HSC Exam Practice

Q1.1 (3 marks)

Back titration: a technique in which a known, measured excess of one reagent is added to the sample, allowed to react completely, and then the unreacted excess is titrated with a standard solution. [1]

Two situations (any two): (i) analyte is insoluble (e.g. CaCO₃); (ii) analyte reacts too slowly for a direct endpoint to be judged; (iii) analyte is volatile or gaseous; (iv) no suitable indicator exists (coloured/turbid solution or weak/weak system). [1 each, max 2]

Q1.2 (2 marks)

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). [1] Mole ratio CaCO₃ : HCl = 1 : 2. [1]

Q1.3 (2 marks)

The acid must be in excess to guarantee that all of the insoluble CaCO₃ dissolves and reacts completely. [1] If the acid were not in excess, unreacted CaCO₃ would remain in the flask, and there would be no leftover acid to titrate in the back-titration, making the calculation impossible and giving no meaningful result. [1]

Q1.4 (3 marks)

The graph is V-shaped (check-mark). [1] Conductance starts high (H⁺ ions, κ≈350), decreases linearly to a minimum as H⁺ is replaced by Na⁺ (κ≈50), then increases linearly after the EP as excess OH⁻ (κ≈198) is added. [1] The equivalence point is the minimum conductance on the curve (the vertex of the V), found at the intersection of the two best-fit straight lines drawn through the pre-EP and post-EP portions. [1]

Q1.5 (3 marks)

Direct titration: one reaction between analyte and titrant; the analyte itself is titrated; the titre gives moles of analyte directly. [1] Back titration: two reactions — (i) analyte + excess standard reagent; (ii) excess reagent + titrant; the excess of the standard reagent is titrated. [1] The titre gives moles of excess reagent; moles of analyte are found by subtraction (n(reacted) = n(total) − n(excess)) and the mole ratio. [1]

Q2.1 (5 marks)

n(NaOH)_total = 0.0800 × 0.02000 = 1.600 × 10⁻³ mol. [1]

n(NaOH)_excess = n(HCl back-titre) = 0.0400 × 0.009650 = 3.860 × 10⁻⁴ mol. [1]

n(NaOH)_reacted = 1.600 × 10⁻³ − 3.860 × 10⁻⁴ = 1.214 × 10⁻³ mol. [1]

Mole ratio aspirin : NaOH = 1 : 2; n(aspirin) = 1.214 × 10⁻³ ÷ 2 = 6.070 × 10⁻⁴ mol. [1]

mass = 6.070 × 10⁻⁴ × 180.2 = 0.1094 g = 109.4 mg. % of label = (109.4/300) × 100 = 36.5%. [1] (Note: this deliberate data set produces a below-label result to illustrate a failed QC check.)

Q2.2 (2 marks)

Aspirin tablets are solid and dissolve slowly in water. [1] Direct titration would be difficult to perform because the endpoint would occur before the tablet has fully dissolved, making the result unreliable; the slow, heterogeneous dissolution means the indicator colour change cannot be judged precisely. [1]

Q2.3 (2 marks)

Assumption: the NaOH standard solution has not absorbed CO₂ from the atmosphere (which converts NaOH to Na₂CO₃, reducing the effective base concentration). [1] If this assumption is invalid, the true concentration of NaOH is lower than the labelled value, causing a systematic underestimate of n(NaOH)_total and therefore an underestimate of n(aspirin) — a systematic error. [1]

Q3.1 (3 marks)

EP volume = 16 mL (read from graph kink). [1] n(NaOH at EP) = 0.100 × 0.01600 = 1.600 × 10⁻³ mol. Since CH₃COOH + NaOH (1:1), n(CH₃COOH) = 1.600 × 10⁻³ mol. [1] c(CH₃COOH) = 1.600 × 10⁻³ ÷ 0.02000 = 0.0800 mol/L. [1]

Q3.2 (3 marks)

Acetic acid is a weak acid; before NaOH is added the solution contains mostly intact CH₃COOH molecules and very few ions — initial conductance is low. [1] As NaOH is added, the neutralisation reaction CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O converts non-conducting molecules into CH₃COO⁻ and Na⁺ ions, increasing the total ion concentration and hence conductance. [1] In HCl + NaOH, the solution starts with high concentrations of H⁺ (κ≈350); each H⁺ is replaced by Na⁺ (κ≈50), reducing conductance. The pre-EP direction therefore differs because HCl provides initially high-mobility H⁺ ions that are lost, whereas acetic acid provides no such high-mobility ions — only gains them as neutralisation proceeds. [1]

Q4.1 (6 marks) — Marking criteria

Band 5–6 (5–6 marks): Accurately describes how conductometric titration works (conductance measured vs volume added; no indicator required); correctly explains that the equivalence point is found from the intersection of two linear segments of the conductance-vs-volume curve; explains why indicator methods fail for coloured effluents (indicator colour masked by sample colour, making the endpoint indistinct); compares the curve shape for strong acid/strong base (V-shape, EP at minimum) versus describes the principle more generally; identifies and explains a genuine limitation (e.g. requires calibrated conductance probe; temperature must be controlled because ion mobility is temperature-dependent; does not give chemical identity of the acid only its quantity); uses precise scientific language throughout.

Band 4–5 (3–4 marks): Describes the principle adequately; correctly identifies the limitation of indicators in coloured solutions; gives EP identification; may lack precision or miss the limitation of conductometric method or the curve-shape explanation.

Band 3–4 (1–2 marks): Some relevant content such as "no indicator is needed" or "measures conductance" but missing key conceptual links (why conductance changes, how EP is found, why indicators fail).

Sample Band 6 response elements (mark allocation):

• Principle: conductometric titration measures the electrical conductance of the solution as titrant is added; conductance depends on the concentration and molar conductivity of ions present — no colour indicator is needed. [1]
• EP identification: the equivalence point is found graphically as the minimum conductance (for strong acid + strong base) or the intersection of two linear segments (slope change) by drawing best-fit lines through the pre- and post-EP portions. [1]
• Indicator limitation: in a coloured industrial effluent, the colour of an acid-base indicator (e.g. phenolphthalein or methyl orange) cannot be distinguished against the sample colour; the endpoint colour change is masked, making direct titration unreliable. [1]
• Curve shape: for a strong acid effluent + NaOH, conductance falls (H⁺, κ≈350, replaced by Na⁺, κ≈50) to a minimum at the EP, then rises (excess OH⁻, κ≈198). [1]
• Advantage summary: conductometric titration works regardless of solution colour or turbidity; the result is objective and does not require subjective colour judgement. [1]
• Limitation: the conductance probe must be calibrated and the temperature controlled (ion mobility increases with temperature); temperature variation during the titration distorts the linearity of the conductance-vs-volume curve and may shift the apparent equivalence point. Also, the method measures total ionic conductance, not specific acid identity; it cannot distinguish between two acids present simultaneously. [1]