0.0768 and thus,

$\begin{array}{l}6.4\times {10}^{-5}=\frac{0.0768\times c_1{\alpha }_1}{{\mu }_1\left(1-{\alpha }_1\right)}\\ 6.4\times {10}^{-5}=0.0768\times {\alpha }_1\\ {\alpha }_1=8.3\times {10}^{-4}\end{array}$

$\begin{array}{l}=0.0768\times 6.4\times {10}^{-5}\\ =6.4\times {10}^{-5}\mathrm{M}\end{array}$