AQA GCSE Chemistry Chemistry Paper 2 (Foundation), June 2024: Question 8
10 marks · Standard Demand difficulty · Short Answer
Describe tests to identify sodium and chloride ions, explain evaporating to constant mass, and calculate the mean concentration of sodium ions from experimental data.
Practise this questionQuestion
Question text
08 A student investigated an aqueous solution of a salt.
The student identified that the salt solution contained only sodium ions and chloride
ions.
08.1 Describe a test to identify sodium ions.
Give the result of the test.
[2 marks]
Test for sodium ions
Result
08.2 Describe a test to identify chloride ions.
Give the result of the test.
[2 marks]
Test for chloride ions
Result
The student determined the concentration of sodium chloride in the salt solution.
This is the method used.
1. Weigh an empty evaporating dish.
2. Add 25.0 cm3 of the salt solution into the evaporating dish.
3. Heat the evaporating dish and contents.
4. Weigh the evaporating dish and contents.
5. Repeat steps 3 to 4 until there is no further change in mass.
6. Repeat steps 1 to 5 three more times.
08.3 Why did the student heat the evaporating dish and contents until the mass did not
change?
[1 mark]
08.4 How did the student calculate the mass of solid sodium chloride remaining after steps
1 to 5?
[1 mark]
Tick ( ) one box.
Mass of 25 cm3 of salt solution + mass of empty evaporating dish
Mass of 25 cm3 of salt solution − mass of empty evaporating dish
Mass of evaporating dish and dry contents + mass of empty
evaporating dish
Mass of evaporating dish and dry contents − mass of empty
evaporating dish 34
08.5 The student calculated the concentration of sodium chloride in the salt solution.
Table 5 shows the results.
Table 5
Concentration of sodium chloride in g/dm3
Trial 1 Trial 2 Trial 3 Trial 4
*33* 35.2 34.6 36.4 33.8
The percentage by mass of sodium ions in sodium chloride is 39.3%.
Calculate the mean concentration of sodium ions in the salt solution.
[4 marks]
Mean concentration = g/dm3
Mark scheme
Show the mark scheme
Question 8
AO /
Question Answers Extra information Mark
Spec. Ref.
08.1 (test) 1 AO1
flame test 4.8.3.1
RPA7
(result) 1
yellow (flame)
OR
(test)
flame emission spectroscopy (1) allow FES
(result)
lines match sodium spectrum (1)
AO /
Spec. Ref.
08.2 (test) AO1
(add acidified) silver nitrate 1 4.8.3.4
(solution) RPA7
(result) 1
white precipitate MP2 is dependent upon the
award of MP1
AO /
Spec. Ref.
08.3 to ensure that all the water has 1 AO3
evaporated 4.10.1.2
– – – RPA8
AO /
Spec. Ref.
08.4 mass of evaporating dish and 1 AO1
dry contents – mass of empty – – – 4.10.1.2
evaporating dish RPA8
AO /
Spec. Ref.
08.5 (mean concentration of NaCl =) 1 AO2
35.2 + 34.6 + 36.4 + 33.8
4.10.1.2
RPA8
or
allow 1 mark for
140 35.2 + 34.6 + 33.8
4 = 34.5
= 35.0 (g/dm3) 1
(mean concentration of Na+ =)
39.3 allow correct use of an 1
35.0 × incorrectly determined mean
concentration of sodium chloride
= 13.8 (g/dm3) allow 13.755 correctly rounded 1
to at least 3 significant figures
alternative approach 1:
(total concentration of NaCl =
35.2 + 34.6 + 36.4 + 33.8 = 140
total concentration of Na+ =) allow 1 mark for
39.3 (35.2 + 34.6 + 33.8 = 103.6)
140 × (1) 39.3
100 103.6 × = 40.71
= 55.02 (g/dm3) (1)
(mean concentration of Na+ =)
55.02 allow correct use of an
(1)
4 incorrectly determined total
concentration of Na+
=13.8 (g/dm3) (1) allow 13.755 correctly rounded
to at least 3 significant figures 23
– – –
alternative approach 2:
(concentrations of Na+ =)
39.3
35.2 ×
39.3
34.6 ×
allow 1 mark if a
39.3 concentration of 36.4 is
36.4 × treated as an anomaly and
not used
39.3
33.8 × (1)
= 13.83 13.60 14.31 13.28 (1)
(mean concentration of Na+ =)
13.83 + 13.60 + 14.31 + 13.28 allow correct use of incorrectly
4 determined concentration(s) of
(1) Na+
=13.8 (g/dm3) (1) allow 13.755 correctly rounded
to at least 3 significant figures
Total Question 8 10
How to answer it
Testing Ions & Measuring Solution Concentration
This question assesses fundamental practical skills and chemical analysis from topics 4.8 (Chemical Analysis) and 4.10 (Using Resources / Required Practical 8):
- Cation Identification: Recalling the flame test method and colour for sodium (Na⁺) ions.
- Anion Identification: Recalling the reagent and positive result for halide ions (chloride, Cl⁻).
- Gravimetric Practical Technique: Understanding "heating to constant mass" to remove all water.
- Experimental Data & Math: Processing repeated measurements, determining mean concentrations, and working with percentages by mass.
Test and Result for Sodium Ions
Qualitative analysis of metal cations (flame tests)
✅ Correct Answer
- Test: Flame test [1 mark]
- Result: Yellow flame [1 mark]
Alternative accepted: Flame emission spectroscopy (FES) [1 mark] ; result: lines match sodium spectrum [1 mark] .
💡 Key Knowledge
Key flame test colours you must know for AQA GCSE:
- Sodium (Na⁺): Yellow
- Lithium (Li⁺): Crimson
- Potassium (K⁺): Lilac
- Calcium (Ca²⁺): Orange-red
- Copper(II) (Cu²⁺): Green
🧠 Exam Technique
Keep your answer concise. The question specifically asks for the "test" and the "result" in separate lines. Writing "flame test" followed by "yellow" scores both marks instantly.
❌ Common Errors
- Writing "orange" instead of "yellow" — GCSE examiners specifically look for "yellow" for sodium. Orange-red is calcium.
- Confusing flame tests with sodium hydroxide solution tests (sodium ions do not form a precipitate with NaOH).
Test and Result for Chloride Ions
Testing for halide anions using silver nitrate solution
✅ Correct Answer
- Test: Add (dilute nitric acid and) silver nitrate solution [1 mark]
- Result: White precipitate [1 mark]
💡 Key Knowledge
Halide tests rely on precipitation reactions with aqueous silver ions (Ag⁺):
- Ag⁺(aq) + Cl⁻(aq) → AgCl(s) (white precipitate)
- Bromide (Br⁻) gives a cream precipitate (AgBr).
- Iodide (I⁻) gives a yellow precipitate (AgI).
- Acidifying with dilute nitric acid (HNO₃) removes carbonate or sulfite impurities that would give false precipitates.
🧠 Exam Technique
Always state the state/appearance clearly: write "white precipitate" or "white solid", not simply "it turns white" or "white solution".
❌ Common Errors
- Confusing reagents: using barium chloride (the test for sulfate ions, SO₄²⁻) instead of silver nitrate.
- Suggesting hydrochloric acid (HCl) to acidify the test — adding HCl adds chloride ions, ruining the test!
Evaporation & Gravimetric Determination
Required Practical 8: Determining dissolved solids by heating to constant mass
✅ 08.3: Why Heat Until Constant Mass?
Answer: To ensure that all the water has evaporated. [1 mark]
Examiner Insight: Saying "to make sure it's completely dry" is acceptable, but the precise scientific reasoning is that water continues to be driven off until only pure anhydrous salt remains.
✅ 08.4: Calculation of Dry Salt Mass
Correct Box to Tick:
☑ Mass of evaporating dish and dry contents − mass of empty evaporating dish [1 mark]
(The 4th option in the list).
💡 Experimental Technique: Constant Mass
When determining the mass of a dissolved residue:
- Initial weighing: Mass of empty clean dry dish (m₁).
- After evaporation: Mass of dish + salt after heating (m₂).
- Heating, cooling, and reweighing is repeated until m₂ remains unchanged between two consecutive weighings.
- Mass of dry salt = m₂ − m₁.
❌ Common Errors in 08.3 & 08.4
- In 08.3: Vague answers such as "to make the test fair" or "to get accurate results" score 0 marks.
- In 08.4: Choosing options that involve the initial 25 cm³ liquid solution rather than the dried dish and contents.
Calculating Mean Concentration of Sodium Ions
Data processing and percentage composition
| Concentration of sodium chloride in g/dm³ | |||
|---|---|---|---|
| Trial 1 | Trial 2 | Trial 3 | Trial 4 |
| 35.2 | 34.6 | 36.4 | 33.8 |
Given: The percentage by mass of sodium ions in sodium chloride is 39.3%.
📐 Step-by-Step Calculation
Sum of trials = 35.2 + 34.6 + 36.4 + 33.8 = 140.0 g/dm³ [1 mark]
Mean concentration = 140.0 / 4 = 35.0 g/dm³ [1 mark]
Mean Na⁺ concentration = 35.0 × (39.3 / 100) [1 mark]
= 35.0 × 0.393 = 13.755 g/dm³
13.8 g/dm³ (or 13.755) [1 mark]
💡 Alternative Valid Approaches
Alternative 1:
Find total Na⁺ first: 140 × (39.3 / 100) = 55.02 g/dm³.
Then divide by 4: 55.02 / 4 = 13.8 g/dm³.
Alternative 2:
Calculate Na⁺ concentration for each individual trial first:
Trial 1: 13.83 | Trial 2: 13.60 | Trial 3: 14.31 | Trial 4: 13.28
Mean = (13.83 + 13.60 + 14.31 + 13.28) / 4 = 13.8 g/dm³.
🧠 Exam Technique & Anomalies
- Look closely at the data: values range from 33.8 to 36.4. None of them are obvious outliers, so all four trials must be included in the mean.
- Notice the mark scheme allows an error-carried-forward (ECF) mark if you made an arithmetic slip in Step 1 but multiplied correctly by 39.3% in Step 2. Always show your working clearly!
❌ Common Errors
- Dividing by 3 instead of 4 (mistakenly treating one result as an anomaly without good reason).
- Inverting the percentage: e.g., dividing by 0.393 instead of multiplying by 0.393.
- Premature rounding: round only your final answer to avoid rounding errors.
Topics
Chemistry · Required Practicals · C8: Chemical Analysis · C10: Using Resources · Required Practicals
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Foundation), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.