Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4 (aq) H2O2 (aq) 3H2SO4 (aq) →3O2 (g) 2MnSO4 (aq) K2SO4 (aq) 4H2O (l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 20.8 mL of the KMnO4 solution

Answer: 0.595 g of H₂O₂ were dissolved.

Explanation:

1) Balanced chemical equation of the titration (given):

2KMnO₄ (aq) + H₂O₂ (aq) + 3H₂SO₄ (aq) →

→ 3O₂ (g) + 2MnSO₄ (aq) + K₂SO₄ (aq) + 4H₂O (l)

2) Mole ratios of KMnO₄ (aq) and H₂O₂ (aq) (the reactants the question deals with):

2 moles 2KMnO₄ (aq) : 1 mol H₂O₂ (aq)

3) Number of moles of KMnO₄ used:

V = 20.8 mL = 0.0208 liter M = 1.68 M

M = n of solute / V of solution in liter ⇒ n of KMnO₄ = M * V = 1.68 M * 0.0208 liter = 0.0349 moles

4) Number of moles of H₂O₂ used:

Set the proportion using the stoichimetric ratio and the actual number of moles of KMnO₄:

2 moles KMnO₄ / 1 mol H₂O₂ = 0.0349 moles KMnO₄ / x

⇒ x = 0.0349 / 2 moles H₂O₂ = 0.0175 moles H₂O₂

5) Mass of H₂O₂ dissolved:

molar mass H₂O₂ = 2 (1.008g/mol) + 2 (15.999 g/mol) = 34.014 g/mol molar mass = mass in grams / number of moles ⇒ ⇒ mass in grams = molar mass * number of moles = 34.01 g/mol * 0.0175 moles = 0.595 g.

So, the answer, using the appropiate number of signficant digits (3) is 0.595 g of H₂O₂.