Chemistry • Year 11 • Module 2 • Lesson 8
Concentration in Context
Apply purity calculations, unit conversions, and contextual reasoning to real-world scenarios involving water quality, pharmaceutical and industrial chemistry.
1. Interpret water-quality data — Australian Drinking Water Guidelines
The table below shows the WHO and Australian Drinking Water Guidelines (ADWG) maximum concentration limits for four contaminants in drinking water. A chemist tests a sample from a rural reservoir. 10 marks
| Contaminant | ADWG limit (mg L−1) | Molar mass (g mol−1) | ADWG limit (mol L−1) calculate |
Reservoir reading (mg L−1) | Exceeds limit? |
|---|---|---|---|---|---|
| Fluoride (F−) | 1.50 | 19.00 | 1.20 | ||
| Lead (Pb2+) | 0.010 | 207.20 | 0.030 | ||
| Nitrate (NO3−) | 50.0 | 62.00 | 55.0 | ||
| Arsenic (As) | 0.010 | 74.92 | 0.008 |
1.1 Complete the “ADWG limit (mol L−1)” column by converting each mg L−1 value to mol L−1. Show one full calculation below. 4 marks (1 per row)
1.2 Complete the “Exceeds limit?” column by comparing the reservoir reading to the ADWG limit. State clearly which contaminants are above the permitted level. 4 marks (1 per row)
1.3 Lead (Pb) and nitrate (NO3−) have very different molar mass values (207 vs 62 g mol−1). Explain why expressing drinking-water limits in mg L−1 rather than mol L−1 can be misleading when comparing the hazard of different contaminants. 2 marks
2. Interpret graph — purity effect on calculated concentration
A student prepares five solutions of NaOH using samples with different percentage purities. Each time she dissolves the same sample mass (5.00 g) in 250 mL of solution. The graph below shows the resulting molar concentration for each purity level. 7 marks
Figure 2.1. Concentration of NaOH solution vs. percentage purity. Sample mass = 5.00 g; solution volume = 250 mL; MM(NaOH) = 39.997 g mol−1. Calculated data.
2.1 Describe the relationship between percentage purity and calculated NaOH concentration shown in the graph. 2 marks
2.2 Using the graph, estimate the NaOH concentration that would result from a sample with 88% purity. Show how you read this from the graph. 2 marks
2.3 A student preparing solutions assumes 100% purity and uses 5.00 g of a sample that is actually 90% pure. Calculate the error in concentration (as a percentage) and explain the consequence of this error in a practical laboratory context. 3 marks
3. Compare forward calculation vs back-calculation across five features
Complete the two-column table below. For each feature, write a concise description that contrasts the two calculation directions. 10 marks (1 per cell)
| Feature | Forward calculation (finding c from impure sample) | Back-calculation (finding m(sample) from target c) |
|---|---|---|
| Starting quantity | ||
| Step where purity is used | ||
| What you calculate first | ||
| Direction through n = m ÷ MM | ||
| Australian lab example |
4. Predict and justify — swimming pool scenario
The Parramatta Aquatic Centre keeps its pool chlorine between 1.0 and 3.0 ppm (mg L−1) of free chlorine (Cl2). A routine test shows the level has dropped to 0.4 ppm. The duty operator adds a 250 g dose of calcium hypochlorite (Ca(ClO)2, MM = 142.98 g mol−1) labelled as 65% pure (by mass) to the 1 500 000 L pool.
5 marks
4.1 Calculate the mass of pure Ca(ClO)2 added to the pool. 1 mark
4.2 Calculate the moles of Ca(ClO)2 added. 1 mark
4.3 Calculate the molar concentration of Ca(ClO)2 added to the pool (assume it dissolves completely in 1 500 000 L = 1.50 × 106 L). 1 mark
4.4 Predict whether this dose alone is likely to bring the pool chlorine back to within the safe range (1.0–3.0 ppm). Justify your answer by converting your molar concentration to mg L−1 of Ca(ClO)2. 2 marks
5. Case study — lead paint contamination in Sydney terrace houses
A 2022 investigation found that topsoil adjacent to pre-1970 terrace houses in Glebe, Sydney contained lead (Pb) at concentrations up to 3 200 mg kg−1 — far above the NEPM guideline of 300 mg kg−1 for residential land. (Note: for soil, mg kg−1 is equivalent to ppm by mass.) A consultant needed to express this in mol kg−1 to compare with international molar toxicology thresholds.
5 marks
5.1 Convert the measured lead concentration of 3 200 mg kg−1 to mol kg−1. (MM(Pb) = 207.20 g mol−1) 2 marks
5.2 Express the NEPM guideline of 300 mg kg−1 in mol kg−1. By what factor does the measured lead level exceed this guideline in molar terms? 2 marks
5.3 Explain why expressing the lead concentration in mol kg−1 rather than mg kg−1 might be more useful to a toxicologist studying the biological effect of lead on human tissue. 1 mark
Q1.1 — ADWG limits converted to mol L−1
F−: 1.50 ÷ 1000 = 1.50 × 10−3 g L−1; c = 1.50 × 10−3 ÷ 19.00 = 7.89 × 10−5 mol L−1.
Pb2+: 0.010 ÷ 1000 = 1.0 × 10−5 g L−1; c = 1.0 × 10−5 ÷ 207.20 = 4.83 × 10−8 mol L−1.
NO3−: 50.0 ÷ 1000 = 0.0500 g L−1; c = 0.0500 ÷ 62.00 = 8.06 × 10−4 mol L−1.
As: 0.010 ÷ 1000 = 1.0 × 10−5 g L−1; c = 1.0 × 10−5 ÷ 74.92 = 1.34 × 10−7 mol L−1.
Q1.2 — Exceeds limit?
F−: 1.20 mg/L < 1.50 mg/L → No. Pb2+: 0.030 mg/L > 0.010 mg/L → Yes, exceeds limit. NO3−: 55.0 mg/L > 50.0 mg/L → Yes, exceeds limit. As: 0.008 mg/L < 0.010 mg/L → No.
Q1.3 — mg/L vs mol/L for hazard comparison (2 marks)
Because molar mass varies greatly between contaminants, the same mass concentration (mg/L) corresponds to very different numbers of particles (moles) for different chemicals. For lead (MM 207) vs nitrate (MM 62), the same mg/L of Pb2+ means far fewer moles than for NO3−. Biological toxicity often correlates with the number of atoms or molecules (moles) interacting with enzymes, not with mass. Expressing limits in mg/L can understate the particle-level threat of heavy metals like Pb with high molar mass.
Q2.1 — Relationship description (2 marks)
The concentration increases linearly (in direct proportion) as percentage purity increases from 80% to 100%. [1] This is because higher purity means more NaOH per gram of sample; since purity is a simple multiplier of mass, the concentration increase is proportional. [1]
Q2.2 — Estimate at 88% purity (2 marks)
Reading from the graph: at 88%, the line falls between the 85% point (0.425 mol/L) and the 90% point (0.450 mol/L). By interpolation at 88%: c ≈ 0.425 + (3/5)(0.450 − 0.425) = 0.425 + 0.015 = 0.440 mol/L. [1 for reading graph; 1 for interpolation method or acceptable approximation in range 0.43–0.44 mol/L.]
Q2.3 — Error in concentration (3 marks)
Assumed (100% pure): c = 5.00 ÷ 39.997 ÷ 0.250 = 0.500 mol/L. [1]
Actual (90% pure): c = 5.00 × 0.90 ÷ 39.997 ÷ 0.250 = 0.450 mol/L. [1]
% error = (0.500 − 0.450) ÷ 0.500 × 100 = 10.0% overestimate. In a practical context (e.g. acid-base titration), using a solution assumed to be 0.500 mol/L when it is actually 0.450 mol/L will cause a 10% systematic error in any results — for example, a titration end-point volume will be reached sooner than expected, overestimating the concentration of the analyte. [1]
Q3 — Forward vs back-calculation comparison
Starting quantity: Forward = known sample mass (g). Back = target concentration (mol/L) and volume.
Purity step: Forward = multiply m(sample) × purity → m(pure), then proceed. Back = find m(pure) first, then divide by purity → m(sample).
First calculation: Forward = m(pure). Back = n = c × V.
Direction through n = m/MM: Forward = divide (m → n). Back = multiply (n → m).
Australian lab example: Forward = a student dissolves industrial NaOH in a beaker, tests the resulting concentration. Back = a chemist at a CSIRO laboratory weighing out impure reagent to prepare an exact standard solution.
Q4.1–4.4 — Swimming pool scenario
4.1 m(pure) = 250 × 0.65 = 162.5 g.
4.2 n = 162.5 ÷ 142.98 = 1.137 mol.
4.3 c = 1.137 ÷ (1.50 × 106) = 7.58 × 10−7 mol L−1.
4.4 Mass concentration = 7.58 × 10−7 × 142.98 × 1000 = 0.108 mg L−1. This is far below the minimum safe level of 1.0 ppm. The dose of 250 g is entirely insufficient to raise a 1.5 million-litre pool to the safe range — a much larger dose would be required. Note: In a real pool, free chlorine from Ca(ClO)2 releases ClO−; the 0.108 mg/L is the Ca(ClO)2 concentration; actual free chlorine released would be even lower in mass terms. Award marks for correct calculation and a reasoned conclusion.
Q5.1–5.3 — Lead paint case study
5.1 3 200 mg/kg ÷ 1000 = 3.200 g/kg; c = 3.200 ÷ 207.20 = 1.545 × 10−2 mol kg−1.
5.2 300 mg/kg ÷ 1000 = 0.300 g/kg; c = 0.300 ÷ 207.20 = 1.448 × 10−3 mol kg−1. Factor = 1.545 × 10−2 ÷ 1.448 × 10−3 ≈ 10.7× (approximately 10 times the guideline level). In mass terms 3200/300 ≈ 10.7× also, since same MM.
5.3 Biological processes (enzyme inhibition, binding to haem) respond to the number of lead atoms per unit of tissue, not their combined mass. Molar concentrations directly represent particle count and thus allow prediction of reaction stoichiometry between Pb atoms and biological molecules.