Chemistry • Year 11 • Module 2 • Lesson 8
Concentration in Context
Build HSC Band 5–6 extended-response technique on purity calculations, multi-step reasoning, and evaluating chemical claims in real-world contexts.
1. Data + scenario: fluoride water treatment at a Queensland plant (Band 5–6)
8 marks Band 5–6
Scenario. A water treatment plant in regional Queensland adjusts the fluoride (F−) concentration in drinking water to the recommended level of 1.00 mg L−1 by adding sodium fluoride (NaF) solution. The plant processes 4.00 × 106 L of water per day. The available NaF powder is 92.0% pure by mass. The plant manager needs to calculate how much NaF powder to add each day to achieve the target concentration. Relevant molar masses: Na = 22.990, F = 18.998.
| Parameter | Value |
|---|---|
| Target [F−] | 1.00 mg L−1 |
| Daily water volume | 4.00 × 106 L |
| NaF purity | 92.0% |
| MM(NaF) | 41.988 g mol−1 |
| WHO guideline (upper limit) | 1.50 mg L−1 |
Q1. Analyse and evaluate the calculation required by the plant manager. In your response you must:
- Convert the target concentration of 1.00 mg L−1 of F− to mol L−1.
- Calculate the total moles of F− (and therefore NaF) required per day.
- Calculate the mass of pure NaF required per day.
- Calculate the mass of impure NaF powder that must be added, accounting for 92.0% purity.
- Evaluate one consequence of an error in this calculation — specifically, what would happen if a technician forgot to account for purity and used the pure-mass figure instead of the impure-sample figure.
- State one assumption in this calculation and one factor that could affect the accuracy of the result in practice.
2. Source critique — evaluate a student’s laboratory report claim (Band 5–6)
7 marks Band 5–6
Student claim. A Year 11 student writes in their laboratory report:
“I dissolved 10.0 g of potassium nitrate (KNO3, 85% pure) in water and made up to 500 mL of solution. The molar concentration is c = 10.0 ÷ 101.10 ÷ 0.500 = 0.198 mol L−1. I also determined that the % purity is 85% because the label says so, and this is confirmed by my calculation.” (MM(KNO3) = 101.10 g mol−1.)
Q2. Critically evaluate the student’s report. In your response you must:
- Identify the specific error in the student’s concentration calculation and explain what correct value they should have obtained.
- Identify the flaw in their claim that their calculation “confirms” the 85% purity, and explain what a purity determination actually requires.
- Show the correct calculation with full working, and state the correct concentration of KNO3 in the solution.
- Describe one experimental method the student could use to independently determine the actual purity of the KNO3 sample.
- State one assumption in the student’s procedure that could introduce error.
Q1 — Sample Band 6 response (8 marks), annotated
Step 1 — Convert [F−] to mol L−1:
1.00 mg L−1 = 1.00 × 10−3 g L−1. [1]
[F−] = 1.00 × 10−3 ÷ 18.998 = 5.264 × 10−5 mol L−1.
Step 2 — Total moles of F− and NaF per day:
n(F−) = c × V = 5.264 × 10−5 × 4.00 × 106 = 210.6 mol. [1]
Since NaF → Na+ + F− (1:1), n(NaF) = 210.6 mol.
Step 3 — Mass of pure NaF:
m(pure NaF) = n × MM = 210.6 × 41.988 = 8844 g ≈ 8.84 kg. [1]
Step 4 — Mass of impure NaF powder:
m(sample) = m(pure) ÷ (purity ÷ 100) = 8844 ÷ 0.920 = 9613 g ≈ 9.61 kg. [1]
Step 5 — Evaluate the consequence of forgetting purity:
If the technician uses m(pure) = 8.84 kg instead of m(sample) = 9.61 kg, they add too little NaF powder. Only 8.84 × 0.920 = 8.13 kg of pure NaF would actually be added (8 844 g × 0.920 = 8 136 g → [F−] = 8136 ÷ 18.998 ÷ 4.00 × 106 = 0.921 × 10−3 g/L = 0.921 mg/L), giving a concentration below the target of 1.00 mg/L. This would reduce the dental health benefit of fluoridation for the population served by the plant. [1]
Step 6 — Assumption and practical factor:
Assumption: all NaF dissolves completely in the 4.00 × 106 L of water (complete dissolution assumed). [1]
Practical factor: the actual purity of each new batch of NaF may vary from 92.0%; the plant should test the purity of each delivery gravimetrically to avoid systematic error. [1]
Precision note: 4.00 × 106 L is a nominal daily average; actual flow varies and the plant would need to adjust the dose rate continuously — a flow-proportional dosing system is used in practice. [1]
Marking criteria summary (8 marks): 1 = correct mg/L to mol/L conversion for F−; 1 = n = c × V with correct V; 1 = correct m(pure NaF); 1 = correct m(sample) applying purity; 1 = identifies the error (too little added) and quantifies or clearly describes the consequence; 1 = names one valid assumption; 1 = names one valid practical factor affecting accuracy; 1 = uses precise chemical terminology and units throughout.
Q2 — Sample Band 6 response (7 marks), annotated
Error in concentration calculation: The student used the total sample mass (10.0 g) in n = m ÷ MM, but this mass includes 15% impurities. The formula n = m ÷ MM requires the mass of the pure substance only. [1] The student’s calculated value of 0.198 mol/L is an overestimate because it attributes the mass of impurities to KNO3. [1]
Correct calculation:
m(pure KNO3) = 10.0 × 0.85 = 8.50 g [1]
n(KNO3) = 8.50 ÷ 101.10 = 0.08407 mol [1]
c = 0.08407 ÷ 0.500 = 0.168 mol L−1 (not 0.198 mol/L as stated). [1]
Flaw in purity “confirmation”: The student’s calculation uses 85% purity as an input, so it cannot independently confirm that purity. A calculation that assumes a value cannot prove it. Actual purity determination requires an independent experimental measurement — for example, dissolving the sample in water, titrating against a known concentration of sodium hydroxide to determine the actual moles of KNO3 present (where the label value is not used), then using % purity = (n × MM ÷ m(sample)) × 100. [1]
Assumption introducing error: The student assumes no NaOH or other dissolved species from the impurity contributes to the volume or reacts with water; if the impurity is KCl or another soluble salt, it will affect the ionic strength but not the KNO3 concentration. The assumption that “impurity = inert material” may not hold for all reagent grades. [1]
Marking criteria summary (7 marks): 1 = identifies the specific error (used total mass, not m(pure)); 1 = states the student’s value is an overestimate with reasoning; 1 = correct m(pure) = 8.50 g; 1 = correct n = 0.0841 mol; 1 = correct c = 0.168 mol/L; 1 = identifies the circular reasoning flaw in claiming the calculation “confirms” purity; 1 = describes a valid independent experimental method to measure purity.