HSC Exam Practice

Sample response. A conjugate acid-base pair consists of two species that interconvert through the gain or loss of exactly one proton (H⁺). The two species must appear on opposite sides of the equilibrium equation and differ by exactly one hydrogen atom and one unit of charge.

Marking notes. 1 mark for defining a conjugate pair as two species interconverting via gain/loss of one proton; 1 mark for stating the one-proton rule (differ by exactly one H⁺, on opposite sides of equation).

Sample response. BL acid = H₂PO₄⁻ (donates H⁺ to OH⁻); BL base = OH⁻ (accepts H⁺). Conjugate pair 1: H₂PO₄⁻ (acid) / HPO₄²⁻ (conjugate base) — differ by one H⁺, opposite sides. Conjugate pair 2: H₂O (conjugate acid) / OH⁻ (base) — differ by one H⁺, opposite sides.

Marking notes. 1 mark BL acid correctly identified with reason; 1 mark BL base correctly identified; 1 mark conjugate pair 1 correct; 1 mark conjugate pair 2 correct.

Sample response. HCO₃⁻ is amphiprotic because it contains one ionisable proton (can donate H⁺ → acts as acid) and has lone pairs/basic sites that can accept a proton (can accept H⁺ → acts as base), depending on its reaction partner. As acid: HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq). As base: HCO₃⁻(aq) + H⁺(aq) ⇌ H₂CO₃(aq).

Marking notes. 1 mark definition (can both donate and accept H⁺); 1 mark acid equation correct; 1 mark base equation correct; 1 mark use of equilibrium arrows.

Sample response. Amphiprotic (Brønsted-Lowry definition) means a substance can both donate and accept a proton. Amphoteric is broader: it means a substance can react with both acids and bases, including via Lewis acid-base mechanisms that do not involve proton transfer. All amphiprotic substances are amphoteric, but not all amphoteric substances are amphiprotic. Example of amphoteric but not amphiprotic: Al₂O₃ (aluminium oxide) — it reacts with acids and bases but contains no ionisable H⁺ to donate, so it cannot be a proton donor.

Marking notes. 1 mark amphiprotic = donate AND accept proton (Brønsted-Lowry); 1 mark amphoteric = broader/includes non-proton mechanisms; 1 mark valid example (Al₂O₃, ZnO, Al(OH)₃, or similar metal oxide/hydroxide).

Sample response. Na₂HPO₄ (sodium hydrogen phosphate) provides HPO₄²⁻ ions. As acid: HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) (K_a3 = 4.8 × 10⁻¹³). As base: HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq).

Marking notes. 1 mark NESA salt name = Na₂HPO₄; 1 mark acid equation correct; 1 mark base equation correct. Both equations must be present for 2 marks on equations.

Sample response. In the autoionisation 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), one water molecule donates a proton (acts as a Brønsted-Lowry acid, producing OH⁻) while the other molecule accepts the proton (acts as a Brønsted-Lowry base, producing H₃O⁺). Because water simultaneously acts as both acid and base, it is amphiprotic. K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C.

Marking notes. 1 mark for identifying one H₂O as acid (donor) and one as base (acceptor) in the same reaction; 1 mark for the conclusion “therefore water is amphiprotic”; 1 mark for K_w expression with correct value.

Part (a) — 3 marks. K_w increases as temperature increases; the rate of increase accelerates at higher temperatures (curve is non-linear/concave up). Water autoionisation is endothermic (ΔH > 0). By Le Chatelier’s Principle, increasing temperature favours the forward reaction (shifts equilibrium right), increasing concentrations of both H₃O⁺ and OH⁻, and therefore K_w. [1 trend + direction; 1 endothermic; 1 Le Chatelier conclusion]

Part (b) — 4 marks. Neutral pH at 37°C: [H₃O⁺] = [OH⁻] = √K_w = √(2.4 × 10⁻¹⁴) = 1.549 × 10⁻⁷ mol L⁻¹; neutral pH = −log(1.549 × 10⁻⁷) = 6.81. Blood pH 7.40 > 6.81 (neutral pH at 37°C), so [H₃O⁺] < [OH⁻] → blood is basic (alkaline) at body temperature. [1 correct calculation of neutral pH; 1 value 6.81; 1 comparison 7.40 vs 6.81; 1 correct classification as basic]

Part (c) — 2 marks. The student’s claim is incorrect. pH + pOH = 14 applies only at 25°C. At 37°C, pK_w = −log(2.4 × 10⁻¹⁴) = 13.62, so pH + pOH = 13.62. If pOH = 6.8, then pH = 13.62 − 6.8 = 6.82. pH 6.82 > 6.81 (neutral pH at 37°C), so the solution is very slightly basic — but the student’s pH value of 7.2 is wrong. [1 identifying pH + pOH ≠ 14 at 37°C; 1 correct calculation pH = 6.82 and correct classification]

Part (a) — 2 marks. [OH⁻] = n/V = 0.010/0.250 = 0.040 mol L⁻¹. [H₃O⁺] = K_w/[OH⁻] = 1.0 × 10⁻¹⁴ / 0.040 = 2.5 × 10⁻¹³ mol L⁻¹. [1 [OH⁻]; 1 [H₃O⁺]]

Part (b) — 2 marks. pH = −log[H₃O⁺] = −log(2.5 × 10⁻¹³) = 12.60. Alternatively: pOH = −log(0.040) = 1.40; pH = 14 − 1.40 = 12.60. [1 method; 1 correct answer]

Part (c) — 1 mark. Assumption: temperature is exactly 25°C (K_w = 1.0 × 10⁻¹⁴ applies). At another temperature K_w would differ and [H₃O⁺] would be different. Error if NaOH 98% pure: [OH⁻] would be 2% lower than calculated (0.040 × 0.98 = 0.0392 mol L⁻¹), so [H₃O⁺] would be slightly higher and pH slightly lower than 12.60. [1 mark for valid assumption AND plausible error]

Sample response. The statement is partially correct but fundamentally flawed: the causal relationship between pH 7 and neutrality is wrong. Neutrality is defined by [H₃O⁺] = [OH⁻], not by the number 7. From K_w = [H₃O⁺][OH⁻], in a neutral solution [H₃O⁺] = [OH⁻] = √K_w, so neutral pH = −log(√K_w). At 25°C, K_w = 1.0 × 10⁻¹⁴ and neutral pH = 7.00. But K_w is temperature-dependent: autoionisation of water is endothermic, and by Le Chatelier’s Principle higher temperatures drive the equilibrium to the right, increasing both [H₃O⁺] and [OH⁻] equally, so K_w increases. At 37°C, K_w = 2.4 × 10⁻¹⁴ and neutral pH = 6.81; at 60°C, neutral pH ≈ 6.51. Pure water at these temperatures is still neutral ([H₃O⁺] = [OH⁻]) but pH < 7. So the statement “pH of 7” does not equal neutrality at temperatures other than 25°C, and must be rejected. The autoionisation 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) also confirms that water is amphiprotic: in the same reaction, one H₂O molecule acts as a Brønsted-Lowry acid (donates H⁺, producing OH⁻) while the other acts as a Brønsted-Lowry base (accepts H⁺, producing H₃O⁺). Because water can both donate and accept a proton, it meets the definition of amphiprotic. The pH of pure water at any temperature is neutral by definition, but that neutral pH value changes with temperature, making the statement “always pH 7” false.

Marking notes. 1 — defines neutrality correctly as [H₃O⁺] = [OH⁻] (not pH 7). 1 — derives neutral pH = −log(√K_w). 1 — identifies K_w as temperature-dependent (K_w increases with T). 1 — links to endothermic autoionisation and Le Chatelier correctly. 1 — provides a numerical example: neutral pH at 37°C = 6.81 or another valid temperature. 1 — explicitly rejects the statement with correct reasoning. 1 — explains autoionisation as evidence of water being amphiprotic (one molecule = acid, one = base). 1 — reaches a sustained, logically coherent conclusion synthesising all elements.