HSC Exam Practice

Sample response. The equivalence point is the calculated stoichiometric point at which moles of acid exactly equal moles of base; it is determined from the volumes and concentrations used. The endpoint is the experimentally observed colour change of the indicator during the titration. A correctly chosen indicator makes these coincide in practice.

Marking notes. 1 mark: equivalence point = stoichiometric / calculated / moles equal. 1 mark: endpoint = indicator colour change / experimentally observed. Both required for full marks.

Sample response. Step 1: State the equivalence point pH and explain why it is above 7 (conjugate base undergoes base hydrolysis, A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻). Step 2: Name the indicator and state its transition range (e.g. phenolphthalein, 8.3–10.0). Step 3: Confirm that the transition range encompasses the EP pH (i.e. the EP pH falls within or is passed through during the sharp pH jump at equivalence).

Marking notes. 1 mark per step. Must include all three for full marks.

Sample response. At the equivalence point of a weak acid + strong base titration, the acid (HA) has been completely converted to its conjugate base (A⁻) and the cation of the strong base (e.g. Na⁺) is a spectator. The conjugate base A⁻ is the anion of a weak acid and accepts a proton from water via base hydrolysis: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq). This produces OH⁻ ions, making the solution basic (pH > 7) at equivalence.

Marking notes. 1 mark: identifying that A⁻ (conjugate base) is the species present at EP. 1 mark: correct ionic equation for base hydrolysis showing OH⁻ produced. 1 mark: linking OH⁻ production to pH > 7.

Sample response. Error: pKa is not read from the equivalence point pH. The equivalence point pH is above 7 and reflects the hydrolysis of the conjugate base — it is a different quantity from pKa. Correct method: the pKa is read from the pH at the half-equivalence point, which is at volume Vₛₖ/2. At this point exactly half the acid has been neutralised, so [HA] = [A⁻], the Henderson–Hasselbalch log term equals zero, and pH = pKa.

Marking notes. 1 mark: identifying the error (pKa ≠ EP pH). 1 mark: stating pKa = pH at Vₛₖ/2 (half-equivalence point). 1 mark: explaining why (at Vₛₖ/2: [HA] = [A⁻] so log([A⁻]/[HA]) = 0 ⇒ pH = pKa).

Sample response. Feature 1: The starting pH is higher than expected for the given concentration — a weak acid is only partially ionised, so [H⁺] < c and pH > −log(c). Feature 2: There is a buffer plateau (gradual rise) before the equivalence point, where HA and A⁻ coexist; a strong acid shows no buffer region because it has no conjugate base with buffering capacity.

Marking notes. 1 mark per feature. Accept: starting pH above what a strong acid would show; presence of buffer plateau before EP; equivalence point pH above 7 (indicates basic salt from weak acid); smaller pH jump compared to strong acid.

Sample response. All three indicators are suitable for a strong acid + strong base titration. This type produces a very large pH jump (approximately pH 4–10) centred at pH 7, so the transition ranges of all three indicators — methyl orange (3.1–4.4), BTB (6.0–7.6), and phenolphthalein (8.3–10.0) — all fall within the jump region. Any of them will give an accurate endpoint.

Marking notes. 1 mark: identifying strong acid + strong base titration. 1 mark: explaining that the large pH jump encompasses all three transition ranges.

(a) Concordant titres. Trials 1, 2, 3: 19.40, 19.42, 19.44 mL — range = 0.04 mL < 0.10 mL → concordant. Trial 4 (20.85 mL) is excluded: it differs from the concordant group by 1.41–1.45 mL, indicating a significant technique error (likely overshot endpoint or burette leak). Mean concordant titre = (19.40 + 19.42 + 19.44) / 3 = 19.42 mL. [2 marks: 1 identification + 1 mean with reason for exclusion]

(b) Concentration of tartaric acid. n(NaOH) = 0.05000 × 0.01942 = 9.710 × 10⁻⁴ mol. Mole ratio 2 NaOH : 1 H₂Tart → n(H₂Tart) = 9.710 × 10⁻⁴ / 2 = 4.855 × 10⁻⁴ mol. c(diluted) = 4.855 × 10⁻⁴ / 0.02000 = 0.02428 mol/L. Dilution factor: 10.00 mL → 100.0 mL, factor = 10.0. c(undiluted wine) = 0.02428 × 10.0 = 0.2428 mol/L. [3 marks: 1 moles NaOH; 1 mole ratio & n(acid); 1 c diluted & c undiluted with dilution factor]

(c) Mass concentration. mass conc. = 0.2428 × 150.09 = 36.4 g/L. [1 mark]

(d) Indicator justification. Step 1: EP pH ≈ 9.3 — above 7 because tartrate ion (C₄H₄O₆²⁻, conjugate base of the diprotic tartaric acid) undergoes base hydrolysis: C₄H₄O₆²⁻ + H₂O ⇌ HC₄H₄O₆⁻ + OH⁻, producing OH⁻. Step 2: Phenolphthalein — transition range pH 8.3–10.0. Step 3: EP pH 9.3 falls within phenolphthalein’s range (8.3–10.0); the indicator transitions within the sharp pH jump at equivalence, giving an accurate endpoint. Methyl orange (3.1–4.4) is completely unsuitable — it would transition far below the EP pH in the early buffer region. [3 marks: 1 per step, must include hydrolysis equation for Step 1]

(a) pKa from curve. Half-EP at V = 12.50 mL. Reading from the curve at 12.50 mL: pH ≈ 3.74. pKa(graph) = 3.74. Verification: pKa(Ka) = −log(1.8 × 10⁻⁴) = −log(1.8) + 4 = 4 − 0.255 = 3.74. Consistent — graph reading matches Ka within graphical reading precision. [2 marks: 1 reading method; 1 verification with Ka]

(b) Concentration of formic acid. Vₛₖ = 25.00 mL (midpoint of steepest section). n(NaOH) = 0.1000 × 0.02500 = 2.500 × 10⁻³ mol. HCOOH + NaOH → HCOONa + H₂O (1:1). n(HCOOH) = 2.500 × 10⁻³ mol. c(HCOOH) = 2.500 × 10⁻³ / 0.02500 = 0.1000 mol/L. [3 marks: 1 Vₛₖ identified; 1 moles with ratio; 1 concentration with correct units]

(c) Percentage error with methyl orange. Reported n(NaOH) at endpoint = 0.1000 × 0.00550 = 5.50 × 10⁻⁴ mol. Reported n(HCOOH) = 5.50 × 10⁻⁴ mol. Reported c = 5.50 × 10⁻⁴ / 0.02500 = 0.02200 mol/L. True c = 0.1000 mol/L. % error = |0.02200 − 0.1000| / 0.1000 × 100% = 78.0%. [3 marks: 1 reported concentration; 1 difference calculation; 1 % error with correct formula and sign]

Sample high-band response. The claim is incorrect on both counts: not all indicators are suitable for all titrations, and not all titrations reach equivalence at pH 7. The EP pH depends on the nature of the salt formed, which in turn depends on the relative strengths of the acid and base used. For a strong acid + strong base titration (e.g. HCl + NaOH), the salt formed is NaCl — Na⁺ and Cl⁻ are both spectator ions with no hydrolysis, so the EP pH = 7.00. All three indicators are valid here because the pH jump spans approximately pH 4–10, encompassing the ranges of methyl orange (3.1–4.4), BTB (6.0–7.6), and phenolphthalein (8.3–10.0). In contrast, for a weak acid + strong base titration (e.g. CH₃COOH + NaOH), the salt formed is sodium acetate; CH₃COO⁻ undergoes base hydrolysis: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), producing OH⁻ and raising EP pH to approximately 8.7. Here the jump is smaller and centred above pH 7; methyl orange (3.1–4.4) transitions far below the EP in the buffer region (where pKa(CH₃COOH) = 4.74), and using it would produce an endpoint at approximately 20–25% of the true titre volume, causing a ~75–80% underestimation of acid concentration. Quantitatively: if the true EP is at 25.00 mL and methyl orange gives an endpoint at 6.25 mL, reported c = (6.25/25.00) × true c = 25% of true c — a 75% underestimation. Phenolphthalein (8.3–10.0) is the only correct choice because its range encompasses the EP pH of ~8.7. For a strong acid + weak base titration (e.g. HCl + NH₃), the EP pH is below 7 because NH₄⁺ undergoes acid hydrolysis, and phenolphthalein would never change colour — only methyl orange or BTB is valid. The claim therefore fails: EP pH depends on salt hydrolysis (not always 7), and only an indicator whose transition range encompasses the EP pH will give an accurate endpoint. Selecting indicators on visual clarity or as a universal rule leads to systematic errors that can reach 75–80% in the reported concentration.

Marking notes. 1 mark — explicitly rejects the claim that EP is always pH 7, with a stated reason. 1 mark — strong + strong titration correctly described (neutral salt, EP pH 7, any indicator valid). 1 mark — weak acid + strong base correctly described (basic salt, EP pH > 7, base hydrolysis equation). 1 mark — strong acid + weak base correctly described (acidic salt, EP pH < 7, acid hydrolysis). 1 mark — phenolphthalein selected and justified for weak acid + strong base with reference to transition range. 1 mark — quantitative example of error caused by wrong indicator (calculation of % error or reporting underestimation). 1 mark — methyl orange correctly identified as unsuitable for weak acid + strong base, with explanation that it transitions in the buffer region. 1 mark — generalisation of indicator selection rule: transition range must encompass EP pH. 2 marks — sustained quality of explanation: correct use of NESA verbs (evaluate, account for), evidence-based judgement, no unsupported assertions. Deduct 1 for significant chemical error (wrong EP direction, wrong hydrolysis product).