Explain the acidic manganate(VII) endpoint, determine iron content from a prepared solution, and trace how practical errors change the calculated answer.
Derive the mole ratio from electrons
Purple MnO₄⁻ is reduced to very pale pink Mn²⁺ in sufficiently acidic solution; Fe²⁺ is oxidised to Fe³⁺. The manganese oxidation number decreases from +7 to +2, so one MnO₄⁻ accepts five electrons. Each Fe²⁺ supplies one, giving the 1:5 ratio. The solution need not be literally colourless before the endpoint: iron ions and tablet ingredients can contribute colour.
Manganate(VII) is self-indicating. Before the endpoint each added drop reacts and loses its purple colour after mixing; just beyond equivalence a slight excess produces the first faint pink colour throughout the solution. Swirl, add dropwise near the endpoint and observe on a white tile. A deep purple endpoint indicates overshooting, not greater certainty.
The acid is part of the reaction conditions
Use the specified excess of dilute sulfuric acid. It supplies H⁺ without introducing a competing oxidant or readily oxidised chloride. Insufficient acidity can form brown MnO₂ instead of Mn²⁺, changing manganese’s electron acceptance from five to three. The colour and stoichiometry are then unsuitable for the intended calculation.
Hydrochloric acid is unsuitable because chloride can be oxidised by acidified manganate(VII), consuming additional titrant and potentially releasing chlorine. Nitric acid can oxidise Fe²⁺ before the titration, leaving less Fe²⁺ for manganate and lowering the titre. The resulting iron calculation is biased high in the first case and low in the second, assuming the usual 1:5 calculation is used.
Do not justify the acid by saying it is only a catalyst. Eight H⁺ appear per MnO₄⁻ in the net ionic equation. Wear eye protection, use a pipette filler and follow the laboratory controls for acid and oxidising solutions; dispose of manganese-containing waste as instructed.
Core Practical 11: prepare the tablet solution quantitatively
Measure the mass of the crushed iron tablets actually transferred, using weighing by difference. Dissolve the available Fe²⁺ in the specified sulfuric acid, stirring thoroughly. Filter insoluble binders into a volumetric flask; rinse the preparation vessel and residue so that soluble iron is transferred. Make up to the mark only after the solution is at room temperature, stopper and invert repeatedly. A volumetric flask gives a known total solution volume, not merely a measured volume of added water.
Rinse the pipette with the prepared solution, then transfer an accurately measured aliquot to a conical flask. Rinse the burette with the standard manganate solution, fill the tip without an air bubble, remove the filling funnel and record initial and final readings consistently at eye level. Use a rough titration to locate the endpoint, then obtain concordant careful titres and average those appropriate results.
For a clean Fe²⁺ solution, use the first faint pink endpoint that persists through mixing for the stated observation time. Pearson’s iron-tablet method notes that this pink may fade later because another tablet ingredient reacts slowly. Do not keep adding titrant to restore that later colour: follow the stated endpoint and timing. This is an example of sample-specific interference, not a reason to ignore all fading during mixing.
The flask may be rinsed with deionised water because extra water changes concentration but not the moles of Fe²⁺ delivered, provided acidity remains sufficient. Water left inside the pipette instead dilutes the measured sample and lowers its moles. The distinction is whether the water is added before or after the fixed aliquot has been measured.
Worked example: iron mass per tablet
Original illustrative data: five tablets are dissolved and made up to 250.0 cm³. A 25.00 cm³ aliquot requires a mean of 23.20 cm³ of 0.00500 mol dm⁻³ MnO₄⁻. Find the mass of iron per tablet, using Ar(Fe) = 55.8.
1. Titrant amount: n(MnO₄⁻) = cV = 0.00500 × 23.20/1000 = 1.160 × 10⁻⁴ mol. This quantity belongs to one aliquot, not all five tablets.
2. Iron amount in the aliquot = 5 × 1.160 × 10⁻⁴ = 5.800 × 10⁻⁴ mol Fe²⁺. Scale to the full preparation: 250.0/25.00 = 10, so total n(Fe) = 5.800 × 10⁻³ mol.
3. Mass in five tablets = 5.800 × 10⁻³ × 55.8 = 0.32364 g. Per tablet = 0.32364/5 = 0.064728 g = 64.7 mg to three significant figures. The factor of five from the redox ratio and the division by five tablets have different meanings even though they happen to cancel numerically.
If the five transferred tablets had mass 1.850 g, mass percentage of iron = 0.32364/1.850 × 100 = 17.5%. This is elemental iron content; using the formula mass of hydrated iron sulfate would answer a different question. State what each intermediate quantity represents.
Separate uncertainty from a systematic chemical error
If each burette reading has uncertainty ±0.05 cm³, a titre found by difference has a maximum absolute uncertainty of ±0.10 cm³. For 23.20 cm³ this is 0.10/23.20 × 100 = 0.431%. If the 25.00 cm³ pipette is ±0.03 cm³ and the 250.0 cm³ flask ±0.12 cm³, their contributions are 0.120% and 0.0480%; the simple maximum percentage sum is 0.599%, before titrant concentration and other uncertainties. These are specified example tolerances, not universal apparatus values.
Increase an unnecessarily small titre, for example by using a larger aliquot or a suitably less concentrated titrant, to reduce relative burette uncertainty. Concordant titres demonstrate precision, not freedom from a shared extraction or calibration error. Comparing with a supplier value requires both uncertainty and a reasoned account of possible interferences. Practical competence must be demonstrated in the laboratory; these notes teach the method and interpretation.
| Problem | Effect | Targeted improvement |
|---|---|---|
| Incomplete extraction or solution left in the vessel/filter | Too little Fe²⁺ reaches the volumetric flask; result low | Allow dissolution, rinse transfers quantitatively and verify extraction. |
| Fe²⁺ oxidised by air before titration | Less Fe²⁺ remains; result low | Prepare fresh acidified solution and titrate promptly. |
| Manganate endpoint overshot | Recorded volume too large; result high | Dropwise addition, thorough swirling and a consistent faint endpoint. |
| Air bubble initially in burette tip | Some recorded delivery fills the tip; apparent titre too high | Run titrant through the tip before the initial reading. |
| Unrinsed water in burette | Titrant diluted; more volume needed and use of nominal c biases result high | Condition the burette with titrant before filling. |
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Why does one mole of acidified MnO₄⁻ react with five moles of Fe²⁺?Show answer
Mn falls from +7 to +2 and accepts five electrons per ion. Each Fe²⁺ becomes Fe³⁺ and supplies one electron. Balancing electrons therefore gives MnO₄⁻:Fe²⁺ = 1:5, independently of the H⁺ coefficient.
Q2. 25.00 cm³ of Fe²⁺ solution requires 18.60 cm³ of 0.0200 mol dm⁻³ MnO₄⁻. Find [Fe²⁺].Show answer
n(MnO₄⁻) = 0.0200 × 0.01860 = 3.720 × 10⁻⁴ mol. n(Fe²⁺) = 5 × this = 1.860 × 10⁻³ mol. Divide by 0.02500 dm³: [Fe²⁺] = 0.0744 mol dm⁻³. Dividing the titrant concentration by five would reverse the reacting ratio.
Q3. Why are sulfuric acid and a faint pink endpoint used?Show answer
Sulfuric acid supplies enough H⁺ for reduction to Mn²⁺ without the competing chloride oxidation or nitric-acid oxidation of Fe²⁺. A slight excess of MnO₄⁻ gives faint pink after all accessible Fe²⁺ has reacted. Follow the stated time criterion, including the tablet-method caveat about later fading.
Q4. A titre is 12.50 cm³ and each burette reading has uncertainty ±0.05 cm³. Find the maximum percentage titre uncertainty and predict it at 25.00 cm³.Show answer
Two readings give ±0.10 cm³. At 12.50 cm³, 0.10/12.50 × 100 = 0.800%. At 25.00 cm³ it is 0.400%. Doubling the useful titre halves this relative contribution, while the individual reading uncertainty stays unchanged.
Q5. Explain why rinsing the conical flask with water is acceptable but leaving water in the sampling pipette biases the result.Show answer
Flask water is added after the aliquot moles have been measured, so those moles are unchanged and the titre remains the same if adequate acidity is retained. Pipette water dilutes the sample before its fixed volume is measured, so less Fe²⁺ is delivered and the result is low.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification — Issue 3 — Topic 14, printed pp.31–32; UK A-Level scope and Core Practicals 10–11.
- Chemrevise — UK Edexcel Topic 14 — Pages 1–11 used as a secondary coverage check. Teaching, examples and questions here are original Finesse material.
- Pearson Core Practical 11 — iron in tablets — Teacher and student sheets, PDF pp.1–4; preparation, endpoint caveat and analysis.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.