A-LevelChemistryAnalytical techniques9701/33 · Oct/Nov 2019 · Q113 Marks

In this experiment you will determine the concentration of a solution of hydrochloric acid by titration with an alkali. FA 1 is a solution containing 6.00 g dm⁻³ of sodium hydroxide, NaOH. FA 2 is hydrochloric acid, HCl. (This solution is also used in Questions 2 and 3.) methyl orange indicator

(a)Method Dilution of FA 2 • Pipette 10.0 cm³ of FA 2 into the 250 cm³ volumetric flask. • Make the solution up to the mark using distilled water. • Shake the solution in the volumetric flask thoroughly. • This solution of hydrochloric acid is FA 3. Label the volumetric flask FA 3. Titration • Fill the burette with FA 1. • Pipette 25.0 cm³ of FA 3 into a conical flask. • Add several drops of methyl orange indicator. • Perform 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 precision of your practical work. • Record in a suitable form below all of your burette readings and the volume of FA 1 added in each accurate titration. I II III IV V VI VII[7]
(b)From your accurate titration results, obtain a suitable value for the volume of FA 1 to be used in your calculations. Show clearly how you obtained this value. 25.0 cm³ of FA 3 required ...................................... cm³ of FA 1. [1][1]
(c)(i)Give your answers to (ii), (iii) and (iv) to the appropriate number of significant figures. [1][1]
(c)(ii)Calculate the number of moles of sodium hydroxide, NaOH, in the volume of FA 1 calculated in (b). moles of NaOH = ...................................... mol [1][1]
(c)(iii)Write the equation for the neutralisation of hydrochloric acid with sodium hydroxide. Include state symbols. Deduce the number of moles of hydrochloric acid that reacted with the sodium hydroxide in (ii). moles of hydrochloric acid =...................................... mol [1][1]
(c)(iv)Calculate the concentration, in moldm⁻³, of hydrochloric acid in FA 2. concentration of HCl in FA 2 = ...................................... moldm⁻³ [2][2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] I All the following data recorded • two burette readings and titre for the rough titration • initial and final burette readings for two (or more) accurate titrations
[1] II Titre values shown, for accurate titrations, and appropriate headings and units in the accurate titration table • initial / start and (burette) reading / volume • final / end and (burette) reading / volume • titre or volume / FB 4 and used / added • unit: / cm³ or (cm³) or in cm³ (for each heading) or cm³ unit given for each volume recorded
[1] III All accurate burette readings are recorded to the nearest .05 cm³.
[1] IV The final accurate titre recorded is within 0.10 cm³ of any other accurate titre.
[1] Award V if δ ≤ 0.80 (cm³) (Where δ is difference to the supervisor's value)
[1] Award VI if δ < 0.50 (cm³)
[1] Award VII if δ < 0.30 (cm³)
[1] Correctly calculates mean titre from two (or more) accurate titres where the total spread is ≤ 0.20 cm³. AND Answer is given to 2 d.p. AND Working must be shown or ticks must be put next to the two (or more) accurate titres selected.
[1] Significant figures All answers in (ii)-(iv) are expressed to 3 or 4 sig. fig.
[1] Correctly calculates no of moles of NaOH used. No. of moles NaOH = 6.00/40 × answer (b)/1000
[1] NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(I) or H+(aq) + OH-(aq) → H2O(I) AND No. of moles of HCl = same as in moles of NaOH in (ii)
[1] Correct expression shown (for the first step) (Two steps) Concentration HCl in FA 3 = ans (c)(iii) × 1000/25 Allow ans (c)(iii) × 40 (Concentration of FA 2 = Concentration of FA 3 × 25) OR moles of FA 3 in 250 cm³ = ans (c)(iii) × 250/25 Allow ans (c)(iii) × 10 (moles of FA 2) in 1 dm³ = ans above × 1000/10)
[1] Answer = (c)(iii) × 1000