OxfordAQA A-Level Chemistry: Reactions of ions in aqueous solution (9620) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover subsection 3.2.6.1 (the only subsection of the aqueous-ions topic) in OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for exams from May/June 2020. All of it is International A2 content, so it is International A-level only. Unit 3 assesses 3.2.4 to 3.2.6, and Unit 5 can draw on any content. Questions 9 and 10 draw on required practical 9; question 3 is synoptic with Ka from section 3.1.10, and question 8 uses pV = nRT from section 3.1.2.

Question 9 also uses the sulfate test from Group 2 and the ammonium test from required practical 3. Course links: hub, printable checklist, free diagnostics.

Questions

1. Ammonium iron(II) sulfate and aluminium nitrate are dissolved separately in water.

(a) For each, give the formula of the metal-aqua ion formed and the colour of the solution. [2] (b) A week later the iron solution, left open to the air, has turned yellow-brown. Name the iron species now present and explain the change. [2]

2. A student writes: “Copper(II) nitrate solution is more acidic than aluminium nitrate solution of the same concentration, because Cu²⁺ is the larger ion.”

Write an equation to show why aluminium nitrate solution is acidic, then correct the student’s statement in terms of charge/size ratio. [4]

3. A 0.200 mol dm⁻³ solution of aluminium nitrate has a pH of 2.85. Treat [Al(H₂O)₆]³⁺ as a weak monoprotic acid and use the usual weak-acid approximation. Calculate Ka for the aqua ion, giving units. [3]

4. State what you would see in each case.

(a) Aqueous ammonia is added in excess to iron(III) nitrate solution. [1] (b) Sodium carbonate solution is added to ammonium iron(II) sulfate solution. [1] (c) Sodium hydroxide solution is added dropwise, then in excess, to copper(II) nitrate solution. [1] (d) Aqueous ammonia is added dropwise, then in excess, to aluminium nitrate solution. [1]

5. This question is about ammonia.

(a) Give a balanced equation for aqueous ammonia reacting with [Fe(H₂O)₆]²⁺, and state the role of the ammonia. [2] (b) Write an equation for the reaction of copper(II) hydroxide, [Cu(H₂O)₄(OH)₂], with excess ammonia, and give the co-ordination number of copper in the product. [2]

6. 20.0 cm³ of a solution containing 0.0800 mol dm⁻³ [Al(H₂O)₆]³⁺ is titrated with 0.400 mol dm⁻³ sodium hydroxide solution.

(a) Calculate the volume of NaOH(aq) added when all the aluminium has just been precipitated. [2] (b) Calculate the total volume of NaOH(aq) added when the precipitate has just redissolved. [1] (c) Write an equation for the redissolving step. [1] (d) Explain why the precipitate would not redissolve if 0.400 mol dm⁻³ aqueous ammonia were used instead. [1]

7. Sodium carbonate solution is added to separate solutions of copper(II) nitrate and aluminium nitrate. For each solution, give the observations and a balanced equation. Then explain why one product is a carbonate and the other is not. [5]

8. Excess sodium carbonate solution is added to 25.0 cm³ of a solution containing [Al(H₂O)₆]³⁺. The carbon dioxide released has a volume of 120 cm³ at 100 kPa and 300 K. (R = 8.31 J K⁻¹ mol⁻¹; Ar: Al = 27.0, O = 16.0, H = 1.0)

(a) Calculate the amount, in moles, of CO₂ released. [1] (b) Calculate the mass of [Al(H₂O)₃(OH)₃] precipitated. [3] (c) Calculate the concentration of Al³⁺ in the original solution. [1]

9. A pale solid, Z, dissolves in water to give a yellow-brown solution that turns blue litmus red. Z is a double salt containing one type of transition metal ion.

(a) A sample of the solution is warmed with excess NaOH(aq). An orange-brown precipitate forms, and the gas released makes damp red litmus paper go blue. Identify the two cations shown, giving the evidence for each. [3] (b) Another sample is acidified and barium chloride solution is added. A white precipitate forms. Identify the anion. [1] (c) Z contains these three ions only. Use charge balance to deduce the simplest formula of anhydrous Z, showing your reasoning. [2] (d) Write an equation for the formation of the orange-brown precipitate in (a). [1] (e) Explain, with an equation, why the solution of Z is acidic. [2]

10. A solution contains either Fe²⁺(aq) or Fe³⁺(aq) at about 0.1 mol dm⁻³; under yellowish lab lighting its colour cannot be judged. Three students suggest adding (i) NaOH(aq), (ii) NH₃(aq) or (iii) Na₂CO₃(aq) to separate samples.

Evaluate each reagent as a way of deciding which iron ion is present. For each, give the expected result for both ions, explain any difference, and state which reagent gives the clearest single observation. [7]

Answers

1. (a) [Fe(H₂O)₆]²⁺, pale green [1]; [Al(H₂O)₆]³⁺, colourless [1] (b) [Fe(H₂O)₆]³⁺ (iron(III)) [1]; oxygen in the air has oxidised iron(II) to iron(III) [1] [4] Examiner insight: Write “colourless” for the aluminium solution, not “clear”, and give each aqua ion with its charge.

2. [Al(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H₃O⁺(aq) [1]. The statement is wrong: the aluminium solution is the more acidic, and a larger ion is less polarising, not more [1]. Al³⁺ has a higher charge and a smaller radius, so a greater charge/size ratio and greater polarising power [1]. It weakens the O–H bonds in its water ligands more, so H⁺ is released more readily and the aluminium solution has the lower pH [1] [4] Examiner insight: Fix both halves of the claim, which solution is more acidic and what ion size does, then link the charge/size ratio to weaker O–H bonds in the water ligands.

3. [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol dm⁻³ [1]. Ka = [H⁺]²/[HA] = (1.41 × 10⁻³)² / 0.200 [1] = 1.0 × 10⁻⁵ mol dm⁻³ [1] [3] Examiner insight: Always give units with Ka when they are asked for; with pH to two decimal places, two significant figures in Ka is the sensible precision.

4. (a) Orange-brown precipitate, which does not dissolve in excess [1] (b) Green precipitate, with no effervescence [1] (c) Blue precipitate, which does not dissolve in excess [1] (d) White precipitate, which does not dissolve in excess [1] [4] Examiner insight: Where the question says “dropwise, then in excess”, the mark needs both the precipitate and what happens in excess; a single word such as “blue” is not enough.

5. (a) [Fe(H₂O)₆]²⁺(aq) + 2NH₃(aq) → Fe(H₂O)₄(OH)₂ + 2NH₄⁺(aq) [1]; ammonia acts as a Brønsted–Lowry base (proton acceptor) [1] (b) Cu(H₂O)₄(OH)₂ + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O(l) + 2OH⁻(aq) [1]; co-ordination number 6 [1] [4] Examiner insight: In (b) the answer of 4 confuses the number of ammonia ligands with the co-ordination number; the two water ligands still count.

6. (a) n(Al³⁺) = 0.0800 × 20.0/1000 = 1.60 × 10⁻³ mol; n(OH⁻) = 3 × 1.60 × 10⁻³ = 4.80 × 10⁻³ mol [1]. V = 4.80 × 10⁻³ / 0.400 = 0.0120 dm³ = 12.0 cm³ [1] (b) n(OH⁻) = 4 × 1.60 × 10⁻³ = 6.40 × 10⁻³ mol, so V = 16.0 cm³ [1] (c) Al(H₂O)₃(OH)₃ + OH⁻(aq) → [Al(H₂O)₂(OH)₄]⁻(aq) + H₂O(l) [1] (d) Ammonia is a weak base; its OH⁻ concentration is too low to remove a fourth proton [1] [5] Examiner insight: Show the amount in moles at each stage so your method is visible, and check that (b) uses the 1 : 4 ratio rather than adding 1 : 3 twice.

7. Copper: blue-green precipitate, no effervescence [1]; [Cu(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → CuCO₃(s) + 6H₂O(l) [1]. Aluminium: white precipitate with effervescence [1]; 2[Al(H₂O)₆]³⁺(aq) + 3CO₃²⁻(aq) → 2Al(H₂O)₃(OH)₃ + 3CO₂(g) + 3H₂O(l) [1]. The 3+ aqua ion is acidic enough (higher charge/size ratio) to protonate carbonate, giving CO₂, so a hydroxide forms; the 2+ ion is not acidic enough, so the metal carbonate precipitates [1] [5] Examiner insight: An equation producing Al₂(CO₃)₃ is a chemical error and gains no credit, however well balanced it is.

8. (a) n(CO₂) = pV/RT = (1.00 × 10⁵ × 1.20 × 10⁻⁴)/(8.31 × 300) = 4.81 × 10⁻³ mol [1] (b) n(Al) = 4.81 × 10⁻³ × 2/3 = 3.21 × 10⁻³ mol [1]. Mr of [Al(H₂O)₃(OH)₃] = 27.0 + 3(18.0) + 3(17.0) = 132.0 [1]. Mass = 3.21 × 10⁻³ × 132.0 = 0.424 g [1] (c) [Al³⁺] = 3.21 × 10⁻³ / 0.0250 = 0.128 mol dm⁻³ [1] [5] Examiner insight: The volume must be in m³ (1.20 × 10⁻⁴) when the pressure is in Pa; leaving it as 120 gives an answer a million times too big.

9. (a) Fe³⁺: orange-brown precipitate with NaOH(aq) [1]. NH₄⁺: on warming with NaOH a gas turns damp red litmus blue, which is ammonia [1]. Both conclusions stated with the matching evidence [1] (b) Sulfate, SO₄²⁻ [1] (c) Positive charge from one NH₄⁺ and one Fe³⁺ is 1 + 3 = 4, so two SO₄²⁻ ions are needed [1]: NH₄Fe(SO₄)₂ [1] (d) [Fe(H₂O)₆]³⁺(aq) + 3OH⁻(aq) → Fe(H₂O)₃(OH)₃ + 3H₂O(l) [1] (e) [Fe(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]²⁺(aq) + H₃O⁺(aq) [1]; the small, highly charged Fe³⁺ ion polarises the water ligands, weakening O–H bonds so that H⁺ is released [1] [9] Examiner insight: In (a) the identification marks need the observation that supports them; naming both ions without saying which result shows which gains less credit.

10. NaOH(aq): Fe²⁺ gives a green precipitate, Fe³⁺ an orange-brown one [1]; but this relies on judging colour under poor lighting, and the green precipitate browns on standing in air, so the two may be confused [1]. NH₃(aq): the same two precipitate colours as with NaOH [1], so it has the same weakness and no advantage [1]. Na₂CO₃(aq): Fe²⁺ gives a green precipitate with no gas; Fe³⁺ gives a precipitate with effervescence [1]. The difference arises because [Fe(H₂O)₆]³⁺ is acidic enough to protonate carbonate, releasing CO₂, while [Fe(H₂O)₆]²⁺ is not [1]. Sodium carbonate gives the clearest single observation, because gas or no gas does not depend on judging a colour [1] [7] Examiner insight: “Evaluate” needs a judgement as well as results; an answer that lists observations for all three reagents but never says which is best cannot gain the final mark.

Where marks are usually lost

• Quoting only the end result in excess and missing the precipitate seen first.
• Writing a metal(III) carbonate as a product or forgetting the effervescence with 3+ ions.
• Using 1 : 3 instead of 1 : 4 between Al³⁺ and OH⁻ when the precipitate has fully redissolved.

Next steps

• Go back over the ion profiles in the revision notes.
• Use the worked examples in the study guide for any question you missed.
• Revisit ligands and colour in transition metals.
• See the OxfordAQA A-Level Chemistry hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. Section 3.2.6 (International A2), the topic on reactions of ions in aqueous solution, subsection 3.2.6.1, including required practical 9.