In acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of KMnO₄ required to reach the end point is 10 mL, the strength of the KMnO₄ solution is:

Solution & Explanation

Concept: use the law of equivalents — at the end point, equivalents of oxidant = equivalents of reductant. In acidic medium MnO₄⁻ gains 5 electrons (Mn⁺⁷ → Mn²⁺), so its n-factor = 5. Oxalic acid is oxidised (C₂O₄²⁻ → 2CO₂), losing 2 electrons, so its n-factor = 2. The balanced reaction confirms this: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Equating milliequivalents (n × M × V): 5 × M(KMnO₄) × 10 = 2 × 0.25 × 10 50 × M(KMnO₄) = 5 M(KMnO₄) = 0.10 M. Hence the strength of KMnO₄ is 0.10 M — answer (A). Trap: do not equate moles directly; the differing n-factors (5 vs 2) must be included.