AQA Chemistry 8462 · Paper 2 · 4.8.3.1, 4.8.3.2, 4.8.3.3, 4.8.3.4, 4.8.3.5, 4.8.3.6, 4.8.3.7
Identifying ions Both tiers Foundation Higher 4 marks Foundation and Higher Chemistry required practical 7 4.8.3.2 Metal hydroxides
Sodium hydroxide solution is added to solutions of metal ions. Some form coloured precipitates.
.1 A solution gives a blue precipitate. Which ion is present?
[1 mark] A. Magnesium, M g 2 + \mathrm{Mg^{2+}} M g 2 + B. Iron(II), F e 2 + \mathrm{Fe^{2+}} F e 2 + C. Copper(II), C u 2 + \mathrm{Cu^{2+}} C u 2 + D. Iron(III), F e 3 + \mathrm{Fe^{3+}} F e 3 + .2 Write the formula of the blue precipitate.
[1 mark] .3 Iron(III) ions form a brown precipitate of iron(III) hydroxide. Write a balanced ionic equation for this reaction.
[2 marks] 3 marks Foundation and Higher Chemistry required practical 7 4.8.3.2 Metal hydroxides
Solutions of aluminium ions, calcium ions and magnesium ions all form a white precipitate when a few drops of sodium hydroxide solution are added.
.1 Describe how a student could use sodium hydroxide solution to tell aluminium ions apart from magnesium ions.
[2 marks] .2 Another solution gives a brown precipitate with sodium hydroxide. Which ion is present?
[1 mark] A. Calcium B. Iron(III) C. Iron(II) D. Copper(II)
3 marks Foundation and Higher Chemistry required practical 7 4.8.3.3 Carbonates 4.8.3.5 Sulfates
A student has a sample of an unknown white solid. It is a single ionic compound.
.1 When dilute hydrochloric acid is added the solid fizzes, and the gas turns limewater milky. Which ion does the solid contain?
[1 mark] A. Chloride B. Carbonate C. Sulfate D. Hydroxide
.2 Describe how the student could test a solution of a different compound for sulfate ions. Give the result if sulfate ions are present.
[2 marks] 4 marks Higher Chemistry required practical 7 4.8.3.5 Sulfates
A student adds dilute hydrochloric acid and then excess barium chloride solution to 50.0 cm³ of a sulfate solution. The white precipitate of barium sulfate is filtered, washed, dried and weighed.
B a 2 + ( a q ) + S O 4 2 − ( a q ) → B a S O 4 ( s ) \mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)} B a 2 + ( aq ) + S O 4 2 − ( aq ) → BaS O 4 ( s )
Relative formula mass
(M r M_r M r of
B a S O 4 \mathrm{BaSO_4} BaS O 4 = 233.
.1 The dry precipitate has a mass of 2.33 g. Calculate the amount, in moles, of sulfate ions in the sample.
[1 mark] .2 Calculate the concentration of sulfate ions in the solution in mol/dm³.
[2 marks] .3 Explain why the barium chloride solution is added in excess.
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