How to use this review
Use this page after the Unit 4 diagnostic, once you have revisited the topics it flagged. Answer all seven questions on paper before opening any answer. Each combines two or more topics.
These are original Marlbridge practice questions, not past exam questions. The samples in Questions 5 and 6 are fictional, with invented data. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score. All sodium, potassium, ammonium and nitrate salts are soluble in water.
Big ideas of the unit
- A chemical change makes new substances. Gas, heat, a precipitate or a new colour are clues; a test for the new substance is proof (Topic 4.1).
- Atoms and charge are conserved, so equations balance in both, and net ionic equations drop spectator ions (Topic 4.2).
- A particle diagram is an equation at the scale of atoms: same atoms, leftovers kept, dissolved salts as ions (Topic 4.3).
- Bonds decide the type of change. Chemical bonds broken or formed: chemical. Only intermolecular attractions changed: physical (Topic 4.4).
- Coefficients are mole ratios. Grams, gas volumes and solution volumes all go through moles; the limiting reactant sets the product (Topic 4.5).
- A titration is stoichiometry at the equivalence point (Topic 4.6).
- Sort reactions by what moves: electrons (redox, including combustion), protons (acid-base) or ions joining into a solid (precipitation) (Topic 4.7).
- Acids donate protons; bases accept them. Each reaction has two conjugate pairs, and water can play either role (Topic 4.8).
- Half-reactions balance electrons, so a redox equation balances charge as well as atoms (Topic 4.9).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Evidence of change | claim, evidence, reasoning about composition | 4.1 |
| Net ionic equation | split strong electrolytes in solution; keep solids, gases, liquids and weak acids or bases whole; cancel spectators | 4.2 |
| Particle diagram | atoms conserved; reacting ratio = coefficients; leftovers stay | 4.3 |
| Physical or chemical | same formula → physical; new formula → chemical | 4.4 |
| Moles | n = m / M; n = M × V(L); n = PV / RT | 4.5 |
| Limiting reactant | smallest value of moles ÷ coefficient | 4.5 |
| Equivalence point | n(analyte) = n(titrant) × coefficient ratio | 4.6 |
| Oxidation numbers | elements 0; ions = charge; sum = overall charge | 4.7 |
| Conjugate pair | differ by one H⁺ and a charge of 1 | 4.8 |
| Half-reactions (acid) | other atoms, O with H₂O, H with H⁺, charge with e⁻; equalise electrons | 4.9 |
Question 1 (multiple choice · mixed)
A student mixes 30.0 mL of 0.200 M silver nitrate, AgNO₃, with 20.0 mL of 0.200 M sodium carbonate, Na₂CO₃. Insoluble silver carbonate, Ag₂CO₃, precipitates. What is the maximum mass of precipitate, and which reacting ion is left over?
- (A) 0.827 g; carbonate ions are left over
- (B) 1.10 g; silver ions are left over
- (C) 1.65 g; carbonate ions are left over
- (D) 0.827 g; silver ions are left over
Answer and explanation
Answer: (A). Net ionic equation: 2Ag⁺(aq) + CO₃²⁻(aq) → Ag₂CO₃(s), with Na⁺ and NO₃⁻ as spectators (Topic 4.2). Moles ÷ coefficients: Ag⁺ 0.00600 ÷ 2 = 0.00300; CO₃²⁻ 0.00400. Ag⁺ is limiting (Topic 4.5): 0.00300 mol × 275.75 g mol⁻¹ = 0.827 g, leaving 0.00100 mol of carbonate.
- (B) takes carbonate as limiting because there is less of it, but each carbonate needs two Ag⁺.
- (C) uses a 1 : 1 ratio of Ag⁺ to Ag₂CO₃.
- (D) The limiting ion is the one used up, not the one left over.
Question 2 (multiple choice · mixed)
Sodium reacts with water: 2Na(s) + 2H₂O(l) → 2Na⁺(aq) + 2OH⁻(aq) + H₂(g). Which statement is correct?
- (A) It is acid-base: water donates a proton to sodium, so OH⁻ is the conjugate base of water.
- (B) It is redox: sodium is oxidised and hydrogen in water is reduced; two electrons are transferred for each H₂ formed.
- (C) It is redox: sodium is oxidised and oxygen is reduced, because OH⁻ carries a negative charge.
- (D) It is redox: sodium is reduced and hydrogen is oxidised, because sodium ends up as an ion.
Answer and explanation
Answer: (B). Check oxidation numbers first (Topic 4.7): Na 0 → +1 (oxidised); H +1 → 0 in H₂ (reduced); O stays −2. Two H atoms each gain one electron, from two Na atoms (Topic 4.9).
- (A) H leaves as H₂, not as H⁺ passed to a base, and sodium cannot accept a proton (Topic 4.8).
- (C) An ion’s charge is not an oxidation number; O is −2 in both H₂O and OH⁻.
- (D) Losing electrons, as sodium does, is oxidation.
Question 3 (multiple choice · mixed)
Hydrogen chloride gas dissolves in water to give hydrochloric acid, a strong acid. Which particle description and classification fit best?
- (A) Separate HCl molecules among the water molecules; physical, like dissolving sugar.
- (B) H₃O⁺ and Cl⁻ ions among the water molecules; best argued as chemical, because H–Cl bonds break and new O–H bonds form.
- (C) H⁺ and Cl⁻ ions, with water unchanged; physical, because no new substance forms.
- (D) H₂ and Cl₂ molecules among the water molecules; chemical.
Answer and explanation
Answer: (B). A strong acid ionises essentially completely, passing its proton to water acting as a base (Topic 4.8): HCl + H₂O → H₃O⁺ + Cl⁻. A covalent bond breaks and a new one forms (Topic 4.4), so the diagram shows ions, not molecules (Topic 4.3).
- (A) fits sugar or a barely ionised weak acid, not a strong acid.
- (C) A proton in water is bonded to a water molecule, so water does change.
- (D) Dissolving does not split HCl into its elements.
Question 4 (constructed response · mixed)
A student heats 2.451 g of potassium chlorate, KClO₃, with a little manganese(IV) oxide catalyst in an open test tube. Bubbles form, and a glowing splint at the mouth of the tube bursts into flame.
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Before heating, the solid dissolved in water gives no precipitate with silver nitrate solution. After heating, the residue gives a white precipitate. (Silver chlorate is soluble; silver chloride is insoluble.) Ignore the catalyst’s mass.
(a) Give two pieces of evidence that a chemical change took place, each linked to a new substance. (b) Write the net ionic equation for the test on the residue, and name the spectator ions. (c) Use oxidation numbers to show that the decomposition is redox, with electrons lost equal to electrons gained. (d) Calculate the maximum mass of oxygen, and its volume at 25.0 °C and 1.00 atm. (e) The tube loses 0.912 g. Calculate the percentage of the KClO₃ that decomposed.
Worked solution and suggested Marlbridge rubric
(a) The relit splint shows oxygen gas, not present as a gas in the solid. The precipitate shows chloride ions, which the starting solid did not contain.
(b) Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Spectators: K⁺ and NO₃⁻.
(c) In KClO₃, Cl = +5 (K +1, three O at −2); in KCl, Cl = −1; in O₂, O = 0. Two Cl atoms each gain 6: 12 electrons gained. Six O atoms each lose 2: 12 electrons lost. Chlorine is reduced and oxygen oxidised.
(d) n(KClO₃) = 2.451 ÷ 122.55 = 0.02000 mol, so n(O₂) = 0.02000 × 3/2 = 0.03000 mol = 0.960 g. V = (0.03000)(0.08206)(298.15) ÷ 1.00 = 0.734 L.
(e) The open tube loses only the oxygen: 0.912 ÷ 0.960 × 100 = 95.0%.
| Point | What earns it |
|---|---|
| 1 | (a) Both observations, each linked to a new substance |
| 1 | (b) Net ionic equation with states, and K⁺, NO₃⁻ as spectators |
| 1 | (c) Cl +5 → −1 and O −2 → 0 |
| 1 | (c) 12 electrons lost = 12 gained |
| 1 | (d) 0.960 g of O₂ from the 3 : 2 ratio |
| 1 | (d) 0.734 L, with T in kelvin |
| 1 | (e) 95.0% (carry forward from (d)) |
Total: 7 points. “It bubbled” alone does not earn the (a) point; the gas must be identified.
Question 5 (constructed response · mixed)
A fictional supplier sells iron(II) sulfate heptahydrate, FeSO₄·7H₂O (20.09% iron by mass when pure), for treating lawns. A student dissolves 1.250 g in dilute sulfuric acid and makes it up to 100.0 mL. She titrates 25.00 mL portions with 0.01000 M potassium dichromate; the mean titre is 18.20 mL. Fe²⁺ is oxidised to Fe³⁺, and Cr₂O₇²⁻ is reduced to Cr³⁺.
(a) Write the two half-reactions in acidic solution. (b) Write the balanced net ionic equation and show that it conserves charge. (c) Calculate the percent by mass of iron in the sample, and compare it with the pure compound. (d) Explain why the sulfuric acid must be in excess.
Worked solution and suggested Marlbridge rubric
(a) Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻. Reduction (Cr +6 → +3): balance Cr, then O with 7H₂O, H with 14H⁺, charge with 6e⁻: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l).
(b) Multiply the oxidation by 6 and add: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Fe²⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l). Charge: left −2 + 14 + 12 = +24; right 6 + 18 = +24.
(c)
- n(Cr₂O₇²⁻) = 0.01000 × 0.01820 = 1.820 × 10⁻⁴ mol.
- n(Fe²⁺) per portion = 6 × 1.820 × 10⁻⁴ = 1.092 × 10⁻³ mol.
- In 100.0 mL: 4 × 1.092 × 10⁻³ = 4.368 × 10⁻³ mol; × 55.85 = 0.2440 g.
- 0.2440 ÷ 1.250 × 100 = 19.52%, below 20.09%, so the batch is not pure: about 19.52 ÷ 20.09 = 97.2% FeSO₄·7H₂O if that is the only iron compound.
(d) H⁺ is a reactant (14 per dichromate, 2.55 × 10⁻³ mol per titration). If it ran short, the reaction would not follow this equation, and the 1 : 6 ratio used in (c) would not hold.
| Point | What earns it |
|---|---|
| 1 | (a) Iron half-reaction |
| 1 | (a) Dichromate half-reaction with 6e⁻ on the left |
| 1 | (b) Electrons cancelled, and +24 on both sides |
| 1 | (c) 1 : 6 ratio gives 1.092 × 10⁻³ mol per portion |
| 1 | (c) Scales by 4; 19.52% |
| 1 | (c) Concludes not pure (about 97%) |
| 1 | (d) H⁺ is a reactant, so it must not limit the reaction |
Total: 7 points. Common slips: a 1 : 1 ratio (3.25%); no scale-up to 100.0 mL (4.88%).
Question 6 (constructed response · mixed)
A fictional cleaner contains ammonia, NH₃, a weak base. A student dilutes 10.00 mL of it to 250.0 mL. A 20.00 mL portion of the diluted solution needs 16.30 mL of 0.1000 M nitric acid to reach the equivalence point.
(a) Write the molecular, complete ionic and net ionic equations for aqueous ammonia reacting with nitric acid. Explain why NH₃ is not split into ions. (b) Rewrite the net ionic equation with H₃O⁺, and label the acid, base, conjugate acid and conjugate base. (c) A particle diagram shows 5 NH₃ molecules mixed with 3 H₃O⁺ and 3 NO₃⁻ ions (other water not shown). Describe the particles after complete reaction, and show that charge balances. (d) Calculate the ammonia concentration in the original cleaner.
Worked solution and suggested Marlbridge rubric
(a) Molecular: NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq). Complete ionic: NH₃(aq) + H⁺(aq) + NO₃⁻(aq) → NH₄⁺(aq) + NO₃⁻(aq). Net ionic: NH₃(aq) + H⁺(aq) → NH₄⁺(aq). NH₃ is a weak, molecular base that stays mostly as molecules, so it is written whole.
(b) NH₃ + H₃O⁺ → NH₄⁺ + H₂O. Acid H₃O⁺, conjugate base H₂O; base NH₃, conjugate acid NH₄⁺. No oxidation number changes (N stays −3), so this is proton transfer, not redox.
(c) 3 H₃O⁺ react with 3 NH₃. After: 3 NH₄⁺, 2 NH₃ and 3 NO₃⁻, all separate (plus 3 new water molecules). Charge: 3(+1) + 3(−1) = 0.
(d) n(HNO₃) = 0.1000 × 0.01630 = 1.630 × 10⁻³ mol = n(NH₃) (1 : 1). Diluted: 1.630 × 10⁻³ ÷ 0.02000 = 0.08150 M. Original: × 250.0/10.00 = 2.04 M.
| Point | What earns it |
|---|---|
| 1 | (a) Three equations, NO₃⁻ cancelled in the net ionic one |
| 1 | (a) NH₃ kept whole as a weak base |
| 1 | (b) All four species labelled |
| 1 | (c) 3 NH₄⁺, 2 NH₃, 3 NO₃⁻, with the charge check |
| 1 | (d) 0.08150 M diluted |
| 1 | (d) 2.04 M after the dilution factor of 25 |
Total: 6 points.
Question 7 (constructed response · mixed)
A sealed steel container holds 2.20 g of propane, C₃H₈, and 6.40 g of oxygen. A spark ignites it. Assume the only products are CO₂ and H₂O.
(a) Write the balanced equation for complete combustion. (b) Identify the limiting reactant. (c) Calculate the mass of each substance after the reaction, and show that mass is conserved. (d) Find the average oxidation number of carbon in propane and in CO₂, and show that electrons lost equal electrons gained. (e) On cooling to 25 °C, droplets form on the walls. Classify this change, explaining in terms of bonds and attractions.
Worked solution and suggested Marlbridge rubric
(a) C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
(b) n(C₃H₈) = 2.20 ÷ 44.094 = 0.04989 mol; n(O₂) = 6.40 ÷ 32.00 = 0.2000 mol, and 0.2000 ÷ 5 = 0.04000. O₂ is limiting, although its mass is larger.
(c) 0.04000 mol of propane reacts. CO₂: 0.1200 mol = 5.281 g. H₂O: 0.1600 mol = 2.883 g. Propane left: 0.00989 mol = 0.436 g. Total after: 8.600 g = 2.20 + 6.40 before. The sealed container keeps every atom.
(d) C₃H₈: 3C + 8 = 0, so C averages −8/3; in CO₂, C is +4. Each propane loses 3 × (4 + 8/3) = 20 electrons; 10 O atoms each gain 2 = 20 electrons.
(e) Physical: water vapour condenses. The H₂O molecules are unchanged (no O–H bonds break); only attractions between molecules increase.
| Point | What earns it |
|---|---|
| 1 | (a) Balanced equation |
| 1 | (b) O₂ limiting, via moles ÷ coefficients |
| 1 | (c) Masses of CO₂, H₂O and leftover propane |
| 1 | (c) 8.60 g before and after |
| 1 | (d) −8/3 and +4, and 20 lost = 20 gained |
| 1 | (e) Physical: same molecules, only intermolecular attractions change |
Total: 6 points. Taking propane as limiting predicts 6.59 g of CO₂, needing more oxygen than is present.
How did you do?
Total: 29 points (3 multiple choice, 26 constructed response). Where you lost points matters more than the total.
- Classifying changes (3, 4(a), 7(e)): revisit Topics 4.1 and 4.4.
- Equations and particle diagrams (1, 4(b), 6(a) and (c)): revisit Topics 4.2 and 4.3.
- Moles and titrations (1, 4(d), 5(c), 6(d), 7(b) and (c)): revisit Topics 4.5 and 4.6.
- Reaction types and redox (2, 5(a) and (b), 6(b), 7(d)): revisit Topics 4.7 to 4.9.
If several topics were weak, return to the Unit 4 diagnostic. Then tick off each topic checklist: