A-LevelChemistryAnalytical techniques9701/31 · May/June 2021 · Q116 Marks
Iron(II) sulfate crystals, FeSO4•xH2O, contain water of crystallisation. You will carry out a titration to determine the value of x in the formula, where x is an integer. A solution containing a known mass of the crystals will be titrated with acidified aqueous potassium manganate(VII) of known concentration. 5Fe2+(aq) + MnO4¯(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(I) FA 1 contains 26.52 gdm³ of hydrated iron(II) sulfate, FeSO4•xH2O. FA 2 is 0.0200 mol dm-³ potassium manganate(VII), KMnO4. FA 3 is dilute sulfuric acid, H2SO4.
- (a)Method • Fill the burette with FA 2. • Pipette 25.0 cm³ of FA 1 into a conical flask. • Use the 25 cm³ measuring cylinder to transfer 25 cm³ of FA 3 into the same conical flask. • Carry out a rough titration and record your burette readings in the space below. The rough titre is ................................... cm³. • Carry out as many accurate titrations as you think necessary to obtain consistent results. • Make sure any recorded results show the accuracy of your practical work. • Record in a suitable form below all your burette readings and the volume of FA 2 added in each accurate titration. Keep FA 1 for use in Question 3.[7]
- (b)From your accurate titration results, obtain a suitable value to be used in your calculations. Show clearly how you obtained this value. 25.0 cm³ of FA 1 required ................................... cm³ of FA 2. [1][1]
- (c)(i)Give your answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures. [1][1]
- (c)(ii)Calculate the number of moles of potassium manganate(VII) present in the volume of FA 2 calculated in (b). moles of KMnO4 = ................................... mol [1][1]
- (c)(iii)Calculate the number of moles of iron(II) sulfate present in 1.00 dm³ of FA 1. moles of FeSO4 = ................................... mol [1][1]
- (c)(iv)Calculate the mass of iron(II) sulfate present in 1.00 dm³ of FA 1. mass of FeSO4 = ................................... g [1][1]
- (c)(v)Calculate the value of x in FeSO4•xH2O. x = ................................... [2][2]
- (d)Iron(II) sulfate in solution is readily oxidised by air to form iron(III) sulfate. State the effect, on the value of x calculated in (c)(v), if some of your sample of FA 1 had oxidised before you carried out the titration. Explain your answer. [2][2]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [1] The following data must be shown • burette readings and titre for rough titration • 2 × 2 'box' showing both accurate burette readings 'Correct' headings and units are not required for this mark
- [1] Headings and units correct for accurate titration table and headings match readings. • initial/start and (burette) reading / volume + unit (allow vol but not V) • final/end and (burette) reading / volume + unit (allow value for reading) • titre or volume/FA 2 and used/added (not 'difference' or 'total' or 'amount') + unit Units: (cm³) or/cm³ or in cm³ or cm³ by every entry
- [1] All accurate burette readings to 0.05 cm³ Do not award this mark if: 50(.00) is used as an initial burette reading; More than one final burette reading is 50(.00); Any burette reading is greater than 50(.00).
- [1] The final accurate titre recorded is within 0.10 cm³ of any other accurate titre. Do not award the mark if any 'accurate' burette readings (apart from initial 0) are given to zero dp.
- [1] Award V if δ < 0.50 cm³
- [1] Award VI if δ < 0.30 cm³
- [1] Award VII if δ < 0.20 cm³
- [1] Candidate must average two (or more) titres that are all within 0.20 cm³. Working must be shown or ticks must be put next to the two (or more) accurate titres selected.
- [1] Answers for (c)(ii), (c)(iii), (c)(iv) to 3–4 sf
- [1] Correctly calculates n(MnO4¯) = 0.02 × (b) / 1000
- [1] Correctly uses n(FeSO4) in 25 cm³ = (c)(ii) × 5 × 40
- [1] Correctly uses mass(FeSO4) in 1 dm³ = (c)(iii) × 151.9
- [1] Correctly uses n(H2O) = [26.52 – (c)(iv)] / 18
- [1] ratio n(H2O) : n((FeSO4) AND x given as integer OR M1: Mr hydrated salt = 26.52 / (c)(iii) M2: (26.52/(c)(iii) – 151.9) / 18 and answer as integer
- [1] (as) mole / amount / volume MnO4 smaller or (as) mass / moles / amount Fe2+ smaller
- [1] mass/moles/amount water larger so (mole) ratio larger so x increases Allow difference in titre/moles of MnO4¯/Fe2+ will be too small to change the (integer) value of x for 1 mark. Allow for 1 mark: less MnO4¯/Fe2+ and x increases
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