Using Oxidised Mohr’s Salt to Standardise KMnO4
A bottle of Mohr’s salt, FeSO4·(NH4)2SO4·6H2O (molar mass 392 g mol−1), has stood open in a school store for several months. The pale green crystals are now coated with a brown solid, believed to be hydrated iron(III) oxide. A student weighs 3.92 g of this old salt, dissolves it in about 50 mL of distilled water, and plans to acidify with concentrated HCl before titrating with approximately 0.0200 M KMnO4 from a 25 mL burette. The stockroom has dilute H2SO4, concentrated HCl, fresh Mohr’s salt, oxalic acid, and K3[Fe(CN)6] solution. The teacher insists that the old sample cannot be used as a primary standard without treatment. The student is asked to calculate a sensible sample mass, choose the acid, and explain the brown coating in terms of Fe2+ chemistry.
1. Why is HCl unsuitable for acidifying the KMnO4 titration? Write the relevant half-equations and identify the correct acid.
Read the worked answer
Permanganate is a strong oxidising agent in acidic solution:
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O E° ≈ +1.51 V.
Chloride can be oxidised to chlorine:
2Cl− → Cl2 + 2e− E° ≈ −1.36 V (as a reduction, Cl2 + 2e− → 2Cl−, E° = +1.36 V).
Because the permanganate couple has the more positive reduction potential, MnO4− can oxidise Cl− to Cl2. The chloride would consume permanganate in addition to the Fe2+, giving a falsely high titre and a calculated KMnO4 concentration that is too high. Chlorine evolution may also be visible, and the solution may smell of chlorine.
Dilute sulfuric acid is the correct acid. Sulfate is not oxidised under these conditions, and the medium keeps Fe2+ and Mn2+ in solution. Nitric acid is also unsuitable because it is itself an oxidising acid and can oxidise Fe2+. Hydrochloric acid must not be used.
2. Calculate the volume of 0.0200 M KMnO4 that 3.92 g of pure Mohr’s salt would require, and recommend a mass suitable for a 25 mL burette (target 20.0–25.0 mL). Show the stoichiometry.
Read the worked answer
The reaction is:
MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O.
So 1 mol MnO4− reacts with 5 mol Fe2+.
For 3.92 g pure Mohr’s salt: n(Fe2+) = 3.92 g / 392 g mol−1 = 0.0100 mol. n(MnO4−) required = 0.0100/5 = 2.00 × 10−3 mol. V = n/C = 2.00 × 10−3 mol / 0.0200 mol L−1 = 0.100 L = 100.0 mL.
This is four times the burette capacity, so the sample is far too large. For a target titre of 20.0–25.0 mL: At 20.0 mL: n(MnO4−) = 0.0200 × 0.0200 = 4.00 × 10−4 mol; n(Fe2+) = 5 × 4.00 × 10−4 = 2.00 × 10−3 mol; mass = 2.00 × 10−3 × 392 = 0.784 g. At 25.0 mL: n(Fe2+) = 5 × (0.0200 × 0.0250) = 2.50 × 10−3 mol; mass = 0.980 g.
A suitable mass is therefore about 0.80–0.98 g, for example 0.90 g weighed accurately. This matches the burette size and keeps the titre on scale.
3. If the student ignores the brown coating and uses the recommended mass of old salt, will the calculated KMnO4 molarity be too high or too low? Explain using the Fe2+ content, and describe a confirmatory test for Fe2+.
Read the worked answer
The brown coating is iron(III) oxide/hydroxide formed by air oxidation, for example:
4Fe2+ + O2 + 4H+ → 4Fe3+ + 2H2O, followed by hydrolysis and precipitation of hydrated Fe2O3/Fe(OH)3.
Iron(III) does not react with permanganate under these conditions, so the old sample contains fewer moles of Fe2+ per gram than pure Mohr’s salt. If the student weighs the recommended mass and assumes it is pure, the expected Fe2+ amount is too high. The actual Fe2+ is lower, so the volume of KMnO4 consumed, V, is smaller than it would be for pure salt.
The student then calculates:
c(KMnO4) = n(Fe2+ assumed)/5V.
Because V is in the denominator and is too small, the calculated concentration is too high. The result overestimates the KMnO4 molarity.
A confirmatory test for Fe2+ is addition of potassium hexacyanoferrate(III), K3[Fe(CN)6], which gives a deep blue precipitate (Turnbull’s blue) with Fe2+. A fresh solution of Mohr’s salt would give this test strongly; the oxidised sample would give a weaker Fe2+ response and may also give a Prussian-blue-type colour from Fe3+. Recrystallisation or use of fresh Mohr’s salt, or standardisation against oxalic acid, is the proper remedy.