AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2021: Question 9
9 marks · Standard Demand difficulty · Short Answer
Explain acid strength and concentration, suggest titration improvements, and calculate required mass and unknown solution concentration.
Practise this questionQuestion
Question text
09 This question is about acids.
Hydrogen chloride and ethanoic acid both dissolve in water.
All hydrogen chloride molecules ionise in water.
Approximately 1% of ethanoic acid molecules ionise in water.
09.1 A solution is made by dissolving 1 g of hydrogen chloride in 1 dm3 of water.
Which is the correct description of this solution?
[1 mark]
Tick ( ) one box.
A concentrated solution of a strong acid
A concentrated solution of a weak acid
A dilute solution of a strong acid
A dilute solution of a weak acid
09.2 Which solution would have the lowest pH?
[1 mark]
Tick ( ) one box.
0.1 mol/dm3 ethanoic acid solution
0.1 mol/dm3 hydrogen chloride solution
1.0 mol/dm3 ethanoic acid solution
1.0 mol/dm3 hydrogen chloride solution27
A student investigated the concentration of a solution of sodium hydroxide by titration
with a 0.0480 mol/dm3 ethanedioic acid solution.
This is the method used.
1. Measure 25.0 cm3 of the sodium hydroxide solution into a conical flask using a
*26* 25.0 cm3pipette.
2. Add two drops of indicator to the sodium hydroxide solution.
3. Fill a burette with the 0.0480 mol/dm3 ethanedioic acid solution to the
0.00 cm3 mark.
4. Add the ethanedioic acid solution to the sodium hydroxide solution until the
indicator changes colour.
5. Read the burette to find the volume of the ethanedioic acid solution used.
09.3 Suggest two improvements to the method that would increase the accuracy of the
result.
[2 marks]
09.4 Ethanedioic acid is a solid at room temperature.
Calculate the mass of ethanedioic acid (H C O ) needed to make 250 cm3 of a
22 4
solution with concentration 0.0480 mol/dm3
Relative formula mass (Mr): H2C2O4 = 90
[2 marks]
Mass = g
*0279.*5 The student found that 25.0 cm3 of the sodium hydroxide solution was neutralised by
15.00 cm3 of the 0.0480 mol/dm3 ethanedioic acid solution.
The equation for the reaction is:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
Calculate the concentration of the sodium hydroxide solution in mol/dm3
[3 marks]
Concentration = mol/dm3
Mark scheme
Show the mark scheme
Question 9
AO /
Question Answers Extra information Mark
Spec. Ref.
09.1 a dilute solution of a strong acid 1 AO2
4.3.2.5
4.4.2.6
AO /
Spec. Ref.
09.2 1.0 mol/dm3 hydrogen chloride 1 AO2
solution 4.4.2.4
4.4.2.6
AO /
Spec. Ref.
09.3 any two from: 2 AO3
• swirl (the solution) 4.4.2.5
• white tile (under the flask) RPA2
• add (ethanedioic) acid
dropwise (near the endpoint) – HEMISTRY – –
• repeat and calculate mean
Question 9 continued
AO /
Spec. Ref.
09.4 (concentration = 90 × 0.0480 =) 1 AO2
4.32 (g/dm3)
4.3.2.1
4.3.2.5
250 allow correct use of an 1 4.3.4
(mass = 4.32 × 1000 ) = 1.08 (g)
incorrectly calculated value of
concentration in g/dm3
alternative approach:
(moles = 0.0480 × 1000 =)
0.012 (mol) (1)
(mass = 0.012 × 90 ) allow correct use of an
= 1.08 (g) (1) incorrectly calculated value of
number of moles
– HEMISTRY – –
Question 9 continued
AO /
Spec. Ref.
09.5 15.0 1 AO2
(moles H2C2O4 = × 0.0480)
1000 4.3.4
= 0.00072 (mol)
4.4.2.5
RPA2
(moles NaOH =
moles H2C2O4 × 2 = )
allow correct use of an 1
0.00144 (mol)
incorrectly calculated value of
number of moles of H2C2O4
0.00144
(concentration= 25.0 × 1000)
= 0.0576 (mol/dm3) allow 0.058 (mol/dm3) 1
allow correct use of an
incorrectly calculated value of
number of moles of NaOH
alternative approach: 25
volume × conc (acid) 1 allow inverse
volume × conc (NaOH) = 2 (1)
(conc NaOH =)
15.0 × 0.0480 allow correct use of incorrect
2 × 25.0 (1)
mole ratio
= 0.0576 (mol/dm3) (1)
Total 9
How to answer it
Acids, Strength vs Concentration & Titrations
This question assesses fundamental and quantitative concepts from AQA GCSE Chemistry Paper 1 (Topic 4 & Topic 5):
- The difference between acid strength (extent of ionisation) and acid concentration (mass/moles per volume).
- How pH relates to hydrogen ion (H⁺) concentration.
- Required Practical 2: Practical techniques for an accurate acid-base titration.
- Quantitative chemistry: calculating masses needed for standard solutions using m = c × V × Mr .
- Titration calculations: determining an unknown concentration using stoichiometric reacting ratios from a balanced equation.
Classifying Acid Solutions
A solution is made by dissolving 1 g of hydrogen chloride in 1 dm³ of water. Which is the correct description?
✅ Correct Answer
A dilute solution of a strong acid
💡 Key Knowledge: Strength vs Concentration
- Strong acid: Completely ionises in aqueous solution (e.g. all HCl molecules split into H⁺ and Cl⁻ ions).
- Dilute solution: Contains a relatively small amount of solute per unit volume (1 g in 1000 cm³ is very small).
❌ Common Errors
Confusing strong with concentrated! Strength refers purely to degree of ionisation in water. Concentration refers to the amount of acid per volume.
Determining Lowest pH
Which solution would have the lowest pH?
✅ Correct Answer
1.0 mol/dm³ hydrogen chloride solution
💡 Key Knowledge: What Determines pH?
- Lower pH = higher concentration of H⁺ ions.
- HCl is a strong acid (100% ionised), so 1.0 mol/dm³ HCl produces 1.0 mol/dm³ H⁺ ions (pH = 0).
- Ethanoic acid is a weak acid (~1% ionised), producing far fewer H⁺ ions at the same concentration.
- Comparing 0.1 mol/dm³ and 1.0 mol/dm³ HCl: the higher concentration (1.0 mol/dm³) produces ten times more H⁺ ions, giving the lowest pH.
🧠 Exam Technique
Break the comparison into two quick decisions: First pick the strong acid (HCl over ethanoic acid). Second, pick the higher concentration (1.0 mol/dm³ over 0.1 mol/dm³).
Titration Method Improvements
Suggest two improvements to the method that would increase the accuracy of the result.
✅ Correct Answers (Choose Any Two)
- Swirl the conical flask during the titration.
- Place a white tile under the conical flask.
- Add the ethanedioic acid dropwise near the endpoint.
- Repeat the titration and calculate a mean (using concordant results).
💡 Why These Increase Accuracy
- White tile: Makes the subtle colour change of the indicator much easier to see clearly.
- Dropwise near endpoint: Prevents overshooting the exact neutralisation volume.
- Swirling: Ensures thorough mixing so reactants combine completely.
❌ Common Errors & Pitfalls
- Writing just "repeat" without stating "and calculate a mean" (or stating "repeat to spot anomalies", which affects reliability rather than accuracy alone).
- Suggesting measuring cylinders instead of pipettes/burettes (which decreases accuracy).
Mass of Solute Calculation
Calculate the mass of ethanedioic acid (H₂C₂O₄) needed to make 250 cm³ of a solution with concentration 0.0480 mol/dm³. Relative formula mass (Mr): H₂C₂O₄ = 90
📐 Step-by-Step Calculation (Method 1)
1 Calculate concentration in g/dm³:
Concentration = Mr × concentration in mol/dm³
Concentration = 90 × 0.0480 = 4.32 g/dm³ [1 mark]
2 Scale to 250 cm³ (0.250 dm³):
Mass = 4.32 × (250 / 1000) = 1.08 g [1 mark]
📐 Step-by-Step Calculation (Method 2)
1 Calculate moles needed:
Moles = concentration × volume (dm³)
Moles = 0.0480 × (250 / 1000) = 0.012 mol [1 mark]
2 Convert moles to mass:
Mass = moles × Mr = 0.012 × 90 = 1.08 g [1 mark]
❌ Calculation Traps
- Volume conversion trap: Forgetting to divide 250 cm³ by 1000 to convert to dm³. Multiplying 0.0480 by 250 gives 12 moles, which yields an impossible mass of 1080 g!
Mark 1: Working out conc in g/dm³ (4.32 g/dm³) OR moles (0.012 mol).
Mark 2: Correct final mass: 1.08 g (allow error carried forward from step 1).
Titration Concentration Calculation
The student found that 25.0 cm³ of NaOH was neutralised by 15.00 cm³ of 0.0480 mol/dm³ ethanedioic acid solution.
Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O
Calculate the concentration of the sodium hydroxide solution in mol/dm³.
📐 Complete Step-by-Step Method
1 Find moles of acid used (H₂C₂O₄):
Moles = (volume / 1000) × concentration
Moles = (15.00 / 1000) × 0.0480 = 0.00072 mol [1 mark]
2 Use the balanced equation to find moles of NaOH:
The mole ratio of H₂C₂O₄ : NaOH is 1 : 2.
Moles of NaOH = 0.00072 × 2 = 0.00144 mol [1 mark]
3 Calculate concentration of NaOH:
Concentration = moles / volume in dm³ = 0.00144 / (25.0 / 1000)
Concentration = (0.00144 / 25.0) × 1000 = 0.0576 mol/dm³ [1 mark]
🧠 Exam Technique: 3-Step Titration Table
Always structure titration calculations using the three golden steps:
- Moles of known: n = c × V
- Mole ratio: Multiply/divide by the balancing numbers from the chemical equation.
- Concentration of unknown: c = n / V
❌ The Stoichiometry Trap
The single most common error is missing the 1 : 2 ratio! Many students assume a 1:1 ratio and divide by 2 or omit multiplying by 2 entirely, leading to 0.0288 mol/dm³.
Mark 1: Moles of acid = 0.00072 mol.
Mark 2: Moles of NaOH = 0.00144 mol (allow ecf from step 1).
Mark 3: Concentration = 0.0576 mol/dm³ (allow 0.058 mol/dm³). Full marks awarded for correct final answer even if working is concise.
Topics
Chemistry · Required Practicals · C3: Quantitative Chemistry · C4: Chemical Changes · Required Practicals
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.