HSC Exam Practice

Sample response. The equivalence point is the theoretical point in a titration at which stoichiometrically equivalent amounts of acid and base have completely reacted according to the balanced equation; it is a chemical property of the system. The end point is the experimental observation — the point at which the indicator changes colour. The two should ideally coincide, but a small discrepancy (titration error) exists if the indicator changes colour slightly before or after the true equivalence point.

Marking notes. 1 mark for a correct definition of equivalence point (stoichiometric completion); 1 mark for a correct definition of end point (indicator colour change); 1 mark for explicitly distinguishing them (equivalence point is theoretical/chemical; end point is experimental/observed; they should ideally coincide).

Sample response. If the burette is rinsed with water before filling, residual water dilutes the titrant solution, reducing its actual concentration below the labelled value. When this diluted titrant is used, a larger volume must be added to reach the equivalence point (because fewer moles per millilitre are being delivered). This overestimates the titre. Using n = c × V with the stated (undiluted) concentration then overestimates n(titrant) and consequently overestimates the concentration of the unknown analyte. The error is systematic (always in the same direction for that burette preparation).

Marking notes. 1 mark for identifying that water dilutes the titrant (lowers its effective concentration); 1 mark for explaining that this causes a larger titre to be recorded; 1 mark for correctly stating the direction of error in c(unknown) — overestimate — with a brief reasoning chain.

Sample response. Step 1: Calculate moles of the known/standard solution using n = c × V (V must be in litres). Step 2: Use the mole ratio from the balanced equation to convert moles of the known substance to moles of the unknown: n(unknown) = n(known) × (coefficient of unknown ÷ coefficient of known). Step 3: Calculate concentration of unknown: c = n ÷ V (V in litres). Step 4 (if required): Convert moles or concentration to mass (m = n × MM) or percentage purity (% = m(pure) ÷ m(sample) × 100).

Marking notes. 1 mark per correctly stated step (formula + what it calculates). Accept equivalent symbolic expressions. Step 2 must mention the mole ratio from the balanced equation, not assume 1:1.

Sample response. Universal indicator changes colour gradually across the entire pH range 1–14. In a titration, the equivalence point is detected by a sharp, sudden colour change within a single drop of titrant. Universal indicator cannot signal this abrupt transition precisely — its gradual colour change makes it impossible to identify the end point to within ±0.05 mL, introducing large titration error. Phenolphthalein is suitable: it is colourless in acidic solution and turns pink (magenta) in alkaline solution at pH 8.3–10. It is used for strong acid–strong base or weak acid–strong base titrations.

Marking notes. 1 mark for explaining that universal indicator changes gradually (cannot pinpoint end point precisely); 1 mark for naming a suitable indicator (phenolphthalein or methyl orange); 1 mark for correctly stating the colour change and an appropriate titration type for the named indicator.

Sample response. Titres 23.10 and 23.08 mL are concordant: |23.10 − 23.08| = 0.02 mL ≤ 0.10 mL. Titre 23.45 mL is NOT concordant: |23.45 − 23.10| = 0.35 mL > 0.10 mL. The student should exclude 23.45 mL and perform an additional accurate run to obtain at least two concordant titres. Correct average = (23.10 + 23.08) ÷ 2 = 23.09 mL.

Marking notes. 1 mark for correctly identifying concordant (23.10, 23.08) and non-concordant (23.45) titres with the difference shown; 1 mark for stating the non-concordant titre must be excluded (and an additional run recommended); 1 mark for the correct average (23.09 mL).

Sample response. The mole ratio comes directly from the stoichiometric coefficients in the balanced equation; different reactions have different ratios. Assuming 1:1 without checking would give the wrong number of moles of the unknown if the true ratio is different. For example, in H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the ratio is 1:2 (H₂SO₄ : NaOH). If 1:1 is wrongly assumed, n(H₂SO₄) = n(NaOH) — this is double the correct value, giving a concentration of H₂SO₄ that is exactly twice too high.

Marking notes. 1 mark for explaining that the mole ratio is determined by the balanced equation coefficients and varies between reactions; 1 mark for a correct named example where the ratio is not 1:1 (H₂SO₄/NaOH, Na₂CO₃/HCl, Mg(OH)₂/HCl, H₃PO₄/NaOH, etc.); 1 mark for stating the consequence of the 1:1 error (answer double or triple the correct value, depending on ratio).

Sample response (a). Titres: 18.20, 18.18, 18.22 mL. Range = 18.22 − 18.18 = 0.04 mL ≤ 0.10 mL → all three concordant. Average = (18.20 + 18.18 + 18.22) ÷ 3 = 18.20 mL.

Sample response (b). Step 1: V(NaOH) = 18.20 mL = 0.01820 L. n(NaOH) = 0.04000 × 0.01820 = 7.280 × 10⁻⁴ mol.
Step 2 (mole ratio): H₂C₂O₄ : NaOH = 1:2. n(H₂C₂O₄) = 7.280 × 10⁻⁴ ÷ 2 = 3.640 × 10⁻⁴ mol.
Step 3: c(H₂C₂O₄) = 3.640 × 10⁻⁴ ÷ 0.02500 = 0.01456 mol L⁻¹ (in aliquot — this step confirmed; n(aliquot) = 3.640 × 10⁻⁴ mol as required).

Sample response (c). Scale-up: n(H₂C₂O₄) in 250.0 mL total = 3.640 × 10⁻⁴ × (250.0 ÷ 25.00) = 3.640 × 10⁻³ mol.
m(H₂C₂O₄) = 3.640 × 10⁻³ × 90.04 = 0.3278 g = 327.8 mg.
% by mass = (0.3278 ÷ 10.00) × 100 = 3.28%.

Sample response (d). One assumption: the extraction was 100% efficient — all oxalic acid in the 10.00 g of rhubarb was dissolved into solution. In reality some oxalic acid may remain bound in plant cell material and not extracted, causing the calculated % to underestimate the true content.

Marking notes. (a) 1 mark concordance check shown; 1 mark correct average. (b) 1 mark n(NaOH); 1 mark mole ratio step explicitly shown; 1 mark n(H₂C₂O₄). (c) 1 mark scale-up; 1 mark mass; 1 mark % by mass. (d) 1 mark valid extraction assumption that would cause underestimate.

Sample response. Acid–base titration is a highly accurate and precise quantitative technique when performed correctly, but its fitness for purpose depends on identifying and minimising the sources of both systematic and random error.

Systematic errors affect accuracy (closeness to the true value). The most significant sources are: (1) rinsing equipment incorrectly — if the burette is rinsed with water rather than titrant, the titrant is diluted, causing a consistently overestimated titre and an overestimated concentration of the unknown; (2) indicator choice — if the indicator’s transition pH range does not overlap the steep section of the titration curve at the equivalence point, the end point will be recorded either before or after the equivalence point, introducing a constant offset; (3) air bubbles in the burette tip — if a bubble dislodges during the titration it is counted as titrant volume, systematically overestimating the titre. These errors are addressed by: rinsing with titrant; selecting an indicator whose transition zone lies within the steep vertical section of the titration curve (for a strong acid–strong base system, both phenolphthalein at pH 8.3–10 and methyl orange at pH 3.1–4.4 work because the pH jump spans ~8 units); and ensuring no air bubbles before starting.

Random errors affect precision (reproducibility). Sources include: parallax when reading the burette meniscus; inconsistent endpoint judgment between runs (the “pink” of phenolphthalein is subjective); and slight variations in room temperature affecting solution volumes. These are minimised by: averaging concordant titres (within ±0.10 mL) — averaging three concordant titres reduces the random error by a factor of approximately √3; reading the burette at eye-level to eliminate parallax; using a white tile for consistent endpoint detection.

Fitness for ±1.5% pharmaceutical tolerance: A burette reads to ±0.05 mL on a 25 mL titre; relative uncertainty = 0.05/25 × 100 = 0.2%. With three concordant titres averaging ±0.05 mL, the relative error in titre is <0.5%. Combined with volumetric glassware uncertainties (pipette ±0.02 mL on 25 mL = 0.08%; volumetric flask ±0.1 mL on 250 mL = 0.04%), total combined uncertainty is well below 1.5%. Titration is therefore fit for purpose in pharmaceutical QA when the procedure is carefully followed. It is used industrially to verify tablet API content in exactly this role (e.g. antacid tablets, aspirin content).

Marking criteria (7 marks). [1] Identifies and correctly explains at least two sources of systematic error (e.g. water rinsing, wrong indicator, air bubble) with direction of error stated.
[1] Identifies and correctly explains at least two sources of random error (e.g. parallax, inconsistent endpoint judgment, temperature variation).
[1] Explains how concordant titres specifically reduce random error (averaging; must state “concordant” criterion of ±0.10 mL).
[1] Correctly links indicator choice to the equivalence point pH and the shape of the titration curve; names at least one indicator with its transition range.
[1] Applies quantitative or semi-quantitative reasoning to assess whether titration uncertainty (<1%) is within the ±1.5% pharmaceutical tolerance.
[1] Explicitly distinguishes systematic error (affects accuracy) from random error (affects precision) using correct scientific vocabulary.
[1] Reaches an explicit, evidence-based evaluative judgement: titration is fit for purpose when correct procedure is followed, with the key conditions stated (correct rinsing, indicator, concordant titres, eye-level reading).