These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data: all sodium, potassium, ammonium and nitrate salts are soluble in water; any other solubility you need is given in the question. H 1.008, C 12.01, O 16.00 g mol⁻¹. You may write H⁺(aq) or H₃O⁺(aq).
Question 1 (multiple choice · foundation)
Aqueous ammonium carbonate is mixed with aqueous calcium chloride, and a white precipitate forms. Which is the net ionic equation?
- (A) Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
- (B) 2NH₄⁺(aq) + 2Cl⁻(aq) → 2NH₄Cl(s)
- (C) (NH₄)₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2NH₄Cl(aq)
- (D) Ca²⁺(aq) + 2Cl⁻(aq) + CO₃²⁻(aq) → CaCO₃(s) + 2Cl⁻(aq)
Answer and explanation
Answer: (A). All ammonium salts are soluble, so ammonium chloride stays dissolved and the precipitate must be calcium carbonate. NH₄⁺ and Cl⁻ are spectators.
- (B) makes ammonium chloride the precipitate, but ammonium salts are soluble.
- (C) is the balanced molecular equation, not the net ionic one.
- (D) still contains the spectator ion Cl⁻ on both sides. It is part of a complete ionic equation (and it also leaves out NH₄⁺).
Question 2 (multiple choice · core)
Colourless solutions of lead(II) nitrate and potassium iodide are mixed. A yellow precipitate of lead(II) iodide forms. Which ions are spectator ions?
- (A) Pb²⁺ and I⁻
- (B) K⁺ and NO₃⁻
- (C) K⁺ only
- (D) Pb²⁺ and NO₃⁻
Answer and explanation
Answer: (B). Molecular equation: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq). Potassium nitrate is soluble, so K⁺ and NO₃⁻ are dissolved before and after: they are spectators. The net ionic equation is Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
- (A) names the ions that do react; they form the precipitate.
- (C) misses nitrate, which is also unchanged (all nitrate salts are soluble).
- (D) Pb²⁺ leaves the solution in the solid, so it is not a spectator.
Question 3 (multiple choice · core)
Aluminium metal reacts with copper(II) ions in solution to form aluminium ions and copper metal. Which equation is correctly balanced?
- (A) Al(s) + Cu²⁺(aq) → Al³⁺(aq) + Cu(s)
- (B) 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
- (C) 3Al(s) + 2Cu²⁺(aq) → 3Al³⁺(aq) + 2Cu(s)
- (D) 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 2Cu(s)
Answer and explanation
Answer: (B). Atoms: 2 Al and 3 Cu on each side. Charge: left 3(+2) = +6; right 2(+3) = +6.
- (A) balances atoms but not charge (+2 on the left, +3 on the right).
- (C) balances atoms but not charge (+4 on the left, +9 on the right).
- (D) balances charge but not atoms: 3 Cu on the left, 2 Cu on the right.
Question 4 (multiple choice · core)
When a complete ionic equation is written, which of these species is written as separate ions?
- (A) CH₃COOH(aq)
- (B) HNO₃(aq)
- (C) CO₂(g)
- (D) BaSO₄(s)
Answer and explanation
Answer: (B). Nitric acid is a strong acid, so in water it is written as H⁺(aq) + NO₃⁻(aq) (or H₃O⁺(aq) + NO₃⁻(aq)).
- (A) Ethanoic acid (acetic acid) is a weak acid. It stays mostly as molecules, so it is written whole.
- (C) A gas is written as one formula.
- (D) A solid (here a precipitate) is written as one formula, even though it is ionic.
Question 5 (constructed response · core)
(a) Liquid methanol, CH₃OH, burns in oxygen gas to form carbon dioxide gas and liquid water. Write the balanced equation with state symbols, and show that mass is conserved using molar masses. (b) Write balanced equations, with state symbols, for these physical changes: (i) solid potassium sulfate dissolves in water; (ii) methanol vapour condenses to a liquid.
Worked solution and suggested Marlbridge rubric
(a) Unbalanced: CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(l). Balance C (1 = 1), then H (4 on the left, so 2H₂O), then O: right has 2 + 2 = 4 O, left has 1 + 2x, giving x = 1.5. Double everything to clear the fraction: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l) Tally: C 2 = 2; H 8 = 8; O 2 + 6 = 8 and 4 + 4 = 8. Mass (keep unrounded molar masses): left 2(32.042) + 3(32.00) = 160.08 g; right 2(44.01) + 4(18.016) = 160.08 g.
(b)(i) K₂SO₄(s) → 2K⁺(aq) + SO₄²⁻(aq) (b)(ii) CH₃OH(g) → CH₃OH(l)
| Point | What earns it |
|---|---|
| 1 | Correct formulas and states in (a) |
| 1 | Coefficients 2, 3, 2, 4 (smallest whole numbers) |
| 1 | Mass check: both sides 160.08 g (or equal atom tallies used to justify conservation) |
| 1 | (b)(i) with 2K⁺ and intact SO₄²⁻ |
| 1 | (b)(ii) with the same formula on both sides and correct state symbols |
Accept H₂O(g) in (a) if the student states the water is formed as vapour. Do not accept equations that keep the fraction 3/2 unless the question allowed it.
Question 6 (constructed response · core)
Aqueous iron(III) chloride is added to aqueous potassium hydroxide. A red-brown precipitate of iron(III) hydroxide forms.
(a) Write the balanced molecular equation. (b) Write the complete ionic equation. (c) Write the net ionic equation and name the spectator ions. (d) Show that the net ionic equation conserves charge.
Worked solution and suggested Marlbridge rubric
(a) FeCl₃(aq) + 3KOH(aq) → Fe(OH)₃(s) + 3KCl(aq) (b) Fe³⁺(aq) + 3Cl⁻(aq) + 3K⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) + 3K⁺(aq) + 3Cl⁻(aq) (c) Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s). Spectator ions: K⁺ and Cl⁻. (d) Left: +3 + 3(−1) = 0. Right: Fe(OH)₃ is a neutral solid, 0. Charge is conserved.
| Point | What earns it |
|---|---|
| 1 | Balanced molecular equation with states |
| 1 | Complete ionic equation with 3Cl⁻, 3K⁺, 3OH⁻ and the solid kept whole |
| 1 | Correct net ionic equation with states |
| 1 | Names both spectators and shows the charge sum equals zero on each side |
Accept carried-forward coefficients only if the equation in (a) is balanced.
Common errors: writing chloride as “Cl₃⁻” instead of 3Cl⁻ (the subscript becomes a coefficient when the ions separate), and splitting the precipitate into Fe³⁺ and OH⁻ on the right. A solid is written as one formula in every form of the equation.
Question 7 (constructed response · stretch)
Sodium carbonate solution is added to dilute nitric acid. Bubbles of carbon dioxide form. Two students write net ionic equations.
- Student P: Na₂CO₃(aq) + 2H⁺(aq) → 2Na⁺(aq) + H₂O(l) + CO₂(g)
- Student Q: CO₃²⁻(aq) + H⁺(aq) → H₂O(l) + CO₂(g)
(a) Identify the error in each student’s equation. (b) Write the correct balanced molecular equation and the correct net ionic equation. (c) Show that your net ionic equation conserves atoms and charge.
Worked solution and suggested Marlbridge rubric
(a) Student P’s equation is balanced, but it is not net ionic: sodium carbonate is a soluble sodium salt and should be split into 2Na⁺ + CO₃²⁻. Then Na⁺ appears on both sides as a spectator and must be removed. Student Q has the right species but it is not balanced: the left has 1 H and charge −1; the right has 2 H and charge 0.
(b) Molecular: Na₂CO₃(aq) + 2HNO₃(aq) → 2NaNO₃(aq) + H₂O(l) + CO₂(g) Net ionic: CO₃²⁻(aq) + 2H⁺(aq) → H₂O(l) + CO₂(g)
(c) Atoms: C 1 = 1; O 3 = 1 + 2; H 2 = 2. Charge: left −2 + 2(+1) = 0; right 0.
| Point | What earns it |
|---|---|
| 1 | Student P: Na₂CO₃(aq) should be split; Na⁺ is a spectator |
| 1 | Student Q: H not balanced and/or charge not balanced |
| 1 | Correct molecular equation with states |
| 1 | Correct net ionic equation with states |
| 1 | Atom and charge checks shown for the net ionic equation |
Accept 2H₃O⁺(aq) in place of 2H⁺(aq), with 3H₂O(l) on the right.
Why this matters: an equation can fail in two different ways. Student P’s equation obeys conservation but shows species that do not change; Student Q’s equation shows the right species but breaks conservation. A correct net ionic equation must pass both tests.
How did you do?
- Q1, Q2 or Q4 wrong: re-read “Three ways to write a reaction in solution” and Figure 1 in the study guide.
- Q3 or Q7 wrong: practise the charge check in “Atoms balanced is not enough”.
- Q5 or Q6 incomplete: follow Worked examples 1 and 2 step by step, including state symbols.
Then tick off the topic checklist.