A mixture of $ ext{H}_2 ext{C}_2 ext{O}_4$ and $ ext{NaHC}_2 ext{O}_4$ weighing $2.02 ext{ g}$

Question

A mixture of $	ext{H}_2	ext{C}_2	ext{O}_4$ and $	ext{NaHC}_2	ext{O}_4$ weighing $2.02	ext{ g}$ was dissolved in water and the solution made up to one litre. $10.0	ext{ mL}$ of this solution required $3.0	ext{ mL}$ of $0.1	ext{ N NaOH}$ solution for complete neutralization. In another experiment, $10.0	ext{ mL}$ of the same solution, in a hot dilute sulphuric acid medium, required $4.0	ext{ mL}$ of $0.1	ext{ N KMnO}_4$ solution for complete reaction. Which of the following statements is/are CORRECT? (Given molar masses: $	ext{H}_2	ext{C}_2	ext{O}_4 = 90	ext{ g/mol}$, $	ext{NaHC}_2	ext{O}_4 = 112	ext{ g/mol}$)

Professor PC Thomas Classes · Answer

Let 1L contain $a$ moles of $	ext{H}_2	ext{C}_2	ext{O}_4$ and $b$ moles of $	ext{NaHC}_2	ext{O}_4$. For acid-base titration, $10	ext{ mL}$ takes $0.3	ext{ meq NaOH}$. So, $1	ext{ L}$ has $30	ext{ meq}$ acidic protons. $2a + b = 0.03$. For redox, $10	ext{ mL}$ takes $0.4	ext{ meq KMnO}_4$. So $1	ext{ L}$ has $40	ext{ meq}$. Since n-factor is 2 for both, $2a + 2b = 0.04 \implies a+b = 0.02$. Solving yields $a = 0.01	ext{ mol}$, $b = 0.01	ext{ mol}$. Mass of $	ext{H}_2	ext{C}_2	ext{O}_4 = 0.01 	imes 90 = 0.90	ext{ g}$ and $	ext{NaHC}_2	ext{O}_4 = 0.01 	imes 112 = 1.12	ext{ g}$. Total reducing equivalents = $0.04$.