Track iodine through an indirect analysis, control the starch endpoint and derive each conversion factor from a balanced equation instead of memorising one overall ratio.
Iodine accepts electrons from thiosulfate
Iodine is reduced to iodide while thiosulfate is oxidised to tetrathionate, S₄O₆²⁻. Two thiosulfate ions together supply the two electrons accepted by one I₂ molecule. Therefore n(S₂O₃²⁻) = 2n(I₂), not the reverse. Iodine is often retained in solution with excess iodide as I₃⁻; the analysis still counts one I₂ equivalent for each I₃⁻.
Place iodine-containing solution in the flask and standard thiosulfate in the burette. The brown iodine colour fades to pale straw. Add a few drops of fresh starch at this stage: the solution becomes blue-black, and further thiosulfate removes the last iodine so the colour disappears. Adding starch while iodine is concentrated can bind iodine strongly and make endpoint equilibration slow or unclear.
Keep the iodine solution cool and titrate promptly in a suitable covered vessel when waiting, because iodine loss lowers the result. Use enough iodide and the specified acidity for the iodine-generating reaction, but avoid unnecessary strongly acidic conditions during the thiosulfate titration: thiosulfate decomposes in acid. Do not transfer the manganate rule ‘excess strong acid’ to every redox analysis.
An indirect titration measures an equivalent amount
An oxidising analyte can first react with excess iodide to liberate iodine. The iodine is then titrated with thiosulfate. The first reaction’s mole ratio links the original analyte to I₂; the second links I₂ to titrant. This is an indirect analysis, and it is not automatically a back titration. A back titration measures a known reagent remaining after reaction, so its calculation requires subtraction.
Original model: 25.00 cm³ of an iodate(V) solution is treated with excess iodide and acid. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. The iodine requires 22.50 cm³ of 0.0800 mol dm⁻³ thiosulfate. First n(S₂O₃²⁻) = 0.0800 × 0.02250 = 1.800 × 10⁻³ mol; then n(I₂) = 9.000 × 10⁻⁴ mol; then n(IO₃⁻) = n(I₂)/3 = 3.000 × 10⁻⁴ mol. Concentration = 3.000 × 10⁻⁴/0.02500 = 0.0120 mol dm⁻³.
The overall conversion is one iodate per six thiosulfates. Deriving this chain protects against a plausible-looking error such as dividing by two and forgetting the three iodines formed. Excess iodide ensures complete conversion; it does not itself define the number of moles of iodine released.
Worked unfamiliar analysis with dilution and mass percentage
Original data: 10.00 cm³ of a bleach sample is diluted to 250.0 cm³. A 25.00 cm³ aliquot is treated with excess iodide and acid, liberating iodine by ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O. It requires 18.40 cm³ of 0.0500 mol dm⁻³ thiosulfate. Find the original NaClO mass concentration, using Mr(NaClO) = 74.5.
n(thiosulfate) = 0.0500 × 18.40/1000 = 9.200 × 10⁻⁴ mol. Hence n(I₂) = 4.600 × 10⁻⁴ mol. The iodine-generating equation gives n(ClO⁻) = n(I₂), so the aliquot contains 4.600 × 10⁻⁴ mol ClO⁻.
The full diluted flask contains 250.0/25.00 = 10 aliquots, or 4.600 × 10⁻³ mol. These moles originally occupied 10.00 cm³, so original concentration = 4.600 × 10⁻³/0.01000 = 0.460 mol dm⁻³. Mass concentration as NaClO = 0.460 × 74.5 = 34.3 g dm⁻³. If the original solution density is 1.06 g cm⁻³, one dm³ has mass 1060 g and mass percentage = 34.27/1060 × 100 = 3.23%.
Label the initial sample, the diluted flask and the aliquot separately before calculating. The dilution factor is 25 for concentration, while the aliquot-to-flask factor is 10 for amount. Both are legitimate, but applying both to a quantity already corrected for dilution double-counts the same change. Bleach–acid chemistry is an analytical laboratory context; use the prescribed reagents and controls because uncontrolled acidification can release hazardous chlorine.
Derive ratios when a different reductant is used
Manganate(VII) can oxidise hydrogen peroxide or ethanedioate. Both supply two electrons per molecule or ion, so each has a 5:2 reductant:MnO₄⁻ mole ratio in acidic solution. Do not carry the iron 5:1 ratio into these reactions. Ethanedioate oxidation can be slow initially; warming to the specified temperature speeds it, and Mn²⁺ formed during reaction acts as an autocatalyst, developed in Topic 15.
For FeC₂O₄, both Fe²⁺ and C₂O₄²⁻ are oxidised: one formula unit supplies one plus two, or three electrons. Five formula units therefore supply 15 electrons to three MnO₄⁻ ions. Waters of crystallisation affect the molar mass, but do not supply extra redox electrons in this calculation.
Original check: 25.00 cm³ of a prepared FeC₂O₄ solution uses 15.00 cm³ of 0.0200 mol dm⁻³ MnO₄⁻. n(MnO₄⁻) = 3.00 × 10⁻⁴ mol; n(FeC₂O₄) = (5/3) × 3.00 × 10⁻⁴ = 5.00 × 10⁻⁴ mol in the aliquot. If diluted to 250.0 cm³, the whole sample contains 5.00 × 10⁻³ mol. The 5:3 ratio follows the total electron donation.
Connect observations, assumptions and the final result
Iodine lost before titration gives too small a thiosulfate titre and underestimates the original oxidant. An iodine-generating reaction that has not finished produces the same directional bias, but requires a different improvement. Overshooting the starch disappearance endpoint gives a titre that is too large. A reagent blank can detect iodine released by background oxidants: subtract a justified blank titre only if the method specifies that correction.
A valid analysis assumes complete selective conversion, known titrant concentration, quantitative transfer and representative sampling. Explain which assumption an error violates. ‘Human error’ is not an evaluation because it neither predicts direction nor identifies a remedy.
June 2023 9CH0/01 Q10(c) used an unfamiliar dissolved-oxygen analysis. The report highlights missed intermediate ratios and confusion between g dm⁻³ and g cm⁻³. The transferable approach is to derive each mole link, retain units and convert to the quantity requested. A number in mol dm⁻³ is unfinished if the question asks for mass concentration or ppm.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. 19.60 cm³ of 0.100 mol dm⁻³ thiosulfate reacts with iodine. Calculate the amount of iodine.Show answer
n(S₂O₃²⁻) = 0.100 × 0.01960 = 1.960 × 10⁻³ mol. n(I₂) = half this = 9.80 × 10⁻⁴ mol. Two thiosulfates supply the two electrons accepted by one iodine molecule.
Q2. Describe the starch endpoint and why starch is added near it.Show answer
Titrate iodine until pale straw, add starch to form a blue-black colour, then add thiosulfate dropwise until that colour disappears after mixing. Late addition avoids strong binding at high iodine concentration and gives a sharper endpoint.
Q3. An oxidant X produces two I₂ molecules per X. Its iodine requires 30.00 cm³ of 0.0400 mol dm⁻³ thiosulfate. Find n(X).Show answer
n(thiosulfate) = 0.0400 × 0.03000 = 1.200 × 10⁻³ mol. n(I₂) = 6.000 × 10⁻⁴ mol. Two I₂ arise per X, so n(X) = 3.00 × 10⁻⁴ mol. The two separate divisions by two represent separate reactions.
Q4. A diluted iodine sample is left warm in an open flask before titration. Predict the bias and one improvement.Show answer
Iodine can escape, so less thiosulfate is required and the inferred original oxidant amount is too low. Keep the solution cool, covered while waiting and titrate promptly. Repeating the same delay would not remove the systematic loss.
Q5. Derive the MnO₄⁻:FeC₂O₄ ratio and explain whether two waters of crystallisation change it.Show answer
Fe²⁺ supplies one electron and C₂O₄²⁻ supplies two on oxidation to two CO₂, making three per FeC₂O₄. MnO₄⁻ accepts five in acid, so three manganates react with five FeC₂O₄. Hydration changes formula mass, not this electron ratio.
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 9CH0/01 mark scheme — June 2023 — Q7(d), Q10(c–d), PDF pp.25, 33–34; conditions, mole ratios and disproportionation.
- Pearson 9CH0/01 examiner report — June 2023 — Q7(d), Q10(c–d), pp.31–32 and 47–51. Advice is specific to those tasks.
- Pearson 9CH0/01 question paper — June 2023 — Q10(c–d), pp.24–26; original question context for the assessment observations, not reproduced exercises.
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