In the reactant of KMnO₄ with an oxalate in acidic medium. ${\mathrm{MnO}}_4^{-}$ is reduced to ${\mathrm{Mn}}^{2+}$ and ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$is oxidized to CO₂. Hence, 50ml of 0.02 M KMnO₄ is equivalent to

Reaction of KMnO₄ with Oxalate in Acidic Medium

Balanced Chemical Equation:
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Redox Changes:

• MnO₄⁻ is reduced from Mn+7 to Mn+2.
• C₂O₄²⁻ is oxidized to CO₂ (carbon from +3 to +4).

Given:

• Volume of KMnO₄ = 50 mL = 0.050 L
• Molarity of KMnO₄ = 0.02 M

Step 1: Moles of KMnO₄
Moles = M × V = 0.02 × 0.050 = 0.001 mol

Step 2: Mole Ratio
From the equation: 2 mol KMnO₄ reacts with 5 mol C₂O₄²⁻ → ratio = 2.5
Therefore: 0.001 mol KMnO₄ × 2.5 = 0.0025 mol C₂O₄²⁻

Final Answer:
50 mL of 0.02 M KMnO₄ is equivalent to 0.0025 moles of oxalate ions (C₂O₄²⁻)