Question

Oxalic acid can donate two protons to water in successive reactions:
(1) $H_2{C}_2{O}_4\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_{3}O+\left(aq\right)+H{C}_2{O}_4^{-}\left(aq\right)$
(2) $H{C}_2{O}_4^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_{3}O+\left(aq\right)+C_2{O}_4^{2-}\left(aq\right)$
If $K_{c1}=5.9\times {10}^{-2}$ and $K_{c2}=6.4\times {10}^{-5}$ at ${25}^{0}C$, what is the value of $K_c$ for reaction (3)?
(3) $H_2{C}_2{O}_4\left(aq\right)+2H_{2}O\left(l\right)\rightleftharpoons 2H_{3}O+\left(aq\right)+C_2{O}_4^{2-}\left(aq\right)$

• A. $3.8\times {10}^{-6}$
• B. $1.1\times {10}^{-3}$
• C. $5.9\times {10}^{-2}$
• D. $9.2\times {10}^2$

Answer 1

The correct option is A $3.8\times {10}^{-6}$

(1) $H_2{C}_2{O}_4\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+{HC}_2{O}_4^{-}{K}_{C1}$

(2) ${HC}_2{O}_4^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C_2{O}_4^{-2}\left(aq\right)K_{C2}$

(3) $H_2{C}_4{O}_4\left(aq\right)+2H_{2}O\left(l\right)\rightleftharpoons {2H}_3{O}^{+}\left(aq\right)+C_2{O}_4^{-2}{K}_{C3}$

$K_{C1}=5.9\times {10}^{-2}{K}_{C2}=6.4\times {10}^{-5}$

We can observe that reaction (3) sum of reaction (1) and reaction (2) $\Rightarrow K^{\prime }S$ will multiply

$\Rightarrow So,K_{C3}=K_{C1}\times K_{C2}$

$K_{C3}=5.9\times {10}^{-2}\times 6.4\times {10}^{-5}$

$\simeq 3.8\times {10}^{-6}$