HSC Exam Practice

Sample response. A back calculation is a multi-step calculation that works backwards from an experimental result (such as a titre volume or precipitate mass) to determine an unknown quantity, such as the concentration of a solution. The three-step titration pathway is: (1) find n(standard) = c(standard) × V(titre, in L); (2) apply the mole ratio from the balanced equation to find n(unknown); (3) calculate c(unknown) = n(unknown) ÷ V(flask, in L).

Marking notes. 1 mark for a correct definition of back calculation (works backwards from experimental data to find an unknown). 1 mark for steps 1 and 3 stated with correct formula and correct identification of which volume belongs to which substance. 1 mark for explicitly stating the mole ratio step (step 2) as the link between n(standard) and n(unknown).

Sample response. Concordant titres are results that agree within 0.10 mL of each other; only concordant titres are averaged for use in the back calculation. The rough titre is the first trial run, during which the student deliberately adds the titrant quickly to locate the approximate endpoint. This overshooting means the rough result is typically larger than the true endpoint volume and is therefore not representative of precise endpoint detection; it is always discarded to prevent biasing the average titre and the calculated concentration.

Marking notes. 1 mark for correctly defining concordant titres (within 0.10 mL, only these are averaged). 1 mark for explaining that the rough titre is discarded because the student overshoots deliberately to locate the endpoint region. 1 mark for explaining the consequence: including the rough titre would bias the average upward, giving a systematic error in c(unknown).

Sample response. The titre volume is the volume of standard solution dispensed from the burette; it is used to calculate n(standard) via n = c(standard) × V(titre). The flask volume is the fixed volume of unknown solution measured into the conical flask by pipette (typically 25.00 mL); it is used to calculate c(unknown) = n(unknown) ÷ V(flask). They cannot be swapped because each volume measures the amount of a different substance: the titre measures how much standard was consumed; the flask volume measures how much unknown was present. Swapping them would calculate the concentration of the standard (using the flask volume) or the moles of the unknown via the wrong formula, giving an answer that is incorrect by the factor V(flask)/V(titre).

Marking notes. 1 mark for stating that titre volume belongs to n(standard) = c × V(titre). 1 mark for stating that flask volume belongs to c(unknown) = n ÷ V(flask). 1 mark for explaining that each volume measures the amount of a different substance (standard vs unknown). 1 mark for identifying the consequence of swapping (answer off by the ratio of the two volumes, or equivalent correct explanation).

Sample response. A primary standard must have high molar mass so that a large, accurately weighable mass (e.g. >0.5 g) corresponds to a relatively small number of moles, reducing the percentage error in weighing. It must have high purity because the calculation uses n = m(weighed) ÷ M_r; any impurity means the effective moles of active standard is less than calculated, and the computed c(unknown) will be systematically too low. Example: if 0.530 g of a primary standard (M_r = 105.99) is weighed but contains 1% impurity, the true mass of active compound = 0.530 × 0.99 = 0.5247 g, giving n = 0.5247 ÷ 105.99 = 4.950 × 10⁻³ mol instead of 5.001 × 10⁻³ mol. Every subsequent step in the back calculation inherits this 1% underestimate.

Marking notes. 1 mark for explaining high molar mass reduces percentage weighing error. 1 mark for explaining that high purity ensures n(active) = n(weighed). 1 mark for a correct quantitative illustration of the impurity effect (showing the error propagation). 1 mark for stating the direction of the error (impurity → c(unknown) too low, or equivalent correct reasoning for a different impurity scenario).

Sample response. Step 1: Find n(precipitate) = m(precipitate) ÷ M_r(precipitate). Step 2: Use the mole ratio from the balanced equation to find n(unknown) = n(precipitate) × (coeff. unknown ÷ coeff. precipitate). Step 3: Calculate c(unknown) = n(unknown) ÷ V(solution, L), where V is the volume of unknown solution that was treated with the excess precipitating reagent.

Marking notes. 1 mark for step 1 (n = m ÷ M_r). 1 mark for step 2 (mole ratio applied correctly). 1 mark for step 3 (c = n ÷ V, with V being the sample volume, not any other volume).

Sample response. The precipitating reagent must be added in excess to ensure that all of the unknown ion reacts completely and is fully precipitated. If it is not in excess, some of the unknown ion remains in solution; the precipitate mass would be less than it should be, n(unknown) would be underestimated, and c(unknown) would be lower than the true value.

Marking notes. 1 mark for stating that excess ensures complete precipitation of all the unknown ion. 1 mark for explaining the consequence of not using excess (some unknown remains in solution → precipitate mass too low → c(unknown) underestimated).

Sample response (a). M_r(BaSO₄) = 137.33 + 32.06 + 4(15.999) = 137.33 + 32.06 + 63.996 = 233.39 g/mol. n(BaSO₄) = 0.4666 ÷ 233.39 = 2.000 × 10⁻³ mol.

Sample response (b). Mole ratio Ba²⁺:SO₄²⁻:BaSO₄ = 1:1:1, so n(SO₄²⁻) = 2.000 × 10⁻³ mol. c(SO₄²⁻) = 2.000 × 10⁻³ ÷ 0.05000 = 0.04000 mol/L.

Sample response (c). Mass of SO₄²⁻ per litre = 0.04000 mol/L × 96.06 g/mol = 3.842 g/L = 3842 mg/L. This exceeds the guideline of 500 mg/L by a factor of approximately 7.7. The river water does not meet the Australian Drinking Water Guideline for sulfate.

Sample response (d). BaCl₂ in excess ensures all SO₄²⁻ ions are completely precipitated; if it were used in stoichiometric quantities, the reaction might not go to completion and some sulfate would remain in solution, underestimating the concentration. One source of error causing a higher-than-true BaSO₄ mass: incomplete washing of the precipitate — other ionic species (e.g. BaCl₂ crystals, BaCO₃ from dissolved CO₂) may co-precipitate with BaSO₄, adding to the measured mass and overestimating c(SO₄²⁻). Accept also: some Ba(NO₃)₂ or other barium salts co-precipitating from impure BaCl₂; or moisture not fully removed from precipitate during drying.

Marking notes. (a): 1 mark M_r(BaSO₄) = 233.39; 1 mark n = 2.000 × 10⁻³ mol. (b): 1 mark n(SO₄²⁻) = 2.000 × 10⁻³ mol via 1:1 ratio; 1 mark c = 0.04000 mol/L. (c): 1 mark correct conversion to mg/L (3842 mg/L); 1 mark correct comparison to guideline (exceeds 500 mg/L). (d): 1 mark excess ensures complete precipitation; 1 mark one valid source of high-error with explanation.

Sample response. Back-calculation titration is a widely used technique in analytical chemistry because it allows the concentration of an unknown solution to be determined with high accuracy and reasonable precision when performed with appropriate precautions. In Australia, titration underpins quality control in food manufacturing (e.g. acidity testing of juices and vinegars under FSANZ standards), water treatment (monitoring NaOH concentrations for pH adjustment at Sydney Water facilities), and pharmaceutical compounding (verification of acid concentration in drug preparation). Its accuracy depends on the precision of the primary standard, the quality of glassware, and the correct identification of concordant titres. The major sources of random error in titration are: (1) endpoint detection — the colour change of an indicator spans a range of pH, so different observers may stop the titration at slightly different points; (2) parallax in burette reading — reading the meniscus at the wrong eye level introduces a systematic offset. Precautions that minimise these include: using concordant titres (within 0.10 mL) to identify and exclude outliers; reading the burette at eye level to the bottom of the meniscus; using a standardised indicator concentration; and employing a white tile beneath the conical flask to improve colour detection. A further source of error is the purity and stability of the primary standard; this is minimised by selecting a primary standard with high M_r, high purity (>99.9%), and low hygroscopicity. Compared to gravimetric analysis, titration offers faster throughput and is better suited to routine high-volume work. Gravimetric analysis generally provides lower relative error for very small analyte concentrations because precipitation can concentrate the analyte into a weighable mass; however, it is more time-consuming (filtering, drying, ignition steps) and subject to co-precipitation errors. For routine quantitative work in industrial and environmental chemistry at typical analyte concentrations (>0.01 mol/L), back-calculation titration with a well-standardised solution offers excellent precision (relative standard deviation <0.5%) and is the preferred method. In summary, back-calculation titration is accurate and reliable when concordant titre practice is followed and a high-quality primary standard is used, but gravimetric analysis is preferred when absolute certainty of complete reaction is needed or at trace concentration levels.

Marking criteria (6 marks). 1 = identifies at least two main sources of error in titration (endpoint detection, parallax, primary standard purity, or equipment calibration) with explanation. 1 = correctly describes at least two precautions that minimise identified errors (concordant titres, eye-level reading, indicator standardisation, or equivalent). 1 = named Australian industrial, regulatory, or environmental context used correctly. 1 = explains how accuracy is different from precision in this context (i.e. systematic vs random errors addressed separately). 1 = makes a valid comparison of titration vs gravimetric analysis in terms of precision, speed, or application suitability. 1 = reaches an explicit evaluative judgement about when back-calculation titration is preferred over gravimetric analysis, with evidence-based reasoning (not a simple preference statement).