The mass percent of MnO2 in a sample of a mineral is determined by reacting it with a measured exces — Stoichiometry / Redox Titration Chemistry Question
Problem Context
The mass percent of MnO2 in a sample of a mineral is determined by reacting it with a measured excess of As2O3 in acid solution, and then titrating the remaining As2O3 with standard KMnO4. A 0.225 g sample of the mineral is ground and boiled with 75.0 mL of 0.0125 M As2O3 solution containing 10 mL of concentrated sulfuric acid. After the reaction is complete, the solution is cooled, diluted with water, and titrated with 2.28 × 10–3 M KMnO4, requiring 16.34 mL to reach the endpoint. Note: 5 mol of As2O3 react with 4 mol of MnO4–.
Write a balanced equation for the reaction of As2O3 with MnO2 in acid solution. The products are Mn2+ and AsO43–.
Model Answer
2MnO2 + As2O3 + H2O → 2Mn2+ + 2AsO43– + 2H+
Calculate the number of moles of As2O3 added initially.
Model Answer
0.0750 L × 0.0125 mol / L = 9.38 × 10−4 mol As2O3
Calculate the number of moles of MnO4– used to titrate the excess As2O3.
Model Answer
0.01634 L × 2.28 × 10–3 mol / L = 3.73 × 10−5 mol MnO4–
3.73 × 10−5 mol MnO4– × (5 mol As2O3 / 4 mol MnO4–) = 4.66 × 10−5 mol As2O3 left
Calculate the number of moles of MnO2 in the sample.
Model Answer
9.38 × 10–4 – 4.66 × 10–5 = 8.91 × 10–4 mol As2O3 react with MnO2
8.91 × 10−4 mol As2O3 × (2 mol MnO2 / 1 mol As2O3) = 1.78 × 10−3 mol MnO2
Determine the mass percent of MnO2 in the sample.
Model Answer
1.78 × 10–3 mol MnO2 × (86.94 g MnO2 / mol MnO2) = 0.155 g MnO2
mass % MnO2 = (0.155 g MnO2 / 0.225 g sample) × 100 = 68.9% MnO2 in sample
Describe how the endpoint is detected in the KMnO4 titration.
Model Answer
The endpoint corresponds to a slight purple (pink) color due to excess MnO4–(aq).