Chemistry 9701/35 — May/June 2017

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

Sulfur forms the peroxodisulfate anion, $\text{S}_2\text{O}_8^{2-}$ . This ion can oxidise iodide ions, $\text{I}^-$ , to iodine, $\text{I}_2$ , as shown in the equation.

2\text{I}^-(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})

You will carry out a series of experiments to investigate how the rate of this reaction is affected by changing the concentration of the solutions.

The rate can be measured by adding thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ , and starch indicator. As the reaction between $\text{S}_2\text{O}_8^{2-}$ and $\text{I}^-$ occurs iodine is produced, but it reacts immediately with the thiosulfate.

\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})

When all the thiosulfate has reacted, the iodine will remain in the mixture and cause the starch indicator to turn blue-black. The rate of reaction may be determined by timing how long it takes the reaction mixture to turn blue-black.

FA 1 is $0.0200\text{ mol dm}^{-3}$ potassium peroxodisulfate, $\text{K}_2\text{S}_2\text{O}_8$ .
FA 2 is $1.00\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FA 3 is $0.00500\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
starch indicator

Read through the instructions carefully and prepare a table for your results on page 4 before starting any practical work.

(a)

Method

Experiment 1

Fill the burette labelled FA 1 with FA 1.
Use the pen to label one of the $100\text{ cm}^3$ beakers 'A' and the other $100\text{ cm}^3$ beaker 'B'.
Run $20.00\text{ cm}^3$ of FA 1 from the burette into beaker A.
Use the measuring cylinder to add $20.0\text{ cm}^3$ of FA 2 into beaker B.
Use the measuring cylinder to add $10.0\text{ cm}^3$ of FA 3 to beaker B.
Add 10 drops of starch indicator to beaker B.
Add the contents of beaker A to beaker B and start timing immediately.
Stir the mixture once and place the beaker on a white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second in your results table.
Wash out both beakers and shake to remove excess water.

Experiment 2

Fill a second burette with distilled water.
Run $10.00\text{ cm}^3$ of FA 1 into beaker A.
Run $10.00\text{ cm}^3$ of distilled water into beaker A.
Use the measuring cylinder to add $20.0\text{ cm}^3$ of FA 2 into beaker B.
Use the measuring cylinder to add $10.0\text{ cm}^3$ of FA 3 to beaker B.
Add 10 drops of starch indicator to beaker B.
Add the contents of beaker A to beaker B and start timing immediately.
Stir the mixture once and place the beaker on a white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second in your results table.
Wash out both beakers and shake to remove excess water.

Experiments 3-5

Carry out three further experiments to investigate how the reaction time changes with different volumes of potassium peroxodisulfate, FA 1.

Note that the combined volume of FA 1 and distilled water must always be $20.00\text{ cm}^3$ .
Do not use a volume of FA 1 that is less than $6.00\text{ cm}^3$ .

Keep FA 1, FA 2, FA 3 and the starch indicator for use in (e).

Calculating the rate of the reaction

The rate of the reaction can be represented by the formula shown.

\text{rate} = \frac{500}{\text{reaction time in seconds}}

Use this formula to calculate the rate for each of your five experiments.

Record all your results in a single table. You should include the volume of FA 1, the volume of distilled water, the reaction time and the reaction rate for each of your five experiments.

10M

DifficultyMedium

Worked solution

Answer

Results Table (representative values shown; candidate records own data)

Volume of FA 1 /cm³	Volume of distilled water /cm³	Reaction time /s	Rate /s⁻¹
20.00	0.00	22	22.7
16.00	4.00	28	17.9
12.00	8.00	37	13.5
10.00	10.00	44	11.4
8.00	12.00	55	9.1

Key requirements for full marks:

Table includes columns for volume of FA 1, volume of distilled water, reaction time, and reaction rate.
Headings include quantity and unit (e.g. Volume of FA 1 /cm³, Reaction time /s, Rate /s⁻¹).
All volumes recorded to 2 decimal places (e.g. 20.00, not 20).
All reaction times recorded to the nearest second (whole numbers).
Three additional volumes chosen with intervals of at least 2.00 cm³, all ≥ 6.00 cm³, and at least one ≤ 8.00 cm³.
Total volume of FA 1 + distilled water = 20.00 cm³ in every experiment.
Rate calculated as $\frac{500}{\text{time}}$ , given to minimum 2 significant figures and 1 decimal place.
Rate units given as s⁻¹.
Time for 20.00 cm³ of FA 1 within ±3 s of supervisor's value (2 marks) or ±5 s (1 mark).
Ratio of time for 10.00 cm³ / time for 20.00 cm³ between 1.8 and 2.2.

Final answer

See working — candidate-dependent table with five experiments, volumes to 2 d.p., times to nearest second, rates calculated as 500/time in s⁻¹

Detailed explanation

Background Concept

This experiment uses the 'clock reaction' method to measure the rate of the reaction between peroxodisulfate ions (S₂O₈²⁻) and iodide ions (I⁻). The principle is that iodine produced in the main reaction is immediately consumed by thiosulfate ions (S₂O₃²⁻) in a fast secondary reaction. Once all the thiosulfate has been used up, free iodine accumulates and reacts with starch to give a characteristic blue-black colour. The time taken for this colour change is inversely proportional to the rate of the main reaction.

The rate is expressed as 500/time (s⁻¹), where the constant 500 is chosen to give convenient numerical values. Since the total volume is kept constant at 50 cm³, the volume of FA 1 used is directly proportional to the initial concentration of S₂O₈²⁻ in the reaction mixture.

Understanding the Question

Part (a) asks the candidate to carry out five experiments (two prescribed, three self-chosen) and record all results in a single table. The mark scheme awards marks for: table construction with correct headings and units, precision of recorded values, appropriate choice of additional volumes, consistency of total volume, comparison with supervisor's values, and correct rate calculations.

The command words here are practical: 'record', 'calculate', 'construct a table'.

Approach

Set up a table with four columns: volume of FA 1, volume of distilled water, reaction time, and rate.
Carry out Experiments 1 and 2 as described, recording times to the nearest second.
Choose three additional volumes of FA 1 that satisfy all constraints: intervals ≥ 2.00 cm³, all volumes ≥ 6.00 cm³, at least one ≤ 8.00 cm³. A good choice would be 16.00, 12.00, and 8.00 cm³.
For each experiment, calculate water volume as (20.00 − volume of FA 1).
Calculate rate = 500/time for each experiment, to at least 2 significant figures and 1 decimal place.

Step-by-Step Reasoning

Table construction (Mark I): The table must have columns for all four quantities. Without a table, no marks are available for the data columns.

Headings and units (Mark II): Each column heading must state the quantity AND its unit. For volumes, acceptable forms are 'Volume of FA 1 /cm³' or 'Volume of FA 1 (cm³)'. All volumes must be given to 2 decimal places (e.g. 16.00 not 16) because they are measured from a burette.

Times to nearest second (Mark III): Reaction times are recorded as whole numbers (e.g. 22, not 22.0 or 22.5).

Choice of additional volumes (Mark IV): Three further experiments are needed. Constraints: intervals between consecutive volumes must be ≥ 2.00 cm³; no volume below 6.00 cm³; at least one volume at or below 8.00 cm³. A valid set is 16.00, 12.00, 8.00 cm³ (intervals of 4.00 cm³, minimum 8.00 cm³).

Total volume consistency (Mark V): FA 1 volume + water volume must always equal 20.00 cm³. This keeps the total reaction volume constant at 50 cm³, ensuring that concentration changes are due only to the FA 1 volume.

Comparison with supervisor (Marks VI-VII): The time for 20.00 cm³ of FA 1 should be close to the supervisor's value. Within ±3 s earns 2 marks; within ±5 s earns 1 mark.

Ratio check (Mark VIII): Since rate is proportional to [S₂O₈²⁻], halving the concentration (20.00 → 10.00 cm³) should double the time. The ratio t(10.00)/t(20.00) should be between 1.8 and 2.2.

Rate calculations (Mark IX): Rate = 500/time. For example, if time = 22 s, rate = 500/22 = 22.7 s⁻¹. Must show at least 2 significant figures and 1 decimal place.

Rate units (Mark X): The unit must be s⁻¹ (or 'per second').

Key Takeaways

In a clock reaction, the measured time is inversely proportional to rate.
Keeping total volume constant while varying one reactant's volume allows direct proportionality between volume used and concentration.
Table construction marks are often the easiest to lose through missing units or inconsistent decimal places.

Common Mistakes

Recording volumes to 1 decimal place (e.g. 20.0) instead of 2 (e.g. 20.00) — burette readings require 2 d.p.
Choosing volumes with intervals less than 2.00 cm³ (e.g. 19.00, 18.00, 17.00).
Forgetting to include at least one volume ≤ 8.00 cm³.
Calculating rates to only 1 significant figure or with no decimal places.
Omitting units from column headings.
Recording times with decimal places (e.g. 22.5 s) instead of nearest second.

Things to Be Careful About

The mark for the ratio (1.8–2.2) depends on your own recorded times, so ensure consistency between Experiments 1 and 2.
Rate must be given to minimum 2 sf AND 1 dp — a value like '11' (no dp) or '11.4' (2 sf, 1 dp ✓) but '11.43' is also acceptable.
The unit s⁻¹ must appear in the heading or next to values; writing just 'rate' without units loses the mark.

Techniques used

construct a results table with headings and unitsrecord volumes to two decimal placesmeasure reaction time to nearest secondcalculate rate using 500/timechoose appropriate range of independent variable

(b)

On the grid on page 5, plot the rate ( $y$ -axis) against the volume of FA 1 ( $x$ -axis). Include the origin in your plot. Draw a straight line of best fit and circle any clearly anomalous points.

Grid 1

DifficultyMedium-Easy

Worked solution

Answer

Plot rate (s⁻¹) on the $y$ -axis against volume of FA 1 (cm³) on the $x$ -axis.

Both axes clearly labelled with quantity and unit.
Linear scales chosen so that plotted points and the origin use more than half of each axis.
All five data points plotted correctly (within half a small square of the correct position).
Include the origin (0, 0) in the plot.
Draw a straight line of best fit through the points (or a smooth curve if appropriate).
Circle any clearly anomalous points that fall significantly off the line.

Final answer

Graph of rate (s⁻¹) vs volume of FA 1 (cm³) showing a straight line through the origin, with all points plotted correctly and a line of best fit drawn

Detailed explanation

Background Concept

Graphical analysis is a key skill in practical chemistry. A well-constructed graph allows visual identification of relationships between variables, extrapolation to unmeasured values, and detection of anomalous results. In this experiment, since rate is proportional to concentration (and hence to volume of FA 1), the expected graph is a straight line through the origin.

Understanding the Question

The candidate must plot their own data from part (a) on the provided grid. The mark scheme checks four aspects: correct axis labelling, appropriate scale, accurate plotting, and a valid line of best fit.

Approach

Label axes: $y$ -axis = 'Rate /s⁻¹', $x$ -axis = 'Volume of FA 1 /cm³'.
Choose scales so that the data range (approximately 0 to 20 cm³ on $x$ , 0 to ~25 s⁻¹ on $y$ ) occupies more than half the grid in each direction.
Plot each point carefully using the calculated rates.
Include the origin (0, 0) since zero volume of FA 1 means zero concentration and hence zero rate.
Draw a single straight line of best fit that passes as close as possible to all points, including through the origin.
If any point is clearly off the line (more than a few small squares), circle it as anomalous.

Step-by-Step Reasoning

Axis labelling (Mark I): Each axis must show both the quantity name and its unit. Writing just 'Rate' without '/s⁻¹' or just 'Volume' without '/cm³' loses this mark. The $y$ -axis must be rate and the $x$ -axis must be volume of FA 1 — reversing them loses the mark.

Scale selection (Mark II): The scale must be linear (equal divisions throughout) and chosen so that plotted points occupy more than 50% of the grid length in both directions. If points cluster in a small area, the scale is inappropriate.

Plotting accuracy (Mark III): Each point must be within half a small square of its correct position. Using the representative data: (20.00, 22.7), (16.00, 17.9), (12.00, 13.5), (10.00, 11.4), (8.00, 9.1). The origin (0, 0) should also be included.

Line of best fit (Mark IV): A single straight line (since the relationship is linear) drawn with a ruler, passing as close as possible to all points. Anomalous points (those far from the line) should be circled rather than included in the line's path.

Key Takeaways

Always label axes with both quantity and unit.
Choose scales that maximise the use of the available grid.
A line of best fit should reflect the overall trend, not simply connect the dots.
Anomalous points are circled, not ignored or forcibly included.

Common Mistakes

Reversing the axes (putting volume on $y$ and rate on $x$ ).
Using a non-linear scale or one that compresses all points into a small region.
Drawing a curve of best fit when a straight line is clearly appropriate.
Not including the origin when instructed.
Failing to circle anomalous points.

Things to Be Careful About

The mark for 'suitable linear scales' requires equal spacing between divisions — do not use a logarithmic or irregular scale.
'More than half of each axis' means the plotted data should span at least 50% of the grid width and height.
If the line does not pass through the origin, this may indicate a systematic error or an incorrect scale choice.

Techniques used

plot points on a graph with labelled axeschoose appropriate scale to fill griddraw line of best fitidentify anomalous points

(c)

The volume of FA 1 is directly related to the concentration of potassium peroxodisulfate.

From your results, what can be stated about the relationship between the rate of reaction and the concentration of potassium peroxodisulfate?

DifficultyEasy

Worked solution

Answer

The rate of reaction is directly proportional to the concentration of potassium peroxodisulfate.

Final answer

Rate is directly proportional to concentration of potassium peroxodisulfate

Detailed explanation

Background Concept

When one variable is directly proportional to another, their graph is a straight line passing through the origin. In kinetics, if doubling the concentration of a reactant doubles the rate, the reaction is first order with respect to that reactant. The volume of FA 1 used is directly proportional to the concentration of S₂O₈²⁻ in the reaction mixture (since total volume is constant), so a linear graph of rate against volume of FA 1 through the origin demonstrates direct proportionality between rate and [S₂O₈²⁻].

Understanding the Question

The command word is 'state' — the candidate must give a concise conclusion about the relationship. The question explicitly notes that volume of FA 1 is directly related to concentration of K₂S₂O₈, so the conclusion should reference concentration, not just volume.

Approach

From the graph in part (b), if the line of best fit is straight and passes through the origin, this indicates a directly proportional relationship. The conclusion should state this in terms of concentration.

Step-by-Step Reasoning

The graph shows rate on the $y$ -axis and volume of FA 1 on the $x$ -axis. Since volume of FA 1 ∝ [S₂O₈²⁻] (total volume constant), and the graph is a straight line through the origin, we conclude:

Rate ∝ [S₂O₈²⁻]

This means the reaction is first order with respect to peroxodisulfate ions. The mark scheme accepts 'rate is directly proportional to concentration' or any comment suitable to the shape of the graph (e.g. 'a straight line through the origin shows the rate increases linearly with concentration').

Key Takeaways

A straight line through the origin on a graph indicates direct proportionality.
In this experiment, volume of FA 1 is a proxy for concentration because total volume is held constant.
This result is consistent with the reaction being first order in S₂O₈²⁻.

Common Mistakes

Saying 'rate increases with concentration' without specifying 'directly proportional' — this is too vague and may not earn the mark.
Referring to volume of FA 1 rather than concentration of peroxodisulfate (though the mark scheme may accept 'comment suitable to shape of graph').
Stating it is 'linear' without connecting to proportionality through the origin.

Things to Be Careful About

The word 'proportional' (or 'directly proportional') is key — 'related' or 'affected by' would not earn the mark.
The conclusion must reference concentration of potassium peroxodisulfate (or S₂O₈²⁻), not just the volume of FA 1.

Techniques used

interpret linear graph through origin to deduce proportionality

(d)

(i)

Use your graph to calculate the reaction time you would expect to measure if you carried out an experiment using $5.00\text{ cm}^3$ of FA 1.

Show your working.

DifficultyMedium-Easy

Worked solution

Working

From the graph, read the rate at volume of FA 1 = 5.00 cm³.

For example, if the line of best fit gives a rate of approximately $5.7\text{ s}^{-1}$ at 5.00 cm³:

\text{time} = \frac{500}{\text{rate}} = \frac{500}{5.7} = 88\text{ s}

Answer

Reaction time $\approx 88$ s (value depends on candidate's graph; must show use of $500/\text{rate}$ )

Final answer

Approximately 88 s (candidate-dependent based on graph reading)

Detailed explanation

Background Concept

The relationship rate = 500/time can be rearranged to time = 500/rate. By reading the rate from the graph at a specific volume of FA 1, the corresponding reaction time can be calculated. This is an example of using a graph for interpolation/extrapolation and then applying a formula.

Understanding the Question

The candidate must use their own graph from part (b) to find the rate at 5.00 cm³ of FA 1, then convert this to a reaction time. The working must show both the graph reading and the calculation using 500/rate.

Approach

Locate 5.00 cm³ on the $x$ -axis.
Read vertically up to the line of best fit.
Read horizontally to the $y$ -axis to obtain the rate.
Calculate time = 500/rate.

Step-by-Step Reasoning

Reading from the graph (Mark 1): At 5.00 cm³, the line of best fit should give a rate of approximately 5.7 s⁻¹ (this depends on the candidate's graph). The reading must be correct to within one small square. The value read from the graph must then be shown in the calculation.

Using 500/rate (Mark 2): The formula must be rearranged and applied:

\text{time} = \frac{500}{\text{rate}}

If rate = 5.7 s⁻¹, then time = 500/5.7 = 87.7 ≈ 88 s.

Note: since volume of FA 1 is halved from 10.00 to 5.00 cm³, the concentration is halved, so the rate is halved and the time doubles. If time at 10.00 cm³ was 44 s, time at 5.00 cm³ should be about 88 s.

Key Takeaways

Graph reading must be accurate to within one small square.
Always show the rearranged formula and the substitution of the read value.
The answer is dependent on the candidate's own graph, so exact values will vary.

Common Mistakes

Reading the volume from the $y$ -axis instead of the $x$ -axis.
Using rate × 500 instead of 500/rate.
Not showing working (the mark requires the number from the graph to be visibly used in the calculation).
Extrapolating beyond the line of best fit rather than reading from it.

Things to Be Careful About

The mark scheme requires the rate value read from the graph to be shown in the calculation — a bare answer without working does not earn the second mark.
The answer should be to a sensible number of significant figures (2 or 3 is appropriate).

Techniques used

read a value from a graphrearrange rate = 500/time to find time

(ii)

Assume that the error in the time measured for each reaction was $\pm 0.5\text{ s}$ in total.

Calculate the maximum percentage error in the reaction time you measured in Experiment 1.

Show your working.

DifficultyEasy

Worked solution

Working

\text{Maximum percentage error} = \frac{0.5}{\text{time for Experiment 1}} \times 100

If time for Experiment 1 = 22 s:

\text{Percentage error} = \frac{0.5}{22} \times 100 = 2.3\%

Answer

$2.3\%$ (value depends on candidate's recorded time for Experiment 1; must be to 2 or more significant figures)

Final answer

2.3% (candidate-dependent)

Detailed explanation

Background Concept

Percentage error expresses the absolute uncertainty as a fraction of the measured value, multiplied by 100. It indicates the precision of a measurement relative to its magnitude. A larger measured value with the same absolute uncertainty gives a smaller percentage error.

Understanding the Question

The absolute error in timing is given as ±0.5 s. The candidate must calculate the percentage error for their own Experiment 1 time. The mark scheme requires the formula $(0.5/\text{time}) \times 100$ to be correctly applied, with the answer to at least 2 significant figures.

Approach

Divide the absolute error (0.5 s) by the measured time, then multiply by 100 to express as a percentage.

Step-by-Step Reasoning

The formula for maximum percentage error is:

\text{Maximum percentage error} = \frac{\text{absolute error}}{\text{measured value}} \times 100 = \frac{0.5}{t_1} \times 100

If $t_1 = 22$ s: percentage error = (0.5/22) × 100 = 2.27% ≈ 2.3%

If $t_1 = 20$ s: percentage error = (0.5/20) × 100 = 2.5%

The answer must be to at least 2 significant figures (e.g. 2.3%, not 2%).

Key Takeaways

Percentage error = (absolute error / measured value) × 100.
The maximum percentage error uses the maximum absolute error (±0.5 s means the error could be +0.5 or −0.5, so 0.5 is used).
Longer reaction times give smaller percentage errors for the same absolute uncertainty.

Common Mistakes

Dividing by 100 instead of multiplying by 100.
Giving the answer to only 1 significant figure (e.g. '2%').
Using the wrong time (e.g. using Experiment 2's time instead of Experiment 1's).
Writing the answer as a decimal (0.023) instead of a percentage (2.3%).

Things to Be Careful About

The mark scheme specifically requires 2 or more significant figures.
Use the candidate's own recorded time for Experiment 1 (the 20.00 cm³ experiment).
The error is ±0.5 s 'in total', meaning this is the combined uncertainty, not per reading.

Techniques used

calculate percentage error from absolute error and measured value

(iii)

A student suggested that this error could be reduced if $0.0100\text{ mol dm}^{-3}$ sodium thiosulfate were used in place of FA 3.

Do you agree with this student? Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Yes, the student is correct. Using $0.0100\text{ mol dm}^{-3}$ sodium thiosulfate (double the concentration of FA 3) would mean more thiosulfate needs to be consumed before the endpoint, so the reaction time would be longer. A longer reaction time with the same absolute error of $\pm 0.5$ s gives a smaller percentage error.

Final answer

Yes — longer reaction time reduces percentage error

Detailed explanation

Background Concept

In the clock reaction, the time measured depends on how much thiosulfate must be consumed before free iodine appears. The moles of thiosulfate determine the 'clock' duration: more thiosulfate means more iodine must be produced before the endpoint, so the time is longer. Percentage error = (absolute error / measured value) × 100, so increasing the measured value (time) while keeping the absolute error constant reduces the percentage error.

Understanding the Question

FA 3 is 0.00500 mol dm⁻³ Na₂S₂O₃. The student suggests using 0.0100 mol dm⁻³ (double the concentration). The question asks whether this reduces the error in timing. The candidate must agree or disagree and explain why.

Approach

Recognise that doubling [S₂O₃²⁻] doubles the moles of thiosulfate (same volume used).
More thiosulfate means more iodine must be generated before the endpoint → longer reaction time.
Longer time with same ±0.5 s error → smaller percentage error.
Therefore the student's suggestion is correct.

Step-by-Step Reasoning

The amount of thiosulfate in 10.0 cm³ of FA 3 (0.00500 mol dm⁻³) is:

n = 0.00500 \times \frac{10.0}{1000} = 5.00 \times 10^{-5}\text{ mol}

With 0.0100 mol dm⁻³ (double concentration), the moles would be $1.00 \times 10^{-4}$ mol — double.

Since the rate of iodine production is unchanged (same [S₂O₈²⁻] and [I⁻]), it takes twice as long to produce enough iodine to consume the doubled thiosulfate. The reaction time approximately doubles.

If original time = 22 s, new time ≈ 44 s.
Percentage error: (0.5/22) × 100 = 2.3% → (0.5/44) × 100 = 1.1%.

The percentage error is halved, confirming the student is correct.

Key Takeaways

Increasing the amount of 'clock' reagent (thiosulfate) increases the measured time.
Longer measurement times reduce percentage error for a fixed absolute uncertainty.
This is a general principle: measure over a larger range to improve precision.

Common Mistakes

Saying the student is incorrect without justification.
Confusing concentration of thiosulfate with concentration of peroxodisulfate (changing FA 3 does not affect the rate of the main reaction, only the time to endpoint).
Not connecting 'longer time' to 'reduced percentage error'.

Things to Be Careful About

The explanation must explicitly state that the reaction time would be longer AND that this reduces the percentage error. Both elements are needed for the mark.
Do not confuse this with changing the rate of reaction — the rate is determined by [S₂O₈²⁻] and [I⁻], not by [S₂O₃²⁻].

Techniques used

evaluate how changing reagent concentration affects reaction time and percentage error

(iv)

A student repeated Experiment 1 but used $0.100\text{ mol dm}^{-3}$ sodium thiosulfate in place of FA 3. The student found that the reaction mixture never turned blue-black.

Explain why.

DifficultyMedium

Worked solution

Answer

With $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, there is so much thiosulfate present that all the peroxodisulfate (and hence all the iodine that can be produced) reacts with the thiosulfate before the thiosulfate is used up. No free iodine remains to react with the starch indicator, so the mixture never turns blue-black.

Final answer

Excess thiosulfate consumes all iodine produced; no free iodine remains to give blue-black colour with starch

Detailed explanation

Background Concept

The clock reaction works because thiosulfate is present in a limited, known amount. Iodine produced by the main reaction is immediately consumed by thiosulfate. Once all thiosulfate is used up, the next molecule of iodine produced remains free and reacts with starch to give the blue-black colour. If the thiosulfate is in excess relative to the maximum iodine that can be produced, the endpoint is never reached.

Understanding the Question

The student used 0.100 mol dm⁻³ Na₂S₂O₃ (20 times the concentration of FA 3) in the same volume (10.0 cm³). The mixture never turned blue-black. The candidate must explain why.

Approach

Calculate the moles of thiosulfate and compare with the maximum moles of iodine that can be produced from the available peroxodisulfate.

Step-by-Step Reasoning

Moles of S₂O₈²⁻ available:

n(\text{S}_2\text{O}_8^{2-}) = 0.0200 \times \frac{20.00}{1000} = 4.00 \times 10^{-4}\text{ mol}

From the stoichiometry: 1 mol S₂O₈²⁻ produces 1 mol I₂.

Maximum moles of I₂ that can be produced = $4.00 \times 10^{-4}$ mol.

Moles of I₂ needed to consume all thiosulfate:
From the equation I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, 1 mol I₂ reacts with 2 mol S₂O₃²⁻.

n(\text{S}_2\text{O}_3^{2-}) = 0.100 \times \frac{10.0}{1000} = 1.00 \times 10^{-3}\text{ mol}

Moles of I₂ needed to consume this thiosulfate = $1.00 \times 10^{-3}/2 = 5.00 \times 10^{-4}$ mol.

Comparison: Maximum I₂ producible = $4.00 \times 10^{-4}$ mol, but I₂ needed to use up all thiosulfate = $5.00 \times 10^{-4}$ mol.

Since $4.00 \times 10^{-4} < 5.00 \times 10^{-4}$ , all the peroxodisulfate is consumed before all the thiosulfate is used up. There is never any excess iodine to turn the starch blue-black.

Key Takeaways

The clock reaction only works when thiosulfate is the limiting reagent relative to the iodine that can be produced.
If thiosulfate is in excess, the endpoint is never reached.
Quantitative comparison of moles is the key reasoning.

Common Mistakes

Saying 'the reaction is too fast' or 'the concentration is too high' without the mole comparison.
Not recognising that the issue is about relative amounts (moles), not just concentration.
Confusing this with the rate of reaction — the rate is unaffected by thiosulfate concentration.

Things to Be Careful About

The mark scheme accepts 'there is so much thiosulfate that all the iodide reacts' — the key idea is that thiosulfate is in excess relative to iodine produced. The exact phrasing may vary but the concept of insufficient iodine to consume all thiosulfate must be clear.

Techniques used

compare moles of thiosulfate with moles of iodine produced to determine if endpoint is reached

(e)

(i)

Using the same method as in (a), carry out an additional experiment to record the reaction time to the nearest second when the following solutions are mixed together.

$10.00\text{ cm}^3$ of FA 1
$20.0\text{ cm}^3$ of FA 2
$5.0\text{ cm}^3$ of FA 3
$15.00\text{ cm}^3$ of distilled water
10 drops of starch indicator

DifficultyMedium-Easy

Worked solution

Answer

Record the reaction time to the nearest second with units of s.

Representative value: approximately 22 s.

This time should be compared with the time recorded in Experiment 1 (20.00 cm³ FA 1). The candidate's time should be within ±3 s of the supervisor's value for this mixture.

Reasoning for expected value: The total volume is 50.0 cm³ (10.00 + 20.0 + 5.0 + 15.00 = 50.0 cm³). The concentration of S₂O₈²⁻ is the same as in Experiment 2 (10.00 cm³ FA 1 in 50 cm³ total), so the rate is the same as Experiment 2. However, only 5.0 cm³ of FA 3 is used (half the thiosulfate of Experiment 2), so the time to endpoint is halved. If Experiment 2 gave ~44 s, this should give ~22 s.

Final answer

Approximately 22 s (candidate-dependent; must be recorded to nearest second with unit s)

Detailed explanation

Background Concept

The reaction time in this clock reaction depends on two factors: (1) the rate of iodine production (determined by [S₂O₈²⁻] and [I⁻]), and (2) the amount of thiosulfate that must be consumed before the endpoint. Changing the volume of FA 3 changes the moles of thiosulfate and hence the time, without affecting the rate.

Understanding the Question

The mixture in part (e)(i) has:

10.00 cm³ FA 1 (same as Experiment 2)
20.0 cm³ FA 2 (same as all experiments)
5.0 cm³ FA 3 (HALF of the 10.0 cm³ used in Experiments 1-5)
15.00 cm³ distilled water (to make total volume 50.0 cm³)
10 drops starch

Total volume = 10.00 + 20.0 + 5.0 + 15.00 = 50.0 cm³ ✓

The concentration of S₂O₈²⁻ is the same as in Experiment 2 (10.00 cm³ in 50 cm³ total), so the rate of iodine production is the same. But only half the thiosulfate is present, so it takes half the time to reach the endpoint.

Approach

Carry out the experiment using the same method.
Record the time to the nearest second with units.
Compare with the supervisor's value (within ±3 s for full marks).

Step-by-Step Reasoning

Recording (Mark 1): Time must be to the nearest second (whole number) with the unit 's' stated.

Comparison (Mark 2): The candidate's time is compared with the supervisor's value for this specific mixture. Within ±3 s earns the mark.

Expected value reasoning:

Experiment 2 used 10.00 cm³ FA 1 + 10.0 cm³ FA 3 in 50 cm³ total → time ≈ 44 s
Part (e)(i) uses 10.00 cm³ FA 1 + 5.0 cm³ FA 3 in 50 cm³ total → same rate but half the thiosulfate → time ≈ 22 s

The time should be approximately half that of Experiment 2, and similar to Experiment 1 (which had 20.00 cm³ FA 1 and 10.0 cm³ FA 3).

Key Takeaways

The clock time depends on both the rate AND the amount of thiosulfate.
Halving the thiosulfate (while keeping rate constant) halves the time.
Total volume must be kept constant to ensure valid comparisons of concentration.

Common Mistakes

Not recording the unit 's' alongside the numerical value.
Recording time with decimal places (e.g. 22.5 s) instead of nearest second.
Getting a time similar to Experiment 2 (~44 s) instead of ~22 s, suggesting the candidate did not notice the reduced thiosulfate volume.

Things to Be Careful About

The mark for comparison is against the supervisor's value for THIS mixture, not against Experiment 1 or 2 directly.
The total volume must be 50.0 cm³ — verify: 10.00 + 20.0 + 5.0 + 15.00 = 50.0 ✓.

Techniques used

carry out experiment and record time to nearest second with unitscompare result with earlier experiment

(ii)

Use your answer to (i) to estimate the reaction time that would be measured if the following solutions were mixed together.

DO NOT CARRY OUT THIS EXPERIMENT

$10.00\text{ cm}^3$ of FA 1
$20.0\text{ cm}^3$ of FA 2
$20.0\text{ cm}^3$ of FA 3
10 drops of starch indicator

Explain your answer.

DifficultyMedium

Worked solution

Working

In part (e)(i), $5.0\text{ cm}^3$ of FA 3 was used. In this new mixture, $20.0\text{ cm}^3$ of FA 3 is used — four times the volume at the same concentration, so four times the moles of thiosulfate.

Since the rate of iodine production is unchanged (same volumes of FA 1 and FA 2, same total volume), the time to consume four times as much thiosulfate is four times longer.

\text{Estimated time} = 4 \times \text{time from (e)(i)} = 4 \times 22 = 88\text{ s}

Answer

Estimated reaction time = $4 \times$ (answer from (e)(i)) $\approx 88$ s.

The time is proportional to the concentration (and amount) of thiosulfate ions. Increasing the volume of FA 3 from $5.0\text{ cm}^3$ to $20.0\text{ cm}^3$ quadruples the moles of $\text{S}_2\text{O}_3^{2-}$ that must be consumed before free iodine appears, so the reaction time is four times longer.

Final answer

4 × (time from e(i)) ≈ 88 s; time proportional to amount of thiosulfate

Detailed explanation

Background Concept

In the clock reaction, the measured time represents how long it takes for the main reaction to produce enough iodine to consume all the thiosulfate present. If the rate of iodine production is constant (determined by [S₂O₈²⁻] and [I⁻]), then the time is directly proportional to the moles of thiosulfate present. Doubling the moles of thiosulfate doubles the time; quadrupling them quadruples the time.

Understanding the Question

The new mixture uses:

10.00 cm³ FA 1 (same as (e)(i))
20.0 cm³ FA 2 (same as (e)(i))
20.0 cm³ FA 3 (four times the 5.0 cm³ used in (e)(i))
10 drops starch
No water added

Total volume = 10.00 + 20.0 + 20.0 = 50.0 cm³ (same as (e)(i): 10.00 + 20.0 + 5.0 + 15.00 = 50.0 cm³)

The rate of iodine production is the same (same [S₂O₈²⁻] and [I⁻] in same total volume), but four times as much thiosulfate must be consumed.

Approach

Compare the moles of thiosulfate in (e)(i) and (e)(ii).
Since rate is unchanged, time is proportional to moles of thiosulfate.
Multiply the (e)(i) time by the ratio of thiosulfate amounts (4).
Explain the reasoning in terms of concentration/amount of FA 3.

Step-by-Step Reasoning

Moles of thiosulfate comparison:

In (e)(i): $n = 0.00500 \times 5.0/1000 = 2.50 \times 10^{-5}$ mol
In (e)(ii): $n = 0.00500 \times 20.0/1000 = 1.00 \times 10^{-4}$ mol

Ratio = $1.00 \times 10^{-4} / 2.50 \times 10^{-5} = 4$

Mark 1 — Estimate as 4 × (e)(i): The time should be four times the answer from (e)(i). If (e)(i) gave 22 s, the estimate is 88 s.

Mark 2 — Explanation: The time (or rate) is related to the concentration/amount of S₂O₃²⁻ (FA 3). Increasing the amount of thiosulfate means more iodine must be produced before the endpoint, so the time increases proportionally. Alternatively: the rate of the clock reaction (as measured by 1/time) is inversely proportional to the amount of thiosulfate.

The mark scheme accepts: 'time/rate related to concentration of S₂O₃²⁻' and 'increased concentration of FA 3 increases time / decreases rate / reaction slower'.

Key Takeaways

The clock time is proportional to the amount of thiosulfate (when rate is constant).
This is a direct application of the principle that time = (amount to consume) / (rate of production).
Scaling arguments from a measured value are a common Paper 3 technique.

Common Mistakes

Multiplying by 2 instead of 4 (not realising 20.0/5.0 = 4, not 2).
Confusing the effect of changing FA 3 volume with changing FA 1 volume (FA 1 affects rate; FA 3 affects time to endpoint at constant rate).
Not explaining WHY the time changes in terms of thiosulfate amount.
Calculating a new total volume that differs from (e)(i), which would change the concentrations.

Things to Be Careful About

The total volume must be checked: 10.00 + 20.0 + 20.0 = 50.0 cm³, same as (e)(i). If it weren't, the concentrations of S₂O₈²⁻ and I⁻ would change, affecting the rate.
The explanation must specifically reference the concentration or amount of thiosulfate (FA 3), not just 'more solution'.
The answer must be expressed as 4 × the candidate's own (e)(i) value, not an absolute number.

Techniques used

scale reaction time proportionally to thiosulfate concentrationexplain relationship between thiosulfate amount and reaction time

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