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Questions on redox titrations: manganate(VII) and thiosulfate

Manganate(VII) titrations, why they are self-indicating and acidified with sulfuric acid rather than hydrochloric acid, the 1 : 5 ratio with iron(II) and the 2 : 5 ratio with ethanedioate and methanoic acid, autocatalysis, iodine–thiosulfate titrations and when to add starch, chlorate(I) in bleach, standardising thiosulfate with potassium iodate(V), primary standards and dichromate(VI) as a titrant, the calculation chain from titre to moles to mass, the mass of iron in a tablet, copper by iodine liberation, and percentage purity.

22 original questions · 74 marks · the redox titrations: manganate(VII) and thiosulfate notes · Redox · Flashcards for this unit

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• 6 marks · Calculate · RTM-Q21

  Calculate the mass of iron in one tablet, in mg, and the percentage of iron by mass in the tablets. Five iron(II) sulfate tablets, total mass 2.85 g, are dissolved in dilute sulfuric acid and made up to 100.0 cm³. 10.0 cm³ portions are titrated with 0.00500 mol dm⁻³ potassium manganate(VII), with the burette readings given in the table. Ar of Fe is 55.8.

  Titration	Rough	1	2	3
  Final reading / cm³	22.05	21.55	43.20	21.60
  Initial reading / cm³	0.00	0.20	21.55	0.15
  Titre / cm³	22.05	21.35	21.65	21.45

  Mark scheme

  1. M1 titres 1 and 3 are concordant, within 0.10 cm³, so the mean titre is (21.35 + 21.45) ÷ 2 = 21.40 cm³; the rough titre and titre 2 are left out
  2. M1 moles of MnO₄⁻ in the mean titre = 0.00500 × 21.40 ÷ 1000 = 1.070 × 10⁻⁴ mol
  3. M1 moles of Fe²⁺ in each 10.0 cm³ portion = 5 × 1.070 × 10⁻⁴ = 5.350 × 10⁻⁴ mol, from the 1 : 5 ratio in the equation
  4. M1 the 100.0 cm³ of solution from five tablets holds ten times as much, 5.350 × 10⁻³ mol, so one tablet holds 1.070 × 10⁻³ mol of iron
  5. A1 mass of iron in one tablet = 1.070 × 10⁻³ × 55.8 = 0.0597 g, which is 59.7 mg to three significant figures
  6. A1 percentage of iron = 5.350 × 10⁻³ × 55.8 ÷ 2.85 × 100 = 10.5%, to three significant figures; error carried forward from the mean titre

• 5 marks · Calculate · RTM-Q7

  Calculate the concentration, in g dm⁻³, of a hydrogen peroxide solution from the results given in the table. The acidified, diluted solution reacts with manganate(VII): 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 5O₂ + 8H₂O.

  Quantity	Value
  Hydrogen peroxide solution diluted	10.0 cm³ made up to 250.0 cm³
  Portion titrated	25.0 cm³
  Concentration of KMnO₄	0.0200 mol dm⁻³
  Mean titre of KMnO₄	21.20 cm³
  Mr of H₂O₂	34.0

  Mark scheme

  1. M1 moles of MnO₄⁻ = 0.0200 × 21.20/1000 = 4.24 × 10⁻⁴ mol
  2. M1 moles of H₂O₂ in 25.0 cm³ = 4.24 × 10⁻⁴ × 5/2 = 1.06 × 10⁻³ mol
  3. M1 moles in 250.0 cm³, which all came from 10.0 cm³ of the original, = 1.06 × 10⁻² mol
  4. M1 concentration of the original = 1.06 × 10⁻² ÷ 0.0100 = 1.06 mol dm⁻³
  5. A1 1.06 × 34.0 = 36.0 g dm⁻³, to three significant figures

• 5 marks · Calculate · RTM-Q15

  Calculate the percentage purity of a sample of copper(II) sulfate crystals, CuSO₄·5H₂O (Mr 249.6). 2.60 g of the sample is dissolved and made up to 250.0 cm³. 25.0 cm³ portions are added to excess potassium iodide, and the iodine liberated is titrated with 0.0500 mol dm⁻³ sodium thiosulfate, with the burette readings given in the table.

  Titration	Rough	1	2	3
  Final reading / cm³	20.90	20.85	41.30	21.40
  Initial reading / cm³	0.00	0.50	20.85	1.00
  Titre / cm³	20.90	20.35	20.45	20.40

  Mark scheme

  1. M1 titres 1, 2 and 3 are concordant, so the mean titre is 20.40 cm³, the rough titre being left out
  2. M1 moles of S₂O₃²⁻ = 0.0500 × 20.40/1000 = 1.02 × 10⁻³ mol
  3. M1 Cu²⁺ : S₂O₃²⁻ is 1 : 1, so 1.02 × 10⁻³ mol of Cu²⁺ in 25.0 cm³ and 1.02 × 10⁻² mol in 250.0 cm³
  4. M1 mass of CuSO₄·5H₂O = 1.02 × 10⁻² × 249.6 = 2.546 g
  5. A1 percentage purity = 2.546/2.60 × 100 = 97.9%, to three significant figures; error carried forward from M1

• 4 marks · Calculate · RTM-Q5

  Calculate the concentration of a potassium manganate(VII) solution from the standardisation against ethanedioic acid described in the table. The ratio of MnO₄⁻ to C₂O₄²⁻ is 2 : 5.

  Quantity	Value
  Mass of H₂C₂O₄·2H₂O (Mr 126.0) weighed out	0.630 g
  Volume of solution made up	250.0 cm³
  Portion titrated, acidified and warmed	25.0 cm³
  Mean titre of manganate(VII)	19.80 cm³

  Mark scheme

  1. M1 moles of acid weighed out = 0.630/126.0 = 5.00 × 10⁻³ mol, so 5.00 × 10⁻⁴ mol in 25.0 cm³
  2. M1 moles of MnO₄⁻ = 5.00 × 10⁻⁴ × 2/5 = 2.00 × 10⁻⁴ mol
  3. M1 concentration = 2.00 × 10⁻⁴ ÷ (19.80/1000)
  4. A1 0.0101 mol dm⁻³, to three significant figures; error carried forward from M2

• 4 marks · Calculate · RTM-Q6

  Calculate the concentration of ClO⁻ in a household bleach from the results given in the table. The diluted bleach is added to excess acidified potassium iodide, ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O, and the iodine is titrated with sodium thiosulfate.

  Quantity	Value
  Bleach diluted	10.0 cm³ made up to 250.0 cm³
  Portion of diluted bleach used	25.0 cm³
  Concentration of sodium thiosulfate	0.100 mol dm⁻³
  Mean titre of sodium thiosulfate	18.40 cm³

  Mark scheme

  1. M1 moles of S₂O₃²⁻ = 0.100 × 18.40/1000 = 1.84 × 10⁻³ mol
  2. M1 moles of I₂ = 1.84 × 10⁻³ ÷ 2 = 9.20 × 10⁻⁴ mol, equal to the moles of ClO⁻ in 25.0 cm³
  3. M1 moles of ClO⁻ in 250.0 cm³, and so in the 10.0 cm³ of bleach, = 9.20 × 10⁻³ mol
  4. A1 concentration = 9.20 × 10⁻³ ÷ 0.0100 = 0.920 mol dm⁻³, to three significant figures

• 4 marks · Describe · RTM-Q8

  Describe how to carry out a titration to find the concentration of an iron(II) solution using standard potassium manganate(VII) solution.

  Mark scheme

  1. B1 rinse and fill the burette with the manganate(VII) solution and read the top of the meniscus
  2. B1 pipette 25.0 cm³ of the iron(II) solution into a conical flask and add an excess of dilute sulfuric acid
  3. B1 run in the manganate with swirling, dropwise near the end, until the first permanent pale pink, with the flask on a white tile
  4. B1 repeat until titres are concordant, within 0.10 cm³, and take the mean of the concordant titres

• 4 marks · Compare · RTM-Q11

  Compare a manganate(VII) titration with an iodine–thiosulfate titration.

  Mark scheme

  1. B1 manganate(VII) is in the burette and measures a reducing agent directly; thiosulfate is in the burette and measures iodine liberated by an oxidising agent
  2. B1 manganate(VII) is self-indicating, whereas the iodine–thiosulfate titration uses starch as an indicator
  3. B1 the manganate end point is colourless to pale pink; the thiosulfate end point is blue-black to colourless
  4. B1 manganate needs excess dilute sulfuric acid; the iodine method needs excess potassium iodide in the flask

• 4 marks · Calculate · RTM-Q18

  Calculate the concentration of a sodium thiosulfate solution from the standardisation against potassium iodate(V) described in the table. Each portion is added to excess acidified potassium iodide: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O.

  Quantity	Value
  Mass of KIO₃ (Mr 214.0) weighed out	0.214 g
  Volume of solution made up	250.0 cm³
  Portion used	25.0 cm³
  Mean titre of sodium thiosulfate	24.00 cm³

  Mark scheme

  1. M1 moles of KIO₃ = 0.214/214.0 = 1.00 × 10⁻³ mol, so 1.00 × 10⁻⁴ mol in 25.0 cm³
  2. M1 one IO₃⁻ gives three I₂, which need six S₂O₃²⁻, so moles of S₂O₃²⁻ = 6 × 1.00 × 10⁻⁴ = 6.00 × 10⁻⁴ mol
  3. M1 concentration = 6.00 × 10⁻⁴ ÷ (24.00/1000)
  4. A1 0.0250 mol dm⁻³, to three significant figures

• 3 marks · Explain · RTM-Q2

  Explain why a manganate(VII) titration is acidified with dilute sulfuric acid and not with dilute hydrochloric acid.

  Mark scheme

  1. B1 manganate(VII) oxidises chloride ions from hydrochloric acid to chlorine
  2. B1 some manganate is used up reacting with chloride as well as with the substance being analysed, so the titre is too large
  3. B1 sulfate ions are not oxidised by manganate(VII), so all the manganate reacts with the analyte and the ratio in the equation holds

• 3 marks · Explain · RTM-Q3

  Explain why starch is added near the end point of an iodine–thiosulfate titration rather than at the start.

  Mark scheme

  1. B1 the iodine colour fades from brown to pale straw, and the last trace of yellow is hard to see, so starch is used to make the end point sharp
  2. B1 starch forms a blue-black complex with iodine, and the end point is when that colour disappears
  3. B1 at high iodine concentration the starch holds iodine and releases it only slowly, so the colour would persist past the end point and the titre would be too large

• 3 marks · Calculate · RTM-Q4

  Calculate the concentration of an iron(II) solution, 25.0 cm³ of which needs 18.75 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) for complete reaction in acid.

  Mark scheme

  1. M1 moles of MnO₄⁻ = 0.0200 × 18.75/1000 = 3.75 × 10⁻⁴ mol
  2. M1 moles of Fe²⁺ = 5 × 3.75 × 10⁻⁴ = 1.875 × 10⁻³ mol, from the 1 : 5 ratio
  3. A1 concentration = 1.875 × 10⁻³ ÷ 0.0250 = 0.0750 mol dm⁻³, to three significant figures

• 3 marks · Suggest · RTM-Q9

  Suggest why, in a warm titration of ethanedioate with manganate(VII), the first few drops take several seconds to lose their colour, but later drops are decolourised at once.

  Mark scheme

  1. B1 the reaction is slow at first because it involves two negative ions, MnO₄⁻ and C₂O₄²⁻, which repel each other, giving a high activation energy
  2. B1 the Mn²⁺ ions produced act as a catalyst for the reaction
  3. B1 as Mn²⁺ builds up the rate increases, so the reaction is autocatalysed

• 3 marks · Suggest · RTM-Q10

  Suggest the identity of the brown precipitate that appears when a manganate(VII) titration is acidified with too little sulfuric acid, and suggest why the titre is then no longer valid.

  Mark scheme

  1. B1 the precipitate is manganese(IV) oxide, MnO₂
  2. B1 without enough H⁺, manganate(VII) is reduced only to +4, gaining three electrons instead of five
  3. B1 the 1 : 5 ratio no longer applies, and the brown solid masks the pale pink end point

• 3 marks · Evaluate · RTM-Q12

  Evaluate this conclusion. A student dissolves an iron tablet, leaves the solution overnight in an open beaker, then titrates it with manganate(VII). The result is 15% below the iron content on the label, and she concludes that the label is wrong.

  Mark scheme

  1. B1 iron(II) in solution is oxidised to iron(III) by oxygen from the air when left standing
  2. B1 manganate(VII) reacts only with iron(II), so iron already oxidised is not counted and the titre is too small
  3. B1 the low result is explained by the method, so it does not show the label is wrong; a freshly made solution should be titrated straight away

• 3 marks · Explain · RTM-Q14

  Explain why one mole of copper(II) ions corresponds to one mole of thiosulfate ions when copper is determined by iodine liberation.

  Mark scheme

  1. B1 2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq), so two moles of Cu²⁺ liberate one mole of I₂
  2. B1 I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq), so one mole of I₂ reacts with two moles of S₂O₃²⁻
  3. B1 two moles of Cu²⁺ therefore match two moles of S₂O₃²⁻, a 1 : 1 ratio

• 3 marks · Calculate · RTM-Q17

  Calculate the volume of 0.0200 mol dm⁻³ potassium dichromate(VI) that reacts with 25.0 cm³ of 0.0750 mol dm⁻³ iron(II) ions in acid. One Cr₂O₇²⁻ ion accepts six electrons.

  Mark scheme

  1. M1 moles of Fe²⁺ = 0.0750 × 25.0/1000 = 1.875 × 10⁻³ mol
  2. M1 each Fe²⁺ gives one electron, so moles of Cr₂O₇²⁻ = 1.875 × 10⁻³ ÷ 6 = 3.125 × 10⁻⁴ mol
  3. A1 volume = 3.125 × 10⁻⁴ ÷ 0.0200 = 0.015625 dm³, so 15.6 cm³ to three significant figures

• 3 marks · Explain · RTM-Q19

  Explain why an iron sample may be dissolved in hydrochloric acid before titration with potassium dichromate(VI), although hydrochloric acid is not used in a titration with potassium manganate(VII).

  Mark scheme

  1. B1 manganate(VII) is a strong enough oxidising agent to oxidise chloride ions to chlorine, which uses up extra manganate and makes the titre too large
  2. B1 dichromate(VI) is a weaker oxidising agent than manganate(VII)
  3. B1 it does not oxidise chloride ions under titration conditions, so all the dichromate reacts with iron(II) and the 1 : 6 ratio holds

• 3 marks · Suggest · RTM-Q20 CAIE

  Suggest the half-equation for the oxidation of methanoic acid, HCOOH, to carbon dioxide in acid, and suggest the mole ratio of manganate(VII) to methanoic acid in an acidified titration.

  Mark scheme

  1. B1 HCOOH(aq) → CO₂(g) + 2H⁺(aq) + 2e⁻
  2. B1 carbon rises from +2 to +4, giving two electrons per molecule, while each MnO₄⁻ reduced to Mn²⁺ takes five
  3. B1 the electrons match at ten, so the ratio of MnO₄⁻ to HCOOH is 2 : 5

• 3 marks · Justify · RTM-Q22

  Justify the choice of dilute sulfuric acid, rather than dilute nitric acid, to acidify a solution of iron(II) ions before it is titrated with potassium manganate(VII).

  Mark scheme

  1. B1 nitric acid is an oxidising agent, so it would oxidise some of the Fe²⁺ to Fe³⁺ before the titration began
  2. B1 less Fe²⁺ would be left to react with the manganate(VII), so the titre would be too small and the calculated concentration too low
  3. B1 sulfate ions are not oxidised by manganate(VII) and sulfuric acid does not oxidise Fe²⁺, so it supplies the H⁺ the half-equation needs without taking part in any other reaction

• 2 marks · State · RTM-Q1

  State why no indicator is added in a manganate(VII) titration of iron(II) ions, and state the colour change at the end point.

  Mark scheme

  1. B1 manganate(VII) is purple and its reduction product Mn²⁺ is almost colourless, so the reagent is self-indicating
  2. B1 the solution in the flask turns from colourless, or very pale green, to the first permanent pale pink

• 2 marks · Give · RTM-Q13

  Give the ionic equation for the reaction of iodine with thiosulfate ions, and the mole ratio it gives.

  Mark scheme

  1. B1 I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)
  2. B1 one mole of iodine reacts with two moles of thiosulfate

• 1 mark · Name · RTM-Q16

  Name the white precipitate formed when copper(II) ions react with iodide ions.

  Mark scheme

  1. B1 the precipitate is copper(I) iodide, CuI

Practise redox titrations: manganate(VII) and thiosulfate one question at a time