HSC Exam Practice

Sample response. Acid strength describes the degree to which an acid ionises in water at equilibrium, quantified by the acid dissociation constant Ka. A strong acid (e.g. HCl, Ka very large) ionises essentially completely; a weak acid (e.g. CH₃COOH, Ka = 1.8 × 10⁻⁵) ionises only partially. Concentration (mol/L) describes the total amount of acid dissolved per litre of solution, regardless of how much has ionised. These two properties are completely independent: a strong acid can be dilute or concentrated, and a weak acid can be dilute or concentrated.

Marking notes. 1 mark for defining acid strength as degree of ionisation / Ka; 1 mark for defining concentration as mol/L / total dissolved amount; 1 mark for explicitly stating they are independent.

Sample response. The six strong acids are: HCl, H₂SO₄ (first ionisation), HNO₃, HClO₄, HBr, HI. Example equations: HCl(aq) → H⁺(aq) + Cl⁻(aq); HNO₃(aq) → H⁺(aq) + NO₃⁻(aq). [Any two from the six list are acceptable.]

Marking notes. 1 mark for all six strong acids listed correctly (allow minor name omissions, accept formulae); 1 mark per correct ionic equation with correct state symbols and single arrow → (2 marks for 2 equations). Equilibrium arrow in any strong acid equation = 0 for that equation.

Sample response. When HF dissolves in water, only a fraction of HF molecules donate their proton to water molecules, establishing a dynamic equilibrium: HF(aq) ⇌ H⁺(aq) + F⁻(aq). The majority of HF molecules remain intact. HF is a weak acid because the H–F bond is unusually strong due to fluorine’s small atomic radius and very high electronegativity, making proton donation energetically unfavourable (bond enthalpy ~570 kJ/mol vs ~432 kJ/mol for H–Cl). Unlike HCl, HBr and HI, whose longer, weaker bonds make proton donation easy (strong acids), HF’s short, strong bond gives it Ka = 6.8 × 10⁻⁴ — partial ionisation only.

Marking notes. 1 mark for correct equation with ⇌; 1 mark for stating most molecules remain intact (partial ionisation); 1 mark for bond strength / electronegativity / Ka explanation.

Sample response. Weak bases use an equilibrium arrow (⇌) in their ionic equations because proton acceptance is partial and reversible. For NH₃: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). The ⇌ communicates that both the forward reaction (NH₃ accepting a proton from water) and the reverse reaction (NH₄⁺ donating a proton back to OH⁻) occur simultaneously at significant rates, establishing a dynamic equilibrium where most NH₃ molecules remain un-protonated.

Marking notes. 1 mark for identifying ⇌ as correct notation for weak base; 1 mark for correct NH₃ equation with state symbols; 1 mark for explaining ⇌ indicates reversible/partial reaction / dynamic equilibrium.

Sample response. Ca(OH)₂ is a strong base and dissociates completely in solution. Its formula contains two OH⁻ groups per formula unit: Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq). Each mole of Ca(OH)₂ that dissolves releases 2 moles of OH⁻, so [OH⁻] = 2 × [Ca(OH)₂]. For a 0.050 mol/L solution: [OH⁻] = 2 × 0.050 = 0.10 mol/L.

Marking notes. 1 mark for correct explanation referencing 2 OH⁻ per formula unit; 1 mark for correct equation with →; 1 mark for correct calculation: [OH⁻] = 0.10 mol/L.

Sample response. 0.10 mol/L HNO₃ has much higher conductivity than 0.10 mol/L HNO₂. HNO₃ is a strong acid (on the strong acid list) and ionises essentially completely: HNO₃(aq) → H⁺(aq) + NO₃⁻(aq). Every molecule dissociates, producing 0.10 mol/L each of H⁺ and NO₃⁻, giving high conductivity (~40 mS/cm). HNO₂ is a weak acid (Ka = 4.5 × 10⁻⁴) and ionises only partially: HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq). Only a small fraction of molecules ionise, so the ion concentration — and thus conductivity — is far lower (~2–3 mS/cm).

Marking notes. 1 mark for correctly stating HNO₃ is strong / HNO₂ is weak; 1 mark for linking the conductivity difference to complete vs partial ionisation; 1 mark for writing correct equations with correct arrow for each.

Part (a) — 2 marks. For 0.10 mol/L HCl and 0.001 mol/L HCl, the measured pH (1.0 and 3.0) matches pH = −log(concentration) exactly, confirming complete ionisation: these are strong acids. For both CH₃COOH entries (rows 2 and 4), the measured pH is higher than the pH expected from complete ionisation: 0.10 mol/L CH₃COOH gives pH 2.9 instead of expected 1.0; 1.0 mol/L CH₃COOH gives pH 2.4 instead of expected 0. This confirms CH₃COOH is a weak acid (partial ionisation). (1 mark for correctly identifying both HCl entries as strong with justification; 1 mark for correctly identifying both CH₃COOH entries as weak with justification.)

Part (b) — 4 marks. The student’s claim is wrong. HCl is always a strong acid with complete ionisation; CH₃COOH is always a weak acid with Ka = 1.8 × 10⁻⁵. The lower pH of 1.0 mol/L CH₃COOH (2.4) compared to 0.001 mol/L HCl (3.0) is caused by the large difference in concentration — CH₃COOH is 1000 times more concentrated than the dilute HCl. Even at 1 mol/L, only ~0.4% of CH₃COOH ionises, giving [H⁺] ≈ 0.004 mol/L (pH ~2.4). HCl at 0.001 mol/L is 100% ionised giving [H⁺] = 0.001 mol/L (pH 3.0). This is precisely the strength vs concentration distinction: strength (Ka) is intrinsic and does not change with concentration; [H⁺] and pH depend on both Ka and concentration. pH alone cannot determine acid strength. (1 mark for rejecting the claim; 1 mark for identifying concentration as the cause of the pH difference in rows 3–4; 1 mark for using degree of ionisation to support argument; 1 mark for concluding pH alone cannot determine strength.)

Part (c) — 2 marks. [H⁺] = 10⁻⁾ ⁺¹·⁸ = 10⁻₂·⁹ = 1.26 × 10⁻³ ≈ 0.0013 mol/L. Degree of ionisation = (0.0013 / 0.10) × 100% = 1.3%. (1 mark for correct [H⁺] calculation; 1 mark for correct degree of ionisation calculation with working.)

Part (a) — 2 marks. NaOH is a strong base; NH₃ is a weak base. The conductivity of NaOH (24.5 mS/cm) is approximately 52 times greater than NH₃ (0.47 mS/cm) at the same concentration (0.10 mol/L). NaOH dissociates completely, producing a high ion concentration (Na⁺ and OH⁻ each at 0.10 mol/L). NH₃ only partially accepts protons from water, so few NH₄⁺ and OH⁻ ions are present, explaining the very low conductivity. (1 mark for correct identification with comparison; 1 mark for linking conductivity difference to complete vs partial ionisation.)

Part (b) — 2 marks. NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). (1 mark for correct formula and products; 1 mark for equilibrium arrow ⇌ and state symbols.)

Sample response. The statement is incorrect as a general rule. While it is true that at the same concentration, a strong acid will always have a lower pH than a weak acid, pH alone cannot identify acid strength because [H⁺] — which determines pH — depends on both strength and concentration together. Acid strength is defined by Ka, an intrinsic property of the acid molecule fixed at a given temperature. A strong acid (e.g. HCl) ionises completely: HCl(aq) → H⁺(aq) + Cl⁻(aq), so [H⁺] = total concentration. A weak acid (e.g. CH₃COOH, Ka = 1.8 × 10⁻⁵) ionises only partially (⇌), so [H⁺] << total concentration. The critical counter-example that falsifies the statement is: 1 mol/L CH₃COOH has pH ≈ 2.4, while 0.001 mol/L HCl has pH = 3.0. Here, the weak acid (CH₃COOH) has a lower pH than the strong acid (HCl) at these concentrations — yet HCl is unambiguously strong (100% ionised at any concentration) and CH₃COOH is unambiguously weak (only ~0.4% ionised at 1 mol/L). The lower pH of CH₃COOH here arises because its far higher concentration compensates for its low Ka, producing more total H⁺ ions than the very dilute HCl. To reliably identify acid strength, one must compare pH at equal concentration, or measure conductivity (which directly reflects ion concentration from ionisation fraction), or consult Ka values. Therefore, the statement should read: “At equal concentration, a lower pH indicates a stronger acid; but pH alone, without knowing concentration, cannot determine acid strength.”

Marking notes. 1 mark — identifies the statement as incorrect / conditionally true. 1 mark — correctly defines acid strength as Ka / degree of ionisation, independent of concentration. 1 mark — explains [H⁺] depends on both Ka and concentration. 1 mark — names a correct counter-example (concentrated weak acid with lower pH than dilute strong acid, e.g. 1 mol/L CH₃COOH pH 2.4 vs 0.001 mol/L HCl pH 3.0). 1 mark — explains the counter-example correctly (high concentration partially compensates for low Ka). 1 mark — identifies conductivity or equal-concentration comparison as the correct way to determine strength. 1 mark — reaches an explicit, evidence-based evaluative conclusion that rejects the statement and restates the correct relationship.