Only 2 mL of KMnO₄ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of KMnO₄ solution should be ________ M.

Solution

Chemistry ma’am · 4 steps

Match electrons, not volumes: each permanganate takes 5 electrons, and each oxalate gives up 2.

1. Electrons per ion. In acidic medium MnO₄⁻ goes to Mn²⁺ (Mn from +7 to +2), so it gains 5 electrons. Oxalate goes to CO₂ (each C from +3 to +4), so one C₂O₄²⁻ loses 2 electrons.

   2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

2. Moles of oxalic acid. 20 mL of 2 M acid holds 0.020 × 2 = 0.04 mol. By the 2 : 5 ratio, the permanganate needed is 0.04 × 2/5 = 0.016 mol.
3. Molarity of KMnO4. This 0.016 mol sits in 2 mL = 0.002 L, so the molarity is 0.016/0.002 = 8 M. As a check, M × 5 × 2 = 2 × 2 × 20 also gives M = 8.
4. So. The KMnO₄ solution is 8 M.

The common slip: 20. Taking a 1 : 1 mole ratio and writing 2 × 20 = M × 2 gives 20, not 8. The n-factors 5 and 2 must go in.

The idea to keep. In a redox titration, multiply each molarity by its n-factor before equating: M₁ n₁ V₁ = M₂ n₂ V₂. For KMnO₄ in acid n = 5, and for oxalic acid n = 2.

Answer from the NTA final answer key. Solution by PrepWiser.