AQA A-Level Chemistry Paper 1, June 2024: Question 2

10 marks · Medium difficulty · Practical Techniques & Data Analysis

Explain experimental steps in determining strontium hydroxide solubility and calculate its solubility using titration data.

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Question

Question text

02 This question is about an experiment to determine the solubility of strontium hydroxide

in water at 20 °C

Strontium hydroxide is slightly soluble in water. Strontium hydroxide solution reacts in

a similar way to calcium hydroxide solution.

• Some solid strontium hydroxide is added to approximately 1 dm3 of distilled water in

a stoppered flask.

• The mixture is kept at 20 °C. Every day, the mixture is checked. If no solid is

present in the flask, more solid strontium hydroxide is added.

• On the day when no more solid needs to be added, the flask is opened and the

mixture is filtered into another flask and stoppered.

• A 25.0 cm3 sample of the filtrate is transferred to a conical flask with a pipette and a

few drops of indicator added.

• This sample is titrated with 0.100 mol dm–3 hydrochloric acid.

• The titration is repeated several times with further samples of the filtrate. The

results are shown in Table 1 on page 8.

02.1 Suggest why the solution is kept until no more solid needs to be added.

[1 mark]

02.2 Suggest why it is important to remove the undissolved strontium hydroxide before the

titration.

[1 mark]

02.3 After the filtration, the solution is stored in a stoppered flask.

Suggest a reason for stoppering the flask.

[1 mark]

02.4 The diagrams in Figure 2 show the part of a pipette with the graduation line.

Which diagram identifies the pipette that is correctly filled?

[1 mark]

Tick ( ) one box.

Figure 2

A B C

02.5 Solubility can be quoted as ‘g of solute per 100 cm3 of solution’.

Table 1 shows the results of the titrations between strontium hydroxide and

hydrochloric acid. These can be used to determine the solubility of

strontium hydroxide.

Table 1

Titration Rough 1 2 3

Final burette reading / cm3 34.40 38.00 41.05 37.00

Initial burette reading / cm3 0.00 5.55 8.05 4.60

Titre / cm3 34.40 32.45 33.00 32.40

Give the equation for the reaction between strontium hydroxide and hydrochloric acid.

Use the results in Table 1 to calculate the mean titre.

Use the mean titre to calculate the solubility of strontium hydroxide, in g per 100 cm3

of solution, at 20 °C 9

[6 marks]

Equation

*08* Mean titre cm3

Solubility of strontium hydroxide g per 100 cm3 solution

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

(To make sure that) as much as possible/maximum amount (of Do not accept reacted

solid) dissolves

Ignore references to right/correct concentration

OR 1

02.1

(1 x AO3)

(To ensure that) the solution/it is saturated

Otherwise, the titre would be larger

Allow (solid) could block pipette

Or Ignore references to changes in concentration

and pH

Would need a larger volume of acid/HCl

02.2

Or (1 x AO3)

Because undissolved strontium hydroxide will react (with the

acid/HCl)

To prevent reaction with carbon dioxide (in the air) Allow so flask can inverted/shaken (to ensure

homogeneous mixture)

02.3 OR

(1 x AO2)

To prevent evaporation (of water/from solution) Ignore contamination

Answer C 1

02.4

(1 x AO3)

M1 Equation

M1 Sr(OH)2 + 2HCl → SrCl2 + 2H2O

3 M2 allow 32.425

M2 32.43 (cm )

M3 n HCl in mean titre = 3.24(3) × 10–3 mol M3 allow 0.1 x M2 ÷ 1000

M4 n Sr(OH) in 25 cm3 = 1.62 × 10–3 mol 6

2 M4 allow M3 ÷ 2

02.5

(6 x AO2)

M5 n Sr(OH) in 100 cm3 of solution = 6.48 × 10–3 mol

2 M5 allow M4 × 4

M6 mass = (6.48×10–3 mol x 121.6) = 0.788 (g per 100 cm3

M6 allow = M5 × 121.6

solution) M6 allow 0.79

allow M5 and M6 in either order

How to answer it

Determining the Solubility of Strontium Hydroxide

📋 Overview

What this question tests

This question assesses experimental design, practical volumetric analysis, and quantitative stoichiometry involving Group 2 hydroxides.

Solubility equilibrium: Preparing a saturated solution and understanding dynamic equilibrium in dissolution.
Practical titration techniques: Rationalising filtration, preventing atmospheric carbonation (CO₂ interference), and reading a volumetric pipette meniscus correctly.
Concordancy & multi-step calculations: Selecting concordant titres, balancing Group 2 neutralisation equations, and calculating concentration in non-standard units (g per 100 cm³).

Part 02.1 • 1 Mark (AO3)

Waiting for Complete Dissolution

Suggest why the solution is kept until no more solid needs to be added.

✅ Acceptable Answers

To ensure that the solution is saturated.
OR To make sure that the maximum amount (of solid) dissolves / as much as possible dissolves.

❌ Common Errors & Trap Warnings

"To make sure it all reacted": Dissolution is a physical equilibrium process, not an acid-base or chemical decomposition reaction. "Reacted" is strictly rejected.
Vague comments: Writing "to get the correct/right concentration" scores zero without specifying saturation.

Examiner Insight: Solubility is defined at equilibrium saturation at a stated temperature (20 °C). The keyword students must hit is either saturated or maximum mass dissolved.

Part 02.2 • 1 Mark (AO3)

Filtration Before Titration

Suggest why it is important to remove undissolved strontium hydroxide before titration.

✅ Acceptable Answers

Because undissolved strontium hydroxide would react with the acid (HCl) during the titration.
OR Otherwise, the titre would be larger / would need a larger volume of acid.
(Also allowed) Undissolved solid could block the pipette.

🧠 Exam Technique

Trace the consequence through to the quantitative outcome! If solid particulates enter the conical flask, as HCl neutralises aqueous OH⁻ ions, Le Chatelier's principle causes undissolved solid to dissolve and neutralise more acid, causing an falsely inflated titre.

Mark Guidance: Mentioning a "change in pH" or general "change in concentration" is too vague to award a mark. Focus specifically on extra reaction or excessive titre volume.

Part 02.3 • 1 Mark (AO2)

Stoppering the Filtrate Flask

Suggest a reason for storing the solution in a stoppered flask after filtration.

✅ Acceptable Answers

To prevent reaction with carbon dioxide (CO₂) from the air.
OR To prevent evaporation of water from the solution (which would alter concentration).
(Also allowed) So the flask can be inverted / shaken safely to ensure a homogeneous solution.

💡 Key Knowledge

Alkaline solutions absorb acidic atmospheric gases:

Sr(OH)₂(aq) + CO₂(g) → SrCO₃(s) + H₂O(l)

This precipitation reaction reduces [OH⁻(aq)], reducing the required titre and underestimating solubility.

Examiner Tip: Do not just say "to stop contamination"—that is far too unspecific for A-Level Chemistry. Name the substance: carbon dioxide or water evaporating.

Part 02.4 • 1 Mark (AO3)

Reading the Pipette Meniscus

Which diagram identifies the pipette that is correctly filled?

✅ Correct Box to Tick

Box C

The bottom of the curved meniscus must rest precisely on the top of the horizontal graduation line at eye level.

❌ Meniscus Misconceptions

A: The liquid level is below the line (underfilled).
B: The top edges of the meniscus touch the line, meaning the bottom is below the line (underfilled).

Part 02.5 • 6 Marks (AO2)

Solubility Calculation & Reaction Stoichiometry

Equation, Mean Titre, and Solubility in g per 100 cm³

📐 Step-by-Step Calculation Breakdown

Step 1: Write the balanced chemical equation (M1)

Sr(OH)₂ + 2HCl → SrCl₂ + 2H₂O

Sr is in Group 2, forming Sr²⁺; two moles of HCl are required to neutralise each mole of Sr(OH)₂.

Step 2: Determine concordant titres & calculate the mean titre (M2)

Titres from Table 1: Rough = 34.40 cm³, Titre 1 = 32.45 cm³, Titre 2 = 33.00 cm³, Titre 3 = 32.40 cm³.

Concordant titres (within 0.10 cm³ of each other) are Titre 1 (32.45 cm³) and Titre 3 (32.40 cm³).
Titre 2 (33.00 cm³) and the rough titration are non-concordant outliers and must be excluded.

Mean titre = (32.45 + 32.40) / 2 = 32.43 cm³ (or 32.425 cm³)

Step 3: Calculate moles of HCl neutralised (M3)

n(HCl) = concentration × volume = 0.100 × (32.425 / 1000) = 3.243 × 10⁻³ mol

Step 4: Use the reacting ratio to find moles of Sr(OH)₂ in 25.0 cm³ (M4)

From the balanced equation, ratio of Sr(OH)₂ : HCl is 1 : 2.

n(Sr(OH)₂ in 25.0 cm³) = (3.243 × 10⁻³) / 2 = 1.621 × 10⁻³ mol

Step 5: Scale to find moles of Sr(OH)₂ in 100 cm³ solution (M5)

The target volume is 100 cm³, which is 4 times the 25.0 cm³ sample (100 / 25.0 = 4):

n(Sr(OH)₂ in 100 cm³) = 1.621 × 10⁻³ × 4 = 6.485 × 10⁻³ mol

Step 6: Convert moles to mass (g per 100 cm³ solution) (M6)

Calculate Mᵣ of Sr(OH)₂: 87.6 + (2 × 16.0) + (2 × 1.0) = 121.6 g mol⁻¹

Mass = 6.485 × 10⁻³ mol × 121.6 g mol⁻¹ = 0.788 g (allow 0.79 g)

❌ Common Calculation Traps

Including Titre 2 in the mean: Averaging all three non-rough titres gives 32.62 cm³, immediately losing M2 and causing consequential carrying error.
Forgetting the 1:2 reacting mole ratio: Skipping division by 2 in Step 4 yields double the true value (1.58 g).
Scaling to 1 dm³ instead of 100 cm³: Multiplying by 40 instead of 4 produces 7.88 g per 100 cm³—read the required units carefully!
Incorrect Mᵣ: Forgetting the second hydroxide group (using SrOH instead of Sr(OH)₂).

🧠 Summary of Final Marks

Equation: Sr(OH)₂ + 2HCl → SrCl₂ + 2H₂O
Mean Titre: 32.43 cm³ (or 32.425 cm³)
Solubility: 0.788 g per 100 cm³ solution (or 0.79)

Full error-carried-forward (ECF) applies through M3, M4, M5, and M6 if an arithmetic mistake or incorrect mean titre is used. Steps M5 and M6 can also be performed in reverse order (converting 25 cm³ to mass first, then multiplying by 4).

Topics

Physical Chemistry · Inorganic Chemistry · Required Practicals · 3.1.2 Amount of Substance · 3.2.2 Group 2, The Alkaline Earth Metals · Required Practical 1: Making up a volumetric solution

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.