0.01 M KMnO₄ solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO₄ solution left in the burette after the end point is _____________ mL. (nearest integer)

Solution

Chemistry ma’am · 4 steps

Match the electrons: each MnO₄⁻ takes 5, each Fe²⁺ gives 1, then subtract from 50 mL.

1. Who reacts. Mohr's salt is FeSO₄.(NH₄)₂SO₄.6H₂O, so the reducing agent is Fe²⁺. In acid, MnO₄⁻ is reduced to Mn²⁺ and gains 5 electrons, while each Fe²⁺ loses 1 electron to become Fe³⁺.

   MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

2. Moles at end point. Fe²⁺ present = 0.05 × 20.0 = 1.0 mmol. From the 1 : 5 ratio, MnO₄⁻ needed = 1.0/5 = 0.20 mmol.
3. Volume used. Volume of KMnO₄ = (0.20 mmol)/(0.01 mmol mL⁻¹) = 20 mL. The burette reading moves from 0 to 20 mL.
4. So. The burette held 50 mL, so the volume left is 50 - 20 = 30 mL.

The common slip: 20. Reporting the volume of KMnO₄ used gives 20, not 30. Subtract it from the 50 mL the burette started with.

The idea to keep. In any redox titration, balance electrons, not moles: M₁ V₁ n₁ = M₂ V₂ n₂. Then read carefully whether the question wants the volume used or the volume left.

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