Question

A sample is analyzed to determine its iron content Fe2+ via a redox titration with potassium chromate as the titrant. In the titration Fe2+ is oxidized to Fe3+ and CrO4 2- is reduced to Cr 3+. What is the percent by mass of iron in the sample if 0.9087g of the sample required 45.68mL of a 0.04322 M K2CrO4 solution to reach the endpoint?

I worked out the balanced equation to be:

8H+ + 3Fe2+ + CrO4 2- = 3Fe3+ + Cr3+ + 4H2O

Answer 1

8H+ + 3Fe2+ + CrO4 2- = 3Fe3+ + Cr3+ + 4H2O

no of moles of K2CrO4 = molarity* volume in L

                                         = 0.04322*0.04568   = 0.00197 moles

K2CrO4 ---------> 2K⁺ + CrO4^2-

0.00197moles              0.00197moles

from balanced equation

1 mole of CrO4^2- react with 3 moles Fe²⁺

0.00197 moles CrO4^2- react with = 3*0.00197/1 = 0.00591moles of Fe⁺²

mass of Fe   = no of moles* gram atomic mass

                    = 0.00591*56 = 0.33096g of Fe

mass percentage of Fe = 0.33096*100/0.9087   = 36.42%