AQA GCSE Chemistry Chemistry Paper 1 (Higher), 2022: Question 8

14 marks · High Demand difficulty · Extended Answer

Explain pH in terms of acid strength and concentration, analyze titration data to calculate concentration, and explain electrical conductivity changes during a titration.

Practise this question

Question

The question is divided into five parts. Part 8.1 asks to explain why the pH of an acid depends on its strength and concentration. Part 8.2 presents a table of five titration volumes of barium hydroxide solution (23.90, 23.45, 23.55, 23.55, 23.45 cm³) used to titrate 25.00 cm³ of hydrochloric acid, and asks why 23.50 cm³ was used in the calculation. Part 8.3 asks to calculate the concentration of hydrochloric acid in mol/dm³ using the reaction equation 2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l). Part 8.4 shows a graph of electrical conductivity versus volume of barium hydroxide added during a titration with sulfuric acid; the conductivity decreases linearly from 2.5 to 0 at 30 cm³. It asks why the conductivity is zero at neutralization. Part 8.5 asks why the conductivity increases when a further 10 cm³ of barium hydroxide is added.
Question text

08 This question is about acids and alkalis.

08.1 Explain why the pH of an acid depends on:

• the strength of the acid

• the concentration of the acid.

[4 marks]

08.2 A student titrated 25.00 cm3 of hydrochloric acid with 0.100 mol/dm3

barium hydroxide solution.

Table 2 shows the results.

Table 2

Titration number 1 2 3 4 5

Volume of barium hydroxide

3 23.90 23.45 23.55 23.55 23.45

solution used in cm

The student calculated the volume of barium hydroxide solution to be used in the

titration calculation as 23.50 cm3.

Explain why the student used a volume of 23.50 cm3 of barium hydroxide solution in

the titration calculation.

[2 marks]

08.3 25.00 cm3 of the hydrochloric acid reacted with 23.50 cm3 of the 0.100 mol/dm3

barium hydroxide solution.

*24* The equation for the reaction is:

2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l)

Calculate the concentration of the hydrochloric acid in mol/dm3.

[4 marks]

26 3

Concentration of the hydrochloric acid = mol/dm

Another student titrated sulfuric acid with barium hydroxide solution.

The equation for the reaction is:

H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(l)

The student measured the electrical conductivity of the mixture during the titration.

The better a conductor, the higher the electrical conductivity value.

Figure 10 shows the results.

Figure 10

08.4 Explain why the electrical conductivity of the mixture was zero when the sulfuric acid

had just been neutralised.

Use the equation for the reaction.

Refer to ions in your answer.

[3 marks]

08.5 The student then added a further 10 cm3 of barium hydroxide solution.

*26* The electrical conductivity of the mixture increased.

Give one reason why.

[1 mark]

Mark scheme

Show the mark scheme The mark scheme provides detailed marking points for each question part. For 8.1, a level-based mark scheme is used, noting that pH depends on hydrogen ion concentration, strength relates to dissociation, and concentration relates to amount of solute per volume. For 8.2, points are awarded for identifying that the mean of concordant values (titrations 2 to 5) was calculated and 23.90 is an anomaly. For 8.3, a step-by-step calculation shows finding moles of barium hydroxide (0.00235), moles of HCl (0.00470), and concentration of HCl (0.188 mol/dm³). For 8.4, marks are given for stating there are no free ions because barium sulfate is insoluble and hydrogen ions reacted with hydroxide ions to form water. For 8.5, one mark is given for stating the mixture now contains free barium and hydroxide ions.

Question 8

AO/

Question Answers Mark

Spec. Ref

08.1 Level 3: Relevant points (reasons / causes) are identified, given 3-4 AO1

in detail and logically linked to form a clear account. 4.3.2.5

4.4.2.4

Level 2: Relevant points (reasons / causes) are identified, and 1-2 4.4.2.6

there are attempts at logical linking. The resulting account is not

fully clear.

No relevant content 0

Indicative content

General principle

• pH depends on H+ ion concentration

• the higher the concentration of H+ ions the lower the pH

Strength

• the stronger an acid the greater the ionisation / dissociation (in

aqueous solution)

• (so) the stronger the acid the lower the pH

Concentration

• the higher the concentration of an acid the more acid / solute in

the same volume (of solution)

• (so) the higher the concentration of the acid the lower the pH

AO /

Question Answers Extra information Mark

Spec. Ref.

allow identification of titration by

08.2 titration number or volume AO3

the mean of titration numbers 2 1 4.4.2.5

to 5 values is calculated RPA2

(because) 23.90 (cm3) is an allow (because) 23.90 (cm3) is 1

anomalous result not concordant

allow (because) 23.90 (cm3) is

too high a value

allow (because) the first titration

is a rough value

allow for 2 marks an answer of

(because) the mean is taken of

the values within 0.10 (cm3)

allow for 2 marks an answer of

(because) the mean is taken of

the concordant values

AO /

Spec. Ref.

08.3 (moles Ba(OH)2 = AO2

23.50 1 4.3.4

× 0.100 ) = 0.00235 4.4.2.5

1000 RPA2

(moles HCl = 0.00235 × 2 =) allow correct use of an 1

0.00470 incorrectly calculated number of

moles of Ba(OH)2

(concentration =) allow correct use of an 1

1000 incorrectly calculated number of

0.00470 × moles of HC

25.0 l

= 0.188 (mol/dm )

alternative approach:

� moles HCl � allow inverted expression

ratio =

moles Ba(OH)2 27

2 25.0 × concentration allow 1 mark for the expression

= (2)

1 23.50 × 0.100 with an incorrect mole ratio

(concentration =)

2 × 23.50 × 0.100 allow correct use of the

(1) expression with an incorrect

25.00

mole ratio

= 0.188 (mol/dm ) (1)

AO /

Spec. Ref.

08.4 there are no ions that are free to allow there are no ions in 1 AO3

move solution 4.2.2.3

allow there are no ions free to 4.4.2.2

carry the charge 4.4.2.4

4.4.2.5

(because) barium sulfate is 1

solid / insoluble

(and) hydrogen ions have allow (and) water is a covalent / 1

reacted with hydroxide ions to molecular substance

produce water

AO /

Spec. Ref.

08.5 the mixture (now) contains allow excess barium hydroxide 1 AO3

barium ions and hydroxide ions solution contains ions 4.2.2.3

that are free to move 4.4.2.2

4.4.2.4

4.4.2.5

Total Question 8 14

How to answer it

Acids, Titrations, and Electrical Conductivity

What this question tests

This question assesses your understanding of acid strength vs concentration, your ability to process titration data (identifying concordant results and calculating concentrations), and your conceptual understanding of ionic conductivity during a neutralisation reaction.

Part 8.1: Acid Strength vs Concentration

Explain why the pH of an acid depends on its strength and concentration. [4 marks]

💡 Key Knowledge

  • General Principle: pH is a measure of the concentration of hydrogen ions ( H⁺ ). The higher the H⁺ concentration, the lower the pH.
  • Acid Strength: Refers to the degree of ionisation/dissociation of acid molecules in aqueous solution.
  • Acid Concentration: Refers to the amount of acid solute dissolved in a given volume of solution.

🧠 Exam Technique

This is a Level-structured question. To get the maximum 4 marks, you must clearly link both strength and concentration back to the overall concentration of H⁺ ions and how that affects pH.

Level 3 (3-4 marks): Logical, fully linked explanation of both factors.

✅ Model Answer

  • General: pH depends on the concentration of hydrogen ( H⁺ ) ions. The higher the concentration of H⁺ ions, the lower the pH. 1 Mark
  • Strength: A stronger acid ionises/dissociates more completely in aqueous solution, releasing more H⁺ ions, which lowers the pH. 1 Mark
  • Concentration: A higher concentration of acid means there is more acid solute dissolved in the same volume of solution, resulting in more H⁺ ions in solution, which lowers the pH. 1 Mark

Part 8.2: Titration Data Analysis

Explain why the student used a volume of 23.50 cm³ of barium hydroxide solution in the titration calculation. [2 marks]

💡 Concordant Results

In titrations, we only use concordant results to calculate the mean. Concordant results are volumes within 0.10 cm³ of each other. This ensures high precision.

❌ Common Errors

Do NOT calculate the mean of all titrations. Titration 1 ( 23.90 cm³ ) is an anomaly (often the rough titration) and must be discarded.

✅ Model Answer

The student calculated the mean using only the concordant runs (Titrations 2, 3, 4, and 5), which are all within 0.10 cm³ of each other. 1 Mark

Titration 1 ( 23.90 cm³ ) was excluded because it is an anomalous result (or a rough titration). 1 Mark

Part 8.3: Titration Calculation

Calculate the concentration of the hydrochloric acid in mol/dm³. [4 marks]

Equation: 2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l)

Given: 25.00 cm³ of HCl reacted with 23.50 cm³ of 0.100 mol/dm³ Ba(OH)₂.

📐 Step-by-Step Calculation

1
Calculate the moles of Ba(OH)₂ used:
Moles = (Volume in cm³ / 1000) × Concentration
Moles of Ba(OH)₂ = (23.50 / 1000) × 0.100 = 0.00235 mol
1 Mark
2
Use the reacting ratio to find moles of HCl:
From the balanced equation, the ratio is 2 HCl : 1 Ba(OH)₂ .
Moles of HCl = 0.00235 × 2 = 0.00470 mol
1 Mark
3
Calculate the concentration of HCl:
Concentration = Moles / (Volume in cm³ / 1000)
Concentration = 0.00470 / (25.00 / 1000) = 0.00470 × (1000 / 25.00)
Concentration = 0.188 mol/dm³
2 Marks (1 for correct expression, 1 for final answer)

⚠️ Calculation Traps

  • The Ratio Trap: Many students divide by 2 instead of multiplying by 2. Always look at the balanced equation! You need twice as much HCl as Ba(OH)₂.
  • Volume Conversion: Always divide cm³ by 1000 to convert to dm³ before using concentration formulas.

Part 8.4: Conductivity at Neutralisation

Explain why the electrical conductivity of the mixture was zero when the sulfuric acid had just been neutralised. [3 marks]

Equation: H₂SO₄(aq) + Ba(OH)₂(aq) → BaSO₄(s) + 2H₂O(l)

💡 Conductivity Requirements

For a solution to conduct electricity, it must contain free-moving ions to carry the electrical charge.

🧠 Look at the State Symbols!

Notice the state symbols in the equation:
• BaSO₄(s) is a solid precipitate.
• H₂O(l) is a covalent liquid.

✅ Model Answer

  • There are no free-moving ions remaining in the mixture to carry the charge. 1 Mark
  • Because barium sulfate ( BaSO₄ ) is an insoluble solid precipitate, so its ions are locked in a lattice and cannot move. 1 Mark
  • The hydrogen ions ( H⁺ ) and hydroxide ions ( OH⁻ ) have reacted to form water ( H₂O ), which is a covalent molecular substance (not ionic). 1 Mark

Part 8.5: Post-Neutralisation Conductivity

The student then added a further 10 cm³ of barium hydroxide solution. The electrical conductivity increased. Give one reason why. [1 mark]

✅ Model Answer

The mixture now contains excess barium ions ( Ba²⁺ ) and hydroxide ions ( OH⁻ ) which are free to move and carry the electrical charge. 1 Mark

Topics

Chemistry · Required Practicals · Required Practicals · C4: Chemical Changes · C3: Quantitative Chemistry · C2: Bonding, Structure and the Properties of Matter

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.