Chemistry 9701/34 — October/November 2023

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

Q1Presentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationFree sample

Quantitative analysis

Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided.

Show the precision of the apparatus you used in the data you record.

Show your working and appropriate significant figures in the answer to each step of your calculations.

Iron(III) ions, $\text{Fe}^{3+}$ , can oxidise iodide ions, $\text{I}^-$ , to iodine, $\text{I}_2$ .

2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})

It is possible to determine the rate of this reaction by measuring the time to produce a certain amount of iodine. To do this, thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ , are added to the reaction mixture. The thiosulfate ions react immediately with the iodine produced by the reaction and convert it back to iodide ions.

\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})

Iodine remains in the solution when all the thiosulfate ions have reacted. The remaining iodine is detected by starch indicator in the reaction mixture, which causes the solution to turn blue-black.

In this experiment you will investigate how the rate of the reaction between iron(III) ions and iodide ions is affected by the concentration of the iron(III) ions.

FB 1 is $0.0500\text{ mol dm}^{-3}$ acidified iron(III) chloride, $\text{FeCl}_3$ .
FB 2 is $0.0500\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FB 3 is $0.00500\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FB 4 is starch indicator.

(a)

Method

Experiment 1

Fill the burette labelled FB 1 with FB 1.
Run $20.00\text{ cm}^3$ of FB 1 into a $100\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ $25 cm^{3}$ measuring cylinder to add the following to the other $100\text{ cm}^3$ $100 cm^{3}$ beaker:
- $10.0\text{ cm}^3$ of FB 2
- $20.0\text{ cm}^3$ of FB 3
- $10.0\text{ cm}^3$ of FB 4.
Add the contents of the first beaker to the second beaker and start timing immediately. Ignore any initial colouration on mixing.
Stir the mixture once and place the beaker on the white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second.
Rinse both beakers and shake dry.
Rinse and dry the glass rod.

Experiment 2

Fill the other burette with distilled water.
Run $10.00\text{ cm}^3$ of FB 1 into a $100\text{ cm}^3$ beaker.
Run $10.00\text{ cm}^3$ of distilled water into the same beaker containing FB 1.
Use the $25\text{ cm}^3$ $25 cm^{3}$ measuring cylinder to add the following to the other $100\text{ cm}^3$ $100 cm^{3}$ beaker:
- $10.0\text{ cm}^3$ of FB 2
- $20.0\text{ cm}^3$ of FB 3
- $10.0\text{ cm}^3$ of FB 4.
Add the contents of the first beaker to the second beaker and start timing immediately. Ignore any initial colouration on mixing.
Stir the mixture once and place the beaker on the white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second.
Rinse both beakers and shake dry.
Rinse and dry the glass rod.

Experiments 3 to 5

Carry out three further experiments to investigate how the reaction time changes with different volumes of FB 1.

The combined volume of FB 1 and distilled water must always be $20.00\text{ cm}^3$ .
Do not use a volume of FB 1 that is less than $5.00\text{ cm}^3$ .

The rate of reaction is given by the following expression.

\text{rate} = \frac{1000}{\text{reaction time in seconds}}

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

Record all your results in a single table. You should include the volume of FB 1, the volume of distilled water, the reaction time and the rate of reaction.

Results

DifficultyMedium

Worked solution

Answer

Volume of FB 1 / cm³	Volume of water / cm³	Time / s	Rate / s⁻¹
20.00	0.00	40	25.0
15.00	5.00	50	20.0
10.00	10.00	60	16.7
7.50	12.50	75	13.3
5.00	15.00	90	11.1

Note: Times and rates are representative values consistent with the mark scheme requirements (ratio of time for Expt 2 to Expt 1 is 1.50, which is between 1.20 and 1.80). All volumes of FB 1 are ≥ 5.00 cm³, intervals are ≥ 2.00 cm³, and total volume of FB 1 + water is 20.00 cm³. Rates are calculated using $\text{rate} = 1000 / \text{time}$ to 3 significant figures.

Final answer

See table above for representative results.

Detailed explanation

Background Concept

In quantitative kinetics experiments, particularly the iodine clock method, the rate of reaction is determined by measuring the time taken for a fixed, small amount of product to form. Here, thiosulfate ions ( $\text{S}_2\text{O}_3^{2-}$ ) are added to the reaction mixture to react immediately with the iodine ( $\text{I}_2$ ) produced. Once the thiosulfate is exhausted, the next increment of iodine reacts with the starch indicator, causing a sudden blue-black colour change. The time taken for this colour change is recorded, and the initial rate of reaction is approximated as $1000 / t$ (in s⁻¹), since the amount of iodine required to trigger the colour change is constant across all experiments.

Understanding the Question

Part (a) asks you to record your experimental results in a single, well-structured table. The mark scheme is highly specific about what constitutes a valid table: correct headings with units, appropriate significant figures for recorded and calculated data, and a valid experimental plan that meets specific constraints (minimum volumes, fixed total volume, minimum intervals between concentrations). Because this is a practical paper, you will have your own actual data. Here, we provide a representative example table that satisfies all marking criteria so you can see exactly how to format and present your results.

Approach

To earn all 9 marks, the table must include:

A single table with four specific column headings and units.
At least 5 experiments (Experiments 1 and 2 are given; you must do 3 more).
Times recorded to the nearest second; volumes to 1 decimal place (or 2 for burette readings like 20.00).
Volumes of FB 1 ≥ 5.00 cm³, with intervals ≥ 2.00 cm³.
The total volume of FB 1 + distilled water must always equal 20.00 cm³.
Rates calculated as $1000 / t$ to 2–4 significant figures.
The ratio of the time for Experiment 2 to Experiment 1 must be between 1.20 and 1.80 (ideally 1.30–1.60).

Step-by-Step Reasoning

1. Table Structure: Create a table with columns for "Volume of FB 1 / cm³", "Volume of water / cm³", "Time / s", and "Rate / s⁻¹". The units must be in the heading, not next to every entry.

2. Experimental Plan:

Experiment 1: 20.00 cm³ FB 1, 0.00 cm³ water.
Experiment 2: 10.00 cm³ FB 1, 10.00 cm³ water.
Experiments 3–5: Choose volumes of FB 1 that are ≥ 5.00 cm³ and spaced by at least 2.00 cm³. For example: 15.00, 7.50, and 5.00 cm³. Calculate the corresponding water volumes to make the total 20.00 cm³.

3. Representative Data:

Assume Experiment 1 takes 40 s. Rate = $1000 / 40 = 25.0$ s⁻¹.
Experiment 2 should take between $1.20 \times 40 = 48$ s and $1.80 \times 40 = 72$ s. Assume it takes 60 s. Rate = $1000 / 60 = 16.7$ s⁻¹. Ratio = $60 / 40 = 1.50$ (valid).
Fill in plausible times for the other experiments that show time increasing as concentration decreases: 50 s (20.0 s⁻¹), 75 s (13.3 s⁻¹), and 90 s (11.1 s⁻¹).

4. Significant Figures:

Volumes from a burette should be recorded to 2 decimal places (e.g., 20.00 cm³), but measuring cylinders to 1 decimal place (e.g., 10.0 cm³). The mark scheme accepts .#0 or .#5 cm³.
Rates should be given to 2–4 significant figures. Ensure consistency (e.g., all to 3 sig figs: 25.0, 20.0, 16.7, 13.3, 11.1).

Key Takeaways

Always include units in the column headings of a results table.
Ensure the independent variable (volume of FB 1) is varied systematically while keeping the total volume constant to maintain a constant overall concentration of other reactants.
Check your data against the mark scheme's implicit constraints (like the time ratio) to ensure your representative table is valid.

Common Mistakes

Missing units in headings: Writing "Time" instead of "Time / s" or "Time / seconds". The mark scheme explicitly requires units in the heading.
Incorrect total volume: Forgetting to add distilled water to make the total volume 20.00 cm³ in the additional experiments. This changes the total volume of the reaction mixture, invalidating the comparison.
Violating volume constraints: Using a volume of FB 1 less than 5.00 cm³, or choosing volumes that are too close together (interval < 2.00 cm³).
Incorrect rate calculation: Using $1 / t$ instead of $1000 / t$ , or failing to apply the correct number of significant figures.

Things to Be Careful About

Significant figures in calculated rates: The mark scheme awards marks if rates are to 2–4 significant figures and either all have the same number of significant figures or the same number of decimal places. Mixing 25.0 (3 sig figs) with 16.666... (4 sig figs) without rounding consistently can cost marks.
Time recording: Times must be recorded to the nearest second. Do not record 40.5 s.
Ratio check: The examiner will calculate the ratio of time for Expt 2 to Expt 1. If your representative data gives a ratio outside 1.20–1.80, it will be flagged as unrealistic and may not earn the final marks for data consistency.

Techniques used

design a results table with correct headings and unitsplan a series of experiments with controlled variablescalculate reaction rates from time measurements

(b)

On the grid opposite, plot the rate (on the $y$ -axis) against the volume of FB 1 (on the $x$ -axis). Include the origin in your plot. Label any points that you consider to be anomalous. Draw the line of best fit.

Fig. 1.1

DifficultyMedium-Easy

Worked solution

Answer

Graph details:

x-axis: Volume of FB 1 / cm³ (linear scale, 0 to 25, origin included).
y-axis: Rate / s⁻¹ (linear scale, 0 to 30, origin included).
Points plotted: (20.00, 25.0), (15.00, 20.0), (10.00, 16.7), (7.50, 13.3), (5.00, 11.1).
Line of best fit: A smooth curve passing through or near all points, starting at the origin (0,0) and curving upwards, showing that rate increases with volume but is not directly proportional (not a straight line).

Final answer

See graph description above.

Detailed explanation

Background Concept

Plotting a graph of rate against a reactant's concentration (or volume, when total volume is constant) is a standard method for determining the order of reaction with respect to that reactant. If the rate is directly proportional to the concentration (first order), the graph will be a straight line through the origin. If the relationship is more complex (e.g., second order or involving a mixed-order rate equation), the graph will be a curve.

Understanding the Question

Part (b) asks you to plot the results from your table on a grid. The y-axis must be rate, and the x-axis must be the volume of FB 1. You must include the origin, use linear scales based on 1, 2, or 5, and draw a smooth curve of best fit. You should also label any anomalous points.

Approach

Axis labels: Clearly label the y-axis as "Rate / s⁻¹" and the x-axis as "Volume of FB 1 / cm³".
Scales: Choose linear scales that allow the point for 20 cm³ FB 1 to be plotted more than halfway along each axis. For example, x-axis: 0 to 25 cm³ (5 cm³ per major division); y-axis: 0 to 30 s⁻¹ (5 s⁻¹ per major division).
Plotting: Accurately plot all 5 data points.
Curve of best fit: Draw a smooth curve through the points. Do NOT join the dots with straight lines. The curve should start at the origin (0,0) because if there is no iron(III) ions, the rate is zero.

Step-by-Step Reasoning

1. Axis Setup:

x-axis: Label "Volume of FB 1 / cm³". Scale from 0 to 25. Mark 0, 5, 10, 15, 20, 25.
y-axis: Label "Rate / s⁻¹". Scale from 0 to 30. Mark 0, 5, 10, 15, 20, 25, 30.
Ensure the origin (0,0) is included.

2. Plotting Points:

(20.00, 25.0)
(15.00, 20.0)
(10.00, 16.7)
(7.50, 13.3)
(5.00, 11.1)
Plot these to within half a small square.

3. Line of Best Fit:

Draw a smooth curve that passes close to all points. The curve will be concave down (gradient decreases as volume increases) or concave up depending on the exact kinetics, but it will definitely NOT be a straight line. In this case, as volume of FB 1 increases, the rate increases, but the rate of increase slows down, forming a curve.

Key Takeaways

Always include the origin if it makes physical sense (zero reactant = zero rate).
Use linear scales based on 1, 2, or 5 to make reading off values easy.
A curve of best fit must be smooth; avoid zig-zag lines connecting the dots.

Common Mistakes

Non-linear scales: Using a scale based on 3 or 4, which makes plotting and reading difficult.
Forgetting the origin: Not including (0,0) on the graph.
Straight line: Drawing a straight line through the points. The mark scheme specifically asks for a "smooth curved line of best fit" because the rate is not proportional to the concentration.
Incorrect axis assignment: Putting volume on the y-axis and rate on the x-axis.

Things to Be Careful About

Scale selection: The mark scheme requires that the point for 20 cm³ FB 1 is more than halfway along the x-axis. If your x-axis only goes to 20, the point will be at the end, which is not allowed. Extend the axis to at least 22 or 25.
Anomalous points: If any of your experimental points are clearly off the curve (e.g., due to a timing error), label them as anomalous and still draw the curve through the other points.

Techniques used

plot a graph of rate against volumechoose appropriate linear scalesdraw a smooth curve of best fit

(c)

In these experiments, the volume of FB 1 is directly related to the concentration of iron(III) ions.

Using your graph, state what conclusion can be drawn about the relationship between the rate of reaction and the concentration of the iron(III) ions.

DifficultyMedium-Easy

Worked solution

Answer

The rate of reaction is not directly proportional to the concentration of iron(III) ions.

(Alternatively: As the concentration of iron(III) ions increases, the rate of reaction increases.)

Final answer

Rate is not proportional to concentration.

Detailed explanation

Background Concept

The order of reaction with respect to a reactant is determined by how the rate changes when the concentration of that reactant changes. If doubling the concentration doubles the rate, the reaction is first order with respect to that reactant, and a graph of rate against concentration will be a straight line passing through the origin. If the graph is a curve, the reaction is not first order (it could be zero order, second order, or involve a more complex rate equation).

Understanding the Question

Part (c) asks for a conclusion about the relationship between the rate of reaction and the concentration of iron(III) ions, based on the graph drawn in part (b). Since the volume of FB 1 is directly proportional to the concentration of iron(III) ions (as the total volume is kept constant), the x-axis effectively represents concentration.

Approach

Look at the shape of the curve of best fit in your graph. Is it a straight line or a curve? A straight line through the origin would mean rate is proportional to concentration. Since the mark scheme asks for a curved line, the relationship is not proportional. State this clearly.

Step-by-Step Reasoning

1. Observe the graph: The graph of rate against volume of FB 1 (and thus concentration) is a smooth curve, not a straight line.

2. Interpret the shape: A curve indicates that the rate does not change linearly with concentration. Therefore, the rate is not directly proportional to the concentration of iron(III) ions.

3. Alternative observation: You can also simply state the trend observed: as the concentration (volume) of FB 1 increases, the rate of reaction increases. This is also a valid conclusion for 1 mark.

Key Takeaways

A straight line through the origin = directly proportional (first order).
A curve = not directly proportional (not first order, or more complex kinetics).
Always base your conclusion on the actual shape of your graph.

Common Mistakes

Saying 'proportional': If the graph is curved, saying the rate is proportional to concentration is incorrect and will score zero.
Stating the order: You cannot determine the exact order (e.g., second order) from just this graph without further calculations (like plotting rate against concentration squared). Stick to what the graph directly shows: 'not proportional' or 'increases'.
Vague statements: Simply saying 'they are related' is too vague. You must specify the nature of the relationship (proportional vs. not proportional, or increasing vs. decreasing).

Things to Be Careful About

Wording: Use the exact phrase 'not proportional' or 'increases'. Avoid technical jargon like 'non-linear order' unless you are sure it's accepted; 'not proportional' is the safest and most direct answer.
Concentration vs Volume: The question states that volume is directly related to concentration. You can refer to either, but it's safer to refer to 'concentration' as the question asks about the relationship with concentration.

Techniques used

interpret a graph to determine reaction orderdistinguish between proportional and non-proportional relationships

(d)

A student wants to increase the concentration of the sodium thiosulfate solution while keeping the rest of the experiment the same. The student realises that the amount of thiosulfate ions must not be too high otherwise there will be no remaining iodine.

You will calculate the concentration of thiosulfate ions that will react with all the iodine produced in Experiment 1.

(i)

Calculate the amount, in mol, of iron(III) ions in the solution at the start of Experiment 1.

\text{amount of Fe}^{3+} = \text{.............................. mol}

DifficultyEasy

Worked solution

Working

\text{amount of Fe}^{3+} = \text{concentration} \times \text{volume}

\text{amount of Fe}^{3+} = 0.0500 \text{ mol dm}^{-3} \times 0.02000 \text{ dm}^3 = 1.00 \times 10^{-3} \text{ mol}

Answer

\text{amount of Fe}^{3+} = 1.00 \times 10^{-3} \text{ mol}

Final answer

1.00 x 10^-3

Detailed explanation

Background Concept

The amount of substance (in moles) can be calculated from the concentration and volume using the equation $n = c \times V$ . It is crucial to ensure that the volume is in the correct units (dm³) to match the concentration units (mol dm⁻³). To convert from cm³ to dm³, divide by 1000.

Understanding the Question

Part (d)(i) asks for the amount of iron(III) ions at the start of Experiment 1. From the method, Experiment 1 uses 20.00 cm³ of FB 1, which is 0.0500 mol dm⁻³ FeCl₃.

Approach

Identify the concentration: $c = 0.0500 \text{ mol dm}^{-3}$ .
Identify the volume: $V = 20.00 \text{ cm}^3 = 20.00 / 1000 = 0.02000 \text{ dm}^3$ .
Calculate moles: $n = c \times V$ .

Step-by-Step Reasoning

\text{amount} = 0.0500 \times 0.02000 = 1.00 \times 10^{-3} \text{ mol}

The answer should be given to 2–4 significant figures. $1.00 \times 10^{-3}$ has 3 significant figures, which is acceptable.

Key Takeaways

Always convert volume to dm³ before calculating moles if concentration is in mol dm⁻³.
Use scientific notation for very small or very large numbers to maintain clear significant figures.

Common Mistakes

Forgetting to convert cm³ to dm³: Calculating $0.0500 \times 20.00 = 1.00$ , which is wrong.
Wrong significant figures: Giving an answer like 0.001 mol (1 sig fig) might be accepted, but $1.00 \times 10^{-3}$ is clearer.

Things to Be Careful About

Concentration: Ensure you use the correct concentration for FB 1 (0.0500 mol dm⁻³), not FB 2 or FB 3.

Techniques used

calculate moles from concentration and volume

(ii)

Calculate the amount, in mol, of iodide ions in the solution at the start of Experiment 1.

\text{amount of I}^- = \text{.............................. mol}

DifficultyEasy

Worked solution

Working

\text{amount of I}^- = \text{concentration} \times \text{volume}

\text{amount of I}^- = 0.0500 \text{ mol dm}^{-3} \times 0.01000 \text{ dm}^3 = 5.00 \times 10^{-4} \text{ mol}

Answer

\text{amount of I}^- = 5.00 \times 10^{-4} \text{ mol}

Final answer

5.00 x 10^-4

Detailed explanation

Background Concept

Same as part (i). Calculate moles using $n = cV$ .

Understanding the Question

Part (d)(ii) asks for the amount of iodide ions at the start of Experiment 1. From the method, Experiment 1 uses 10.0 cm³ of FB 2, which is 0.0500 mol dm⁻³ KI.

Approach

Identify the concentration: $c = 0.0500 \text{ mol dm}^{-3}$ .
Identify the volume: $V = 10.0 \text{ cm}^3 = 10.0 / 1000 = 0.01000 \text{ dm}^3$ .
Calculate moles: $n = c \times V$ .

Step-by-Step Reasoning

\text{amount} = 0.0500 \times 0.01000 = 5.00 \times 10^{-4} \text{ mol}

Key Takeaways

Consistent application of $n = cV$ .
Pay attention to the volume given in the method (10.0 cm³, not 20.0 cm³).

Common Mistakes

Using the wrong volume: Using 20.00 cm³ (the volume of FB 1) instead of 10.0 cm³ (the volume of FB 2).

Things to Be Careful About

Significant figures: 10.0 cm³ has 3 sig figs, 0.0500 has 3 sig figs. The answer should be to 3 sig figs: $5.00 \times 10^{-4}$ .

Techniques used

calculate moles from concentration and volume

(iii)

Use the equation to determine the maximum amount, in mol, of iodine that can be made during this reaction.

2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})

\text{amount of I}_2 = \text{.............................. mol}

DifficultyMedium-Easy

Worked solution

Working

From the balanced equation:

2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})

The molar ratio of $\text{Fe}^{3+}$ to $\text{I}^-$ is 2:2, or 1:1.

Amount of $\text{Fe}^{3+}$ = $1.00 \times 10^{-3} \text{ mol}$
Amount of $\text{I}^-$ = $5.00 \times 10^{-4} \text{ mol}$

Since $5.00 \times 10^{-4} < 1.00 \times 10^{-3}$ , $\text{I}^-$ is the limiting reagent.

From the equation, 2 moles of $\text{I}^-$ produce 1 mole of $\text{I}_2$ .

\text{amount of I}_2 = \frac{\text{amount of I}^-}{2} = \frac{5.00 \times 10^{-4}}{2} = 2.50 \times 10^{-4} \text{ mol}

Answer

\text{amount of I}_2 = 2.50 \times 10^{-4} \text{ mol}

Final answer

2.50 x 10^-4

Detailed explanation

Background Concept

In a chemical reaction, the limiting reagent is the reactant that is completely consumed first and determines the maximum amount of product that can be formed. To find the limiting reagent, compare the mole ratio of the reactants available to the mole ratio required by the balanced equation.

Understanding the Question

Part (d)(iii) asks for the maximum amount of iodine that can be produced in Experiment 1. You have calculated the initial amounts of $\text{Fe}^{3+}$ and $\text{I}^-$ . You must use the stoichiometry of the reaction to find the maximum $\text{I}_2$ .

Approach

Compare moles of $\text{Fe}^{3+}$ and $\text{I}^-$ using the 1:1 ratio from the equation.
Identify the limiting reagent ( $\text{I}^-$ ).
Use the molar ratio between $\text{I}^-$ and $\text{I}_2$ (2:1) to calculate the maximum moles of $\text{I}_2$ .

Step-by-Step Reasoning

1. Mole comparison:

Required ratio $\text{Fe}^{3+} : \text{I}^- = 2:2 = 1:1$ .
Available $\text{Fe}^{3+} = 1.00 \times 10^{-3} \text{ mol}$ .
Available $\text{I}^- = 5.00 \times 10^{-4} \text{ mol}$ .
Since $5.00 \times 10^{-4} < 1.00 \times 10^{-3}$ , $\text{I}^-$ is in deficit. $\text{I}^-$ is the limiting reagent.

2. Calculate $\text{I}_2$ :

Ratio $\text{I}^- : \text{I}_2 = 2:1$ .
Moles of $\text{I}_2 = \frac{1}{2} \times \text{moles of I}^- = \frac{5.00 \times 10^{-4}}{2} = 2.50 \times 10^{-4} \text{ mol}$ .

Key Takeaways

Always check for the limiting reagent before calculating product amounts.
Pay close attention to the stoichiometric coefficients in the balanced equation.

Common Mistakes

Assuming $\text{Fe}^{3+}$ is limiting: Because it has a larger volume, students might incorrectly assume it is limiting. Always compare moles.
Wrong ratio: Using a 1:1 ratio between $\text{I}^-$ and $\text{I}_2$ instead of the correct 2:1 ratio from the equation.

Things to Be Careful About

Equation balancing: The equation is given as $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ . The coefficient for $\text{I}_2$ is 1, while for $\text{I}^-$ it is 2.

Techniques used

determine limiting reagentuse stoichiometric ratios from balanced equation

(iv)

Use the equation to determine the concentration, in $\text{mol dm}^{-3}$ , of sodium thiosulfate solution that will react with all the iodine produced in Experiment 1.
Show your working.

\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})

\text{concentration} = \text{.............................. mol dm}^{-3}

DifficultyMedium-Easy

Worked solution

Working

From the balanced equation for the thiosulfate reaction:

\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})

The molar ratio of $\text{I}_2$ to $\text{S}_2\text{O}_3^{2-}$ is 1:2.

Amount of $\text{I}_2$ (from part iii) = $2.50 \times 10^{-4} \text{ mol}$

\text{amount of S}_2\text{O}_3^{2-} = 2 \times \text{amount of I}_2 = 2 \times 2.50 \times 10^{-4} = 5.00 \times 10^{-4} \text{ mol}

Volume of $\text{Na}_2\text{S}_2\text{O}_3$ solution = $20.0 \text{ cm}^3 = 0.0200 \text{ dm}^3$

\text{concentration} = \frac{\text{amount}}{\text{volume}} = \frac{5.00 \times 10^{-4} \text{ mol}}{0.0200 \text{ dm}^3} = 0.0250 \text{ mol dm}^{-3}

Answer

\text{concentration} = 0.0250 \text{ mol dm}^{-3}

Final answer

0.0250

Detailed explanation

Background Concept

This part asks you to find the concentration of a thiosulfate solution that would react completely with all the iodine produced. This is essentially a stoichiometry calculation in reverse: you know the amount of product ( $\text{I}_2$ ) and the volume of the reactant ( $\text{S}_2\text{O}_3^{2-}$ ), and you need to find the concentration.

Understanding the Question

Part (d)(iv) asks for the concentration of sodium thiosulfate that will react with all the iodine produced in Experiment 1. The volume of thiosulfate solution used is 20.0 cm³ (from FB 3 in the method).

Approach

Use the moles of $\text{I}_2$ calculated in part (iii).
Use the 1:2 molar ratio from the thiosulfate equation to find moles of $\text{S}_2\text{O}_3^{2-}$ .
Calculate concentration using $c = n / V$ .

Step-by-Step Reasoning

1. Moles of thiosulfate:

Ratio $\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2$ .
Moles of $\text{S}_2\text{O}_3^{2-} = 2 \times 2.50 \times 10^{-4} = 5.00 \times 10^{-4} \text{ mol}$ .

2. Concentration:

Volume = $20.0 \text{ cm}^3 = 0.0200 \text{ dm}^3$ .
$c = \frac{5.00 \times 10^{-4}}{0.0200} = 0.0250 \text{ mol dm}^{-3}$ .

Key Takeaways

Chain calculations carefully: use the result from the previous part directly.
Ensure units are consistent (convert cm³ to dm³).

Common Mistakes

Wrong ratio: Using a 1:1 ratio between $\text{I}_2$ and $\text{S}_2\text{O}_3^{2-}$ instead of 1:2.
Wrong volume: Using the total volume of the mixture (50 cm³) instead of the volume of the thiosulfate solution alone (20 cm³) to calculate the concentration of the thiosulfate solution.

Things to Be Careful About

Significant figures: The answer should be to 2–4 significant figures. $0.0250$ has 3 significant figures, which is correct.
Error carried forward: If you made a mistake in part (iii), you can still earn marks here if you use your wrong answer from (iii) correctly in the subsequent steps (ecf).

Techniques used

calculate concentration from moles and volumeuse stoichiometric ratios from balanced equation

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2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation8M
Q3Qualitative Analysis · Manipulation, Measurement and Observation13M

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