AQA GCSE Chemistry Chemistry Paper 1 (Higher), June 2025: Question 8

13 marks · Standard Demand difficulty · Short Answer

Identify the alkali ion, suggest improvements to a titration plan, read a burette, calculate concentration from titration data, and compare strong and weak acid solutions.

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Question

Question 8 includes six parts on acids, alkalis, and titrations. 08.1 asks for the formula of the ion all alkalis contain in solution. 08.2 lists a 9-step titration plan using sulfuric acid, universal indicator, and sodium hydroxide, asking for two improvements and reasons. 08.3 shows a magnified burette reading between 24 and 25 cm³. 08.4 gives a neutralisation equation between NaOH and H2SO4 with volumes and concentration to calculate NaOH concentration. 08.5 and 08.6 present Table 4 with concentrations and ionization percentages of HCl and HCN, asking for a correct comparative statement and an explanation for why HCl has a lower pH.
Question text

08 This question is about alkalis and acids.

08.1 What is the formula of the ion that all alkalis contain when in solution?

[1 mark]

A student planned a titration experiment to determine the concentration of a

sodium hydroxide solution.

This is the student’s plan.

1. Measure 25.0 cm3 of sulfuric acid of known concentration using a pipette.

2. Add the sulfuric acid to a conical flask.

3. Add 3 drops of universal indicator to the conical flask.

4. Place the conical flask on a white tile.

5. Fill a burette with sodium hydroxide solution to the 0.00 cm3 mark.

6. Add sodium hydroxide solution to the conical flask and swirl the flask.

7. Stop adding sodium hydroxide solution when there is a colour change.

8. Record the volume of sodium hydroxide solution used.

9. Repeat steps 1 to 8 until consistent results are obtained.

08.2 Suggest two improvements to the plan.

Give one reason for each improvement.

[4 marks]

Improvement 1

Reason

Improvement 2

Reason

08.3 Figure 9 shows the burette after one titration.

Figure 9

What is the reading on the burette?

[1 mark]

25cm3

08.4 In a different titration, 25.0 cm3 of 0.216 mol/dm3 sulfuric acid needed 11.25 cm3 of

sodium hydroxide solution for neutralisation.

The equation for the reaction is:

2 NaOH + H2SO4 ⟶ Na2SO4 + 2 H2O

Calculate the concentration of the sodium hydroxide solution.

[4 marks]

26 3

Concentration = mol/dm

Table 4 gives information about solutions of two acids.

Table 4

Concentration

Formula of Percentage (%) of acid

of solution in

acid 3 molecules that are ionised

mol/dm

HCl 0.1 100

HCN 2.0 0.01

08.5 Which is a correct statement about the solutions of HCl and HCN in Table 4?

[1 mark]

Tick ( ) one box.

The HCl is a more concentrated and stronger acid than the HCN.

The HCl is a more concentrated and weaker acid than the HCN.

The HCl is a more dilute and stronger acid than the HCN.

The HCl is a more dilute and weaker acid than the HCN.

08.6 In Table 4, the solution of HCl has a lower pH than the solution of HCN.

Explain why.

Use the data in Table 4.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 8 shows: 8.1 accepts OH⁻ (1 mark). 8.2 awards up to 4 marks for improvements with linked reasons, such as using phenolphthalein/methyl orange instead of universal indicator for a sharp colour change, swirling continuously, adding drop by drop near endpoint, or reading at eye level to avoid parallax. 8.3 accepts 24.70 cm³ or 24.7 cm³ (1 mark). 8.4 gives 4 marks for moles of acid (0.00540), moles of alkali (0.0108), and final concentration 0.960 mol/dm³. 8.5 marks 'The HCl is a more dilute and stronger acid than the HCN' (1 mark). 8.6 awards 2 marks for stating HCl has a higher concentration of hydrogen ions because HCl has 0.1 mol/dm³ H⁺ compared to 0.0002 mol/dm³ for HCN.

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 OH– 1 AO1

4.4.2.4

AO /

Spec. Ref.

08.2 any two from: 1 mark for improvement, 1 mark AO3

for reason 4 4.4.2.5

reason must be linked to the RPA 2

corresponding improvement

(improvement)

use phenolphthalein (instead of allow methyl orange for

universal indicator) phenolphthalein

allow any named single change

indicator for phenolphthalein

(reason)

so that the colour change is allow universal indicator

sharp changes colour too gradually

allow universal indicator has too

many colour changes

ignore specific colour changes

or

(improvement)

use a pH probe

(reason)

so the pH probe shows a

sharper endpoint (than universal

indicator)

OR

(improvement)

add the sodium hydroxide allow add the alkali drop by drop

solution drop by drop (near the (near the end point)

end point)

(reason)

the volume (of sodium hydroxide allow so the end point is more

solution) used is more accurate accurate

OR

(improvement)

read burette at eye level allow read (volume) at the

bottom of the meniscus

(reason)

the volume (of sodium hydroxide allow to reduce parallax error

solution) used is more accurate

OR

(improvement)

rinse burette with sodium

hydroxide and pipette with acid

(reason)

remove contaminants

AO /

Spec. Ref.

08.3 24.70 (cm3) allow 24.7 (cm3) 1 AO2

4.4.2.5

RPA 2

AO /

Spec. Ref.

08.4 25.0 × 0.216 AO2

�moles acid = =� 4.3.2.5

1000

1 4.4.2.5

0.00540 RPA 2

(moles alkali = 0.00540 × 2 =)

0.0108 allow correct use of incorrectly 1

determined number of moles of

acid

(concentration =)

1000 allow correct use of incorrectly 1

0.0108 × determined number of moles of

11.25

alkali

= 0.960 (mol/dm3) allow = 0.96 (mol/dm3) 1

alternative approach:

moles acid

� =�

moles alkali

2 11.25 × conc allow inverted expression

= (2)

1 25.0 × 0.216

allow 1 mark for the expression

with an incorrect mole ratio

(concentration =)

2 × 25.0 × 0.216 allow correct use of the

(1) expression with incorrect mole

11.25

ratio

= 0.960 (mol/dm3) (1) allow = 0.96 (mol/dm3)

AO /

26Question Answers Extra information Mark

Spec. Ref.

08.5 the HCl is a more dilute and 1 AO2

stronger acid than the HCN 4.4.2.6

AO /

Spec. Ref.

08.6 (HCl has a) higher concentration allow HCN has a lower 1 AO2

of hydrogen ions concentration of hydrogen ions 4.3.2.5

4.4.2.6

(because)

concentration of hydrogen ions 1

in HCl is 0.1 (mol/dm3)

and

concentration of hydrogen ions

in HCN is 0.0002 (mol/dm3)

Total Question 8 13

How to answer it

Acids, Alkalis, Titrations & Acid Strength

What this question tests

This question assesses practical and theoretical aspects of neutralisation and quantitative acid-base chemistry from AQA Topic 4 (Chemical Changes) and Topic 3 (Quantitative Chemistry):

  • Recall: The characteristic ion responsible for alkalinity in aqueous solutions.
  • Required Practical 2 (Titration): Evaluating experimental procedures, identifying errors, suggesting valid procedural improvements, and reading burette scales.
  • Reacting Mole Calculations: Multi-step solution titration stoichiometry linking volume, concentration, mole ratios, and units ( mol/dm³ ).
  • Strong vs. Weak & Concentrated vs. Dilute: Distinguishing acid strength (degree of ionisation) from concentration (amount per volume), and linking H⁺ ion concentration to pH.
Question 08.1 • 1 Mark

Hydroxide Ions in Alkaline Solutions

Identifying the universal ion present in all aqueous alkalis

✅ Correct Answer

OH⁻

[1 mark] formula must have correct capitalisation and negative charge.

💡 Key Knowledge

  • Acids produce aqueous hydrogen ions: H⁺(aq)
  • Alkalis are soluble bases that produce aqueous hydroxide ions: OH⁻(aq)
  • Neutralisation equation: H⁺(aq) + OH⁻(aq) → H₂O(l)
Question 08.2 • 4 Marks

Evaluating and Improving the Titration Plan

Suggesting two improvements and providing a linked reason for each

✅ Accepted Pairs (Pick Any TWO)

  • Improvement 1: Use a single-change indicator (e.g., phenolphthalein or methyl orange) instead of universal indicator.
    Reason: Universal indicator changes colour too gradually; a single indicator gives a sharp, distinct end-point.
  • Improvement 2: Add the sodium hydroxide drop by drop near the end point.
    Reason: Prevents overshooting the end-point so the volume recorded is accurate.
  • Alternative: Read burette at eye level / to the bottom of the meniscus.
    Reason: Avoids parallax error / gives more accurate volume.
  • Alternative: Rinse the burette with NaOH solution and pipette with acid before filling.
    Reason: Removes residual water or contaminants that could dilute the solutions.
[4 marks: 1 mark for each improvement + 1 linked mark for its correct reason]

❌ Common Errors

  • Naming an indicator without a reason: Saying "use phenolphthalein" earns 1 mark, but stating "because it is pink" scores 0 for the reason. The reason must focus on the sharp colour change.
  • Vague comments: Saying "repeat the experiment" is already step 9 in the student's plan and gains zero marks.
  • Mismatched reasons: Giving a valid improvement but linking it to the wrong justification (e.g. saying "use a pipette filler to prevent parallax error").
Question 08.3 • 1 Mark

Reading a Burette Scale

Determining liquid level using the meniscus

✅ Correct Answer

24.70 cm³

(Acceptable: 24.7 cm³ )

[1 mark] Value must be read from the bottom of the curve.

🧠 Exam Technique: Downward Burette Scales

  • Burettes have 0.00 cm³ at the top and increase downwards!
  • Between 24 and 25, the scale counts downwards: each small division is 0.1 cm³ .
  • The bottom of the curved liquid surface (meniscus) rests precisely on the 7th tick below 24, giving 24.70 cm³ .
  • Watch out: Misreading from the bottom up gives 25.30 cm³ —a classic trap!
Question 08.4 • 4 Marks

Titration Concentration Calculation

Calculating concentration from reacting volumes and stoichiometry

📐 Step-by-Step Calculation

Reaction: 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O

Step 1: Calculate moles of known acid (H₂SO₄)

Volume = 25.0 cm³ = 0.0250 dm³ | Concentration = 0.216 mol/dm³

Moles of H₂SO₄ = (25.0 × 0.216) / 1000 = 0.00540 mol

[1 mark] Correct moles of acid.
Step 2: Use stoichiometric mole ratio to find moles of alkali (NaOH)

From equation: 1 mole H₂SO₄ reacts with 2 moles NaOH (1 : 2 ratio).

Moles of NaOH = 0.00540 × 2 = 0.0108 mol

[1 mark] Multiplying acid moles by 2.
Step 3: Convert volume of NaOH to dm³ and calculate concentration

Volume of NaOH used = 11.25 cm³

Concentration = moles / volume (dm³) = 0.0108 / (11.25 / 1000)

Concentration = (0.0108 × 1000) / 11.25

[1 mark] Dividing alkali moles by volume in dm³.
Step 4: Final Answer

0.960 mol/dm³ (or 0.96)

[1 mark] Final numerical value.

❌ Calculation Traps to Avoid

  • Forgetting cm³ to dm³ conversion: Always divide volume in cm³ by 1000 before using conc = moles / volume .
  • Ignoring the 2:1 ratio: Sulfuric acid is diprotic ( H₂SO₄ ). 1 mole of acid neutralises 2 moles of NaOH. Many students divide by 2 instead of multiplying by 2.
Question 08.5 • 1 Mark

Strong vs. Weak and Concentrated vs. Dilute

Interpreting acid data from Table 4

✅ Correct Answer

Tick Box 3:

The HCl is a more dilute and stronger acid than the HCN.

[1 mark] Exact match required.

💡 Strength vs. Concentration Rules

  • Concentration (mol/dm³): Measures how many solute particles are dissolved per unit volume.
    • HCl (0.1 mol/dm³) < HCN (2.0 mol/dm³), so HCl is more dilute.
  • Acid Strength (% ionised): Measures the fraction of acid molecules that dissociate into H⁺ ions.
    • HCl is 100% ionised (strong acid); HCN is 0.01% ionised (weak acid), so HCl is stronger.
Question 08.6 • 2 Marks

Explaining Differences in pH Using Data

Connecting degree of ionisation, concentration, and pH

✅ Correct Answer & Mark Breakdown

  • Mark 1: HCl has a higher concentration of hydrogen ions (H⁺) than HCN (or HCN has a lower H⁺ concentration).
  • Mark 2: Quantitative justification from table:
    • [H⁺] in HCl = 0.1 mol/dm³ (100% of 0.1)
    • [H⁺] in HCN = 0.0002 mol/dm³ (0.01% of 2.0 = 2.0 × 0.0001)
[2 marks] 1 mark for stating higher [H⁺], 1 mark for supporting data values.

🧠 Exam Technique: "Use the Data"

  • Whenever a question explicitly commands "Use the data in Table 4", you must calculate or quote numbers derived from the table.
  • Simply stating "HCl is strong and HCN is weak" will not score the data mark! Show the calculation: 2.0 × (0.01 / 100) = 0.0002 mol/dm³ .
  • Remember: Lower pH = higher concentration of H⁺ ions.

Topics

Chemistry · Required Practicals · C4: Chemical Changes · C3: Quantitative Chemistry · Required Practicals

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.