Question 116 marks
Question
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(l) FA 1 contains 26.52 g dm–3 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. I II III IV V VI VII [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] (c) Calculations (i) Give your answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures. [1] (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] (iii) Calculate the number of moles of iron(II) sulfate present in 1.00 dm3 of FA 1. moles of FeSO4 = …… mol [1] (iv) Calculate the mass of iron(II) sulfate present in 1.00 dm3 of FA 1. mass of FeSO4 = …… g [1] (v) Calculate the value of x in FeSO4•xH2O. x = …… [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] [Total: 16] Question 2 starts on the next page.
Mark scheme
1(a) I The following data must be shown 1 • 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 II Headings and units correct for accurate titration table and headings match readings. 1 • 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 III All accurate burette readings to 0.05 cm3 1 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). IV The final accurate titre recorded is within 0.10 cm3 of any other accurate titre. 1 Do not award the mark if any ‘accurate’ burette readings (apart from initial 0) are given to zero dp. For assessment of accuracy (Q) marks, the Examiner should round any burette readings to the nearest 0.05 cm3. Check 1 and correct subtractions. Then select the ‘best’ titres using the hierarchy: • two (or more) accurate identical titres (ignoring any that are labelled ‘rough’), then • two (or more) accurate titres within 0.05 cm3, then • two (or more) accurate titres within 0.10 cm3, etc. These best titres should be used to calculate the mean titre, expressed to nearest 0.01 cm3. Calculate the difference (δ) between the candidate’s titre and the supervisor’s titre. Award the accuracy (Q) marks as shown below. Award V if δ ⩽ 0.50 cm3 1 Award VI if δ ⩽ 0.30 cm3 1 Award VII if δ ⩽ 0.20 cm3 1(b) Candidate must average two (or more) titres that are all within 0.20 cm3. 1 Working must be shown or ticks must be put next to the two (or more) accurate titres selected. 1(c)(i) Answers for (c)(ii), (c)(iii), (c)(iv) to 3–4 sf 1 1(c)(ii) Correctly calculates 1 n(MnO4−) = 0.02 × (b) / 1000 1(c)(iii) Correctly uses 1 n(FeSO4) in 25 cm3 = (c)(ii) × 5 × 40 1(c)(iv) Correctly uses 1 mass(FeSO4) in 1 dm3 = (c)(iii) × 151.9 1(c)(v) Correctly uses 1 n(H2O) = [26.52 – (c)(iv)] / 18 ratio n(H2O) : n((FeSO4) 1 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(d) (as) mole / amount / volume MnO4− smaller 1 or (as) mass / moles / amount Fe2+ smaller mass/moles/amount water larger so (mole) ratio larger so x increases 1 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