Iron in a tablet by permanganate titration

UKChO Round 1 · unreviewed

Fe²⁺ titrated with acidified permanganate; one portion of a made-up solution titrated.

Worked example

An iron tablet of mass 1.076 g was dissolved in dilute sulfuric acid and made up to 250.0 cm³. A 25.00 cm³ portion needed 31.60 cm³ of 0.00200 mol dm⁻³ potassium manganate(VII) to reach the end point. Calculate the percentage by mass of iron in the tablet. M(Fe) = 55.85 g mol⁻¹.

1. Concentration from moles and volume: solve for n(MnO₄⁻)
$n = cV$
= 6.320 × 10⁻⁵ mol
2. Mole ratio from the balanced equation: solve for n(Fe²⁺) in the portion
$n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$
= 3.160 × 10⁻⁴ mol
3. Divide made up to ÷ portion titrated → Flask volume ÷ portion
= 10.00
4. Multiply n(Fe²⁺) in the portion × (Flask volume ÷ portion) → n(Fe²⁺) in the tablet
= 0.003160 mol
5. Moles from mass: solve for m(Fe)
$m = nM$
= 0.1765 g
6. Divide m(Fe) ÷ m(tablet) → Fe by mass
= 0.1640

Answer: Fe by mass = 0.164

Method

n(MnO₄⁻) = cV. The half-equations (MnO₄⁻ gains 5e⁻, Fe²⁺ loses 1e⁻) give n(Fe²⁺) = 5n(MnO₄⁻) in the portion. Multiply by flask volume ÷ portion volume for the whole tablet, then convert to a mass and a percentage.

Common mistakes

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O: each permanganate oxidises five Fe²⁺.
Only one portion of the solution was titrated: scale up by the flask volume ÷ the portion volume.

Formulas used

Concentration from moles and volume
$c = \dfrac{n}{V}$
Mole ratio from the balanced equation
$n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$
Moles from mass
$n = \dfrac{m}{M}$

Practise this with fresh numbers

Iron in a tablet by permanganate titration

Worked example

Method

Common mistakes

Formulas used

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