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A-LevelChemistryAtoms, molecules and stoichiometry9701/51 · Oct/Nov 2019 · Q113 Marks

Yttrium barium copper oxide, YBa₂Cu3O7, is a crystalline compound. You are to design an experiment in which YBa₂Cu3O7 is first synthesised and then analysed by titration.

  1. (a)(i)YBa₂Cu3O7 can be synthesised by reacting Y2O3, BaCO3 and CuO using the following method. • Place solid Y2O3, BaCO3 and CuO together in a mortar and grind the mixture well with a pestle. • Transfer the mixture to a porcelain crucible and place this in an oven set at 920°C. • Heat the mixture for 12 hours, then allow the crucible and its contents to cool slowly in the oven to below 100°C before removing it. The equation for the reaction is given. Y2O3 + 4BaCO3 + 6CuO + ½O2 → 2YBa₂Cu3O7 + 4CO2 YBa₂Cu3O7 contains Y, Ba and Cu in the molar ratio of 1:2:3. Calculate the minimum masses of BaCO3 and CuO that are needed to react with 0.750g of Y2O3, to give a Y: Ba: Cu ratio of 1:2:3. [Ar: Y, 88.9; Ba, 137.3; Cu, 63.5; O, 16.0; C, 12.0] mass of BaCO3 = __________ g mass of CuO = __________ g[3]
  2. (a)(ii)State what should be done once the solid product has cooled to ensure that the highest possible yield of YBa₂Cu3O7 has been produced.[1]
  3. (b)(i)YBa₂Cu3O7 contains some copper ions in the unusual +3 oxidation state. The proportion of Cu³+ in YBa₂Cu3O7 can be determined by titration. • Step 1 A sample of YBa2Cu3O7 is reacted with an excess of concentrated aqueous HBr. Cu³+ ions are reduced to Cu2+ ions and Br3¯ ions are formed. 2Cu³+(s) + 3Br¯(aq) → 2Cu²+(aq) + Br3¯(aq) • Step 2 A solution of 1.0 moldm⁻³ sodium citrate is added to the mixture from Step 1. The resulting mixture is then neutralised with a minimum volume of concentrated NH3(aq). • Step 3 Excess I⁻ is added which reacts with Br3¯ to form I2. Br3⁻(aq) + 2I⁻(aq) → 3Br⁻(aq) + I2(aq) • Step 4 The I₂ is titrated with a standard solution of S₂O₃²⁻ and starch solution as an indicator. 2S2O3²⁻(aq) + I2(aq) → S4O6²⁻(aq) + 2I⁻(aq) The concentration of I2(aq) can therefore be determined and hence the concentration of Br3¯(aq). From this the amount of Cu³+(s) can be determined. The table gives some electrochemical data. reduction process | E°/V ---|--- I₂ + 2e⁻ ⇌ 2I⁻ | +0.54 Cu²⁺ + I⁻ + e⁻ ⇌ CuI | +0.86 O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O | +1.23 Use these data and the information given above to answer the following questions. The citrate anion forms an insoluble complex with Cu2+ and so removes Cu2+ from solution. Explain why this is necessary.[1]
  4. (b)(ii)Explain why it is necessary to neutralise the mixture in Step 2.[1]
  5. (b)(iii)When starch indicator is added in Step 4, the mixture turns blue-black due to the presence of I2(aq). The end-point of the titration with S2O3²⁻(aq) is a colourless solution. The number of moles of S₂O₃²⁻(aq) needed for complete reaction with I₂(aq) can be calculated from the mean titre value. Hence the moles of I₂(aq) can be determined. State the expression for the moles of Cu³+ in the sample of YBa₂Cu3O7. Use A to represent the number of moles of I₂(aq) in Step 4. moles Cu³+ = __________ mol[1]
  6. (c)(i)Calculate the mass of hydrated sodium citrate, Na3C6H5O7•2H₂O, that would be required for the preparation of 250.0 cm³ of a solution of 1.0 moldm⁻³ citrate ions, C6H5O7³⁻. [Mr: Na3C6H5O7•2H₂O, 294.0] mass of Na3C6H5O7•2H2O = __________ g[1]
  7. (c)(ii)A student places the mass of Na3C6H5O7•2H2O calculated in (c)(i) into a beaker. Describe how the student can prepare exactly 250.0 cm³ of a solution of 1.0 mol dm⁻³ citrate ions from the sample in the beaker. Give the name and capacity, in cm³, of any apparatus used.[3]
  8. (d)A different student records the following titration data in Step 4. experiment | rough | 1 | 2 ---|---|---|--- final reading/cm³ | 21.20 | 24.60 | 47.75 initial reading/cm³ | 0.00 | 3.10 | 25.30 titre/cm³ | 21.20 | 21.50 | 22.45 Identify the problem with the student's titration method and suggest how it could be improved.[2]

Mark scheme

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

  1. [3] M1: moles of Y2O3 = 0.750 ÷ (2 × 88.9 + 3 × 16.0) = 0.750 ÷ 225.8 = 3.32 × 10⁻³ (mol) 3.3215235 × 10⁻³ M2: mass of BaCO3 = 4 x 3.32 × 10⁻³ × (137.3 + 12.0 + 3 × 16.0) = 4 x 3.32 × 10⁻³ × 197.3 = 2.62 (g) 2.6213463 M3: mass of CuO = 6 x 3.32 × 10⁻³ × (63.5 + 16.0) = 6 x 3.32 × 10⁻³ × 79.5 = 1.58 (g) 1.5813667
  2. [1] heat solid again (and allow to cool) AND (to) constant mass
  3. [1] (Prevents) reaction of Cu2+ with I- OR (prevents) formation of CuI / Cu+ / copper(I) OR (prevents) oxidation of I- (to I2) by Cu2+
  4. [1] I- is oxidised (to I2) in acidic solution
  5. [1] 2A
  6. [1] 1.0 × 250.0/1000 × 294.0 = 73.5 (g)
  7. [3] M1: Dissolve / make a solution in (beaker) in (small volume of distilled water) M2: Add / transfer solution to a 250 cm³ volumetric flask M3: Make to mark of (volumetric) flask with distilled water and the washings
  8. [2] M1: titres are not concordant M2: repeat titration until concordant titres are obtained OR improved valid experimental technique

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About this question

Topic
Atoms, molecules and stoichiometry, A-Level Chemistry (syllabus 9701). Structured question, 13 marks.
Source
A-Level Chemistry, Oct/Nov 2019 Paper 5 Variant 1, question 1.

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