Topic 16 · page 14 of 19

Core Practical 13a: iron(II) with manganate(VII)

Titrate iron(II) ions against potassium manganate(VII) in acid. The manganate(VII) is its own indicator, and the end point is the first permanent pale pink.

13 min read16.17(i)16.15

Read first: Redox titration reactions and ratios

Method 16.17(i)

Aim: to find the concentration of an iron(II) solution by titrating it against potassium manganate(VII) of known concentration.

MnOX4X−(aq)+8 HX+(aq)+5 FeX2+(aq)→MnX2+(aq)+4 HX2O(l)+5 FeX3+(aq) \begin{aligned} &\ce{MnO4-(aq) + 8H+(aq) + 5Fe^2+(aq)} \\ &\ce{-> Mn^2+(aq) + 4H2O(l) + 5Fe^3+(aq)} \end{aligned}

PracticalTitrating iron(II) with manganate(VII)

  1. Pipette 25.0 cm³ of the iron(II) solution into a conical flask.
  2. Add an excess of dilute sulfuric acid (about 10 cm³).
  3. Fill the burette with potassium manganate(VII) solution. Read the top of the meniscus, because the solution is too dark to see the bottom.
  4. Titrate, swirling, until the first permanent pale pink colour appears.
  5. Repeat until you have concordant titres, that is, within 0.10 cm³ of each other.

The first titration is a rough one, run in larger steps to find the approximate end point. Do not use it in the mean. Make iron(II) solutions fresh, because Fe²⁺ is slowly oxidised by air.

Three conical flasks: pale green at the start, pale yellow with purple drops fading as they mix in, and permanent pale pink at the end point.
The colour sequence for iron(II) with manganate(VII). The colours are drawn to show the sequence.

Explaining the method 16.17(i)

Safety. Potassium manganate(VII) is an oxidising agent and stains skin and clothes. Wear eye protection.

Recording and using results 16.17(i)16.15

Worked exampleConcentration of an iron(II) solution (illustrative results)

25.0 cm³ portions of Fe²⁺(aq) are titrated against 0.0200 mol dm⁻³ KMnO₄. Burette readings give titres of 23.15 (rough), 22.40 and 22.30 cm³.

  1. The rough titre is not used. The other two are within 0.10 cm³, so they are concordant. Mean titre = (22.40 + 22.30) ÷ 2 = 22.35 cm³.
  2. n(MnO₄⁻) = 0.0200 × 22.35 ÷ 1000 = 4.470 × 10⁻⁴ mol
  3. n(Fe²⁺) = 5 × 4.470 × 10⁻⁴ = 2.235 × 10⁻³ mol in 25.0 cm³
  4. [Fe²⁺] = 2.235 × 10⁻³ ÷ 0.0250 = 0.0894 mol dm⁻³

Check yourself

  1. Why is the top of the meniscus read in this titration?
  2. Why is an excess of sulfuric acid added to the flask?
  3. Why is hydrochloric acid unsuitable for acidifying the mixture?
  4. State the colour change at the end point and explain why no indicator is needed.
  5. What would you see if too little acid were used, and why does it matter?
  6. 25.0 cm³ of Fe²⁺(aq) needs a mean titre of 21.80 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate [Fe²⁺] to 3 significant figures.

Answers cover these until you have tried the questions

  1. The manganate(VII) solution is a dark purple, so the bottom of the meniscus cannot be seen.
  2. The reaction uses H⁺ ions, and the acid supplies them. An excess ensures the manganate(VII) is reduced to Mn²⁺ and not to brown MnO₂.
  3. MnO₄⁻ oxidises Cl⁻ (Ecell⊖E^\ominus_\text{cell} = +0.15 V). Extra manganate(VII) is used up, so the titre is too large.
  4. Colourless (pale green) to the first permanent pale pink. MnO₄⁻ is purple and Mn²⁺ is almost colourless, so the first excess drop gives the colour: the titration is self-indicating.
  5. A brown precipitate of MnO₂ would form. It hides the end point and means the reduction of MnO₄⁻ is incomplete, so the amount of manganate(VII) used does not match the ratio 1 : 5.
  6. n(MnO₄⁻) = 0.0200 × 21.80 ÷ 1000 = 4.360 × 10⁻⁴ mol. n(Fe²⁺) = 5 × 4.360 × 10⁻⁴ = 2.180 × 10⁻³ mol. [Fe²⁺] = 2.180 × 10⁻³ ÷ 0.0250 = 0.0872 mol dm⁻³.