AQA A-Level Chemistry Paper 3, June 2018: Question 31
1 mark · Medium difficulty · Multiple Choice
Identify which pair of equimolar solutions produces the greatest mass of solid precipitate when equal volumes are mixed.
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Question text
31 Some 1.0 mol dm–3 solutions were mixed using equal volumes of each solution.
Which pair of solutions would give the greatest mass of solid?
[1 mark]
A Ba(OH)2 and MgCl2
B Ba(OH)2 and MgSO4
C Ba(OH)2 and NaCl
D Ba(OH)2 and Na2SO4
Mark scheme
Show the mark scheme
31 B
How to answer it
Precipitation and Group 2 Solubility Trends
- Group 2 Solubility Trends: Hydroxide solubility increases down Group 2 (Mg(OH)₂ is sparingly soluble/insoluble; Ba(OH)₂ is soluble).
- Sulfate Solubility Trends: Sulfate solubility decreases down Group 2 (MgSO₄ is soluble; BaSO₄ is completely insoluble).
- Stoichiometry & Mass of Precipitates: Deducing the total mass of solid products formed from equimolar mixtures of ionic compounds.
Question 31 Analysis
Equimolar mixtures of Group 2 and ionic solutions
✅ Correct Answer: B
Ba(OH)₂ and MgSO₄ produce two separate insoluble precipitates:
Ba(OH)₂(aq) + MgSO₄(aq) → BaSO₄(s) + Mg(OH)₂(s)
Because both products are insoluble solids, the total mass of solid produced is the sum of both: 1 mol of BaSO₄ (233.4 g) plus 1 mol of Mg(OH)₂ (58.3 g) = 291.7 g.
💡 Key Knowledge: Group 2 Solubility
- Hydroxides: Solubility increases down the group.
• Mg(OH)₂ → Sparingly soluble (forms a white solid precipitate)
• Ba(OH)₂ → Soluble - Sulfates: Solubility decreases down the group.
• MgSO₄ → Soluble
• BaSO₄ → Completely insoluble (used in barium meals / sulfate tests) - Group 1 salts and chlorides: NaCl and BaCl₂ are highly soluble.
📐 Step-by-Step Comparison (Assuming 1 dm³ of each 1.0 mol dm⁻³ solution)
Equal volumes of 1.0 mol dm⁻³ mean exactly 1.0 mol of each reactant is available.
- Option A: Ba(OH)₂ + MgCl₂
Reaction: Ba(OH)₂(aq) + MgCl₂(aq) → Mg(OH)₂(s) + BaCl₂(aq)
Solid formed: 1.0 mol of Mg(OH)₂
Mass = 1.0 × (24.3 + 34.0) = 58.3 g - Option B: Ba(OH)₂ + MgSO₄
Reaction: Ba(OH)₂(aq) + MgSO₄(aq) → BaSO₄(s) + Mg(OH)₂(s)
Solid formed: 1.0 mol BaSO₄ + 1.0 mol Mg(OH)₂
Mass = (137.3 + 32.1 + 64.0) + (24.3 + 34.0) = 233.4 g + 58.3 g = 291.7 g - Option C: Ba(OH)₂ + NaCl
Reaction: No insoluble combination (BaCl₂ and NaOH are both soluble).
Solid formed: 0 g - Option D: Ba(OH)₂ + Na₂SO₄
Reaction: Ba(OH)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaOH(aq)
Solid formed: 1.0 mol BaSO₄
Mass = 1.0 × 233.4 = 233.4 g
❌ Common Errors & Pitfalls
- Picking Option D: Students often recognize BaSO₄ as a heavy insoluble solid (Mr = 233.4) and choose D without noticing that in B, Mg(OH)₂ also precipitates.
- Mixing up solubility trends: Confusing the trends for Group 2 sulfates and hydroxides (e.g., thinking MgSO₄ is insoluble or Ba(OH)₂ is insoluble).
- Assuming only one precipitate forms: Overlooking that both cation-anion swap products in a double displacement can be insoluble at the same time.
🧠 Exam Technique: MCQ Elimination
- Quick Rule: When asked for the "greatest mass of solid", immediately check if any reaction produces two precipitates instead of just one.
- Check spectator ions: Na⁺ and Cl⁻ ions almost always remain dissolved in aqueous solution, so combinations involving them (C and D) will never produce more than one precipitate.
Topics
Inorganic Chemistry · Physical Chemistry · 3.2.2 Group 2, The Alkaline Earth Metals · 3.1.2 Amount of Substance
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.