AQA A-Level Chemistry Paper 1, 2023: Question 6
10 marks · Medium difficulty · Practical Techniques & Data Analysis
Calculate masses, concentrations, and percentage errors, and describe the preparation and practical techniques involved in a titration between standard barium hydroxide and hydrochloric acid.
Practise this questionQuestion
Question text
06 The concentration of dilute hydrochloric acid can be found by titration using a
standard solution of barium hydroxide.
06.1 Calculate the mass, in g, of solid barium hydroxide (Mr = 171.3) needed to prepare
250 cm3 of 0.100 mol dm–3 barium hydroxide solution.
[1 mark]
Mass g
06.2 The mass of barium hydroxide from Question 06.1 is dissolved in a beaker containing
150 cm3 of distilled water.
Describe how this solution is used to make 250 cm3 of the
0.100 mol dm–3 barium hydroxide solution.
[3 marks]
06.3 Before the first titration, the 25 cm3 pipette is rinsed with a small volume of the
0.100 mol dm–3 barium hydroxide solution.
State why it is good practice to rinse the pipette in this way.
[1 mark]
06.4 Hydrochloric acid is added to the burette using a funnel.
State why it is good practice to remove the funnel from the burette before the titration.
[1 mark]
06.5 In a different experiment, 0.952 g of solid barium hydroxide is used to make 250 cm3
of standard barium hydroxide solution.
25.0 cm3 of this barium hydroxide solution reacts with exactly
24.50 cm3 of hydrochloric acid.
Calculate the concentration of the hydrochloric acid.
[3 marks]
Concentration mol dm–3
06.6 The uncertainty in the 25.0 cm3 of solution from the pipette is ±0.05 cm3
The total uncertainty in the 24.50 cm3 of solution from the burette is ±0.15 cm3
Calculate the total percentage error in using the pipette and burette.
[1 mark]
Percentage error
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark
0.100 × 250 × 171.3 = 4.28 g 1
06.1 Allow 4.3 g
1000 (AO2)
M1 Transfers the solution to a volumetric/graduated flask
M2 Add washings using distilled water and make up to the
06.2 3 (3 x AO2)
graduation mark/250 cm
M3 Invert many times / shake to mix
So that the titration is done with known concentration of Ba(OH)2
Allow so that water does not dilute the solution 1
06.3
Allow remove water/prevent contamination (AO3)
Drops of acid could fall into the burette (so no longer know how Ignore changes the titre 1
much has been added from burette)
Do not accept increases titre (AO1)
06.4
OR
Decreases titre
3 0.952 –4
M1 n Ba(OH)2 in 25 cm = = 5.56 x 10 mol
171.3 𝐱𝐱𝐱𝐱𝐱𝐱
M2 n HCl in 24.5 cm3 = 2 x 5.56 x 10–4 = 0.00111 mol 3
06.5 M2 = M1 x 2
(3 x AO2)
0.00111 × 1000 –3
M3 Concentration of HCl = = 0.045 mol dm
24.50 M2 × 1000
M3 =
24.50
0.15 + 0.05 x 100% = 0.8(1)% 1
06.6
24.50 25.00
(AO2)
How to answer it
Standard Solutions, Titration Practicalities & Volumetric Analysis
This question assesses Required Practical 1 competencies and quantitative chemistry: calculating the exact mass required to prepare a primary standard, describing the standard procedure for making a volumetric solution, identifying critical errors and precautions in titration apparatus handling (rinsing pipettes and removing funnels), carrying out a multi-step back/direct titration calculation involving a 1:2 reacting mole ratio, and calculating total combined percentage uncertainties.
Calculating Mass Needed for a Standard Solution
Calculate mass of solid Ba(OH)₂ (Mr = 171.3) needed for 250 cm³ of 0.100 mol dm⁻³
📐 Step-by-Step Calculation
- Find moles of Ba(OH)₂ needed:
n = c × V
n = 0.100 × (250 / 1000) = 0.0250 mol - Calculate mass:
m = n × Mr = 0.0250 × 171.3 = 4.2825 g - Final Answer: 4.28 g (or 4.3 g )
✅ Mark Scheme Breakdown
- 4.28 g [1 mark]
- Allow rounded to 4.3 g.
Method: Preparing a 250 cm³ Standard Solution
Describe how the solution is transferred and made up to 250 cm³
✅ 3-Step Standard Method
- Transfer: Pour the dissolved solution from the beaker into a 250 cm³ volumetric (or graduated) flask (ideally using a funnel).
- Washings & Dilution: Rinse the beaker and funnel with distilled water and add all washings to the flask. Make up to the graduation mark until the bottom of the meniscus touches the line.
- Homogenisation: Stopper the flask and invert multiple times / shake thoroughly to ensure a uniform concentration throughout.
❌ Common Errors & Lost Marks
- Missing "washings": Simply stating "pour into flask and fill to line" loses Mark 2. Examiners insist on mentioning rinsing apparatus and adding washings.
- Wrong water type: Mentioning tap water or not specifying distilled/deionised water.
- Forgetting to invert: Leaving out inversion/mixing loses Mark 3. Solutions do not mix by diffusion alone; the heavier solute settles at the bottom.
Apparatus Technique: Rinsing the Pipette
Why rinse the 25 cm³ pipette with 0.100 mol dm⁻³ Ba(OH)₂ solution?
💡 Key Knowledge
If residual water droplets remain inside the pipette from prior cleaning, filling it directly will dilute the reagent being transferred. Rinsing with the actual solution ensures that the concentration transferred into the conical flask is exactly the intended standard concentration.
✅ Accepted Mark Scheme Answers
- So that the titration is done with a known concentration of Ba(OH)₂.
- Allow: To ensure water does not dilute the solution.
- Allow: To remove residual water / prevent contamination.
Apparatus Technique: Removing the Funnel
Why remove the funnel from the burette before titration?
🧠 Exam Technique & Logic
If the funnel remains in place, residual drops of acid hanging on the stem may drop into the burette mid-titration.
- Additional acid enters the burette, keeping the liquid level higher than it should be.
- The final burette reading will be too high (less volume appears to have been used).
- Therefore, this decreases the measured titre!
❌ Examiner Pitfall
- Do NOT accept "increases titre": Acid dripping in reduces the apparent volume delivered.
- "Changes titre" is too vague and receives 0 marks. You must specify that it decreases the titre or that drops could fall in making the delivered volume uncertain.
Titration Calculation: Hydrochloric Acid Concentration
0.952 g Ba(OH)₂ made to 250 cm³. 25.0 cm³ reacts with 24.50 cm³ HCl.
📐 Step-by-Step Calculation
- Balanced Equation & Ratio:
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
Mole ratio is 1 Ba(OH)₂ : 2 HCl - Moles of Ba(OH)₂ in 25.0 cm³:
Total moles in 250 cm³ = 0.952 / 171.3 = 5.5575 × 10⁻³ mol
Moles in 25.0 cm³ sample = 5.5575 × 10⁻³ / 10 = 5.56 × 10⁻⁴ mol [M1] - Moles of HCl reacted:
n(HCl) = 2 × 5.5575 × 10⁻⁴ = 1.111 × 10⁻³ mol [M2] - Concentration of HCl:
c = n / V = (1.111 × 10⁻³ × 1000) / 24.50 = 0.0453 mol dm⁻³ [M3]
(0.045 mol dm⁻³ based on 2 s.f.)
❌ Two Fatal Traps
- Forgetting the Aliquot Factor (÷10): Only 25.0 cm³ of the 250 cm³ solution is used per titration. Failing to divide by 10 is an automatic error.
- Ignoring the 1:2 Stoichiometry: Ba(OH)₂ has two OH⁻ ions per formula unit, so it neutralises 2 moles of HCl. Forgetting to multiply by 2 forfeits M2.
Apparatus Uncertainty Calculation
Pipette error: ±0.05 cm³ on 25.0 cm³ | Burette error: ±0.15 cm³ on 24.50 cm³
📐 Percentage Uncertainty Calculation
- Pipette % uncertainty:
(0.05 / 25.0) × 100% = 0.200% - Burette % uncertainty:
(0.15 / 24.50) × 100% = 0.612% - Total combined percentage error:
0.200% + 0.612% = 0.812% ≈ 0.81% (or 0.8% )
🧠 Exam Technique: Adding Uncertainties
When multiple measuring instruments are used in an experiment to obtain data, the combined experimental percentage uncertainty is simply the sum of the individual percentage uncertainties.
Topics
Physical Chemistry · Required Practicals · 3.1.2 Amount of Substance · Required Practical 1: Making up a volumetric solution
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.