Chemistry • Year 11 • Module 1 • Lesson 5
IQ1 Consolidation — Classification and Separation
Build HSC Band 5–6 extended-response technique on gravimetric analysis calculations, technique evaluation, and experimental design in real-world contexts.
1. Multi-step calculation — chloride in Sydney drinking water (Band 4–5)
8 marks Band 4–5
Scenario. The NSW EPA requires that drinking water contain no more than 250 mg/L of chloride ions (Cl−). A water quality officer tests a 200 mL sample of tap water from the Blue Mountains. She adds excess silver nitrate (AgNO3) solution, forming a white silver chloride (AgCl) precipitate. After filtering, washing and drying to constant mass, the dry precipitate weighs 0.1434 g.
Molar masses: Ag = 107.9 g mol−1; Cl = 35.5 g mol−1; N = 14.0 g mol−1; O = 16.0 g mol−1. So M(AgCl) = 143.4 g mol−1; M(Cl−) = 35.5 g mol−1.
(a) Write the net ionic equation for the precipitation reaction. (1 mark)
(b) Calculate the number of moles of AgCl precipitate collected. Show all working. (1 mark)
(c) Apply the molar ratio to find the moles of Cl− in the 200 mL sample. (1 mark)
(d) Calculate the mass of Cl− in the 200 mL sample in milligrams, then convert to a concentration in mg/L. Show full working. (2 marks)
(e) State whether the sample complies with the NSW EPA limit of 250 mg/L and justify your answer using your calculated value. (1 mark)
(f) The officer is concerned that bromide ions (Br−) may also be present in the water. Br− also reacts with Ag+ to form the insoluble precipitate AgBr. Explain how co-precipitation of AgBr would affect the calculated Cl− concentration, and state whether it would cause an overestimate or underestimate. (2 marks)
2. Scenario + evaluation — separating components of sea water at Port Kembla (Band 5–6)
8 marks Band 5–6
Scenario. A Year 11 student at Port Kembla collects a 500 mL sample of sea water. She observes that the water contains: dissolved NaCl (~35 g/L), dissolved MgCl2 (~5 g/L), fine suspended sand and algae, and traces of immiscible oil on the surface. The student must: (A) obtain the dry sand; (B) obtain pure NaCl crystals; (C) determine the exact concentration of Cl− ions; (D) explain why simple distillation is not the best technique for obtaining pure NaCl from sea water. The table below shows solubility data for NaCl and MgCl2 at different temperatures.
| Temperature (°C) | Solubility of NaCl (g per 100 mL water) | Solubility of MgCl2 (g per 100 mL water) |
|---|---|---|
| 20 | 35.9 | 54.6 |
| 40 | 36.4 | 57.5 |
| 60 | 37.1 | 61.0 |
| 80 | 38.0 | 66.8 |
| 100 | 39.2 | 73.3 |
| 0 (ice water) | 35.6 | 52.8 |
Solubility data: CRC Handbook of Chemistry and Physics, 95th edition (illustrative values).
Q2. Analyse, justify and evaluate the techniques required for tasks A–D. In your response you must:
- For (A): identify the correct technique, state the key property exploited, and describe the procedure in at least two steps.
- For (B): identify the correct technique, use the solubility data to justify why it works, and describe the procedure including a sequential step that must happen first.
- For (C): identify gravimetric analysis as the correct technique and outline the full procedure, including why excess reagent is used and why ashless filter paper is specified.
- For (D): use the solubility data and/or boiling point reasoning to explain why simple distillation is inappropriate for obtaining pure solid NaCl (two reasons).
- State one limitation of gravimetric analysis in this specific sea water context and how it could be addressed.
Q1 — Multi-step calculation (8 marks)
(a) Net ionic equation: Ag+(aq) + Cl−(aq) → AgCl(s). [1 mark — correct ions, correct state symbols, arrow not ⇌]
(b) n(AgCl) = mass ÷ M = 0.1434 ÷ 143.4 = 1.000 × 10−3 mol = 0.001000 mol. [1 mark for correct calculation]
(c) From the 1:1 molar ratio in the equation: n(Cl−) = n(AgCl) = 0.001000 mol. [1 mark]
(d) m(Cl−) = n × M = 0.001000 × 35.5 = 0.0355 g = 35.5 mg [1 mark for correct mass]. Concentration = mass ÷ volume = 35.5 mg ÷ 0.200 L = 177.5 mg/L ≈ 178 mg/L [1 mark for correct conversion and concentration].
(e) The sample complies with the NSW EPA limit: 178 mg/L < 250 mg/L. [1 mark]
(f) If AgBr also precipitates, the total mass of precipitate collected will be higher than the mass of AgCl alone. Since the calculated result uses the total precipitate mass as if it were all AgCl, the calculation will attribute the extra mass to Cl−. This leads to an overestimate of the Cl− concentration [1 mark]. To address this: pre-treat the sample to remove Br− before the AgNO3 addition, or use a technique specific for Cl− only (e.g. potentiometric titration with AgNO3 and a Cl−-selective electrode) [1 mark].
Marking criteria summary: (a) 1; (b) 1; (c) 1; (d) 2 (mass calculation + unit conversion); (e) 1; (f) 2 (identifies overestimate with reasoning + addresses the problem). Total 8 marks.
Q2 — Band 6 sample response (8 marks)
(A) Obtaining dry sand: Technique = filtration. Key property: particle size — sand is insoluble and the particles are large enough to be retained by filter paper. Procedure: (1) Pour the sea water through a funnel fitted with filter paper into a conical flask; sand is collected as residue on the filter paper. (2) Wash the sand on the filter paper with distilled water to remove residual salt, then transfer to a watch glass and dry in an oven. [1 mark — correct technique + property + at least 2 procedure steps]
(B) Pure NaCl crystals: First, filter the sea water to remove sand (sequential step), so that the crystals are not contaminated. Technique = crystallisation [1 mark]. Justification using solubility data: NaCl solubility changes by only 35.9 → 35.6 g/100 mL from 20 °C to 0 °C — a very small change. However, by concentrating the solution first (heating to reduce water volume), the NaCl concentration can exceed its solubility and it will crystallise. MgCl2 is more soluble (52.8 g/100 mL at 0 °C) and remains in solution, so NaCl crystals can be collected with reduced MgCl2 contamination. Repeat crystallisation (recrystallisation) to improve purity [1 mark for reasoning using data].
(C) Gravimetric analysis for Cl−: Procedure: (1) Add excess AgNO3 to the sea water sample — excess ensures all Cl− (and Br−) precipitate; insufficient reagent causes incomplete precipitation and underestimates the true Cl−. (2) Filter using ashless filter paper — ordinary filter paper contains cellulose that would add carbon mass when the paper is ignited during the gravimetric procedure; ashless paper burns to leave no residue. (3) Wash, dry to constant mass, weigh, use stoichiometry with the balanced equation Ag+(aq) + Cl−(aq) → AgCl(s) and molar masses to calculate [Cl−]. [2 marks: 1 for excess reagent justification + ashless paper reason; 1 for full procedural sequence with stoichiometry step]
(D) Why simple distillation is inappropriate for pure solid NaCl: Reason 1 — Simple distillation removes water as vapour and leaves the dissolved solids behind; this would concentrate NaCl and MgCl2 together in the residue, not separate them. Since both NaCl and MgCl2 are non-volatile, they both remain in the flask — the technique produces a mixed salt residue, not pure NaCl [1 mark]. Reason 2 — Simple distillation requires a large boiling point difference between the components to separate them; here, the “component” to remove is water (b.p. 100 °C) versus solid NaCl (m.p. 801 °C, non-volatile), so technically boiling off water does separate water from salt — but it does not separate NaCl from MgCl2, which is the purity problem. Additionally, boiling away large volumes of sea water uses enormous energy, making it impractical compared to crystallisation [1 mark].
Limitation of gravimetric analysis in sea water: Sea water contains multiple halide ions (Cl−, Br−, I−), all of which precipitate as insoluble silver salts with AgNO3, causing co-precipitation and overestimation of Cl− concentration. This can be addressed by adding a mask for Br−/I− (e.g. excess AgF, which specifically reacts with Br− and I− to convert them to AgBr/AgI before the gravimetric step), or by using an instrumental technique such as ion chromatography which can distinguish all halide ions individually [1 mark].
Marking criteria summary (8 marks): (A) technique + property + 2 steps [1]; (B) technique + solubility data justification + sequential filtration step [2]; (C) excess reagent + ashless paper reason + full procedure [2]; (D) two valid reasons why distillation fails [2]; limitation + solution [1]. Total 8 marks.