Question

Back-Titration of Ferrous Ion with Permanganate

Pyrolusite samples are analyzed by, first, allowing the MnO2 to react with excess Fe2+.

MnO2 + 2Fe2+ + 4H+ → Mn2+ + 2Fe3+ + 2H2O

Subsequently, the remaining ferrous ion is back-titrated in acidified solution by permanganate.

MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

A 0.150-g sample, containing pyrolusite, is treated with 60.0 mL of 0.0438-M Fe(NH4)2(SO4)2. The solution is acidified and the excess ferrous ion is titrated with 21.0 mL of 0.0175-M KMnO4.

Determine the managanese content as percent Mn2O3 in the original sample.

Answer 1

Let us go backwards

Total Fe(NH4)2(SO4)2 used

mmol = MV = 0.0438*60 = 2.628 mmol of  Fe(NH4)2(SO4)2

1 mmol of  Fe(NH4)2(SO4)2 = 1 mmol of Fe+2

mmol of Fe+2 = 2.628

Backtitration

mmol of KMnO4 used = MV = 21*0.0175 = 0.3675 mmol of KMnO4

5 mmol of Fe+2 = 1 mmol of MnO4-

mmol of Fe+2 titrated = 1/5*0.3675 = 0.0735 mmol of Fe+2

mmol of of Fe+2 reacted in MnO2 = 2.628 - 0.0735

mmol of Fe+2 reacted with MnO2 = 2.5545

1 mmol of MnO2 ---> 2 mmol of Fe+2

x mmol of MnO2 = 2.5545 mmol of Fe+2

x = 2.5545 /2 = 1.27725 mmol of MnO2

mass o fMnO2 = mmol*MW = 1.27725*86.9368 = 111.04 mg = 0.11104 g

% sample = 0.11104 /0.150 *100 = 74.026 % of MnO2