AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2020: Question 9
18 marks · Standard Demand difficulty · Short Answer
Investigate temperature changes in an endothermic neutralization reaction and perform titration calculations with citric acid and sodium hydroxide.
Practise this questionQuestion
Question text
09 This question is about citric acid (C6H8O7).
Citric acid is a solid.
A student investigated the temperature change during the reaction between citric acid
and sodium hydrogencarbonate solution.
This is the method used.
1. Pour 25 cm3 of sodium hydrogencarbonate solution into a polystyrene cup.
2. Measure the temperature of the sodium hydrogencarbonate solution.
3. Add 0.20 g of citric acid to the polystyrene cup.
4. Stir the solution.
5. Measure the temperature of the solution.
6. Repeat steps 3 to 5 until a total of 2.00 g of citric acid has been added.
The student plotted the results on a graph.
Figure 6 shows the student’s graph.
Figure 6
09.1 Figure 6 shows an anomalous point when 0.60 g of citric acid was added.
This was caused by the student making an error.
The student correctly:
• measured the mass of the citric acid
*26* • read the thermometer
• plotted the point.
Suggest one reason for the anomalous point.
[1 mark]
09.2 Explain the shape of the graph in terms of the energy transfers taking place.
You should use data from Figure 6 in your answer.
[3 marks]
09.3 A second student repeated the investigation using a metal container instead of the
polystyrene cup. The container and the cup were the same size and shape.
Sketch a line on Figure 6 to show the second student’s results until 1.00 g of
citric acid had been added. The starting temperature of the solution was the same.
Explain your answer.
[3 marks]
The student used a solution of citric acid to determine the concentration of a solution
of sodium hydroxide by titration.
09.4 The student made 250 cm3 of a solution of citric acid of concentration 0.0500 mol/dm3
Calculate the mass of citric acid (C6H8O7) required.
Relative atomic masses (Ar): H = 1 C = 12 O = 16
[3 marks]
Mass = g
This is part of the method the student used for the titration.
1. Measure 25.0 cm3 of the sodium hydroxide solution into a conical flask
using a pipette.
2. Add a few drops of indicator to the flask.
3. Fill a burette with citric acid solution.
09.5 Describe how the student would complete the titration.
[3 marks]
09.6 Give two reasons why a burette is used for the citric acid solution.
[2 marks]
09.7 13.3 cm3 of 0.0500 mol/dm3 citric acid solution was needed to neutralise
25.0 cm3 of sodium hydroxide solution.
The equation for the reaction is:
3NaOH + C6H8O7 → C6H5O7Na3 + 3H2O
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 didn’t stir (the solution enough) allow measured the temperature 1 AO3
before the temperature stopped 4.5.1.1
falling RPA 4
allow measured the temperature
too soon
09.2 the temperature decreases allow temperature decreases 1 AO2
(initially) because energy is (initially) because the reaction is AO3
taken in (by the reaction from endothermic 4.5.1.1
the solution) RPA 4
when 1.5 g (of citric acid) is allow when the temperature 1
added the sodium reaches 11.6 °C the sodium
hydrogencarbonate has all hydrogencarbonate has all
reacted reacted
or
from 1.5 g the citric acid is in allow after the temperature
excess reaches 11.6 °C the citric acid is
in excess
or
when 1.5 g (of citric acid) is allow when the temperature
added the reaction is complete reaches 11.6 °C the reaction is
complete
(so) the temperature increases allow (so) the temperature 1
as energy is transferred from the increases as energy is
room to the solution transferred from the excess
citric acid to the solution
Question 9 continued
AO /
Spec. Ref.
09.3 less steep line starting at ignore any part of the line drawn 1 AO3
16.8 °C and reaching 1.00 g (of beyond 1.00 g 4.2.2.8
citric acid) 4.5.1.1
27 RPA 4
(as) metal is a better conductor allow (as) polystyrene is a better 1
insulator
(so) more energy is absorbed allow (so) more heat is 1
(from the surroundings) absorbed (from the
surroundings)
09.4 (Mr citric acid =) 192 1 AO2
4.3.2.5
250 4.3.4
(moles = × 0.0500)
1000
= 0.0125 1
(mass = 0.0125 × 192 =) 2.4 (g) allow correct use of an 1
incorrectly calculated Mr
allow correct use of an
incorrectly calculated number of
moles
alternative approach:
(Mr citric acid =) 192 (1)
(concentration = 0.0500 × 192)
= 9.6 (g/dm3) (1) allow correct use of an
incorrectly calculated Mr
250 allow correct use of an
(mass = × 9.6 =) 2.4 (g) (1)
1000 incorrectly calculated
concentration in g/dm3
Question 9 continued
AO /
Spec. Ref.
09.5 add the citric acid (to the flask) ignore colours of indicator 1 AO1
until there is a (permanent) 4.4.2.5
colour change RPA 2
measure / record the volume (of allow take the final (and initial) 1
citric acid) added burette reading
any one from: 1
• swirl
• use a white tile
• add the citric acid dropwise allow add the citric acid slowly
(near the end-point) (near the end-point)
• repeat and calculate a mean
09.6 any two from: 2 AO1
4.4.2.5
• can add (the citric acid) in allow can add (the citric acid) RPA 2
small increments drop by drop
allow can add (the citric acid)
slowly
• can measure variable volumes allow has a scale
• more accurate than a
measuring cylinder
Question 9 continued
AO /
Spec. Ref.
09.7 (moles citric acid = AO2
13.3 4.3.4
× 0.0500 )
1000 1 4.4.2.5
= 0.000665 RPA 2
(moles NaOH = 3 × 0.000665 ) allow correct use of an 1
= 0.001995 incorrectly calculated number of
moles of citric acid
1000
(conc = × 0.001995)
25 1
= 0.0798 (mol/dm3)
allow 0.08 or 0.080 (mol/dm3)
allow correct use of an
incorrectly calculated number of
moles of NaOH
alternative approach:
25.0 × conc NaOH 3 13.3 ×0.0500 1
= (1) allow =
13.3 × 0.0500 1 25.0 ×conc NaOH 3
13.3 × 0.0500
(conc NaOH =) 3 ×
25.0
(1)
= 0.0798 (mol/dm3) (1) allow 0.08 or 0.080 (mol/dm3)
Total 18
How to answer it
Citric Acid: Thermochemistry & Acid-Base Titrations
This question assesses fundamental practical and analytical skills across two major GCSE Chemistry topics:
- Required Practical 4 (Temperature Changes): Identifying procedural experimental anomalies, linking temperature decreases to endothermic reactions, using graph endpoints to identify limiting reactants, and predicting heat loss effects from apparatus changes.
- Required Practical 2 & Quantitative Chemistry (Titrations): Describing titration methodology, identifying advantages of specific volumetric apparatus (burette vs cylinder), and multi-step stoichiometric molarity calculations (including finding Mᵣ , calculating mole ratios from balanced equations, and unit conversions from cm³ to dm³).
Identifying an Experimental Anomaly
Suggest one reason for the anomalous point at 0.60 g [1 mark]
✅ Acceptable Answers (Any 1)
- Did not stir the solution (or did not stir enough).
- Measured the temperature too soon / before the temperature stopped falling.
❌ Common Errors
- "Misread the thermometer" or "Misweighed citric acid" — the stem clearly states the student correctly weighed, read, and plotted!
- "Heat was lost to the surroundings" — this reaction is endothermic; heat is absorbed from the surroundings.
Explaining the Graph Using Energy Transfers
Explain the shape of the graph using data from Figure 6 [3 marks]
✅ 3-Mark Complete Model Answer
- Initial fall (0.0 g to 1.5 g): The temperature decreases because the reaction is endothermic (energy is taken in from the solution). [1 mark]
- Turning point: At 1.5 g (or when temperature reaches 11.6 °C), the sodium hydrogencarbonate has completely reacted / citric acid is now in excess. [1 mark]
- Subsequent rise (above 1.5 g): The temperature increases as energy is transferred from the warmer room/surroundings into the cooler solution (or from the added room-temperature acid). [1 mark]
🧠 Exam Technique
- Must quote data: You cannot get full marks without stating either the mass at the minimum (1.5 g) or the minimum temperature reached (11.6 °C).
- Address both sections: Describe why it goes down, what happens at the vertex (reaction stops), and why it creeps back up.
Impact of Replacing Insulation with Metal
Sketch line on graph to 1.00 g and explain your answer [3 marks]
✏️ What to Draw on Figure 6
Draw a straight line that:
- Starts at the exact same starting point: (0.0 g, 16.8 °C).
- Is less steep than the original line (lies entirely above the original line).
- Ends cleanly at 1.00 g.
✅ Explanation Points
- Metal is a better thermal conductor than polystyrene (or polystyrene is a better insulator). [1 mark]
- Therefore, more energy/heat is absorbed from the surroundings into the cold mixture, keeping its temperature higher than in the cup. [1 mark]
Calculation: Preparing a Standard Solution
Calculate mass of citric acid required for 250 cm³ of 0.0500 mol/dm³ [3 marks]
📐 Step-by-Step Calculation
- Step 1: Calculate Relative Formula Mass (Mᵣ) of C₆H₈O₇
Mᵣ = (6 × 12) + (8 × 1) + (7 × 16) = 72 + 8 + 112 = 192 [1 mark] - Step 2: Calculate moles of citric acid required
Volume in dm³ = 250 / 1000 = 0.250 dm³
Moles = Concentration × Volume = 0.0500 mol dm⁻³ × 0.250 dm³ = 0.0125 mol [1 mark] - Step 3: Calculate required mass
Mass = Moles × Mᵣ = 0.0125 mol × 192 g/mol = 2.4 g [1 mark]
❌ Common Trap
Forgetting to divide 250 cm³ by 1000. Using 250 directly gives 12.5 moles and a massive mass of 2400 g!
Completing a Titration Method
Describe how the student would complete the titration [3 marks]
✅ 3 Essential Marking Points
- Mark 1: Add citric acid to the conical flask until there is a permanent colour change.
- Mark 2: Measure / record the final volume (or initial and final burette reading) to determine volume added.
- Mark 3 (Any 1 procedural refinement):
- Swirl the flask during addition
- Place the flask on a white tile
- Add dropwise near the end-point
- Repeat titration until concordant results are obtained and calculate a mean
❌ Common Errors
- Specifying exact colour changes without naming the indicator (e.g. stating "turns red" without context is ignored).
- Saying "pour" instead of adding carefully via the burette tap.
Apparatus: Advantages of a Burette
Give two reasons why a burette is used for the citric acid solution [2 marks]
✅ Acceptable Reasons (Any 2)
- Can add the acid drop by drop / in small increments / slowly.
- Can measure variable volumes (unlike a fixed-volume pipette).
- Has a scale / graduations to deliver flexible volumes accurately.
- More accurate / precise than a measuring cylinder.
Titration Calculation: Reacting Ratios
Calculate the concentration of the sodium hydroxide solution [3 marks]
3 NaOH + C₆H₈O₇ → C₆H₅O₇Na₃ + 3 H₂O
📐 Step-by-Step Calculation
- Step 1: Calculate moles of citric acid used
Volume = 13.3 cm³ = 13.3 / 1000 = 0.0133 dm³
Moles of C₆H₈O₇ = 0.0133 dm³ × 0.0500 mol dm⁻³ = 0.000665 mol [1 mark] - Step 2: Determine moles of NaOH using reacting ratio
Ratio is 3 NaOH : 1 C₆H₈O₇
Moles of NaOH = 0.000665 mol × 3 = 0.001995 mol [1 mark] - Step 3: Calculate concentration of NaOH
Volume of NaOH = 25.0 cm³ = 0.0250 dm³
Concentration = Moles / Volume = 0.001995 mol / 0.0250 dm³ = 0.0798 mol dm⁻³ [1 mark]
(Accept 0.08 or 0.080 mol dm⁻³)
❌ The #1 Stoichiometry Mistake
Dividing by 3 instead of multiplying by 3 in Step 2. Look at the equation: 3 moles of NaOH react with every 1 mole of citric acid, so there must be 3 times as many moles of NaOH!
Topics
Chemistry · Required Practicals · Required Practicals · C3: Quantitative Chemistry · C4: Chemical Changes · C5: Energy Changes
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.