A 20.0 mL solution containing 0.2 g impure H₂O₂ reacts completely with 0.316 g of KMnO₄ in acid solution. The purity of H₂O₂ (in %) is _____________
(mol. wt. of H₂O₂ = 34; mol. wt. of KMnO₄ = 158)

Solution

Chemistry ma’am · 4 steps

Find the moles of permanganate, then use the 2 : 5 ratio to get how much real H₂O₂ is there.

1. Balanced equation. In acid, each MnO₄⁻ takes 5 electrons to become Mn²⁺. Each H₂O₂ gives 2 electrons and becomes O₂. Matching the electrons gives the equation below.

   2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

2. Moles of KMnO4. 0.316/158 = 0.002 mol of KMnO₄ is used. By the 2 : 5 ratio, it reacts with 0.002 × 5/2 = 0.005 mol of H₂O₂.
3. Mass of pure H2O2. 0.005 × 34 = 0.17 g of the 0.2 g sample is real H₂O₂. The 20.0 mL volume is not needed.
4. So. Purity = 0.17/0.2 × 100 = 85%.

The common slip: 34. Taking KMnO₄ and H₂O₂ as reacting 1 : 1 gives only 0.068 g of H₂O₂ and a purity of 34%, not 85%.

The idea to keep. In any redox titration, balance by electrons first: MnO₄⁻ takes 5 in acid and H₂O₂ gives 2. That electron count fixes the mole ratio, and the ratio gives the pure mass.

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