HSC Exam Practice

Sample response. A primary standard is a substance of very high purity, known molar mass, and stability in air that is used to prepare a solution of precisely known concentration. A standard solution is a solution whose concentration is known precisely. The relationship is that a standard solution is prepared from a primary standard: the mass of the primary standard is weighed, dissolved, and made up to an exact volume in a volumetric flask, giving a solution with a precisely calculable concentration.

Marking notes. 1 mark for a correct definition of primary standard (substance of high purity/stability used to make a standard solution); 1 mark for a correct definition of standard solution (precisely known concentration); 1 mark for correctly stating the relationship (standard solution is prepared from a primary standard / the primary standard is weighed and dissolved to make a standard solution).

Sample response. (1) High purity — impurities alter the actual moles present, making the calculated concentration incorrect. (2) Stable in air (not hygroscopic, not reactive with O₂, CO₂, or H₂O) — the composition must not change between weighing and dissolving. (3) High molar mass — more grams per mole are needed, so a small balance error is a smaller percentage of the total mass. (4) Readily soluble in water — must dissolve completely to give a homogeneous solution of defined concentration.

Marking notes. 1 mark per property correctly named. Accept minor variations in wording. A property that is listed without any qualifier (e.g. just “pure” without “high” or “very”) may receive the mark if context makes it clear the student means analytical-grade purity.

Sample response. When a solution is diluted, only solvent (water) is added — no solute is added or removed. Therefore the number of moles of solute before dilution (n₁) equals the number of moles after dilution (n₂): n₁ = n₂. Since n = cV for any solution, this becomes c₁V₁ = c₂V₂. The concentration decreases because the same number of moles is now spread through a larger volume.

Marking notes. 1 mark for stating that only water is added so moles of solute are unchanged (n₁ = n₂); 1 mark for substituting n = cV on each side to derive c₁V₁ = c₂V₂; 1 mark for explaining that concentration decreases because the same moles are distributed in a larger volume.

Sample response. V₂ is the total final volume of the diluted solution, whereas the “volume of water added” is only the volume of solvent added. V₂ equals V₁ plus the volume of water added. In the example: V₁ = 30.0 mL; volume of water added = 120 mL; V₂ = 30.0 + 120 = 150 mL (not 120 mL). Using 120 mL as V₂ would give an incorrect (too high) final concentration.

Marking notes. 1 mark for defining V₂ as total final volume (not volume of water added); 1 mark for correctly calculating V₂ = 150 mL using the given example; 1 mark for explaining the consequence of the error (using V_water instead of V₂ gives an overestimate of the final concentration).

Sample response. Concentrated HCl cannot be used as a primary standard for at least two reasons. First, it fails the purity criterion: concentrated HCl is a solution of HCl gas in water whose concentration varies with temperature, age, and storage; the mass you weigh out does not correspond to a fixed number of moles of HCl. Second, it fails the stability criterion: HCl is volatile and loses HCl gas when opened, continuously changing its concentration. Third, it fails the high molar mass criterion: MM = 36.46 g mol⁻¹, so very small masses would be required, greatly amplifying percentage weighing error. (Any two of these three, well explained, earn full marks.)

Marking notes. 1 mark for each correctly named property that HCl fails, with a specific explanation of why. Maximum 2 marks for properties. 1 additional mark for precise language (e.g. “variable concentration,” “volatile,” “percentage weighing error”). Do not award marks for listing a property without explaining why HCl fails it.

Sample response. A volumetric flask is used because it is calibrated to contain a precise volume (±0.1 mL or better) at a specific temperature. Its long, narrow neck allows the bottom of the meniscus to be read very accurately when making up to the calibration mark. A measuring cylinder is accurate only to ±1–2 mL — insufficient for preparing a standard solution.

Calculation: n = m ÷ MM = 2.140 ÷ 214.00 = 0.01000 mol. V = 500.0 mL = 0.5000 L. c = n ÷ V = 0.01000 ÷ 0.5000 = 0.02000 mol L⁻¹.

Marking notes. 1 mark for explaining volumetric flask accuracy (calibrated, narrow neck, precise meniscus reading); 1 mark for contrasting with measuring cylinder (±1–2 mL); 1 mark for correct calculation of n (0.01000 mol); 1 mark for correct c (0.02000 mol L⁻¹).

Sample response (a) — concentrations.

Step 1 (full working): c₁ = 0.500; V₁ = 25.0 mL; V₂ = 250 mL. c₂ = (0.500 × 25.0) ÷ 250 = 0.0500 mol L⁻¹. [1 mark for working; 1 mark for correct answer]

Step 2: c = (0.0500 × 25.0) ÷ 250 = 0.00500 mol L⁻¹. [0.5 mark]

Step 3: c = (0.00500 × 25.0) ÷ 250 = 5.00 × 10⁻⁴ mol L⁻¹. [0.5 mark] (Accept 0.000500 mol L⁻¹.)

Sample response (b) — moles in 10.0 mL of Step 3.

n = c × V = 5.00 × 10⁻⁴ × 0.0100 = 5.00 × 10⁻⁶ mol. [1 mark]

Sample response (c) — evaluation of the student’s claim.

The student’s claim is partially correct but imprecise. The concentration of paracetamol at Step 3 (5.00 × 10⁻⁴ mol L⁻¹) is much less than at Step 0 (0.500 mol L⁻¹), so in that sense there is less paracetamol per litre of solution [1]. However, if the claim means there are fewer moles of paracetamol in the entire 250 mL flask, this is also technically true — the 25 mL aliquot transferred at each step carries only a fraction of the previous flask’s moles [1]. The key principle is that moles are conserved within the aliquot transferred: the moles entering each flask equal the moles in the 25 mL portion used. No paracetamol is destroyed during dilution — it is simply spread through larger and larger volumes [1].

Marking notes. Part (a): 1 mark for Step 1 with correct c₁V₁ = c₂V₂ working; 1 mark for correct Step 1 answer; 1 mark for both Step 2 and Step 3 answers correct. Part (b): 1 mark for correct n. Part (c): 3 marks as annotated above.

Sample response.

Calculations. Each step halves the concentration (takes 125 mL of current solution and makes to 250 mL). Using c₁V₁ = c₂V₂:

Step 1: V₁ = (0.500 × 250) ÷ 1.00 = 125 mL of 1.00 mol L⁻¹ stock → 250 mL at 0.500 mol L⁻¹. Step 2: V₁ = (0.250 × 250) ÷ 0.500 = 125 mL of Step 1 → 250 mL at 0.250 mol L⁻¹. Step 3: 125 mL of Step 2 → 250 mL at 0.125 mol L⁻¹. Step 4: 125 mL of Step 3 → 250 mL at 0.0625 mol L⁻¹. [2 marks for full, correct set of calculations; 1 mark if 3 out of 4 are correct]

Procedure for accuracy at each step. (1) Rinse the pipette with the source solution before use so no carry-over contamination dilutes the aliquot. (2) Use a calibrated volumetric pipette (not measuring cylinder) to transfer exactly 125 mL. (3) Transfer to a clean, labelled 250 mL volumetric flask. (4) Add distilled water to just below the calibration mark, then use a dropper pipette to bring the meniscus to the mark. (5) Stopper and invert at least 10 times to ensure uniform mixing before the next step. [2 marks for ≥4 specific accuracy measures; 1 mark for 2–3 measures]

Error accumulation. Each pipetting step introduces a small random error (e.g. ±0.1 mL in a 125 mL transfer = ±0.08%). Because Step 2 uses the solution from Step 1, any error in Step 1 is compounded into Step 2, and so on. By Step 4, the error in concentration is approximately the sum of four independent pipetting errors. The implication for the forensic laboratory is that the concentration of the 0.0625 mol L⁻¹ standard is less precisely known than the 0.500 mol L⁻¹ standard, which could affect calibration accuracy for low-concentration breathalyser readings. [2 marks: 1 for explaining cumulative error mechanism; 1 for implication for precision of final solution]

Comparison with independent preparation. Preparing each solution independently from the 1.00 mol L⁻¹ stock requires four separate c₁V₁ = c₂V₂ calculations and four pipetting operations directly from the same stock; errors in each do not carry over to the others. This means each standard is independently accurate — an error preparing the 0.500 mol L⁻¹ solution does not affect the 0.0625 mol L⁻¹ solution. The disadvantage is that more distilled water and pipetting time is required, and the calculations are slightly more complex for each step. Serial dilution is faster and more efficient when the same dilution factor is applied at each step; independent preparation gives higher independent precision for each standard, which is preferable for critical forensic applications. [2 marks: 1 for comparing error independence; 1 for evaluative judgement with reasoning about which method is preferable for this context]

Quality of language. [1 mark if response uses precise chemical terminology throughout: c₁V₁ = c₂V₂, meniscus, calibration mark, aliquot, volumetric pipette, moles conserved, cumulative error, precision, standard solution.]

Marking criteria summary (9 marks): 2 = full, correct set of four calculations (1 if 3/4 correct); 2 = four or more specific procedural accuracy measures described; 2 = cumulative error mechanism identified and implication for forensic precision discussed; 2 = comparison of serial dilution vs independent preparation including evaluative judgement; 1 = precise chemical terminology throughout.