Question

$K_1$ and $K_2$ for oxalic acid 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. $9.6\times {10}^{-6}$
• B. $1.4\times {10}^{-1}$
• C. $1.3\times {10}^{-6}$
• D. $1.3\times {10}^{-8}$

Answer 1

The correct option is C $1.3\times {10}^{-6}$

The hydrolysis of $C_2{O}_4^{2-}$ is represented by:
$C_2{O}_4^{2-}+H_{2}O\rightleftharpoons H{C}_2{O}_4^{-}+O{H}^{-}$

$K_h=\frac{K_W}{K_2}=\frac{1.0\times {10}^{-14}}{6.1\times {10}^{-5}}=\frac{x^2}{0.01}$

$\therefore x=1.3\times {10}^{-6}$