What volume (in mL) of a 0.1 M solution is needed to completely oxidize 144 mg of ferrous oxalate () — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Moles of $FeC_2O_4 = 144 \times 10^{-3} / 144 = 10^{-3}$ mol. The n-factor of $FeC_2O_4$ is 3 (Fe changes +2 $\rightarrow$ +3; two C atoms change +3 $\rightarrow$ +4). Equivalents of $FeC_2O_4 = 3 \times 10^{-3}$ eq = 3 meq. For $KMnO_4$ in acidic medium, n-factor = 5. Milli-equivalents of $KMnO_4 = 0.1 \times V \times 5 = 0.5V$. Therefore, $0.5V = 3 \Rightarrow V = 6.0$ mL.