HSC Exam Practice

Sample response. Ka is the equilibrium constant for the ionisation of a weak acid in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). Ka = [H⁺][A⁻] / [HA], where all concentrations are equilibrium values. A larger Ka means more ionisation at equilibrium and a stronger acid.

Marking notes. 1 mark — correct equilibrium equation with reversible arrow; 1 mark — correct Ka expression with product concentrations in numerator and reactant in denominator.

Sample response. Ka is an intrinsic molecular property — it depends only on the identity of the acid and temperature, not on concentration. pH = −log[H⁺] measures the concentration of H⁺ions in a specific solution; it depends on both Ka and the acid's concentration. Two acids with different Ka values can have identical pH at different concentrations, so pH cannot be used to compare inherent acid strengths unless concentrations are equal. Only Ka provides a concentration-independent measure of acid strength.

Marking notes. 1 mark — Ka is concentration-independent / intrinsic; 1 mark — pH depends on both Ka and concentration; 1 mark — different Ka acids can have same pH at different concentrations (or equivalent statement that Ka is the correct comparison tool).

Sample response. Ranking from strongest to weakest (using Ka data from the data sheet): HF (6.8 × 10⁻⁴) > CH₃COOH (1.8 × 10⁻⁵) > H₂CO₃ (4.3 × 10⁻⁷) > HCN (6.2 × 10⁻¹⁰) > H₂O (1 × 10⁻¹⁶). Justification: larger Ka means more ionisation at equilibrium = stronger acid.

Marking notes. 1 mark — all five acids in correct order; 1 mark — explicit justification using Ka (larger Ka = stronger).

Sample response. In the first ionisation, a proton H⁺ is removed from the neutral H₂CO₃ molecule — there is no net charge opposing removal. The first ionisation produces HCO₃⁻ (charge −1). The second ionisation requires removing H⁺ from this negatively charged species: the proton is electrostatically attracted to the negative ion, making removal much harder. Greater electrostatic opposition to proton donation results in a much smaller Ka2. Ka1/Ka2 = 4.3 × 10⁻⁷ / 4.7 × 10⁻¹¹ ≈ 9,100 — Ka1 is ≈10,000 times larger.

Marking notes. 1 mark — identifies removal from neutral vs negatively charged species; 1 mark — electrostatic attraction between H⁺ (positive) and anion (negative) opposes removal; 1 mark — Ka ratio cited OR statement that magnitude difference is ≈10⁴. Award all 3 if all three elements present without using the word "electrostatic" — the mechanism must be clear.

Sample response. Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C is valid only for a conjugate acid–base pair: Ka(HA) × Kb(A⁻) = Kw, where A⁻ is formed by HA donating one proton. If Ka increases (stronger acid), then Kb = Kw/Ka decreases — the conjugate base becomes weaker. Conjugate bases of strong acids have very small Kb values.

Marking notes. 1 mark — condition: applies only to a conjugate pair; 1 mark — inverse relationship correctly stated (if Ka increases, Kb decreases).

Sample response. Ka(HF) = 6.8 × 10⁻⁴; Ka(CH₃COOH) = 1.8 × 10⁻⁵. pKa(HF) = −log(6.8 × 10⁻⁴) = 4 − log(6.8) = 4 − 0.833 = 3.17. pKa(CH₃COOH) = −log(1.8 × 10⁻⁵) = 5 − log(1.8) = 5 − 0.26 = 4.74. HF is the stronger acid (lower pKa, larger Ka). At the same concentration, HF ionises to a greater extent than CH₃COOH: for 0.1 mol/L, [H⁺](HF) = √(6.8 × 10⁻⁵) = 8.2 × 10⁻³ mol/L vs [H⁺](CH₃COOH) = √(1.8 × 10⁻⁶) = 1.3 × 10⁻³ mol/L — HF produces ≈6.3 times more H⁺ ions per litre.

Marking notes. 1 mark — Ka values stated; 1 mark each for correct pKa of HF and CH₃COOH (to 2 d.p.); 1 mark — correct conclusion that HF is stronger and explanation using degree of ionisation or [H⁺] comparison. [4 marks total — deduct 1 if pKa wrong or degree of ionisation not mentioned]

Sample response. The longest bar on the chart is HCN (pKa 9.21), meaning it has the largest pKa and therefore the smallest Ka — hence HCN is the weakest acid. Its conjugate base CN⁻ is therefore the strongest conjugate base in the set. Kb(CN⁻) = Kw / Ka(HCN) = 1.0 × 10⁻¹⁴ / 6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵.

Marking notes. 1 mark — identifies HCN as weakest acid (largest pKa) → CN⁻ is strongest base; 1 mark — Kb = Kw/Ka(HCN) correctly set up; 1 mark — correct numerical answer.

Sample response. Ka2 is much smaller than Ka1 because the second proton must be removed from HCO₃⁻, which carries a net negative charge. The electrostatic attraction between the positively charged H⁺ and the negatively charged HCO₃⁻ is much stronger than for the neutral H₂CO₃ in the first ionisation, requiring far more energy to overcome — hence Ka2 is ≈10,000 times smaller. Ka1 dominates ocean acidification chemistry: Ka1/Ka2 ≈ 9,100 means the first ionisation produces ≈9,100 times more H⁺ than the second at any given H₂CO₃ concentration, so the second ionisation is negligible for pH calculations and ocean pH modelling.

Marking notes. 1 mark — electrostatic explanation (removal from negatively charged HCO₃⁻ vs neutral H₂CO₃); 1 mark — Ka1 identified as dominant (Ka1/Ka2 ≈10⁴ cited or equivalent); 1 mark — conclusion that Ka1 governs ocean acidification / Ka2 negligible.

Sample response. Error: the student has confused bar length (pKa magnitude) with acid strength. A longer bar means a larger pKa, which means a smaller Ka — therefore HCN is the weakest acid, not the strongest. Corrected statement: HCN has the longest bar because it has the largest pKa (9.21) and therefore the smallest Ka (6.2 × 10⁻¹⁰), making it the weakest acid shown. HF (shortest bar, pKa 3.17) is the strongest.

Marking notes. 1 mark — identifies the error (conflating bar length / large pKa with strong acid); 1 mark — corrected statement explicitly linking large pKa to small Ka to weak acid.

Sample response.
HCN(aq) ⇌ H⁺(aq) + CN⁻(aq)    Ka = 6.2 × 10⁻¹⁰
ICE: I: 0.100, 0, 0  | C: −x, +x, +x  | E: 0.100−x, x, x
Assume x « 0.100: x² = Ka × 0.100 = 6.2 × 10⁻¹¹
x = √(6.2 × 10⁻¹¹) = 7.87 × 10⁻⁶ mol/L
Check: 7.87 × 10⁻⁶ / 0.100 = 0.0079% « 5% — assumption valid.
[H⁺] = 7.87 × 10⁻⁶ mol/L  •  pH = −log(7.87 × 10⁻⁶) = 6 − log(7.87) = 6 − 0.896 = 5.10

Marking notes. 1 mark — correct ICE table with correct expressions; 1 mark — correct [H⁺] calculation; 1 mark — correct pH value and 5% check shown.

Sample response. Kb(CN⁻) = Kw / Ka(HCN) = 1.0 × 10⁻¹⁴ / 6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵. A 0.100 mol/L NaCN solution is basic: Na⁺ is a spectator ion, but CN⁻ has a moderate Kb and hydrolyses water (CN⁻ + H₂O ⇌ HCN + OH⁻), producing excess OH⁻ and raising pH above 7.

Marking notes. 1 mark — Kb calculated correctly; 1 mark — correctly predicts basic solution with valid reasoning (CN⁻ hydrolyses / produces OH⁻ because HCN is a weak acid).

Sample response. Assumption: the change in [HCN] due to ionisation (x) is negligible compared to the initial concentration (x « 0.100), so [HCN] at equilibrium ≈ 0.100 mol/L. Assessment: the check gives 0.0079% ionisation — well below the 5% threshold, so the assumption is valid and does not significantly affect the accuracy of the result. The error introduced would be less than 0.04% in [H⁺].

Marking notes. 1 mark — correctly identifies and states the 5% approximation assumption; 1 mark — correctly assesses it as valid (references the «5% check value) and concludes it does not significantly affect accuracy. Accept "x « initial concentration" phrasing.

Sample response. Dental erosion occurs when the pH in the mouth falls below 5.5 — the critical pH at which hydroxyapatite, the main mineral of tooth enamel, dissolves. Ka values determine how much [H⁺] an acid generates at a given concentration and therefore predict how far pH can fall. For carbonic acid in sparkling water, Ka1(H₂CO₃) = 4.3 × 10⁻⁷. At a typical carbonation level of ~0.001 mol/L, [H⁺] ≈ √(4.3 × 10⁻⁷ × 10⁻³) ≈ 6.6 × 10⁻⁶ mol/L, giving pH ≈ 5.2 — marginally below the erosion threshold of 5.5, so sparkling water does carry some erosion risk. Cola, however, contains phosphoric acid (H₃PO₄, Ka1 = 7.5 × 10⁻³ — approximately 10⁴ times larger than Ka1 of H₂CO₃) and citric acid (Ka1 = 7.4 × 10⁻⁴). Phosphoric acid ionises to a far greater degree at equivalent concentrations, producing vastly more H⁺ and driving cola pH to 3.0–3.5 — well below 5.5. This explains why cola causes far more aggressive erosion: the higher Ka of phosphoric acid, combined with the presence of multiple acids, produces [H⁺] orders of magnitude greater than carbonic acid alone. Both contain H₂CO₃, but cola's additional acids (especially H₃PO₄ with its much larger Ka) account for the pH difference. Acetic acid (Ka = 1.8 × 10⁻⁵) in vinegar similarly produces pH around 2.4 at 1 mol/L, which is extremely corrosive to enamel — again explained by its Ka being much larger than Ka(H₂CO₃). In summary, Ka is the predictor of dental risk: acids with larger Ka values ionise more, produce higher [H⁺], and push mouth pH further below 5.5, causing more rapid enamel dissolution.

Marking notes. 1 mark — correctly links dental erosion to the pH 5.5 threshold and hydroxyapatite dissolution; 1 mark — uses Ka of H₂CO₃ to explain why sparkling water has moderate risk (pH near but below 5.5); 1 mark — identifies Ka(H₃PO₄) as much larger than Ka(H₂CO₃) (ratio ≈10⁴) and uses this to explain cola's lower pH; 1 mark — explicitly states that degree of ionisation is proportional to Ka, providing mechanism; 1 mark — uses Ka data for at least three named acids (H₂CO₃, H₃PO₄, and at least one other: citric, acetic, or tartaric); 1 mark — correctly evaluates why sparkling water and cola differ despite both containing H₂CO₃ (presence of additional stronger acids); 1 mark — reaches an explicit analytical conclusion linking Ka magnitude to dental erosion severity. Band 5–6 responses name three or more acids with Ka values and provide quantitative reasoning or calculation supporting their argument.