Edexcel A-Level Chemistry AS Paper 2, June 2024: Question 6

7 marks · Medium difficulty · Practical Techniques and Data Analysis

Analyse the disappearing cross reaction between sodium thiosulfate and hydrochloric acid to determine reaction rate and the effect of concentration.

Practise this question

Question

Question 6 describes an experiment investigating the effect of sodium thiosulfate concentration on the rate of reaction with hydrochloric acid by observing a disappearing cross. It includes a diagram of a conical flask placed over a cross on paper viewed from above, a table detailing five experiments with varying volumes of thiosulfate, water, and acid, and a graph plotting 1/time against concentration of sodium thiosulfate. Parts (a) to (e) ask for reasons for cloudiness, initial concentration calculation, purpose of added water, graphical determination of reaction time, interpretation of graph linearity, and evaluation of experimental accuracy and flask volume changes.
Question text

6 Sodium thiosulfate solution reacts with aqueous hydrochloric acid as shown.

Na2S2O3 + 2HCl → S + SO2 + H2O + 2NaCl

During the reaction the mixture becomes cloudy.

A student carried out an investigation to determine the effect of the concentration of

sodium thiosulfate on the rate of the reaction.

Procedure

Step 1 Place 10 cm3 of a solution of sodium thiosulfate and 40 cm3 of deionised water

in a clean 200 cm3 conical flask.

Step 2 Place the flask on a piece of paper with a black cross marked on it.

Step 3 Add 20 cm3 of hydrochloric acid (an excess) to the flask, swirl the solution and

start a timer.

Step 4 Look down through the solution at the black cross and record the time taken

for the cross to no longer be visible through the solution.

Step 5 Calculate 1/time to find the average rate of reaction.

Step 6 Change the concentration of sodium thiosulfate by repeating

Steps 1–5 using different volumes of the sodium thiosulfate solution and

deionised water.

Apparatus

reaction mixture

paper marked with cross

(a) State why the reaction mixture becomes cloudy.

(1)

(b) A student carried out the investigation using five different concentrations of

sodium thiosulfate solution.

20 Experiment 1 2 3 4 5

Volume of Na2S2O3*P76894A02032*solution added in Step1/ cm31020 30 40 50

Volume of water added in Step 1 / cm3 40 30 20 10 0

Volume of hydrochloric acid added in Step 3 / cm3 20 20 20 20 20

Concentration of Na2S2O3 immediately after

–3 0.03 0.06 0.09 0.12 0.15

adding the acid in Step 3 / mol dm

(i) What was the concentration, in mol dm–3, of the original solution of

sodium thiosulfate?

(1)

A 0.03

B 0.15

C 0.21

D 0.71

(ii) State why water is added in Experiments 1 to 4, but not in Experiment 5.

(1)

(c) The student plotted a graph of 1/time against concentration of

sodium thiosulfate.

0.05

0.04

0.03

1/time

/ s–1

0.02

0.01 *P76894A02132*

0.00

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

Concentration of Na S2O3 in the reaction mixture / mol dm–3

(i) Calculate, using the graph, the time taken for the cross to be obscured using

a concentration of sodium thiosulfate of 0.10 mol dm–3.

(1)

(ii) State and justify the relationship between the rate of reaction and the

concentration of sodium thiosulfate as shown by the graph.

(1)

(d) Which will decrease the accuracy of the experiment?

(1)

22 A rinsing the flask with deionised water before each new experiment

B stirring the solution throughout each experiment*P76894A02232*

C using a different 50 cm3 measuring cylinder for each solution

D using the same piece of paper in each experiment

(e) Experiment 1 is repeated at the same temperature, but using a 100 cm3

conical flask in place of the 200 cm3 flask.

Which statement is correct about the repeated experiment?

(1)

A it is not possible to predict how the time taken will be affected

B the time taken will decrease using the 100 cm3 flask

C the time taken will increase using the 100 cm3 flask

D the time taken will be the same using the 100 cm3 flask

(Total for Question 6 = 7 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 6 outlining the answers for 7 marks: 6(a) sulfur is insoluble in water or precipitates; 6(b)(i) option C (0.21); 6(b)(ii) to keep total volume constant; 6(c)(i) reads 1/time = 0.03 s^-1 giving time = 33.3 s; 6(c)(ii) directly proportional because the line is straight and passes through the origin; 6(d) option A; 6(e) option B.

Question

Answer Additional Guidance Mark

Number

6(a) An answer that makes reference to the following point: (1)

• sulfur is insoluble in water / precipitates Allow sulfur forms a yellow solid / comes out of solution /

does not dissolve

Allow just ‘sulfur is formed’

Question

Answer Mark

Number

6(b)(i) The only correct answer is C (0.21) (1)

A is not correct because this is the concentration in the first experiment which is diluted with water and acid

B is not correct because this is the concentration in the most concentrated of the experiments and is diluted with acid

D is not correct because this is the ratio of the volumes of thiosulfate and total volume in the reaction = 5/7

Question

Answer Additional Guidance Mark

Number

6(b)(ii) An answer that makes reference to the following point: (1)

• so that the total volume of the reaction mixture Accept the volume of sodium thiosulfate and water is kept

is kept constant / 70 cm3 constant / 50 cm3

Allow the same volume of 70 cm3 / 50 cm3

Question

Answer Additional Guidance Mark

Number

6(c)(i) Example of calculation (1)

• find the value of 1/time for the concentration of 1/time = 0.03 (s‒1)

0.10 mol dm‒3

and

calculates time time = 1 ÷ 0.03 = 33.3 / 33.333 (s)

Ignore SF except 1SF

Allow ± ½ small square so range can be up to

time = 1 ÷ 0.0305 = 32.79 / 32.8 (s)

Question

Answer Additional Guidance Mark

Number

6(c)(ii) An answer that makes reference to the following point: (1)

• (directly) proportional Allow just directly proportional

and Allow as concentration doubles rate doubles

because the graph is a straight line (which passes Allow linear

through the origin) Ignore increasing concentration increases rate

Question

Answer Mark

Number

6(d) The only correct answer is A (rinsing the flask with deionised water before each new experiment) (1)

B is not correct because any effect on the rate of the reaction will be equal in each case

C is not correct because each cylinder is the same size so will measure with the same accuracy

D is not correct because this will not affect the rate of the reaction

Question

Answer Mark

Number

6(e) The only correct answer is B (the time taken will decrease using the 100 cm3 flask) (1)

A is not correct because the depth of the solution has increased so more solid is between the eye and the cross

C is not correct because the depth of the solution has increased so more solid is between the eye and the cross

D is not correct because the depth of the solution has increased so more solid is between the eye and the cross

(Total for Question 6 = 7 marks)

How to answer it

Rates of Reaction: Sodium Thiosulfate & Hydrochloric Acid

What this question tests

This question assesses practical and mathematical competencies in chemical kinetics, including:

  • Identifying the chemical product responsible for turbidity (precipitate formation).
  • Dilution calculations using volume and concentration relationships ( C₁V₁ = C₂V₂ ).
  • Understanding control variables in reaction rate experiments (maintaining constant depth/volume).
  • Interpreting graphical rate data ( 1/time against concentration) and justifying mathematical relationships.
  • Evaluating procedural errors and apparatus modifications (path length and light obstruction).
Part (a)

Why the Reaction Mixture Becomes Cloudy

Identifying products and state changes [1 Mark]

Na₂S₂O₃(aq) + 2HCl(aq) → S(s) + SO₂(aq) + H₂O(l) + 2NaCl(aq)

✅ Correct Answer

Sulfur is formed, which is insoluble in water / precipitates as a solid.

💡 Key Knowledge

Of all four reaction products, only elemental sulfur ( S ) is insoluble in water. It forms a pale yellow precipitate that suspended in solution causes light scattering (cloudiness).

❌ Common Errors

Stating that "sulfur dioxide gas causes bubbles/cloudiness". SO₂ is a soluble toxic gas; only the solid sulfur precipitate causes turbidity.

Mark scheme guidance: Allow "sulfur forms a solid / precipitate / does not dissolve" or simply "sulfur is formed".
Part (b)(i)

Original Concentration of Sodium Thiosulfate

Dilution calculation from reaction mixture data [1 Mark]

📐 Step-by-Step Calculation

  1. Identify total volume in any experiment:
    For Exp 5: V(thiosulfate) = 50 cm³, V(water) = 0 cm³, V(acid) = 20 cm³.
    Total volume = 50 + 0 + 20 = 70 cm³.
  2. Use the dilution equation:
    C₁ × V₁ = C₂ × V₂
    C(original) × 50 cm³ = 0.15 mol dm⁻³ × 70 cm³
  3. Solve for initial concentration:
    C(original) = (0.15 × 70) / 50 = 10.5 / 50 = 0.21 mol dm⁻³

✅ Correct Answer

C (0.21 mol dm⁻³)

❌ Common Distractor Traps

  • A (0.03): Concentration of Exp 1 after dilution by both water and acid.
  • B (0.15): Concentration of Exp 5 after adding the 20 cm³ acid. Students forget the acid also dilutes the stock solution!
  • D (0.71): Incorrect volume ratio inversion ( 50 / 70 = 0.71 ).
Mark scheme guidance: 1 mark for option C.
Part (b)(ii)

Purpose of Adding Water in Experiments 1 to 4

Controlling experimental variables [1 Mark]

✅ Correct Answer

To keep the total volume of the reaction mixture constant (at 70 cm³) across all experiments.

🧠 Exam Technique

Always state what is being kept constant and why. If the volume varied, the concentration of the acid and the depth of the liquid would also change, introducing multiple uncontrolled variables.

❌ Common Errors

Vague answers such as "to dilute the solution" or "to make it a fair test" without specifying that volume is the variable being kept constant.

Mark scheme guidance: Accept "volume of sodium thiosulfate + water is kept constant / 50 cm³" or "to keep total volume constant at 70 cm³".
Part (c)(i)

Calculate Time Taken from the Rate Graph

Graph reading and reciprocal relationship [1 Mark]

📐 Step-by-Step Calculation

  1. Locate 0.10 mol dm⁻³ on the x-axis:
    Read vertically up to the line of best fit.
  2. Read the corresponding y-axis value:
    1/time = 0.030 s⁻¹ (allow ±½ square: 0.0300 to 0.0305 s⁻¹).
  3. Calculate the time taken (t):
    time = 1 / (1/time) = 1 / 0.030 = 33.3 s
    (Range: 32.8 s to 33.3 s).

✅ Correct Answer

33.3 s (accept 33 s, 33.33 s, or range 32.8 to 33.3 s)

❌ Common Calculation Traps

  • Forgetting to invert: Quoting the y-axis reading 0.03 directly as the time!
  • Rounding incorrectly: Truncating to 1 significant figure ( 30 s ) loses the mark. Standard guidance permits 2 or 3 sig figs.
Mark scheme guidance: Ignore SF except 1 SF. Reading of 0.03 gives 33.3 s; reading of 0.0305 gives 32.8 s.
Part (c)(ii)

Relationship Between Rate and Concentration

Mathematical interpretation of the graph [1 Mark]

✅ Correct Answer

Rate is directly proportional to concentration because the graph is a straight line passing through the origin (0,0).

🧠 Exam Technique: "State and Justify"

Two essential elements are required for a complete answer:

  • State: "Directly proportional" (first order).
  • Justify: Must mention both that it is a straight line AND it passes through the origin.

❌ Insufficient Responses

Writing "as concentration increases, rate increases" only describes a positive correlation, NOT direct proportionality. This scores 0 marks.

Mark scheme guidance: Allow "directly proportional" on its own, or "as concentration doubles, rate doubles", or "linear through the origin". Ignore "increasing concentration increases rate".
Part (d)

Source of Experimental Inaccuracy

Evaluating laboratory procedure [1 Mark]

✅ Correct Answer

A (rinsing the flask with deionised water before each new experiment)

💡 Why Option A Decreases Accuracy

Rinsing leaves residual water droplets inside the flask. This unmeasured extra water dilutes the reaction mixture in subsequent runs, altering the intended concentrations and decreasing accuracy.

❌ Why Other Options Are Incorrect

  • B (stirring continuously): Improves mixing consistency and ensures homogeneity across all runs.
  • C (different cylinders): Prevents cross-contamination between reagents.
  • D (same paper): Ensures the darkness/thickness of the cross remains a strictly controlled visual standard.
Mark scheme guidance: 1 mark for option A.
Part (e)

Effect of Flask Size on Reaction Time

Apparatus geometry and optical depth [1 Mark]

✅ Correct Answer

B (the time taken will decrease using the 100 cm³ flask)

💡 The Physics of the "Disappearing Cross"

A smaller conical flask (100 cm³ vs 200 cm³) has a narrower base.

  • The same total volume of liquid (70 cm³) will reach a greater depth (height).
  • Looking down from above, the light must pass through a longer path length of solution.
  • Fewer sulfur particles per unit volume are required to obscure the cross, so the cross disappears sooner (shorter time).

❌ Common Misconception

Thinking that because the quantities/concentrations are identical, the time taken must be the same (Option D). Flask geometry directly dictates liquid column height!

Mark scheme guidance: 1 mark for option B. Depth increases, meaning more solid is in the path between the eye and the cross.

Topics

Physical Chemistry · Topic 9: Kinetics I · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.