A-LevelChemistryAnalytical techniques9701/52 · Oct/Nov 2024 · Q116 Marks

Solution X is dilute hydrochloric acid of unknown concentration. A student uses titration with aqueous sodium carbonate, Na2CO3(aq), to determine the concentration of solution X. The student plans to prepare 250.0 cm³ of 0.500 moldm-3 aqueous sodium carbonate, Na2CO3(aq).

(a)(i)Calculate the mass of Na2CO3(s) needed to make 250.0 cm³ of 0.500 moldm-3 Na2CO3(aq), using a two decimal place balance. mass of Na2CO3(s) = .............................................. g [1][1]
(a)(ii)Describe how the student should make 250.0 cm³ of 0.500 moldm-3 Na2CO3(aq) starting from the mass of Na2CO3(s) calculated in (i) supplied in a 50 cm³ beaker. Give the name and size of any key apparatus to be used. Write your answer using a series of numbered steps. [3][3]
(b)(i)In step 1, 25.0 cm³ of solution Y is transferred using a volumetric pipette. State what the volumetric pipette should be rinsed with before carrying out step 1. [1][1]
(b)(ii)Describe how the student detects the end-point of the titration in step 2. [1][1]
(b)(iii)Explain why the evaporating basin is not heated directly with a Bunsen burner in step 7. [1][1]
(b)(iv)Suggest what the student should do to ensure all of the water has evaporated from the residue before completing step 8. [1][1]
(c)(i)Use Fig. 1.1 to complete Table 1.1. Table 1.1 burette reading (final)/cm³ burette reading (initial)/cm³ volume of solution X added/cm³ [2][2]
(c)(ii)Calculate the percentage error in the volume of solution X calculated in (i). Show your working. percentage error = ......................................... [1][1]
(d)A second student repeats the experiment in (b) using solution X and 25.0 cm³ of solution Y. The results are shown in Table 1.2. Table 1.2 volume of solution X added in titration/cm³ 13.35 mass of dry evaporating basin/g 44.52 mass of evaporating basin and solid residue/g 45.69 The reaction that takes place in the titration is: 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) Using the results in Table 1.2, calculate the following: • the amount, in mol, of sodium chloride amount of sodium chloride = .............................................. mol • the concentration of solution X and the concentration of solution Y. concentration of solution X = .............................................. mol dm-3 concentration of solution Y = .............................................. moldm-3 [3][3]
(e)State what happens to the value obtained for the concentration of solution Y if not all the water is evaporated in step 7. Explain your answer. [2][2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

[1] 13.25 g
[1] M1 dissolve the solid / sodium carbonate (in the beaker) (using a small volume of) (distilled) water
[1] M2 transfer / add to a (250 cm³ volumetric) flask and rinse (with distilled water)
[1] M3 top the 250 cm³ volumetric flask up to mark (with distilled water)
[1] solution Y
[1] indicator / methyl orange changes colour
[1] to avoid spitting (of solid when it forms)
[1] heat to constant mass
[1] M1 burette readings 14.85 (cm³) and 1.35 (cm³)
[1] M2 volume of solution X added = 13.50 (cm³)
[1] working must be shown 2×0.05 / 13.50 × 100 = 0.741
[1] M1 Mr(NaCl) = 58.5 n(NaCl) = 1.17 g / 58.5 = 0.0200
[1] M2 n(HCl) = 0.02 c(HCl) (X) = 0.02/0.01335 = 1.50 (mol dm-³)
[1] M3 n(Na2CO3) = 0.02 / 2 = 0.01 c(Na2CO3) (Y) = 0.01 / 0.025 = 0.400 (mol dm-³)
[1] M1 calculated concentration of Y will be higher than the true value.
[1] M2 mass / amount of NaCl (in step 7) will be too high.