In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of K

NEET Chemistry PYQ solved by Nitish Sir 🔥 A Redox Titration numerical on KMnO₄ vs oxalic acid, n-factors, and the normality equation. 📚 Chapter: Redox Reactions (Class 11 NCERT) | Topic: Redox titrations, equivalent concept and n-factor 📚 Cross-link: d- and f-Block Elements (Class 12) — KMnO₄ as an oxidising agent in acidic medium ❓ QUESTION In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of KMnO₄ solution required to reach the end point is 10 mL, the strength of the KMnO₄ solution is (1) 0.25 M (2) 0.15 M (3) 0.10 M (4) 0.20 M ✅ Correct answer: (3) 0.10 M The balanced equation → 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O Manganese falls from +7 in MnO₄⁻ to +2 in Mn²⁺, a gain of 5 electrons, so the n-factor of KMnO₄ in acidic medium is 5. Each carbon in oxalate rises from +3 to +4, and there are two carbons, so oxalic acid loses 2 electrons and its n-factor is 2. Method 1, mole ratio → Moles of oxalic acid = 0.25 × 10/1000 = 2.5 × 10⁻³ mol. From the equation, 5 mol of oxalate need 2 mol of MnO₄⁻, so moles of KMnO₄ = (2/5) × 2.5 × 10⁻³ = 1.0 × 10⁻³ mol. This sits in 10 mL, so molarity = 1.0 × 10⁻³ / (10/1000) = 0.10 M. Method 2, normality equation → At the end point the equivalents are equal, so N₁V₁ = N₂V₂. For oxalic acid, N = 0.25 × 2 = 0.5 N, giving 0.5 × 10 = 5 milliequivalents. For KMnO₄, N = 5/10 = 0.5 N, and M = N / n-factor = 0.5/5 = 0.10 M. Both routes agree, so option (3). The trap to avoid → The volumes are equal, which tempts students to conclude the concentrations are equal and pick 0.25 M. That only works when the n-factors match. Here the 5:2 electron ratio is what fixes the answer, so always write the n-factor before touching the numbers. Also remember the n-factor of KMnO₄ changes with medium: 5 in acidic, 3 in neutral or faintly alkaline, and 1 in strongly alkaline medium. Bonus → This titration is self-indicating. The end point is the first permanent pale pink tinge from a trace of excess MnO₄⁻, and the mixture is warmed to about 60 °C because the reaction is slow to start until the Mn²⁺ produced begins to autocatalyse it. 👨‍🏫 ABOUT NITISH SIR B.E (Hons) BITS Pilani • AIR-16 GATE 2012 • AIR-732 JEE Main 2008 • 15+ years teaching. 🎯 FREE for re-NEET aspirants → https://neetsambhav.in 🔔 SUBSCRIBE to Chemistry Forum for daily NEET Chemistry PYQs & live doubt sessions. 💬 Telegram: https://t.me/ChemistryForumNitishSir 🌐 Website: https://neetsambhav.in #NEET #NEETChemistry #RedoxReactions #Titration #Stoichiometry NEET chemistry, redox titration KMnO4 oxalic acid, n factor of KMnO4, normality equation N1V1 N2V2, permanganate titration acidic medium, equivalent concept redox, molarity to normality conversion, self indicator KMnO4, redox reactions class 11 NEET, NEET chemistry PYQ solution, Nitish Sir chemistry, NEET Sambhav, Chemistry Forum