oxalate analysis (oxidation and reduction)

Question

A 20.00 ml sample of 0.520 M oxalic acid is titrated with 26.50 of an unknown KMnO4 solution.

a) How many moles of oxalic acid were present in this titration?

b.)How many moles of KMnO4 were needed for the reaction?

c.) What is the molarity of the KMnO4 solution?

Answer 1

a) moles of oxalic acid = 0.02000 L x 0.520 mol/L = 0.0104 moles oxalic acid

b) 2MnO₄^- + 16H⁺ + 5C2O₄^2- ===> 2Mn²⁺ + 8H₂O + 10CO₂ balanced equation for the reaction

moles KMnO₄ = 0.0104 moles oxalic acid x 2 mol KMnO₄/5 moles O.A. = 0.00416 moles KMnO₄

c) molarity of KMnO₄ = 0.00416 moles/0.02650 L = 0.1570 M