Question

A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate: 5H2O2(aq) + 2KMnO4(aq) + 6H⁺(aq). 8H2O(l) + 5O2(g) + 2K⁺(aq) + 2Mn²⁺(aq). What is the mass percent of H2O2 in solution if 55.7g of solution required 38.9 ML of 0.534 M KMnO4 for it's titration?

Answer 1

Ans. Moles of KMnO₄ consumed = Molarity x Volume of solution in liters

                                                = 0.534 M x 0.0389 L

                                                = 0.0207726 mol

# Balanced reaction:

5 H₂O₂(aq) + 2 KMnO₄(aq) + 6 H⁺(aq)------> 8 H₂O(l) + 5 O₂(g) + 2 K⁺(aq) + 2 Mn²⁺(aq).

According to the stoichiometry of balanced reaction, 2 mol KMnO₄ neutralizes 5 mol H₂O₂.

So,

            Moles of H₂O₂ in sample = (5/2) x moles of KMnO₄ consumed

                                                = (5/2) x 0.0207726 mol

                                                = 0.0519315 mol

Now,

            Mass of Moles of H₂O₂ in sample = Moles of H₂O₂ x Molar mass

                                                = 0.0519315 mol x (34.01468 g /mol)

                                                = 34.01468 g

# Mass % of H₂O₂ in sample = (Mass of H₂O₂ / Mass of sample) x 100

                                                = (34.01468 g / 55.7 g) x 100

                                                = 61.07 %