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 $\left[H{O}^{-}\right]$ in a $0.01M$ 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 reaction for the hydrolysis of the oxalate ion is $\underset{0.01}{C_2{O}_4^{2-}}+H_{2}O\rightleftharpoons \underset{x}{H{C}_2{O}_4^{-}}+\underset{x}{O{H}^{-}}$.
The expression for the hydrolysis constant is, $K_h=\frac{K_w}{K_2}=\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-5}}$.
But $K_h=\frac{\left[H{C}_2{O}_4^{-}\right]\left[O{H}^{-}\right]}{\left[C_2{O}_4^{2-}\right]}=\frac{x^2}{0.01}$.
Hence, $\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-5}}=\frac{x^2}{0.01}$ or $x=1.3\times {10}^{-6}$.