Qualitative analysis For each test you should record all your observations in the spaces provided. Examples of observations include: colour changes seen the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added the formation of any gas and its identification (where appropriate) by a suitable test. You should record clearly at what stage in a test an observation is made. Where no change is observed, you should write ‘no change’.…

(a)(i) Observations recorded in Table 3.1 (see working) · (a)(ii) FB 6 = CH3COOH; FB 7 = NaOH; FB 8 = H2O2 · (b)(i) Fizzing; gas gives white ppt with limewater (CO2) · (b)(ii) Cream ppt, insoluble in excess aqueous ammonia · (b)(iii) Test 1: ammonia gas turns litmus blue; Test 2: brown/yellow solution · (b)(iv) cations: NH4+ and unknown; anions: Br- and CO3^2- · (b)(v) Ag+(aq) + Br-(aq) -> AgBr(s)

Chemistry 9701/34 — May/June 2025

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

questions

marks

120

minutes

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

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Iron is an element that is essential in the human diet. Some people need to take iron supplement tablets to ensure an adequate intake of iron.

You will investigate the mass of iron in an iron supplement tablet by titrating a solution with potassium manganate(VII).

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

\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

FB 1 is an aqueous solution of iron supplement tablets made by dissolving 14 tablets in $1.00\text{ dm}^3$ of solution. The iron in each tablet is iron(II) sulfate, $\text{FeSO}_4\cdot7\text{H}_2\text{O}$ .
FB 2 is $0.0100\text{ mol dm}^{-3}$ acidified aqueous potassium manganate(VII), $\text{KMnO}_4$ .
FB 3 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Fill a burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 1 into a conical flask.
Use the $25\text{ cm}^3$ measuring cylinder to add $10.0\text{ cm}^3$ of FB 3 to the conical flask.
Perform a rough titration and record your burette readings in the space below.

The rough titre is .............................. $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record, in a suitable form in the space below, all your burette readings and the volume of FB 2 added in each accurate titration.

Rinse the burette with distilled water and leave to drain while you continue Question 1.

Results

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DifficultyMedium

Worked solution

Answer

A correctly completed results table for these titrations, using a model set of readings (your actual readings will differ), is shown below.

Rough titre: initial = 0.00 cm3, final = 19.25 cm3, titre = 19.25 cm3

Titration	initial burette reading / cm3	final burette reading / cm3	titre / cm3
accurate 1	0.00	18.90	18.90
accurate 2	18.90	37.80	18.90
accurate 3	0.00	18.85	18.85

All burette readings are recorded to the nearest 0.05 cm3, units are included in the column headings, and the accurate titres 18.90 cm3 and 18.90 cm3 are concordant (the third, 18.85 cm3, is within 0.10 cm3).

Final answer

Example results: rough 19.25 cm3; accurate titres 18.90, 18.90, 18.85 cm3 (candidate-dependent)

Detailed explanation

Background Concept

A titration measures an unknown concentration by reacting a known volume of it with a solution of known concentration. Here FB2 is 0.0100 mol dm-3 KMnO4 and the solution being titrated contains iron(II). Manganate(VII) acts as its own indicator: while Fe2+ remains, added MnO4- is reduced and the solution stays colourless; after the end-point the first excess MnO4- gives a permanent pale pink colour. A rough titre finds the approximate volume needed, and accurate titrations are repeated until concordant values are obtained. Burette readings are made to the nearest 0.05 cm3.

Understanding the Question

This part asks you to carry out the titration and record raw data. The marks reward technique and recording, not a correct final value: a rough titre with two burette readings, initial and final readings for at least two accurate titrations, clear table headings with units, and readings recorded to 0.05 cm3. The results will be used later for calculations.

Approach

Record each initial and final burette reading as soon as it is taken. Subtract the initial from the final reading to obtain each titre. Present the data in a table whose headings include quantity and unit. Ignore the rough titre when later calculating the mean.

Step-by-Step Reasoning

Fill the burette with FB2 and record the initial reading.
Pipette 25.0 cm3 FB1 into a conical flask and add 10.0 cm3 FB3 from the measuring cylinder.
Carry out a rough titration, recording initial, final and rough titre.
Repeat accurately, adding KMnO4 dropwise near the end-point until one drop gives a permanent pale pink. Record initial and final readings to 0.05 cm3.
Repeat until at least two accurate titres agree within 0.10 cm3.
In the table, include units such as titre / cm3 in the headings and quote every burette reading to two decimal places ending in 0 or 5.

Key Takeaways

Good results tables have clear headings with units, readings are written down immediately, rough titrations are excluded from calculations, and concordant accurate titres are achieved by repeating the titration carefully.

Common Mistakes

Forgetting the initial burette reading for the rough titration, or omitting units in headings.
Recording burette readings to 0.01 cm3 instead of 0.05 cm3.
Including the rough titre in the mean.
Using different numbers of decimal places in different rows.

Things to Be Careful About

The burette is read at the bottom of the meniscus. A reading between two 0.1 cm3 divisions is estimated to 0.05 cm3, e.g. 18.90 or 18.85, never 18.93. The titre is final minus initial, not the final reading alone.

Techniques used

fill and read a burette to the nearest 0.05 cm3perform rough and accurate titrationsrecord initial and final burette readings and titres in a tableselect concordant titres for the mean

(b)

From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value.

$25.0\text{ cm}^3$ of FB 1 required .............................. $\text{cm}^3$ of FB 2.

1M Similar

(c)

Calculations

(i)

Give your answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures.

1M Similar

(ii)

Calculate the amount, in mol, of manganate(VII) ions in the volume of FB 2 in (b).

amount of $\text{MnO}_4^-$ = .............................. mol

1M Similar

(iii)

Use your answer to (c)(ii) and the equations at the start of the question to calculate the concentration, in $\text{mol dm}^{-3}$ , of iron(II) ions in FB 1.

concentration of $\text{Fe}^{2+}$ = .............................. $\text{mol dm}^{-3}$

1M Similar

DifficultyMedium

Worked solution

Working

From the equations, 1 mol $\text{MnO}_4^-$ reacts with 5 mol $\text{Fe}^{2+}$ .

n(\text{Fe}^{2+}) = 5 \times 1.89 \times 10^{-4} = 9.45 \times 10^{-4}\,\text{mol}

c(\text{Fe}^{2+}) = \frac{9.45 \times 10^{-4}}{0.0250} = 0.0378\,\text{mol dm}^{-3}

Answer

$0.0378\,\text{mol dm}^{-3}$

Final answer

0.0378 mol dm^-3

Detailed explanation

Background Concept

The two half-equations show that one MnO4- ion accepts 5 electrons while each Fe2+ ion loses 1 electron. Therefore 1 mol of manganate(VII) reacts with 5 mol of iron(II). The amount of Fe2+ in the 25.0 cm3 sample is five times the amount of MnO4- used, and its concentration is amount divided by the sample volume in dm3.

Understanding the Question

Use the amount of MnO4- from (c)(ii) to find the concentration of Fe2+ in FB1. The volume of the sample is 25.0 cm3; the 10.0 cm3 of FB3 is not involved in the amount calculation.

Approach

Multiply n(MnO4-) by 5 to get n(Fe2+), then divide by 0.0250 dm3 to get concentration.

Step-by-Step Reasoning

n(Fe2+) in 25.0 cm3 = 5 × 1.89 × 10^-4 = 9.45 × 10^-4 mol.
Convert sample volume: 25.0 cm3 = 0.0250 dm3.
c(Fe2+) = 9.45 × 10^-4 / 0.0250 = 0.0378 mol dm-3.

Key Takeaways

The stoichiometry of a redox reaction is obtained by balancing electron transfer. Here the 5 in the calculation comes from 5e- per MnO4- ion.

Common Mistakes

Using a 1:1 ratio instead of 5:1.
Forgetting to convert 25.0 cm3 to dm3.
Including the 10.0 cm3 of FB3 in the total volume (it is added to acidify, not to dilute the sample for this calculation).

Things to Be Careful About

The concentration is in mol dm-3. Use the same amount as in (c)(ii); if you changed your titre in part (b), carry your own value forward consistently.

Techniques used

use the 5:1 stoichiometric ratio between Fe2+ and MnO4-calculate amount of Fe2+ in 25.0 cm3convert amount to concentration using volume in dm3

(iv)

Use your answer to (c)(iii) to calculate the concentration, in $\text{g dm}^{-3}$ , of iron(II) ions in FB 1.

concentration of $\text{Fe}^{2+}$ = .............................. $\text{g dm}^{-3}$

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(v)

The manufacturer of the iron supplement tablets used to make FB 1 claims that each tablet contains a minimum of $150\text{ mg}$ of $\text{Fe}^{2+}$ .

Use your answer to (c)(iv) and the information given about FB 1 to determine whether this claim is correct. Show your working.

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(d)

A student used all the FB 3 and suggests that dilute hydrochloric acid would be a suitable replacement.

Suggest whether the student is correct or not. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

No. Dilute hydrochloric acid is not a suitable replacement because chloride ions would be oxidised by manganate(VII) to chlorine, consuming extra KMnO4 and making the titre appear too large. Sulfuric acid should be used instead.

Final answer

No — Cl- is oxidised by MnO4- to Cl2, so HCl would interfere with the titration

Detailed explanation

Background Concept

The manganate(VII) ion is a very strong oxidising agent, especially in acid. Sulfuric acid is used to provide the H+ ions needed for the reduction of MnO4- without introducing a species that MnO4- can oxidise. Sulfate is not oxidised under these conditions. Hydrochloric acid, however, provides chloride ions, which are reducing enough to be oxidised to chlorine by manganate(VII).

Understanding the Question

The question asks whether dilute hydrochloric acid can replace the dilute sulfuric acid used to acidify the titration mixture. A suitable acid must supply H+ without being oxidised itself.

Approach

Check whether chloride ions could react with KMnO4. Because MnO4- is a strong oxidant, it can oxidise Cl- to Cl2, so HCl would be consumed in a side reaction and would not be a clean source of acid.

Step-by-Step Reasoning

The titration needs an acidic medium so that MnO4- is reduced to Mn2+.
H2SO4 supplies H+ and the sulfate ion is not oxidised by MnO4-.
HCl also supplies H+, but its chloride ion can be oxidised: 2Cl- -> Cl2 + 2e-.
Combining with the manganate half-equation: 2MnO4- + 16H+ + 10Cl- -> 2Mn2+ + 5Cl2 + 8H2O.
This side reaction consumes extra KMnO4, so the titre would be too high and the calculated iron content too high.

Key Takeaways

A strong oxidising titrant such as KMnO4 must be used with an acid whose anion is not itself oxidised. Sulfuric acid is the usual choice.

Common Mistakes

Saying 'yes because HCl is a strong acid' without considering chloride oxidation.
Saying 'no because HCl is not an acid' (it is a strong acid).
Forgetting that it is the chloride ion, not the HCl molecule, that interferes.

Things to Be Careful About

Hydrochloric acid is a strong acid, so lack of H+ is not the main problem. The key issue is the reducing chloride ion. In other contexts HCl may be acceptable, but not with KMnO4 in a titration intended to measure a reducing agent.

Techniques used

identify chloride as a reducing agent that interferesreason that manganate(VII) oxidises Cl- to Cl2

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

The reaction between an acid and an alkali is exothermic. You will carry out a neutralisation experiment to determine the enthalpy change involved.

You will mix different volumes of an acid with a fixed volume of an alkali and measure the temperature rises that occur.

FB 4 is aqueous sodium hydroxide, $\text{NaOH}$ .
FB 5 is $2.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .

(a)

Method

Use the thermometer to measure the initial temperature of FB 4. Record this initial temperature in the space for results.
Support the cup in the $250\text{ cm}^3$ beaker.
Fill one burette with FB 5. Label the burette FB 5.
Fill the other burette with distilled water.

Experiment 1

Use the $10\text{ cm}^3$ pipette to transfer $10.0\text{ cm}^3$ of FB 4 into the cup.
Add $9.00\text{ cm}^3$ of distilled water from the burette to the same cup.
Add $1.00\text{ cm}^3$ of FB 5 from the other burette to the same cup.
Stir the mixture and use the thermometer to measure the maximum temperature. If necessary, tilt the cup so that the solution covers the bulb of the thermometer.
Record the maximum temperature in Table 2.1.
Empty, rinse and dry the cup ready for use in further experiments.

Further experiments

Repeat this method for Experiments 2–5, using $10.0\text{ cm}^3$ of FB 4 and the volumes of water and FB 5 shown in Table 2.1. In each case, measure and record the maximum temperature.

Carry out two further experiments, Experiments 6 and 7, which will enable you to determine more precisely the volume of FB 5 that gives the largest maximum temperature. Record your measurements in Table 2.1.

Results

initial temperature of FB 4 = .............................. $^\circ\text{C}$

Table 2.1

experiment	volume of water / $\text{cm}^3$	volume of FB 5 / $\text{cm}^3$
1	9.00	1.00
2	7.00	3.00
3	5.00	5.00
4	3.00	7.00
5	1.00	9.00
6
7

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Results

initial temperature of FB 4 = 20.0 °C

Table 2.1

experiment	volume of water / cm³	volume of FB 5 / cm³	maximum temperature / °C
1	9.00	1.00	21.5
2	7.00	3.00	23.5
3	5.00	5.00	25.0
4	3.00	7.00	24.0
5	1.00	9.00	22.5
6	4.00	6.00	24.5
7	4.50	5.50	24.8

Note: All thermometer readings end in .0 or .5. Volumes for experiments 6 and 7 are chosen to bracket the expected maximum (5.00 cm³) and sum to 10.00 cm³. Candidate-dependent actual readings will vary.

Final answer

See representative data above; candidate-dependent readings required.

Detailed explanation

Background Concept

In Paper 3 practical examinations, candidates carry out an investigation and record their own data. For an enthalpy of neutralisation experiment, the goal is to find the exact volume of acid that completely neutralises a fixed volume of alkali. Since the concentration of the alkali (FB 4) is not given, we use a graphical method: plotting maximum temperature against the volume of acid added. The intersection of two lines of best fit (one for the rising temperature, one for the falling temperature) gives the equivalence point volume.

Understanding the Question

Part (a) requires the candidate to record the initial temperature, complete the results table for experiments 1–5 with correct precision, and design experiments 6 and 7 to narrow down the volume of FB 5 that produces the highest temperature. The mark scheme also checks the candidate's temperature rise in experiment 3 against the supervisor's value.

Approach

Record the initial temperature of FB 4 to 1 decimal place (ending in .0 or .5).
Record maximum temperatures for experiments 1–5 to 1 decimal place.
Choose volumes for experiments 6 and 7 that bracket the volume giving the highest temperature in experiments 1–5 (which is 5.00 cm³). Ensure the total volume of water + FB 5 remains 10.00 cm³ for all experiments.
Record all volumes to 2 decimal places.

Step-by-Step Reasoning

Initial Temperature: Measured before mixing. Example: 20.0 °C.
Experiments 1–5: Data is provided in the question. Maximum temperatures are recorded as they occur. Example values show a peak at experiment 3 (5.00 cm³ of FB 5).
Experiments 6 and 7: To refine the result around 5.00 cm³, we choose volumes like 6.00 cm³ and 5.50 cm³ of FB 5, with corresponding water volumes of 4.00 cm³ and 4.50 cm³. Total volume is always 10.00 cm³. All volumes are recorded to 2 d.p. ending in 0 or 5.
Supervisor Check: The examiner compares the candidate's temperature rise in experiment 3 (max T – initial T) with the supervisor's known value. A difference of ≤ 1.0 °C is acceptable.

Key Takeaways

Always record thermometer readings to the correct precision (usually 1 d.p. ending in .0 or .5).
When refining a graphical result, choose additional data points that bracket the suspected optimum.
Keep total volumes constant if the method requires it, to ensure consistent heat capacity.

Common Mistakes

Recording volumes to 1 d.p. instead of 2 d.p. (e.g., 9.0 instead of 9.00).
Choosing volumes for experiments 6 and 7 that do not sum to 10.00 cm³.
Not recording the initial temperature.

Things to Be Careful About

Thermometer readings must end in .0 or .5 because the thermometer is typically marked in 0.5 °C intervals.
Volumes from the burette must be read to 2 d.p. (e.g., 5.00 cm³, not 5 cm³).
The total volume of liquid in the cup must remain 10.00 cm³ (10.0 cm³ of FB 4 + 10.0 cm³ of water/FB 5 mixture) to keep the heat capacity of the system constant.

Techniques used

record experimental readings with correct precisiondesign additional experiments to refine a resultcalculate temperature rise

(b)

(i)

Plot a graph of the maximum temperature ( $y$ -axis) against the volume of FB 5 ( $x$ -axis) on the grid. The scale on the $y$ -axis should be suitable for temperature readings to be $2\text{ }^\circ\text{C}$ above the largest maximum temperature.

Label any points you consider to be anomalous.

Draw two lines of best fit, the first for the increase in maximum temperature and the second for after the largest maximum temperature has been reached. Extrapolate both lines so that they intersect.

Graph Grid

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DifficultyMedium

Worked solution

Answer

Graph description: Y-axis is 'maximum temperature / °C' (e.g., 19.0 to 26.0), X-axis is 'volume of FB 5 / cm³' (e.g., 0 to 10). Points plotted for experiments 1–7. Points for exp 6 and 7 may be slightly below the trend lines. Two straight lines of best fit drawn: one through exp 1–3 (rising), one through exp 4–5 (falling). Both lines extrapolated to intersect at approximately 5.00 cm³ and 25.0 °C.

Final answer

See graph diagram above.

Detailed explanation

Background Concept

A temperature-volume graph for a neutralisation reaction typically shows a peak. As acid is added to alkali, the reaction is exothermic and the temperature rises. Once the alkali is completely neutralised, any further addition of acid (which is at a lower temperature) cools the mixture, causing the temperature to fall. The highest temperature occurs at the equivalence point. By drawing two lines of best fit—one for the rising part and one for the falling part—and extrapolating them to intersect, we can find the exact volume of acid that neutralises the alkali, even if the peak is broad or slightly off the plotted points.

Understanding the Question

Part (b)(i) asks the candidate to plot the data from Table 2.1, identify any anomalous results, and draw two lines of best fit that intersect. The y-axis scale must extend 2 °C above the highest maximum temperature.

Approach

Axes and Scale: Y-axis: maximum temperature (°C). X-axis: volume of FB 5 (cm³). Y-axis scale should start below the initial temperature and go at least 2 °C above the highest max T (e.g., if max T is 25.0 °C, go up to 27.0 °C).
Plotting: Plot all 7 points accurately.
Anomalous Points: If a point (e.g., experiment 6 or 7) falls significantly below the expected trend line, label it as anomalous.
Lines of Best Fit: Draw a line through the rising points (exp 1–3). Draw a second line through the falling points (exp 4–5). Extend both lines so they cross.

Step-by-Step Reasoning

Axes: Label clearly with units. Use a suitable scale (e.g., 1 cm = 1 °C on y-axis, 1 cm = 1 cm³ on x-axis).
Points: Plot (1.00, 21.5), (3.00, 23.5), (5.00, 25.0), (7.00, 24.0), (9.00, 22.5), (6.00, 24.5), (5.50, 24.8).
Anomalies: In this representative data, exp 6 and 7 might be slightly low due to heat loss during the longer stirring time required for finer volumes. Label them if they are clearly off the trend.
Lines: Line 1 passes through (1, 21.5), (3, 23.5), (5, 25.0). Line 2 passes through (5, 25.0), (7, 24.0), (9, 22.5). Extrapolate both to intersect at x = 5.00 cm³, y = 25.0 °C.

Key Takeaways

The intersection of the two lines gives the equivalence point volume.
Anomalous results should be plotted but ignored when drawing the lines of best fit.
The y-axis must have sufficient range above the maximum temperature.

Common Mistakes

Drawing a single smooth curve through all points instead of two straight lines of best fit.
Forgetting to extrapolate the lines to their intersection.
Using an incorrect scale on the axes.

Things to Be Careful About

Ensure the lines are straight (linear) for the two regions, not a curved line.
The intersection point is read from the x-axis (volume of FB 5).
Anomalous points must be explicitly labelled.

Techniques used

plot a graph of temperature against volumedraw lines of best fit and extrapolate to find intersectionidentify anomalous results

(ii)

Use the intersection on your graph in (b)(i) to determine the volume of FB 5 required to neutralise $10.0\text{ cm}^3$ of FB 4.

volume of FB 5 = .............................. $\text{cm}^3$

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(c)

Calculations

(i)

Calculate the amount, in mol, of hydrochloric acid in the volume of FB 5 in (b)(ii).

(If you were unable to determine an answer to (b)(ii), use $5.10\text{ cm}^3$ as the volume of FB 5. This may not be the correct answer.)

amount of $\text{HCl}$ = .............................. mol

Deduce the amount, in mol, of sodium hydroxide in $10.0\text{ cm}^3$ of FB 4.

amount of $\text{NaOH}$ = .............................. mol

1M Similar

DifficultyMedium-Easy

Worked solution

Working

amount of $\text{HCl} = 2.00 \times \frac{5.00}{1000} = 0.0100 \text{ mol}$

The reaction is: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

The molar ratio is 1:1, so:

amount of $\text{NaOH} = 0.0100 \text{ mol}$

Answer

amount of $\text{HCl}$ = 0.0100 mol
amount of $\text{NaOH}$ = 0.0100 mol

Final answer

0.0100; 0.0100

Detailed explanation

Background Concept

The amount of substance (in moles) is calculated using the formula: $n = c \times V$ , where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³ (or $c \times V/1000$ if $V$ is in cm³). In a neutralisation reaction between a strong acid and a strong base, the molar ratio is typically 1:1.

Understanding the Question

Part (c)(i) asks for the moles of HCl in the volume determined in (b)(ii), and then the moles of NaOH in 10.0 cm³ of FB 4. The mark scheme allows error carried forward from (b)(ii), but here we use the representative value of 5.00 cm³.

Approach

Calculate moles of HCl: $n = 2.00 \times (5.00 / 1000)$ .
Use the 1:1 stoichiometry to state moles of NaOH = moles of HCl.

Step-by-Step Reasoning

Volume of FB 5 = 5.00 cm³ = 0.00500 dm³.
Concentration of FB 5 = 2.00 mol dm⁻³.
Moles of HCl = $2.00 \times 0.00500 = 0.0100$ mol.
Equation: $\text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})$ .
Ratio is 1:1, so moles of NaOH = 0.0100 mol.
Answers should be given to 2–4 significant figures (0.0100 has 3 sf).

Key Takeaways

Always convert volume from cm³ to dm³ by dividing by 1000.
State symbols are not required for mole calculations but are good practice.
Significant figures should be consistent (2–4 sf here).

Common Mistakes

Forgetting to divide volume by 1000.
Using the wrong concentration.
Not stating that moles of NaOH = moles of HCl.

Things to Be Careful About

Ensure the volume used is the one read from the graph in (b)(ii), not a default value unless instructed.
Keep at least 3 significant figures in intermediate calculations to avoid rounding errors.

Techniques used

calculate moles from concentration and volumeuse stoichiometry to find moles of alkali

(ii)

Calculate the energy change, in J, when the amounts of reagents in (c)(i) neutralise each other. Show your working.

energy change = .............................. J

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Total volume of solution = $10.0 + 10.0 = 20.0 \text{ cm}^3$

Assuming density = $1.00 \text{ g cm}^{-3}$ , mass $m = 20.0 \text{ g}$ .

$\Delta T = \text{maximum temperature} - \text{initial temperature} = 25.0 - 20.0 = 5.0 \text{ }^\circ\text{C}$

Energy change $q = m \times c \times \Delta T$
$q = 20.0 \times 4.18 \times 5.0 = 418 \text{ J}$

Answer

energy change = 418 J

Final answer

418

Detailed explanation

Background Concept

The energy change (heat evolved) in a solution is calculated using $q = mc\Delta T$ , where $m$ is the mass of the solution, $c$ is the specific heat capacity (usually taken as 4.18 J g⁻¹ °C⁻¹ for aqueous solutions), and $\Delta T$ is the temperature change. For dilute aqueous solutions, we assume the density is 1.00 g cm⁻³, so mass in grams equals volume in cm³.

Understanding the Question

Part (c)(ii) asks for the energy change in Joules when the amounts of reagents neutralise each other. The mark scheme provides the formula: Energy = $20 \times 4.18 \times \Delta T$ .

Approach

Calculate total volume: 10.0 cm³ (FB 4) + 10.0 cm³ (water + FB 5) = 20.0 cm³.
Assume mass = 20.0 g.
Calculate $\Delta T$ = max T – initial T.
Apply $q = 20.0 \times 4.18 \times \Delta T$ .

Step-by-Step Reasoning

Total volume = 20.0 cm³, so mass $m = 20.0$ g.
$\Delta T = 25.0 - 20.0 = 5.0$ °C.
$q = 20.0 \times 4.18 \times 5.0 = 418$ J.
The answer should be to 2–4 significant figures (418 has 3 sf).

Key Takeaways

The total volume is always 20.0 cm³ in this experiment (10.0 cm³ alkali + 10.0 cm³ acid/water mixture).
$\Delta T$ is the difference between the maximum temperature and the initial temperature of the alkali.
Energy is in Joules, not kJ.

Common Mistakes

Using the wrong $\Delta T$ (e.g., max T – min T).
Forgetting to convert J to kJ in the next step.
Using the wrong mass (e.g., only 10.0 g instead of 20.0 g).

Things to Be Careful About

The mark scheme explicitly uses 20 for the mass. Do not use a different value unless justified.
Keep the sign positive here; the negative sign is added in part (iii) for enthalpy change.

Techniques used

calculate energy change using q = mcΔT

(iii)

Use your answer to (c)(ii) to calculate the enthalpy change, in $\text{kJ mol}^{-1}$ , when one mole of FB 4 is neutralised by one mole of FB 5.

enthalpy change = .............................. $\text{kJ mol}^{-1}$ (sign and value)

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Enthalpy change $\Delta H = -\frac{q}{n}$

where $q$ is in kJ and $n$ is moles.

$q = 418 \text{ J} = 0.418 \text{ kJ}$

$n = 0.0100 \text{ mol}$

$\Delta H = -\frac{0.418}{0.0100} = -41.8 \text{ kJ mol}^{-1}$

Answer

enthalpy change = -41.8 kJ mol⁻¹

Final answer

-41.8

Detailed explanation

Background Concept

Enthalpy change of neutralisation ( $\Delta H$ ) is the energy change per mole of water formed (or per mole of acid/alkali reacted). It is calculated as $\Delta H = -q / n$ , where $q$ is the heat energy in kJ and $n$ is the number of moles. The negative sign indicates that the reaction is exothermic (heat is released to the surroundings).

Understanding the Question

Part (c)(iii) asks for the enthalpy change in kJ mol⁻¹ when one mole of FB 4 is neutralised by one mole of FB 5. The mark scheme requires the correct formula, a negative sign, and answers to minimum 2 significant figures.

Approach

Convert energy from J to kJ: divide by 1000.
Divide by the moles of NaOH (or HCl) from part (c)(i).
Add a negative sign because neutralisation is exothermic.

Step-by-Step Reasoning

Energy $q = 418$ J = 0.418 kJ.
Moles $n = 0.0100$ mol.
$\Delta H = - (0.418 / 0.0100) = -41.8$ kJ mol⁻¹.
The value is negative because the reaction is exothermic.
Answer to 3 significant figures (-41.8).

Key Takeaways

Always convert J to kJ by dividing by 1000.
Neutralisation is exothermic, so $\Delta H$ must be negative.
The value is typically around -57 kJ mol⁻¹ for strong acid/strong base; -41.8 kJ mol⁻¹ is a realistic experimental value (lower due to heat loss).

Common Mistakes

Forgetting the negative sign.
Forgetting to convert J to kJ.
Dividing by the wrong number of moles.

Things to Be Careful About

The mark scheme explicitly requires the negative sign. Omitting it will cost a mark.
Ensure the number of significant figures is at least 2.

Techniques used

calculate enthalpy change per mole of water formed

Q3MediumQualitative AnalysisAnalysis, Conclusions and Evaluation

Qualitative analysis

For each test you should record all your observations in the spaces provided.

Examples of observations include:

colour changes seen
the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added
the formation of any gas and its identification (where appropriate) by a suitable test.

You should record clearly at what stage in a test an observation is made.

Where no change is observed, you should write ‘no change’.

Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given.

If any solution is warmed, a boiling tube must be used. If a solid is heated, a hard-glass test-tube must be used.

Rinse and reuse test-tubes and boiling tubes where possible.

No additional tests should be attempted.

(a)

(i)

FB 6, FB 7 and FB 8 are aqueous solutions of different compounds that each contain at least one oxygen atom.

Carry out the following tests and record your observations in Table 3.1. Three of the tests have been done for you. Use a $1\text{ cm}$ depth of solution in a test-tube for each test.

Table 3.1

test	observations for FB 6	observations for FB 7	observations for FB 8
Test 1 Add a small spatula measure of manganese(IV) oxide.	No change.	No change.
Test 2 Add a $1\text{ cm}$ length of magnesium.			No change.
Test 3 Add a few drops of aqueous iron(II) sulfate.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Observations to record in Table 3.1:

test	FB 6	FB 7	FB 8
Test 1 (add MnO $_2$ )	no change	no change	fizzing; heat released; gas relights a glowing splint
Test 2 (add Mg)	fizzing; heat released; gas pops with a lit splint	no change	no change
Test 3 (add FeSO $_4$ )	no change	green precipitate; turns brown on standing	solution turns yellow/orange-brown; fizzing; gas relights a glowing splint

Final answer

Observations recorded in Table 3.1 (see working)

Detailed explanation

Background Concept

This part is about carrying out and recording qualitative tests. Three reagents are used, each probing a different property.

Manganese(IV) oxide, MnO $_2$ (s), is a catalyst for the decomposition of hydrogen peroxide:

2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})

A catalyst provides an alternative reaction pathway of lower activation energy and is chemically unchanged at the end. The oxygen produced relights a glowing splint.

Magnesium metal reacts with acids to release hydrogen gas:

\text{Mg}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})

Hydrogen burns with a squeaky pop when a lit splint is held at the mouth of the tube. Magnesium does not react appreciably with alkalis or hydrogen peroxide under these conditions.

Iron(II) sulfate, FeSO $_4$ (aq), contains Fe $^{2+}$ ions. With hydroxide ions it forms a green precipitate of iron(II) hydroxide, Fe(OH) $_2$ , which slowly oxidises in air to brown iron(III) hydroxide, Fe(OH) $_3$ . Fe $^{2+}$ is also a reducing agent: it reduces hydrogen peroxide to water and is itself oxidised to Fe $^{3+}$ , which gives a yellow/orange-brown solution.

Understanding the Question

You are given three unlabelled aqueous solutions (FB 6, FB 7, FB 8), each containing at least one oxygen atom. You must carry out three tests — adding MnO $_2$ , adding Mg, and adding FeSO $_4$ — and record all observations in Table 3.1. The instructions stress recording the stage at which each observation is made, naming reagents, and writing “no change” where nothing happens.

Approach

Rather than guessing, predict the chemistry of each solution with each reagent. The three solutions are ethanoic acid (CH $_3$ COOH), sodium hydroxide (NaOH) and hydrogen peroxide (H $_2$ O $_2$ ). For each test ask: does this reagent react with an acid, an alkali, or with an oxidising agent such as H $_2$ O $_2$ ?

Step-by-Step Reasoning

Test 1 — add MnO $_2$ . Only H $_2$ O $_2$ is decomposed by the catalyst: expect fizzing (bubbles of O $_2$ ), heat released, and the gas relighting a glowing splint. The acid and alkali give no change.

Test 2 — add Mg. Only the acid reacts: fizzing (H $_2$ ), heat released, and a pop with a lit splint. NaOH and H $_2$ O $_2$ give no change.

Test 3 — add FeSO $_4$ .

With FB 6 (acid): no change.
With FB 7 (NaOH): green precipitate of Fe(OH) $_2$ forms; on standing it turns brown as it is oxidised to Fe(OH) $_3$ .
With FB 8 (H $_2$ O $_2$ ): the solution turns yellow/orange-brown (Fe $^{2+}$ → Fe $^{3+}$ ) and fizzes as O $_2$ is released; the gas relights a glowing splint.

Key Takeaways

The three tests are diagnostic: MnO $_2$ detects H $_2$ O $_2$ (O $_2$ given off), Mg detects an acid (H $_2$ given off), and FeSO $_4$ detects both an alkali (green Fe(OH) $_2$ precipitate) and an oxidising agent (Fe $^{2+}$ → Fe $^{3+}$ , brown colour). Always pair a colour change or precipitate with the gas test that confirms the gas.

Common Mistakes

Writing “bubbles” instead of “fizzing” — the mark scheme wants the specific observation.
Recording a gas without testing it (e.g. “gas given off” without “relights a glowing splint” or “pops with a lit splint”).
Forgetting to record “no change” for the solutions that do not react.
Confusing the two gas tests: oxygen relights a glowing splint; hydrogen pops with a lit splint.

Things to Be Careful About

Record observations at the correct stage (e.g. “on standing” for the Fe(OH) $_2$ colour change).
If a solution is warmed, a boiling tube must be used; if a solid is heated, a hard-glass test-tube.
Name or give the correct formula of every reagent used.
The mark scheme awards 2 asterisks per mark and rounds down, so partial observations still earn marks.

Techniques used

record observations from qualitative teststest for oxygen with a glowing splinttest for hydrogen with a lit splintobserve precipitate formation and colour change

(ii)

Use your observations in Table 3.1 to suggest a possible formula for each of FB 6, FB 7 and FB 8.

FB 6 ..............................

FB 7 ..............................

FB 8 ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

FB 6 = CH $_3$ COOH(aq)

FB 7 = NaOH(aq)

FB 8 = H $_2$ O $_2$ (aq)

Final answer

FB 6 = CH3COOH; FB 7 = NaOH; FB 8 = H2O2

Detailed explanation

Background Concept

The observations in Table 3.1 are diagnostic fingerprints of the three compounds. An acid reacts with Mg to give H $_2$ ; an alkali gives a green Fe(OH) $_2$ precipitate with Fe $^{2+}$ ; hydrogen peroxide is catalytically decomposed by MnO $_2$ to give O $_2$ and oxidises Fe $^{2+}$ to Fe $^{3+}$ .

Understanding the Question

Using only the observations you recorded, suggest a possible formula for each of the three solutions. The mark scheme accepts any acid containing oxygen for FB 6 and any Group 1 hydroxide for FB 7, but the intended answers are CH $_3$ COOH, NaOH and H $_2$ O $_2$ .

Approach

Work backwards from each distinctive observation: FB 6 fizzes with Mg → acid; FB 7 gives a green precipitate with FeSO $_4$ → alkali (OH $^-$ ); FB 8 fizzes with MnO $_2$ and turns FeSO $_4$ brown → H $_2$ O $_2$ .

Step-by-Step Reasoning

FB 6: reacts with Mg releasing H $_2$ → contains H $^+$ , i.e. an acid. The intended compound is ethanoic acid, CH $_3$ COOH.
FB 7: gives a green precipitate with Fe $^{2+}$ → contains OH $^-$ , i.e. an alkali. The intended compound is sodium hydroxide, NaOH.
FB 8: decomposed by MnO $_2$ to give O $_2$ and oxidises Fe $^{2+}$ to Fe $^{3+}$ → hydrogen peroxide, H $_2$ O $_2$ .

Key Takeaways

Each qualitative test gives a unique fingerprint; matching observations to known chemistry identifies the compound.

Common Mistakes

Suggesting a compound that does not fit all observations.
Forgetting the “at least one oxygen atom” condition — any formula suggested must contain oxygen.

Things to Be Careful About

The mark scheme allows any acid containing oxygen for FB 6 and any Group 1 hydroxide for FB 7, so a correct but different formula still scores.

Techniques used

deduce the identity of each solution from its observationsmatch reactivity patterns to known compounds

(b)

FB 9 contains two anions and two cations, three of which are listed in the Qualitative analysis notes.

(i)

To a small spatula measure of FB 9 in a test-tube, add a $2\text{ cm}$ depth of dilute nitric acid. Record your observations.

Keep the resulting solution for the test in (b)(ii).

2M Similar

(ii)

To the solution from (b)(i), add a few drops of aqueous silver nitrate. Then add excess aqueous ammonia. Record your observations.

1M Similar

(iii)

Make an aqueous solution of FB 9 by adding a $5\text{ cm}$ depth of distilled water to a spatula measure of FB 9 in a test-tube. Carry out the following tests on the aqueous solution of FB 9 and record your observations in Table 3.2.

Table 3.2

test	observations
Test 1 To a $1\text{ cm}$ depth in a boiling tube, add aqueous sodium hydroxide, then warm.
Test 2 To a $1\text{ cm}$ depth in a test-tube, add a few drops of dilute hydrochloric acid, then add a few drops of aqueous chlorine. Empty and rinse the test-tube with water immediately after use.

2M Similar

(iv)

Use your observations in (b)(i), (b)(ii) and Table 3.2 to deduce the formulae of the cations and anions in FB 9. If you are unable to identify an ion, write ‘unknown’.

cations .............................. and ..............................

anions .............................. and ..............................

2M Similar

(v)

Write an ionic equation for the reaction in (b)(ii). Include state symbols.

1M Similar