AQA GCSE Chemistry Chemistry Paper 2 (Higher), 2022: Question 8
17 marks · High Demand difficulty · Short Answer
Analyze the reaction between sodium thiosulfate and hydrochloric acid by calculating reaction rates from a tangent, explaining concentration effects, sketching predicted curves, and determining reactant volume ratios from graphs.
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Question text
08 This question is about the reaction between sodium thiosulfate solution and
hydrochloric acid.
When hydrochloric acid is added to sodium thiosulfate solution, the mixture gradually
becomes cloudy.
The equation for the reaction is:
Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s)
08.1 Sulfur is produced in the reaction.
Why does the mixture become cloudy?
[1 mark]
A student investigated the effect of changing the concentration of sodium thiosulfate
solution on the rate of the reaction.
Figure 6 shows the apparatus used.
Figure 6
A smaller percentage of light from the light source reaches the light sensor as the
mixture becomes more cloudy.
This is the method used.
1. Measure 50 cm3 of 0.10 mol/dm3 sodium thiosulfate solution into the beaker.
2. Add 10 cm3 of hydrochloric acid to the sodium thiosulfate solution.
3. Immediately start a timer.
4. Record the percentage of light from the light source that reaches the light sensor
every 20 seconds for 120 seconds.
30 3
5. Repeat steps 1 to 4 using 0.20 mol/dm sodium thiosulfate solution.
*28* Figure 7 shows the results for 0.10 mol/dm3 sodium thiosulfate solution.
Figure 7
08.2 The percentage of light reaching the light sensor decreases by 1% when 7.1 × 10−5
moles of sulfur is produced.
Determine the rate of reaction in mol/s for the production of sulfur at 30 seconds.
You should draw a tangent on Figure 7.
[5 marks]
Rate = mol/s
08.3 Explain why the rate of reaction changes between 0 and 60 seconds.
Answer in terms of concentration.
Use Figure 7.
[2 marks]
Figure 8 is a repeat of Figure 7.
Figure 8
Figure 8 shows the results for 0.10 mol/dm3 sodium thiosulfate solution.
Sodium thiosulfate solution was in excess in the investigation.
08.4 The line of best fit on Figure 8 is horizontal between 80 and 120 seconds because the
reaction stopped.
Why did the reaction stop?
[1 mark]
08.5 Sketch a line on Figure 8 to show the results you would predict for 0.20 mol/dm3
sodium thiosulfate solution. 33
[2 marks]
The same student did the investigation again the next day.
The student found that the same method produced different results for the percentage
*32* of light reaching the light sensor.
08.6 How could the student improve the method so that the same percentages of light
reached the light sensor?
[1 mark]
Tick ( ) one box.
Record the percentage of light every 10 seconds.
Stop light from other sources reaching the light sensor.
Use a larger volume of sodium thiosulfate solution.
Use a more sensitive light sensor.
08.7 The student improved the method so that similar results were obtained on
different days.
What name is given to similar results obtained on different days under the same
conditions by the same student?
[1 mark]
Tick ( ) one box.
Anomalous
Precise
Repeatable
Reproducible 34
Figure 9 shows the volumes of:
• sodium thiosulfate solution of concentration 0.10 mol/dm3
• hydrochloric acid of concentration 0.05 mol/dm3
which completely react to produce different masses of sulfur.
Figure 9
08.8 Which expression represents the relationship between the volume (V) of sodium
thiosulfate solution used and the mass (m) of sulfur produced?
Use Figure 9.
[1 mark]
Tick ( ) one box.
V ∝ m
V ~ m
V << m
V = m
08.9 Determine the simplest whole number ratio of the volumes of
sodium thiosulfate solution : hydrochloric acid
which completely react with each other.
Use Figure 9.
[3 marks]
Simplest whole number ratio = :
Mark scheme
Show the mark scheme
Question 8
AO /
Question Answers Extra information Mark
Spec. Ref.
08.1 (sulfur is a) precipitate / solid 1 AO2
or 4.6.1.2
(sulfur is an) insoluble RPA5
substance
AO /
Spec. Ref.
08.2 correctly drawn tangent at 30 s 1 AO2
4.6.1.1
View with
RPA5
Figure 7
correct values for x step and y allow correct use of an 1
step from tangent incorrectly drawn tangent
allow a tolerance of ± ½ a small
square for each coordinate
value for y step allow correct use of incorrectly 1
(ratio =) value for x step determined values from tangent
for x step and/or y step
correct calculation of ratio 1
(conversion allow correct use of an 1
rate = ratio × 7.1 × 10-5 ) incorrectly calculated ratio
correct evaluation of rate (mol/s)
AO /
Spec. Ref.
08.3 rate decreases allow the collision frequency 1 AO2
decreases 4.6.1.1
4.6.1.2
(because) concentration of 1 RPA5
reactants decreases
24 alternative approach:
greatest rate at start (1) allow the collision frequency is
highest at the start
(because) greatest
concentration of reactants at
start (1)
AO /
Spec. Ref.
08.4 (hydrochloric) acid is used up allow (hydrochloric) acid is the 1 AO3
limiting reactant 4.3.2.4
4.6.1.2
ignore reactants used up RPA5
AO /
Spec. Ref.
08.5 decreasing curve starting at 1 AO2
(0,95) and steeper initially than 4.6.1.2
View with 3 RPA5
curve for 0.10 mol/dm sodium
Figure 8 thiosulfate solution
levelling at 24% 1
AO /
Spec. Ref.
08.6 stop light from other sources 1 AO3
reaching the light sensor 4.6.1.2
RPA5
AO /
Spec. Ref.
08.7 repeatable 1 AO3
4.6.1.2
RPA5
AO /
Spec. Ref.
08.8 V ∝ m 1 AO2
4.6.1.2
AO /
Spec. Ref.
08.9 volume of sodium thiosulfate allow a tolerance of ± ½ a small 1 AO2
solution and volume of square for volume readings 4.6.1.2
View with
hydrochloric acid at any fixed
Figure 9 mass
allow
volume of Na2S2O3 solution volume of hydrochloric acid 1
� =� � =�
volume of hydrochloric acid volume of Na2S2O3 solution
0.25 4
allow correct use of incorrectly
determined volumes
1 : 4 1
Total Question 8 17
How to answer it
Rates of Reaction & Stoichiometry
AQA GCSE Chemistry Required Practical 5 (Disappearing Cross / Turbidity)
What this question tests
This question assesses your understanding of factors affecting the rate of chemical reactions, specifically concentration. It tests your ability to interpret experimental data, draw tangents to curves to calculate instantaneous rates, explain rate changes using collision theory, identify limiting reactants, and determine reacting ratios from graphical data.
Part 8.1: Why the Mixture Becomes Cloudy
Correct Answer
Sulfur is produced as a precipitate (or a solid / insoluble substance).
Key Knowledge
Look at the state symbols in the equation:
S(s) indicates that sulfur is a solid. Because it is insoluble in water, it forms a suspension that blocks light.
Part 8.2: Calculating Rate of Reaction Using a Tangent
Step-by-Step Calculation
- Draw the Tangent: Place a ruler flat against the curve at exactly 30 seconds on Figure 7. Draw a straight line that matches the slope of the curve at that exact point.
- Find the Gradient (y-step / x-step): Pick two easy-to-read points on your tangent line.
• Let's say your tangent line starts at (0 s, 80%) and ends at (70 s, 10%) .
• y-step (change in % light) = 80 - 10 = 70%
• x-step (change in time) = 70 - 0 = 70 s - Calculate the Gradient Ratio:
Ratio = y-step / x-step = 70 / 70 = 1.0% per second (Note: Your exact values will depend on your drawn tangent, but the method is identical). - Convert to mol/s: The question states that a 1% decrease in light corresponds to 7.1 × 10⁻⁵ moles of sulfur.
Rate = Gradient Ratio × (7.1 × 10⁻⁵)
Rate = 1.0 × 7.1 × 10⁻⁵ = 7.1 × 10⁻⁵ mol/s
Exam Technique
Always draw your tangent line long enough to span across several grid lines. This makes reading the coordinates much easier and significantly reduces your margin of error!
Common Errors
Do not calculate the gradient using points directly from the curve. You must draw a straight tangent line and use points from that line.
Part 8.3: Explaining Rate Changes
Correct Answer
- The rate of reaction decreases 1 Mark
- Because the concentration of reactants decreases (which reduces the frequency of collisions) 1 Mark
Collision Theory Link
As reactant particles react to form products, there are fewer reactant particles left in the same volume. This means they collide less frequently, slowing down the rate.
Part 8.4: Why Did the Reaction Stop?
Correct Answer
The hydrochloric acid is used up (or hydrochloric acid is the limiting reactant).
Common Pitfalls
Do not just write "the reactants are used up". The question states that sodium thiosulfate was in excess, so you must specify that it is the acid that ran out.
Part 8.5: Sketching the Curve for Higher Concentration
How to Draw the Curve on Figure 8
To get both marks, your sketched line must have these two distinct features:
- Feature 1 (Steeper Start): Start the curve at (0, 95) but make it drop much more steeply than the original curve. This shows that doubling the concentration increases the initial rate of reaction. 1 Mark
- Feature 2 (Same Endpoint): Level the curve off horizontally at exactly 24% (the same level as the original curve). 1 Mark
Why does it level off at the same height?
Even though we doubled the concentration of sodium thiosulfate, it was already in excess. The amount of product (sulfur) is entirely controlled by the limiting reactant (hydrochloric acid), which did not change. Therefore, the same total mass of sulfur is made, and the light sensor levels off at the same percentage.
Parts 8.6 & 8.7: Experimental Design & Terminology
8.6 Correct Option
☑ Stop light from other sources reaching the light sensor.
Why? External room light acts as a systematic error. Shielding the apparatus ensures only light from the source is measured.
8.7 Correct Option
☑ Repeatable
Why? "Repeatable" means the same student gets similar results using the same method and equipment.
Part 8.8: Graphical Relationships
Correct Option
☑ V ∝ m
This symbol (∝) means directly proportional. Because the graph in Figure 9 is a straight line passing through the origin (0,0), volume and mass are directly proportional.
Part 8.9: Determining the Simplest Reacting Volume Ratio
Step-by-Step Ratio Calculation
- Pick a fixed mass of sulfur on the y-axis of Figure 9:
Let's choose 0.20 g of sulfur because it aligns perfectly with the grid lines. - Read the volumes required for this mass:
• Volume of 0.10 mol/dm³ sodium thiosulfate (dashed line) = 50 cm³ 1 Mark
• Volume of 0.05 mol/dm³ hydrochloric acid (solid line) = 200 cm³ - Set up the ratio:
Sodium Thiosulfate : Hydrochloric Acid
50 : 200 1 Mark - Simplify to the simplest whole numbers:
Divide both sides by 50:
1 : 4 1 Mark
Alternative Check
You can pick any mass! For example, at 0.30 g of sulfur:
• Thiosulfate = 75 cm³
• Acid = 300 cm³
Ratio = 75 : 300 = 1 : 4. The ratio remains constant!
Common Trap
Be careful not to write the ratio backwards (4 : 1). Always double-check which reactant is listed first in the question: sodium thiosulfate : hydrochloric acid.
Topics
Chemistry · Required Practicals · Required Practicals · C6: The Rate and Extent of Chemical Change · C3: Quantitative Chemistry
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.