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A-LevelChemistryAnalytical techniques9701/52 · Feb/Mar 2020 · Q119 Marks

Brass is an alloy of copper and zinc. Typical copper concentrations vary from 50% to 85%, depending upon the properties needed in the alloy. There may be small amounts of other metals present. A student found a method to determine the percentage of copper in a sample of brass. A known mass of brass powder is reacted with excess concentrated nitric acid. Both the copper and the zinc and any other metals present are oxidised into aqueous ions by the nitric acid. The amount of Cu²⁺(aq) ions present can be determined by a titration technique. step 1 Use a weighing boat to accurately weigh by difference approximately 2 g of brass powder and place the brass into a small glass beaker. step 2 In a fume cupboard add approximately 20 cm³ of concentrated nitric acid to the brass in the beaker. Allow the brass to completely react to form solution A. The equation for the reaction is shown. Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l) step 3 Dilute all of solution A to form exactly 250.0 cm³ of solution B. step 4 Place 25.00 cm³ of solution B into a conical flask. step 5 Use a dropping pipette to add aqueous sodium carbonate, Na₂CO₃(aq), to solution B in the conical flask until there is no more acid present. step 6 Add approximately 20cm³ of aqueous potassium iodide, KI(aq), to the conical flask. A white precipitate forms as well as a brown solution of aqueous iodine, I₂(aq). step 7 Fill a burette with 0.100 moldm⁻³ sodium thiosulfate solution, Na₂S₂O₃(aq), so it is ready for the titration in step 8. step 8 Carry out a titration of the aqueous iodine produced in the conical flask against the 0.100 mol dm⁻³ Na₂S₂O₃(aq).

  1. (a)Outline how the student should accurately weigh by difference in step 1 in order that the exact mass of brass transferred into the small glass beaker is known. Include a results table, with appropriate headings, ready for the student to fill in.[2]
  2. (b)Suggest why it is necessary to do step 2 in a fume cupboard.[1]
  3. (c)Outline how the student should carry out step 3. Include the name and capacity of the suitable piece of apparatus in which solution B should be prepared.[2]
  4. (d)Name the apparatus needed to transfer solution B into the conical flask in step 4.[1]
  5. (e)State how the student would know there was no more acid present in the mixture in step 5.[1]
  6. (f)The student is given 200 cm³ of 0.100 moldm⁻³ Na₂S₂O₃(aq). Outline how the student should use this solution to fill the burette in step 7 so it is ready for titration. Include any relevant procedures the student should follow to ensure the burette is correctly filled before any readings are taken.[2]
  7. (g)(i)The titration table the student used is shown. Complete the table and calculate the mean titre to be used in calculating the percentage of copper in brass. Show your working. titration number | rough | 1 | 2 | 3 ---|---|---|---|--- final burette reading/cm³ | 20.50 | 40.25 | 19.90 | 39.65 initial burette reading/cm³ | 0.00 | 20.60 | 0.00 | 19.90 titre/cm³ | | | | mean titre = ...................................... cm³[2]
  8. (g)(ii)The burette used by the student has graduations of 0.10cm³. Determine the percentage error in the titre measured in titration number 2. Show your working. percentage error = ......................................[1]
  9. (g)(iii)Other than a change in apparatus, suggest one change to the experiment which would lead to a reduction in the percentage error in a measured titre.[1]
  10. (h)(i)Steps 1–8 were repeated, this time using 1.88 g of brass. The end-point of the titration was found to be 16.50 cm³. The equations for the reactions occurring are shown. equation 1 2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq) (step 6) equation 2 I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq) (step 8) Determine the number of moles of I₂ formed when excess KI(aq) was added to 25.00 cm³ of solution B in step 6. Use the data from the repeated experiment in your calculations. moles of I₂ = ......................................[2]
  11. (h)(ii)Use your answer to (h)(i) to determine the mass of Cu²⁺ ions in solution A and therefore the percentage by mass of copper in this sample of brass. If you were unable to obtain an answer to (h)(i), assume the number of moles of I₂ to be 8.85 × 10⁻⁴ mol. This is not the correct value. [Aᵣ: Cu, 63.5] percentage by mass of copper in the sample of brass = ......................................[3]
  12. (i)A small percentage of silver is sometimes found in some brass alloys. In step 2, when concentrated nitric acid is added, silver metal is oxidised to silver ions, Ag⁺(aq). At the end of step 6 the Ag⁺(aq) ions no longer remain in solution. Explain why.[1]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

  1. [2] (a) M1 Order of weighing • boat + brass is weighed • (brass transferred) • empty boat reweighed M2 table + units /g Mass of boat + brass before transfer Mass of boat after transfer Mass of brass transferred
  2. [1] (b) toxic / poisonous gas given off
  3. [2] (c) M1 transfer of solution A into a 250 cm³ volumetric flask and rinsing of beaker M2 top up to mark of (250 cm³) volumetric flask using distilled water
  4. [1] (d) (25 cm³) pipette
  5. [1] (e) no more effervescence is seen
  6. [2] (f) M1 rinse burette with Na2S2O3 M2 (idea of) run some Na2S2O3 through the tap / remove air from below tap
  7. [2] (g)(i) M1 all titres recorded to 2 dp: 20.50; 19.65; 19.90; 19.75 M2 (titres 1 and 3 averaged and) answer given as 19.7(0)
  8. [1] (g)(ii) (0.05×2)/19.90 ×100=0.503% working must be shown
  9. [1] (g)(iii) increase mass of brass OR decrease concentration of Na2S2O3(aq)
  10. [2] (h)(i) M1 mol of thio used = 0.1×16.50/1000 =1.65×10⁻³ mol M2 mol of I2 produced = 1.65×10⁻³/2 = 8.25×10⁻⁴ mol
  11. [3] (h)(ii) M1 mol of Cu2+ produced from brass in 25.00 cm³ = 8.25 × 10⁻⁴ × 2 = 1.65 × 10⁻³ mol M2 mass of Cu in brass in 250.0 cm³ = 1.65×10⁻³ x 250/25 x 63.5 = 1.04(775) g M3 percentage of Cu in brass = [1.04(775)/1.88] x100 = 55.7% OR M1 8.85 × 10⁻⁴ × 2 = 1.77 × 10⁻³ mol M2 M1 x 250/25 x 63.5 OR 1.77 × 10⁻² × 63.5 = 1.12(395) g M3 [M2/1.88] ×100 or [1.12(395)/1.88] ×100 = 59.8%
  12. [1] (i) (Ag+) react with I⁻/ iodide (ions) to form a precipitate / solid

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About this question

Topic
Analytical techniques, A-Level Chemistry (syllabus 9701). Structured question, 19 marks.
Source
A-Level Chemistry, Feb/Mar 2020 Paper 5 Variant 2, question 1.

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