Question

A solution of $KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O$ is $0.2N$ as an acid. Therefore, it is :

• A. $0.267N$ as a reducing agent
• B. $0.6M$ as an acid
• C. $0.067M$ as an acid
• D. $0.067M$ as a reducing agent

Answer 1

Answer: (A), (C)

The correct options are
A $0.067M$ as an acid
C $0.267N$ as a reducing agent

(i) Three ionizable protons are present in $KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O$.
$0.2N$ solution will be $\frac{0.2}{3}=0.067M$ solution

(ii) Two ${C_2{O}_4}^{2-}$ ions are oxidized to $C{O}_2$.
${2C_2{O}_4}^{2-}\to 4C{O}_2+4e^{-}$

Hence, $0.067M=0.267N$

Hence, both the options A and C are correct.