HSC Exam Practice

Sample response. pKa = −log₁₀(Ka), where Ka is the acid dissociation constant for the equilibrium HA + H₂O ⇌ A⁻ + H₃O⁺. A lower pKa value indicates a larger Ka — the equilibrium lies further to the right and more H₃O⁺ is produced — so the compound is a stronger acid. Each unit decrease in pKa represents a 10-fold increase in acid strength.

Marking notes. 1 mark for the definition (pKa = −log₁₀Ka or equivalent). 1 mark for the interpretation (lower pKa = stronger acid, with reason: larger Ka or more ionisation).

Sample response. Brønsted–Lowry acid: CH₃COOH (ethanoic acid) — it donates a proton to water. Brønsted–Lowry base: H₂O — it accepts the proton from ethanoic acid. Conjugate base: CH₃COO⁻ (ethanoate / acetate ion) — the species remaining after ethanoic acid donates its proton.

Marking notes. 1 mark for identifying CH₃COOH as the Brønsted–Lowry acid. 1 mark for identifying H₂O as the Brønsted–Lowry base. 1 mark for naming CH₃COO⁻ as the conjugate base (ethanoate or acetate accepted).

Sample response. Test 1 — NaHCO₃: Add sodium hydrogen carbonate solution. Propanoic acid (pKa 4.87 < 6.4) reacts to produce CO₂ gas — effervescence observed; phenol (pKa 10 > 6.4) produces no CO₂. Reasoning: reaction proceeds only when the acid on the left is stronger than the acid formed (H₂CO₃, pKa 6.4). Test 2 — NaOH: Add sodium hydroxide solution. Both compounds react with NaOH (strong base deprotonates both), so this test alone does not distinguish them. However, if used after Test 1, phenol (unidentified residue) reacts with NaOH to form sodium phenoxide; propanoic acid has already been identified. A better second test is Na metal: both produce H₂ gas — cannot distinguish. Therefore the NaHCO₃ test alone unambiguously identifies propanoic acid (CO₂) and by elimination, the other sample is phenol.

Marking notes. 1 mark for NaHCO₃ test — correct observation (CO₂ from propanoic acid, not from phenol). 1 mark for correct chemical reasoning for NaHCO₃ test (pKa 4.87 < 6.4; pKa 10 > 6.4). 1 mark for identifying a second valid distinguishing test (NaOH: phenol reacts → sodium phenoxide; propanoic acid has already been identified — or a temperature/solubility test). 1 mark for correct observation/reasoning for the second test. Award 2 marks if the student correctly concludes NaHCO₃ alone unambiguously distinguishes both with valid pKa reasoning.

Sample response. When ethanoic acid loses its proton, the carboxylate ion (CH₃COO⁻) is formed. The negative charge is delocalised by resonance across both equivalent oxygen atoms of the –COO group, distributing it over two atoms and making the conjugate base very stable [1]. When ethanol loses its proton, the alkoxide ion (CH₃CH₂O⁻) is formed. There is no adjacent pi system, so no resonance is possible — the negative charge remains localised on one oxygen atom; the alkoxide is a stronger, less stable base [1]. A more stable conjugate base is a weaker base; the equilibrium HA + H₂O ⇌ A⁻ + H₃O⁺ therefore lies much further right for ethanoic acid — its Ka is much larger and its pKa (4.76) is much lower than ethanol's (~16) [1].

Marking notes. 1 mark for describing resonance stabilisation in carboxylate (delocalised across 2 equivalent O). 1 mark for contrasting with alkoxide (no resonance, charge localised on 1 O). 1 mark for linking to Ka/pKa (more stable conjugate base → weaker base → equilibrium further right → stronger acid).

Sample response. Decreasing base strength: ethylamine > aniline > ethanamide. Ethylamine (Kb ~4 × 10⁻⁴): nitrogen lone pair is in a freely available sp³ orbital with no adjacent pi system; the ethyl group donates electron density inductively, making the lone pair more electron-rich and more available to accept a proton [1]. Aniline (Kb ~4 × 10⁻¹⁰): lone pair is partially delocalised into the benzene ring's pi system via resonance — three resonance structures place electron density on the ring carbons, reducing lone pair availability [1]. Ethanamide (Kb ~10⁻¹⁵): lone pair is fully delocalised into the adjacent carbonyl C=O via resonance (amide resonance) — essentially no lone pair is available for proton acceptance; ethanamide is essentially not basic [1].

Marking notes. 1 mark each for the correct structural feature of the lone pair for each compound, in the correct decreasing order.

Sample response. The chlorine atom in chloroacetic acid is highly electronegative and withdraws electron density inductively through the sigma bonds of the C–C chain toward the –COO⁻ group [1]. This inductive withdrawal stabilises the conjugate base (ClCH₂COO⁻) — the electron-withdrawing effect partially disperses the negative charge on the carboxylate anion, making it a weaker base than CH₃COO⁻ [1]. A more stable conjugate base corresponds to a stronger acid (the ionisation equilibrium lies further right; Ka is larger); hence chloroacetic acid (pKa 2.86) is stronger than ethanoic acid (pKa 4.76). The ~100-fold difference (10^(4.76−2.86) ≈ 79) reflects how effectively one Cl atom stabilises the carboxylate [1].

Marking notes. 1 mark for identifying Cl as electron-withdrawing by induction. 1 mark for explaining how this stabilises the conjugate base ClCH₂COO⁻. 1 mark for linking to lower pKa / stronger acid (Ka relationship).

Part (a) — Ka ratio (3 marks). Ka(aspirin) = 3.16 × 10⁻⁴ (from table). Ka(ibuprofen) = 1.23 × 10⁻⁵ (from table). Ratio = Ka(aspirin) / Ka(ibuprofen) = (3.16 × 10⁻⁴) / (1.23 × 10⁻⁵) = 25.7 [1 mark for each Ka correctly read, 1 mark for correct ratio, with working shown; accept 25–26]. Interpretation: aspirin is approximately 26 times more acidic than ibuprofen — at equivalent concentration, aspirin produces ~26 times more H₃O⁺ at equilibrium [1].

Part (b) — Predominantly HA form explanation (4 marks). All three drugs have pKa values in the range 3.5–4.91. At stomach pH 2.0, the pH is lower than all three pKa values (pH 2.0 < pKa for all three) [1]. By the Brønsted–Lowry principle, when pH < pKa, the acid equilibrium HA ⇌ A⁻ + H⁺ lies to the left — the high H₃O⁺ concentration in the acidic stomach suppresses ionisation, so the neutral (HA) form predominates [1]. This is confirmed by the % ionised data: even aspirin (strongest acid, pKa 3.5) is only ~3.1% ionised at pH 2.0 [1]. Pharmacological consequence: the neutral HA form is lipid-soluble (uncharged) and can pass through the lipid bilayer of the stomach lining cells by passive diffusion; the ionised A⁻ form is hydrophilic and cannot readily cross the membrane, so drugs are absorbed in the stomach because the acidic environment maintains the neutral form [1].

Part (a) — pKa calculation (2 marks). Ka(ethylammonium) = Kw / Kb = (1.0 × 10⁻¹⁴) / (4.3 × 10⁻⁴) = 2.33 × 10⁻¹¹ [1]. pKa = −log₁₀(2.33 × 10⁻¹¹) = 10.63 ≈ 10.6 [1].

Part (b) — Reaction with ethanoic acid (3 marks). For the reaction to proceed, pKa(acid on left) must be < pKa(conjugate acid of amine) = 10.6. Ethanoic acid pKa = 4.76 < 10.6 → the reaction proceeds strongly to the right [1]. Equation: CH₃COOH + C₂H₅NH₂ → CH₃COO⁻ + C₂H₅NH₃⁺ (ethylammonium ethanoate salt) [1]. The large difference (ΔpKa = 10.6 − 4.76 = 5.84 ≈ 10^5.84 ≈ 7 × 10⁵ in favour of products) indicates the reaction is essentially complete [1].

Part (c) — Reaction with ethanol (1 mark). Ethanol pKa ≈ 16 > pKa of conjugate acid of ethylamine (10.6). The reaction would need pKa(ethanol) < 10.6, but 16 >> 10.6 — the reaction does NOT proceed to any significant extent. Ethanol is too weak an acid to protonate ethylamine under standard aqueous conditions [1].

Sample Band 6 response summary. The claim is partially supported — pKa values quantitatively describe differences in acid strength, but they do not by themselves explain why those differences exist. An explanation requires understanding of resonance stabilisation and the inductive effect.

pKa values correctly rank the three classes: carboxylic acids (~4–5) are the strongest organic acid class, phenols (~10) intermediate, and alcohols (~16) the weakest. Each unit decrease in pKa represents a 10-fold increase in Ka. For example, ethanoic acid (pKa 4.76) vs phenol (pKa 10) → Ka ratio = 10^(10−4.76) ≈ 1.7 × 10⁵, confirming carboxylic acids are ~170 000 times more acidic than phenol. So pKa values are useful descriptors and allow quantitative comparison.

However, pKa values alone do not explain the underlying cause. The mechanistic explanation requires resonance: the carboxylate (RCOO⁻) is stabilised by delocalisation of the negative charge across two equivalent oxygen atoms (high resonance stability → weak base → strong acid → low pKa). Phenoxide (C₆H₅O⁻) has partial resonance into the benzene ring (ortho and para carbons carry δ⁻) — intermediate stability → pKa ~10. Alkoxide (RO⁻) has no resonance — charge localised on one O — least stable conjugate base → strongest base → weakest acid (pKa ~16).

The inductive effect further modifies pKa within the carboxylic acid class: chloroacetic acid (pKa 2.86) is stronger than ethanoic acid (pKa 4.76) because the Cl atom withdraws electron density inductively, further stabilising the conjugate base beyond what resonance alone provides. This ~100-fold difference in Ka cannot be predicted from pKa alone — it requires the inductive effect concept.

Conclusion: pKa values are necessary (they quantify strength) but not sufficient (they don't explain mechanism). A complete account of acid strength differences requires both pKa data AND the structural reasoning of resonance stabilisation and inductive effects. The claim is partially supported: pKa describes differences but does not explain them.

Marking criteria (8 marks):

• 1 mark — Clear evaluative judgement stated upfront (e.g. "partially supported — describes but does not explain").
• 1 mark — Correctly states what pKa values do measure/describe (quantitative ranking, 10× per unit).
• 1 mark — Uses at least two specific pKa values in a valid quantitative comparison (e.g. ethanoic acid 4.76 vs phenol 10; chloroacetic acid 2.86 vs ethanoic 4.76).
• 1 mark — Describes resonance stabilisation of carboxylate (RCOO⁻) correctly — negative charge delocalised across two equivalent O atoms.
• 1 mark — Correctly describes why phenoxide (partial ring resonance) and alkoxide (no resonance) differ in stability, linked to their pKa rankings.
• 1 mark — Introduces the inductive effect with a specific example (e.g. ClCH₂COOH vs CH₃COOH) showing pKa alone does not explain within-class variation.
• 1 mark — Explicitly links conjugate base stability → base strength → Ka/pKa (the three-component argument) for at least one compound class.
• 1 mark — Justified, precise conclusion that pKa describes but does not explain, and that resonance + inductive effect are needed for complete explanation; does not overstate or understate the claim.