Aim: To determine the amount of Fe2+ in a dissolved wire solution.
Method: 25cm^3 of a sol. Of iron wire in dilute sulphuric acid(some in the form of Fe3+ and some in the form of Fe2+) was pipetted and placed into a conical flask to which 10cm^3 of dil. H2SO4 was added. The contents were titrated using Potassium Manganate (3.16 gdm^-3 KmnO4) until a pale pink colour was observed.
(1M H2SO4 was used)
Calculations:
Title value =22.73
Calculate the concentration in mol/dm^3 Potassium Manganate, KMnO4
Mass ok KmnO4 = 3.16 gdm^-3
Molecular mass of KMnO4 = 158g
Concentration = Mass/Mol. Mass
= 3.16/158
= 0.02 mol dm^-3
What is the concentration in mol/dm^3 of MnO4- ions in the sol.
Mass of MnO4/mass of KMnO4 x conc. Of KMnO4
= 119/158 x 0.02
= 0.015 mol dm^-3
Calculate the no. of moles of MnO4- ions run from the burette into the titration flask
If in 1000cm^3 of KMnO4 sol, there is 0.015 mol dm^-3 of MnO4-
Then in 22.7 cm^3 (title value), there will be: 1000/22.73 x 0.015 mol dm^-3
= 0.0003 mol
I have to write half equations for the reaction:
MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l)
and Fe2+(aq) -----> Fe3+(aq) + e-
To combine the no. of electrons used must be equal to the no. formed so the second equation needs to be multiplied by five:
MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l)
5Fe2+(aq) -----> 5Fe3+(aq) + 5e-
Then I add them together to give:
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Find the no. of moles Fe2+ ions present in the titration flask:
Ratio MnO4- : 5Fe2+
So no. of moles Fe2+ present = 0.0003 x 5
= 0.0015 mol
Calculate the mass of Fe2+ ions in 1 dm^3:
If in 25 cm^3, there was 0.0015 mol
Then in 1000cm^3, there will be: 1000/25 x 0.0015
= 0.06 mol dm^-3
PLEASE CORRECT ME!
Quick Reply
