Question 4.17: Oxidation–Reduction Titration Analysis Iron ores often invol...
Oxidation–Reduction Titration Analysis
Iron ores often involve a mixture of oxides and contain both [latex] Fe^{2+} and Fe^{3+} ions. Such an ore can be analyzed for its iron content by dissolving it in acidic solution, reducing all the iron to Fe^{2+} ions, and then titrating with a standard solution of potassium permanganate. In the resulting solution, MnO_{4} ^{-} is reduced to Mn^{2+}, and Fe^{2+} is oxidized to Fe^{3+}. A sample of iron ore weighing 0.3500 g was dissolved in acidic solution, and all the iron was reduced to Fe^{2+}. Then the solution was titrated with a 1.621 × 10 ^{-2} M KMn O_{4} [/latex] solution. The titration required 41.56 mL of the permanganate solution to reach the light purple (pink) endpoint. Determine the mass percent of iron in the iron ore.
Where are we going?
We are asked to determine the mass percent of iron in an iron ore.
What do we know?
We know the mass of the iron ore and the volume and concentration of the KMnO_{4} solution used to titrate the iron in the ore.
How do we get there?
Mass percent of iron = \frac{mass of iron }{mass of iron ore} × 100 %
We know that the mass of the mixture is 0.3500 g, so we change the question to “What is the mass of the iron present in the ore?”
All of the iron metal is converted to Fe^{2+}, which is reacted with a known volume and molarity of Mn O_{4} ^{-} . From volume and molarity we can get moles, and by using the mole ratio in a balanced equation, we can determine the moles of iron. We convert from moles to mass using the atomic mass of iron.
From the problem it is obvious that this is a redox reaction, so we will need to balance the equation accordingly.
First, we write the unbalanced equation for the reaction:
H^{+}(aq) + Mn O_{4} ^{-} (aq) + Fe^{2+} (aq) → Fe^{3+} (aq) + Mn^{2+} (aq) + H_{2}O (l)
Using the half-reaction method, we balance the equation:
8H^{+} (aq) + Mn O_{4} ^{-} (aq) + 5 Fe^{2+}(aq) → 5 Fe^{3+}(aq) + Mn^{2+} (aq) +4 H_{2}O (l)
The number of moles of Mn O_{4} ^{-} ion required in the titration is found from the volume and concentration of permanganate solution used:
41.56 mL ×\frac{1 L}{1000 mL}× \frac{1.621 × 10 ^{-2} mol Mn O_{4} ^{-} }{L} = 6.737 × 10 ^{-4} mol Mn O_{4} ^{-}
The balanced equation shows that five times as much Fe^{2+} as Mn O_{4} ^{-} is required:
6.737 × 10 ^{-4} mol Mn O_{4} ^{-} ×\frac{5 mol Fe^{2+}}{1 mol Mn O_{4} ^{-}} = 3.368 × 10 ^{-3} mol Fe^{2+}
Thus the 0.3500-g sample of iron ore contained 3.368 × 10 ^{-3} mol of iron. The mass of iron present is
3.368 × 10 ^{-3} mol Fe ×\frac{55.85 g Fe}{1 mol Fe} = 0.1881 g Fe
The mass percent of iron in the iron ore is
\frac{ 0.1881 g }{0.3500 g} × 100% = 53.74 %