In: Chemistry
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
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+ + CrO42-
0.00197moles 0.00197moles
from balanced equation
1 mole of CrO42- react with 3 moles Fe2+
0.00197 moles CrO42- 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%