A-LevelChemistryAnalytical techniques9701/32 · May/June 2022 · Q112 Marks
A bottle containing the acid salt sodium hydrogen sulfate, NaHSO4, has been contaminated. You will determine the percentage purity by mass of the sodium hydrogen sulfate by titrating a solution of the acid salt against a known concentration of sodium hydroxide. NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H₂O(I) The impurity in the sodium hydrogen sulfate does not react with aqueous sodium hydroxide under the conditions of the titration. FB 1 is 0.100 mol dm-³ sodium hydroxide, NaOH. FB 2 is 12.53 gdm⁻³ impure sodium hydrogen sulfate. FB 3 is thymol blue indicator.
- (a)Method • Fill a burette with FB 1. • Pipette 25.0 cm³ of FB 2 into a conical flask. • Add approximately 10 drops of FB 3. • Perform a rough titration and record your burette readings in the space below. The end-point is shown by the appearance of a permanent blue colour. The rough titre is ...................................... cm³. • Carry out as many titrations as you think necessary to obtain consistent results. • Make certain any recorded results show the precision of your practical work. • Record, in a suitable form below, all your burette readings and the volume of FB 1 added in each accurate titration. I II III IV V VI VII[7]
- (b)From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value. 25.0 cm³ of FB 2 required ................................................. cm³ of FB 1.[1]
- (c)(i)Give all your answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures.[1]
- (c)(ii)Use your answer to (b) to calculate the amount, in mol, of sodium hydroxide, FB 1, titrated. amount of NaOH = .............................................. mol Hence, deduce the amount, in mol, of sodium hydrogen sulfate present in 25.0 cm³ of FB 2. amount of NaHSO4 = .............................................. mol[1]
- (c)(iii)Use your final answer to (c)(ii) to calculate the mass of sodium hydrogen sulfate present in 1.00 dm³ of FB 2. mass of NaHSO4 = .............................................. g[1]
- (c)(iv)Use your answer to (c)(iii) and the information on page 2 to calculate the percentage purity by mass of the sodium hydrogen sulfate. percentage purity = .............................................. %[1]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- I All the following data are recorded: • two burette readings and titre for the rough titration • initial and final burette readings for two (or more) accurate titrations
- II Correct headings and units in the accurate titration table and titre values recorded for accurate titrations • initial / start and (burette) reading / volume + unit • final / end and (burette) reading / volume + unit • titre + unit or volume / FB 1 and used / added + unit
- III All accurate burette readings are recorded to the nearest 0.05 cm³
- IV The final accurate titre recorded is within 0.10 cm³ of any other accurate titre
- For assessment of accuracy marks, round all burette readings to the nearest 0.05 cm³. Check 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. Write the candidate's mean titre on each script. Write the Supervisor's [corrected] mean titre in a ring on each candidate script. Calculate the difference (δ) between the candidate's mean titre and the supervisor's mean titre. Write the value of δ on each script. Award the accuracy marks as shown below. Award V if δ ≤ 0.50 (cm³) Award VI if δ < 0.30 Award VII if δ < 0.20
- [1] Correctly calculates the mean titre correct to 2 dp from accurate titres that are within 0.20 cm³ total spread
- [1] All final answers for (c)(ii), (c)(iii), and (c)(iv) are to 3–4 sf
- [1] Correctly calculates n(NaOH) = 0.1 × (b) / 1000 AND n(NaHSO4) = same answer
- [1] Correctly uses [NaHSO4] = final answer (c)(ii) × 1000 / 25 (= (c)(ii) × 40) AND answer above × 120.1
- [1] Correctly uses % NaHSO4 = (final answer (c)(ii) / 12.53) × 100
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