Question

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 1

Answer: 0.595 g of H₂O₂ were dissolved.

Step-by-step 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.