HSC Exam Practice

Sample response. A standard solution is a solution of accurately known concentration, prepared from a precisely weighed primary standard and used as the reference reagent in a titration to determine the concentration of an unknown solution.

Marking notes. 1 mark for “accurately known concentration”; 1 mark for linking it to quantitative analysis / use as a reference in titration. Do not accept “exact” without “known” or “accurately”.

Sample response. A primary standard must be: (1) high purity (≥99.9%); (2) chemically stable (does not decompose or react with air); (3) non-hygroscopic (does not absorb water vapour); (4) high molar mass (to minimise percentage error from weighing); and (5) readily soluble in water. Any four of these five criteria for full marks.

Marking notes. 1 mark per criterion correctly stated. Accept slight paraphrasing. “Pure” alone without a quantitative threshold or reason is acceptable for 1 mark. Do not accept “strong acid/base” as a criterion.

Sample response. NaOH cannot be a primary standard for three reasons: (i) it reacts with CO₂ in air (NaOH + CO₂ → Na₂CO₃ + H₂O), reducing its purity and changing its effective concentration; (ii) it is hygroscopic, absorbing water vapour from air, which increases its mass without increasing the moles of NaOH, making the weighed mass an overestimate of the moles available; (iii) it has a low molar mass (M = 40.0 g/mol), so small weighing errors represent a large percentage error in the moles calculated. Because these properties violate the primary standard criteria, NaOH solutions must be standardised against a suitable primary standard before use.

Marking notes. 1 mark for CO₂ absorption / purity loss (with equation or specific chemical statement); 1 mark for hygroscopicity / water vapour absorption; 1 mark for a third valid reason (low molar mass, or a clear consequence statement that the concentration is unreliable). Do not award marks for “NaOH is a strong base” as a reason.

Sample response. The equivalence point is the theoretical point in a titration at which stoichiometrically equivalent amounts of acid and base have completely reacted, as calculated from the balanced equation. The endpoint is the experimentally observed point where the indicator changes colour, signalling the approximate completion of the reaction. When the indicator is correctly chosen (its pK_In lies within 1 unit of the equivalence-point pH), the endpoint coincides with the equivalence point; if the wrong indicator is used, these differ, introducing a systematic error.

Marking notes. 1 mark for equivalence point defined as stoichiometric completion (theoretical); 1 mark for endpoint defined as observed indicator colour change; 1 mark for explaining the relationship between them and when they coincide / when they differ.

Sample response. (1) Rinse the burette with the titrant solution before filling: residual distilled water from washing would dilute the titrant, reducing its effective concentration and causing the titre to be too large, overestimating the unknown concentration. (2) Read the burette at the bottom of the meniscus with the eye level with the liquid surface: reading from above or below introduces parallax error, causing a systematic over- or underestimate in all volume readings and introducing a systematic error into the titre.

Marking notes. For each technique: 1 mark for correctly describing what must be done; 1 mark for correctly explaining why (quantitative reason linked to the error that would result if ignored). 2 × 2 = 4 marks. Accept any two of: rinsing with titrant, removing air bubbles from the tip, reading at the bottom of the meniscus, recording to two decimal places.

Sample response. Step 1: Write the balanced equation and identify the mole ratio of the known to the unknown substance. Step 2: Calculate the moles of the known (standard) solution: n(known) = c(known) × V(known), where V must be converted to litres. Step 3: Use the mole ratio to find the moles of the unknown: n(unknown) = n(known) × (coefficient of unknown / coefficient of known). Step 4: Calculate the concentration of the unknown: c(unknown) = n(unknown) / V(unknown), where V is again in litres.

Marking notes. 1 mark per step correctly outlined. Must mention: Step 2 — V in litres; Step 3 — mole ratio from balanced equation; Step 4 — divides by V. Do not award Step 1 mark if student states “ratio is always 1:1”.

Sample response (a) — titres and concordance. T₁ = 20.65 − 0.10 = 20.55 mL; T₂ = 20.70 − 0.05 = 20.65 mL; T₃ = 21.25 − 0.20 = 21.05 mL. Concordance check: T₂ − T₁ = 0.10 mL ≤ 0.10 mL (concordant); T₃ − T₁ = 0.50 mL > 0.10 mL; T₃ − T₂ = 0.40 mL > 0.10 mL ⇒ T₃ is non-concordant. Concordant: T₁ and T₂. Average = (20.55 + 20.65)/2 = 20.60 mL = 0.02060 L.

Marking notes (a). 1 mark per correctly calculated titre (3 marks); 1 mark for correctly identifying non-concordant T₃ with supporting arithmetic; 1 mark for correct average of T₁ and T₂.

Sample response (b) — four-step calculation. n(Na₂CO₃) total = 1.060/106.0 = 1.000×10⁻² mol in 200.0 mL. c(Na₂CO₃) = 1.000×10⁻²/0.2000 = 0.05000 mol/L. Step 1: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; ratio Na₂CO₃:HCl = 1:2. Step 2: n(Na₂CO₃) = 0.05000 × 0.02500 = 1.250×10⁻³ mol. Step 3: n(HCl) = 1.250×10⁻³ × 2 = 2.500×10⁻³ mol. Step 4: c(HCl) = 2.500×10⁻³ / 0.02060 = 0.1214 mol/L.

Marking notes (b). 1 mark for correct c(Na₂CO₃); 1 mark for correct mole ratio applied in Step 3 (1:2, n(HCl) = 2 × n(Na₂CO₃)); 1 mark for correct final c(HCl) using the concordant average titre. Mark consequentially if concordant average from (a) is different.

Sample response (a) — c(CH₃COOH). Step 1: CH₃COOH + NaOH → CH₃COONa + H₂O; mole ratio 1:1. Step 2: n(NaOH) = 0.5000 × 0.01720 = 8.600×10⁻³ mol. Step 3: n(CH₃COOH) = 8.600×10⁻³ mol (ratio 1:1). Step 4: c(CH₃COOH) = 8.600×10⁻³ / 0.01000 = 0.8600 mol/L. [3 marks: 1 for balanced equation and ratio, 1 for n(NaOH) with correct unit conversion, 1 for correct c(CH₃COOH).]

Sample response (b) — percentage by mass. mass(CH₃COOH) = 8.600×10⁻³ × 60.06 = 0.5165 g. mass(vinegar sample) = 10.00 × 1.004 = 10.04 g. % by mass = (0.5165/10.04) × 100 = 5.14%. The result of 5.14% is above the label claim of 4.8%, so the vinegar is more acidic than labelled (or the result is inconsistent with the label). [3 marks: 1 for correct mass(CH₃COOH); 1 for correct mass(vinegar) using density; 1 for correct % and comparison statement.]

Sample response (c) — systematic error direction. If the conical flask was rinsed with the vinegar (analyte) before adding the pipetted volume, extra moles of CH₃COOH would remain on the flask walls in addition to the pipetted 10.00 mL. This extra acid would require more NaOH to neutralise, so the titre would be too large → n(NaOH) calculated in Step 2 is too large → n(CH₃COOH) in Step 3 is too large → c(CH₃COOH) in Step 4 is overestimated → mass(CH₃COOH) is overestimated → percentage by mass is too high. [2 marks: 1 for correct specific error named; 1 for complete directional chain from error to overestimated %.]

Sample response. Accurate titration results require two interlocking foundations: a reliable starting concentration (established through primary standard selection) and correct procedural technique throughout the experiment. A primary standard must satisfy five criteria: high purity, chemical stability (must not decompose or react with air), non-hygroscopic, high molar mass, and ready solubility. Anhydrous Na₂CO₃ (M = 106.0 g/mol) satisfies all five and is used to standardise HCl; oxalic acid dihydrate (M = 126.1 g/mol) satisfies all five and is used to standardise NaOH. NaOH cannot itself be a primary standard because it absorbs both CO₂ (reducing its purity) and water vapour (increasing its apparent mass without increasing moles), violating the purity and stability criteria. If the NaOH concentration is unknown or unstable, every subsequent calculation is unreliable regardless of how carefully the titration is performed.

Correct titration technique addresses three key areas. First, the burette must be rinsed with the titrant before filling: if residual water remains, the titrant is diluted below its nominal concentration, the titre becomes too large, and the calculated concentration of the unknown is overestimated. Second, air bubbles in the burette tip must be eliminated before starting: a bubble released during the titration adds to the apparent volume delivered, inflating the titre and overestimating the unknown concentration. Third, the burette must be read at the bottom of the meniscus with the eye level with the surface to eliminate parallax error — reading from above causes the meniscus to appear lower, underestimating the volume; reading from below causes the opposite. Each of these errors propagates systematically through the four-step calculation: an erroneously large titre increases n(standard) in Step 2, inflates n(unknown) in Step 3 via the mole ratio, and inflates c(unknown) in Step 4. Concordant results (within ±0.10 mL) provide a check that random technique errors are not dominating the data. Overall, primary standard selection establishes the accuracy of the reference concentration, and correct technique ensures the delivered volume accurately reflects the moles of standard used — both are essential and neither alone is sufficient.

Marking notes. 1 mark — names a specific primary standard with at least two criteria. 1 mark — explains why NaOH fails (CO₂ absorption or hygroscopicity with correct reasoning). 1 mark — links primary standard quality to the reliability of all subsequent calculations. 1 mark — first technique error named with correct direction of effect on c(unknown). 1 mark — second technique error named with correct direction of effect on c(unknown). 1 mark — traces the error through the four-step calculation (titre → n → n(unknown) → c). 1 mark — explains the role of concordant results in identifying technique errors. 1 mark — reaches a substantive evaluative conclusion that integrates both dimensions (primary standard selection + technique) as jointly necessary for accuracy.