Redox Chemistry

Question

What is the percent by mass of oxalate ion in a 0.278 g sample of the iron salt that took 17.64 mL of 0.02085 M potassium permanganate to titrate to a light pink endpoint.

Answer 1

The balanced equation for this titration is as follows:

2MnO₄^- + 5C₂O₄^2- + 16H⁺ ==> 10CO₂ + 2Mn²⁺ + 8H₂O

To find % by mass of oxalate ion (C₂O₄^2-), we need to find the mass of this ion and then express it as a percent of the total mass, 0.278 g.

To find the mass if C₂O₄^2-, we use the balanced titration equation and first find the moles of MnO₄^- used. Then find moles of C₂O₄^2- and finally convert those moles to grams using the molar mass.

Moles MnO₄^- used = 17.64 ml x 1 L/1000 ml x 0.02085 mol/L = 3.6779x10^-4 moles

Moles C₂O₄^2- present = 3.6779x10^-4 moles MnO₄^- x 5 mols C₂O₄^2-/2 mols MnO₄^- = 9.1949x10^-4 moles

Mass C₂O₄^2- = 9.1949x10^-4 mole C₂O₄^2- x 88.02 g/mole = 0.08093 g

Percent oxalate by mass = 0.08093 g oxalate/0.278 g (x100%) = 29.1% (to 3 significant figures)