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) Purple crystals crackle on heating, forming a black powder/residue; gas evolved relights a glowing splint (oxygen). · (a)(ii) Test 1: purple solution turns yellow. Test 2: purple decolourises with effervescence; gas relights a glowing splint (O2). · (a)(iii) Dark green solution · (b) Test 1: no change with HNO3 and with AgNO3. Test 2: white ppt insoluble in HNO3 (sulfate). Test 3: pale brown ppt darkening, turning black with H2O2, with effervescence (O2). · (c) Test…

Chemistry 9701/35 — October/November 2024

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

questions

marks

120

minutes

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

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

The neutralisation of an acid by an alkali is an exothermic reaction. The concentration of an acid can be found by measuring the temperature change when the acid reacts with an alkali.

You will determine the concentration of sulfuric acid by adding aqueous sodium hydroxide of known concentration to the sulfuric acid and measuring the temperature change.

FA 1 is $2.01 \text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FA 2 is sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Support the cup in the $250 \text{ cm}^3$ beaker.
Pipette $25.0 \text{ cm}^3$ of FA 1 into the cup.
Place the thermometer into the FA 1 in the cup. Tilt the cup if necessary to ensure the bulb of the thermometer is fully covered. Record the temperature of FA 1 in Table 1.1. This is the temperature when the volume of FA 2 is $0.00 \text{ cm}^3$ .
Fill the burette with FA 2.
Run $5.00 \text{ cm}^3$ of FA 2 into the cup containing FA 1.
Stir the mixture and record the maximum temperature in Table 1.1.
Run further $5.00 \text{ cm}^3$ portions of FA 2 into the same cup.
After each addition of FA 2 stir the contents of the cup. Record the maximum temperature for each addition.

Table 1.1

total volume of FA 2 / $\text{cm}^3$	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / °C

Keep the rest of FA 2 for use in Question 2.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

The candidate records thermometer readings for all nine volumes of FA 2 added, with each reading ending in .0 or .5 °C (i.e. recorded to the nearest 0.5 °C or 0.1 °C depending on thermometer resolution). The maximum temperature is recorded after each addition following stirring.

Representative example data (candidate-dependent):

total volume of FA 2 / cm³	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / °C	18.5	20.5	22.5	24.5	26.0	27.0	26.0	24.5	23.5

The maximum temperature ( $T_{\text{max}}$ ) is 27.0 °C at 25.00 cm³ of FA 2 added.

Final answer

Candidate-dependent; representative $T_{\text{max}} = 27.0$ °C at 25.00 cm³

Detailed explanation

Background Concept

Neutralisation between a strong acid and a strong base is exothermic: $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O(l)}$ , $\Delta H = -57.1 \text{ kJ mol}^{-1}$ . When measuring temperature change to find an equivalence point, the principle is that the temperature rises as long as reaction occurs (heat is released), reaches a maximum when the reaction is complete (all limiting reagent consumed), and then falls as the added solution is at room temperature and dilutes the mixture while heat is lost to surroundings.

Understanding the Question

This part requires the candidate to carry out the experiment and record nine temperature readings in Table 1.1. The command word is implicit: "Record the maximum temperature for each addition." The mark scheme rewards: (I) readings for all 9 volumes, (II) all readings ending in .0 or .5 (consistent precision), and (III)/(IV) closeness of $T_{\text{max}}$ to the supervisor's value (within 2.0 °C or 1.0 °C respectively).

Approach

The candidate must: pipette exactly 25.0 cm³ of FA 1 into the polystyrene cup, record the initial temperature, then add FA 2 in 5.00 cm³ portions from the burette, stirring after each addition and recording the maximum temperature reached. The cup is supported in a beaker to reduce heat loss.

Step-by-Step Reasoning

Initial reading (0.00 cm³ FA 2): Record the temperature of FA 1 before any acid is added. This is the starting point and should be close to room temperature (e.g. 18.5 °C).
Subsequent additions (5.00 to 40.00 cm³): After each 5.00 cm³ portion, stir thoroughly and record the maximum temperature. The temperature rises as neutralisation proceeds, peaks near the equivalence point, then falls as excess cold acid is added and heat dissipates.
Precision requirement: All readings must end in .0 or .5 °C, indicating the thermometer is read to the nearest 0.5 °C (or the candidate rounds consistently). This is marked as requirement II.
Accuracy against supervisor: The supervisor's $T_{\text{max}}$ is the reference. Marks III and IV are awarded based on how close the candidate's $T_{\text{max}}$ is to the supervisor's (within 2.0 °C and 1.0 °C respectively).

Key Takeaways

In a thermometric titration, the maximum temperature corresponds to the point of complete neutralisation.
Consistent recording precision (all readings to the same number of decimal places) is essential.
Stirring ensures uniform temperature throughout the solution before reading.

Common Mistakes

Recording temperatures to inconsistent precision (e.g. mixing 22.5 with 23.7).
Not waiting for the maximum temperature to stabilise before recording.
Failing to stir after each addition, giving a lower apparent maximum.
Recording the temperature immediately after adding acid rather than the peak temperature.

Things to Be Careful About

The thermometer bulb must be fully immersed in the solution throughout.
The cup should be supported in the beaker to minimise heat loss to the bench.
Each reading must be the maximum temperature reached after that addition, not the temperature at a fixed time.

Techniques used

record thermometer readings to 0.1 °C precisionstir mixture after each additionrecord maximum temperature for each volume added

(b)

Plot a graph of temperature of solution ( $y$ -axis) against total volume of FA 2 added ( $x$ -axis) on the grid. Select a scale for the $y$ -axis to include a value $3 \text{ }^{\circ}\text{C}$ above your maximum temperature reading. Label any points you consider to be anomalous.

Draw two lines of best fit through the points on your graph. Draw the first line for the increase in temperature and the second line after the maximum temperature was reached. Extrapolate the lines so they intersect. This intersection corresponds to the volume of FA 2 needed to neutralise the FA 1 in your experiment in (a).

Fig. 1.1

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

5M Similar

DifficultyMedium

Worked solution

Answer

Graph construction:

$x$ -axis: total volume of FA 2 / cm³ (scale: e.g. 5 cm³ per 10 small squares, range 0–45 cm³)
$y$ -axis: temperature of solution / °C (scale chosen so the top of the axis is at least 3 °C above the maximum recorded temperature; e.g. 2 °C per 10 small squares, range 16–30 °C)
Plot all nine points accurately (within half a small square)
Draw a smooth curve through the rising points (0–25 cm³) and a straight line or smooth curve through the falling points (25–40 cm³)
Extrapolate both lines so they intersect at or above the highest recorded temperature
Read the volume at the intersection to 1 decimal place

Representative result:

25.0 cm³ of FA 1 required 25.0 cm³ of FA 2.

Final answer

25.0 cm³ (candidate-dependent; read to 1 dp from graph intersection)

Detailed explanation

Background Concept

In a thermometric titration, the temperature of the solution rises linearly (approximately) as acid is added and reacts with the base, reaching a maximum at the equivalence point. Beyond this point, additional acid does not react but simply dilutes the warm solution and absorbs heat, causing the temperature to fall. The intersection of the extrapolated rising and falling lines gives the volume at which neutralisation was complete — the equivalence point volume.

This method works because the heat released is proportional to the amount of reaction occurring, and the maximum temperature corresponds to the point where all the limiting reagent has been consumed.

Understanding the Question

The candidate must plot their data from part (a) on the provided grid (Fig. 1.1), with temperature on the $y$ -axis and volume of FA 2 on the $x$ -axis. The key requirements are:

A $y$ -axis scale that extends at least 3 °C above the maximum reading (to allow extrapolation to be visible)
Two separate lines of best fit: one for the rising portion, one for the falling portion
Both lines extrapolated to intersect
The intersection volume read to 1 decimal place

Approach

Choose scales that use at least half the grid in each direction.
Plot points carefully — the intersection region is the most critical.
The rising line should be a smooth curve (temperature increase is not perfectly linear due to increasing total volume and heat loss).
The falling line may be straight or a gentle curve.
Extrapolate both beyond the data range until they meet.
Drop a vertical line from the intersection to the $x$ -axis and read the volume.

Step-by-Step Reasoning

Mark I — Axes and scale: Both axes must be clearly labelled with quantity and unit. The $y$ -axis must extend at least 3 °C above $T_{\text{max}}$ so the extrapolation is visible within the grid. Scales should be linear with sensible divisions (1, 2, or 5 per 10 small squares).

Mark II — Plotting: All nine points must be plotted to within half a small square. Anomalous points (if any) should be circled and labelled.

Mark III — Lines of best fit: Two distinct lines are required. The rising line must be a smooth curve (not straight, because the rate of temperature rise changes as total volume increases). The falling line may be straight or curved.

Mark IV — Extrapolation: Both lines must be extended beyond the last data points until they intersect. The intersection must occur at or above the highest recorded temperature (i.e. the lines cross in the extrapolation region, not within the data range).

Mark V — Reading: The volume at the intersection is read from the $x$ -axis to 1 decimal place (e.g. 25.0 cm³, not 25 cm³ or 25.00 cm³).

Key Takeaways

The intersection of extrapolated lines gives the equivalence point volume more accurately than simply reading the maximum temperature point, because it corrects for heat loss during the experiment.
The rising line is curved because each successive addition of acid is diluted by a larger total volume, so the temperature increment per 5 cm³ decreases.
Reading to 1 dp is required because the grid allows this precision.

Common Mistakes

Drawing only one line through all points instead of two separate lines.
Making the rising line straight rather than a smooth curve.
Not extrapolating far enough for the lines to intersect (or the intersection falling below $T_{\text{max}}$ ).
Choosing a $y$ -axis scale that is too compressed, making the 3 °C headroom requirement impossible to meet.
Reading the intersection volume to 0 dp or 2 dp instead of 1 dp.
Plotting points inaccurately, especially near the maximum.

Things to Be Careful About

The $y$ -axis scale selection is critical: if $T_{\text{max}} = 27.0$ °C, the axis must go to at least 30.0 °C.
The intersection volume is used in part (c)(ii), so accuracy here directly affects the final concentration calculation (error carried forward applies).
Label any anomalous points with a circle and the word "anomalous" or a cross.

Techniques used

plot points accurately on a griddraw two lines of best fit through rising and falling sectionsextrapolate lines to find intersectionread volume at intersection to 1 dp

(c)

(i)

Calculate the amount, in mol, of sodium hydroxide, FA 1, pipetted into the cup.

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

1M Similar

DifficultyEasy

Worked solution

Working

n = \frac{c \times V}{1000} = \frac{2.01 \times 25.0}{1000} = 0.05025 \text{ mol}

Answer

amount of NaOH = $5.025 \times 10^{-2}$ mol (or $5.03 \times 10^{-2}$ mol)

Final answer

5.025 × 10⁻² mol

Detailed explanation

Background Concept

The amount (number of moles) of a solute in solution is calculated from $n = c \times V$ where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³. Since volumes are typically given in cm³, the conversion $V(\text{dm}^3) = V(\text{cm}^3)/1000$ gives $n = cV/1000$ .

Understanding the Question

The question asks for the amount of NaOH in the 25.0 cm³ pipetted into the cup. The concentration of FA 1 is given as 2.01 mol dm⁻³. This is a direct substitution into the mole formula.

Approach

Use $n = cV/1000$ with $c = 2.01$ mol dm⁻³ and $V = 25.0$ cm³.

Step-by-Step Reasoning

Identify the known values: $c = 2.01$ mol dm⁻³, $V = 25.0$ cm³.
Apply the formula: $n = \frac{2.01 \times 25.0}{1000}$ .
Calculate: $n = \frac{50.25}{1000} = 0.05025$ mol.
Express to 3 or 4 significant figures: $5.025 \times 10^{-2}$ mol or $5.03 \times 10^{-2}$ mol.

The mark scheme requires 3 or 4 significant figures. Both $5.025 \times 10^{-2}$ and $5.03 \times 10^{-2}$ are accepted.

Key Takeaways

Always convert cm³ to dm³ by dividing by 1000 when using $n = cV$ .
The answer must be given to 3 or 4 significant figures as specified.
This value is used in part (c)(ii), so accuracy here affects the final answer (but ecf applies).

Common Mistakes

Forgetting to divide by 1000, giving 50.25 mol instead of 0.05025 mol.
Giving the answer to only 2 significant figures (e.g. $5.0 \times 10^{-2}$ ), which loses the mark.
Confusing the volume of FA 1 (25.0 cm³) with the volume of FA 2 from part (b).

Things to Be Careful About

The mark scheme specifies 3 or 4 significant figures — 2 sf would be rejected.
Units must be mol.

Techniques used

calculate moles from concentration and volume using n = cV/1000

(ii)

The equation for this neutralisation reaction is shown.

2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}

Calculate the concentration, in $\text{mol dm}^{-3}$ , of sulfuric acid in FA 2.

Show your working.

concentration of $\text{H}_2\text{SO}_4$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation: $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

n(\text{H}_2\text{SO}_4) = \frac{n(\text{NaOH})}{2} = \frac{5.025 \times 10^{-2}}{2} = 2.513 \times 10^{-2} \text{ mol}

Using the volume of FA 2 from part (b) = 25.0 cm³:

c(\text{H}_2\text{SO}_4) = \frac{n \times 1000}{V} = \frac{2.513 \times 10^{-2} \times 1000}{25.0} = 1.005 \text{ mol dm}^{-3}

Answer

concentration of H₂SO₄ = 1.01 mol dm⁻³ (3 sf)

Final answer

1.01 mol dm⁻³

Detailed explanation

Background Concept

Stoichiometric calculations from a balanced equation allow conversion between amounts of reactants and products. The equation $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ shows that 2 moles of NaOH react with 1 mole of H₂SO₄. The concentration of the unknown acid is then found from $c = n \times 1000/V$ , where $V$ is the volume of acid at the equivalence point (read from the graph in part b).

Understanding the Question

The candidate must use:

The moles of NaOH calculated in (c)(i)
The 2:1 stoichiometric ratio from the given equation
The volume of FA 2 at the equivalence point from the graph in (b)

to find the concentration of H₂SO₄.

Approach

Divide moles of NaOH by 2 to get moles of H₂SO₄ (M1).
Divide moles of H₂SO₄ by the volume of FA 2 (in dm³) to get concentration (M2).

Step-by-Step Reasoning

M1 — Moles of H₂SO₄:

From the balanced equation, 2 mol NaOH reacts with 1 mol H₂SO₄. Therefore:
$n(\text{H}_2\text{SO}_4) = \frac{n(\text{NaOH})}{2} = \frac{5.025 \times 10^{-2}}{2} = 2.513 \times 10^{-2} \text{ mol}$

This is M1. The mark is awarded for correctly halving the answer from (c)(i), even if (c)(i) was wrong (ecf applies).

M2 — Concentration of H₂SO₄:

The volume of FA 2 at the equivalence point is read from the graph in part (b). Using the representative value of 25.0 cm³:
$c = \frac{n \times 1000}{V} = \frac{2.513 \times 10^{-2} \times 1000}{25.0} = 1.005 \text{ mol dm}^{-3}$

Rounded to 3 sf: $1.01$ mol dm⁻³.

M2 requires some working shown AND the final answer to 3–4 sf. If the candidate's volume from (b) differs, the concentration will differ accordingly (ecf from part b).

Key Takeaways

The stoichiometric ratio (2:1) must be correctly applied — it is easy to forget to divide by 2.
The volume used in the concentration calculation comes from the graph (part b), not from any other source.
Error carried forward applies: if (c)(i) or (b) is wrong but used consistently here, method marks can still be awarded.
Final answer must be to 3 or 4 significant figures.

Common Mistakes

Forgetting to divide by 2 (using moles of NaOH directly as moles of H₂SO₄), giving double the correct concentration.
Using 25.0 cm³ (the volume of FA 1) instead of the volume of FA 2 from the graph.
Not showing working (M2 requires "some working shown").
Giving the answer to only 2 significant figures.
Dividing by 2 the wrong way (multiplying by 2 instead of dividing).

Things to Be Careful About

The mark scheme explicitly requires the final answer to 3–4 sf.
ECF: if the volume from (b) is incorrect but used correctly here, M2 can still be awarded.
The units must be mol dm⁻³.
Ensure the volume from (b) is in cm³ when using $c = n \times 1000/V$ .

Techniques used

apply stoichiometric ratio from balanced equationcalculate concentration from moles and volume using c = n×1000/V

Q2MediumPresentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

Acids react with carbonates to produce carbon dioxide gas.

\text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

This reaction can be used to determine the concentration of acid, using the mass of carbon dioxide released.

You will determine the concentration of sulfuric acid in FA 2.

FA 2 is the solution used in Question 1.
FA 3 is sodium carbonate, $\text{Na}_2\text{CO}_3$ .

(a)

Method

Use the $25 \text{ cm}^3$ measuring cylinder to transfer $25.0 \text{ cm}^3$ of FA 2 into the conical flask.
Weigh the flask with the acid. Record the mass.
Weigh the container with FA 3. Record the mass.
Carefully tip all of FA 3 into the acid in the conical flask. Swirl the contents of the flask and leave the flask to stand with occasional swirling until the fizzing stops.
Weigh the container with any residual FA 3. Record the mass.
Calculate and record the mass of FA 3 added to the flask.
Calculate and record the total mass of flask + acid + FA 3.
Weigh the flask and contents when the fizzing has stopped. Record the mass.
Calculate and record the mass of carbon dioxide given off during the experiment.

Results

(Prepare a table for your results in the space below)

5M Similar

DifficultyMedium-Easy

Worked solution

Answer

Set up a results table with these headings, each followed by the unit /g:

mass of flask + acid (FA 2)
mass of container + FA 3
mass of container (empty / with residue)
mass of FA 3 added = (mass of container + FA 3) – (mass of container)
mass of flask + acid + FA 3 (initial total)
mass of flask and contents after reaction
mass of carbon dioxide given off = initial total – final mass

Record every balance reading to the same precision (2 or 3 decimal places) and put the unit beside every entry and calculated value.

Then substitute your own measured masses into the calculations to complete the table.

Final answer

Results table with correct mass headings, units in g, and consistent decimal places (candidate readings)

Detailed explanation

Background Concept

This experiment uses the mass lost as carbon dioxide gas to find how much acid reacted. When sodium carbonate is added to sulfuric acid, the reaction produces carbon dioxide which escapes from the open flask, so the total mass of the flask and contents decreases. The decrease in mass is exactly the mass of CO2 released. Once the mass (and hence the amount) of CO2 is known, the amount and concentration of acid can be found from the 1:1 stoichiometry of the equation. A well-organised results table is needed because the method depends on several masses being measured and combined.

Understanding the Question

Part (a) asks you to prepare the results table for the experiment, not to give a numerical answer. The marks come from the structure of the table: correct headings, units, consistent precision, and the correct calculated quantities. The mark scheme especially rewards showing all four balance readings to the same degree of precision and using the right arithmetic to obtain the mass of FA 3 and the mass of CO2.

Approach

Think of the table as having two kinds of rows: direct readings and calculated values. The direct readings are the actual balance readings; the calculated values are obtained by subtraction. Make sure the heading states what is being measured and the unit is shown for every row.

Step-by-Step Reasoning

Record the mass of the conical flask plus the acid: this is one balance reading.
Record the mass of the container plus FA 3 before transfer.
After tipping FA 3 into the acidchers, reweigh the container with any residue: this second mass of the container lets you find how much FA 3 was actually transferred.
Calculate mass of FA 3 added = mass(container + FA 3) – mass(container with residue).
Calculate initial total mass = mass(flask + acid) + mass of FA 3 added.
After fizzing stops, weigh the flask and its contents and record this final mass.
Calculate mass of CO2 = initial total mass – final mass of flask and contents.

Every calculated row should be shown in the table, with the same units and the same number of decimal places as the readings.

Key Takeaways

A mass-loss titration/gas-evolution experiment depends on carefully recording all masses.
The mass of a transferred solid is found by weighing the container before and after transfer.
The mass of gas released is found by the difference between the initial total mass and the final mass.
A results table must have clear headings, units, and consistent precision.

Common Mistakes

Including no unit or putting the unit only in the first row.
Mixing 2-decimal-place and 3-decimal-place readings in the same table.
Forgetting to reweigh the container after transferring FA 3.
Calculating the mass of CO2 directly as mass of FA 3 instead of using the total mass difference.
Labelling a value with the wrong quantity, e.g. calling the mass loss 'mass of FA 3'.

Things to Be Careful About

The mark scheme requires all four specified balance readings to be shown and all readings to be given to either 2 dp or 3 dp.
The unit must be cited as /g or g in every relevant column or value.
The mass of CO2 must come only from the mass loss, not from any other mass in the table.

Techniques used

design a results table with headings and unitsrecord balance readings to consistent decimal placescalculate masses by differenceobtain the mass of carbon dioxide by mass loss

(b)

(i)

Calculate the amount, in mol, of carbon dioxide given off in the reaction.

amount of $\text{CO}_2$ = .............................. mol

1M Similar

DifficultyEasy

Worked solution

Working

\text{amount of CO}_2 = \frac{\text{mass of CO}_2}{M_r(\text{CO}_2)} = \frac{m(\text{CO}_2)}{44} \text{ mol}

where $m(\text{CO}_2)$ is the mass of carbon dioxide found in part (a).

Substitute your recorded mass and give the answer to 2 – 4 significant figures.

Answer

\frac{m(\text{CO}_2)}{44} \text{ mol}

Final answer

m(CO2) / 44 mol

Detailed explanation

Background Concept

The amount of a substance is related to its mass by $n = m/M$ , where $M$ is the molar mass in g mol^-1. For carbon dioxide, $M_r = 12 + 2(16) = 44$ , so its molar mass is 44 g mol^-1.

Understanding the Question

Part (b)(i) asks for the amount of CO2 produced. You have the mass of CO2 from the mass loss in part (a), so you simply divide that mass by 44.

Approach

Use the relationship amount = mass / molar mass. No stoichiometric ratio is needed here; the division by 44 is the key step.

Step-by-Step Reasoning

Write the mass of CO2 you recorded from part (a), e.g. as $m(\text{CO}_2)$ .
Calculate $M_r(\text{CO}_2) = 12 + 16 + 16 = 44$ .
Substitute into $n = m/M$ to get $n(\text{CO}_2) = m(\text{CO}_2)/44$ .
Round the final value to 2 – 4 significant figures, because the balance readings are only quoted to that precision.

Key Takeaways

The molar mass of carbon dioxide is 44 g mol^-1.
Moles of a gas can be obtained from the mass of gas released if the gas is collected as a mass loss.
The final numerical value should be quoted to a sensible number of significant figures.

Common Mistakes

Using $M = 44$ with a unit of dm^3 or thinking that CO2 occupies 24 dm^3 mol^-1 at r.t.p.; this part is about mass, not volume.
Forgetting to use the mass loss from part (a) and instead using the mass of FA 3.
Quoting a value with too many decimal places or using the calculator's unrounded answer slavishly when a 2 – 4 sf value is requested.

Things to Be Careful About

Answers in this paper are candidate-dependent; replace $m(\text{CO}_2)$ with your own measured mass.
The unit is mol, and it must be shown.
The mark scheme requires the final value to be given to 2 – 4 sf.

Techniques used

calculate amount using n = m/Mapply the molar mass of carbon dioxidegive the answer to an appropriate number of significant figures

(ii)

The sodium carbonate, FA 3, was in excess in the reaction with sulfuric acid. Show by calculation that the sodium carbonate was in excess. Use your answer to (b)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{amount of Na}_2\text{CO}_3 \text{ used} = \frac{\text{mass of FA 3}}{M_r(\text{Na}_2\text{CO}_3)} = \frac{m(\text{FA 3})}{106} \text{ mol}

From the equation, 1 mol $\text{Na}_2\text{CO}_3$ produces 1 mol $\text{CO}_2$ . Hence the amount of $\text{Na}_2\text{CO}_3$ needed is exactly the amount of $\text{CO}_2$ calculated in part (b)(i).

Compare the amounts:

\frac{m(\text{FA 3})}{106} > \text{amount of CO}_2

Therefore some sodium carbonate remains unreacted, so FA 3 was in excess.

Answer

Moles of Na2CO3 used = m(FA 3)/106; this is greater than the moles of CO2 formed, so Na2CO3 is in excess.

Final answer

m(FA3)/106 > amount of CO2, so Na2CO3 is in excess

Detailed explanation

Background Concept

A reagent is in excess when more moles of it are present than are required by the stoichiometry of the reaction. In the equation Na2CO3 + H2SO4 -> Na2SO4 + CO2 + H2O, one mole of sodium carbonate reacts with one mole of sulfuric acid and produces one mole of carbon dioxide.

Understanding the Question

Part (b)(ii) asks you to prove, using your own mass data, that the sodium carbonate was in excess. The calculation must use the mass of FA 3 from part (a) and the amount of CO2 from part (b)(i).

Approach

Convert the mass of FA 3 into moles using the molar mass of sodium carbonate. Since the reaction is 1:1, compare those moles with the moles of CO2 already found. If moles Na2CO3 > moles CO2, the sodium carbonate is in excess.

Step-by-Step Reasoning

Work out $M_r(\text{Na}_2\text{CO}_3) = 2(23) + 12 + 3(16) = 106$ .
Convert the mass of FA 3 used to moles: $n = m/106$ .
Use part (b)(i) to write the amount of CO2: this is also the amount of Na2CO3 needed, because one mole of Na2CO3 gives one mole of CO2.
Compare the two amounts and state that there is enough sodium carbonate, and some remains unreacted, so it is in excess.

Key Takeaways

Molar mass of sodium carbonate is 106 g mol^-1.
The stoichiometric ratio between Na2CO3 and CO2 is 1:1 here.
To prove a reagent is in excess, compare the actual moles present with the moles required by the reaction.

Common Mistakes

Using the wrong molar mass, e.g. forgetting the waters of hydration for a carbonate that is actually anhydrous Na2CO3.
Comparing masses instead of moles.
Stating that FA 3 is in excess without showing the comparison of amounts.
Thinking that the acid must be in excess because all the solid disappears; here the solid carbonate is explicitly stated to be in excess.

Things to Be Careful About

The mark scheme gives M1 for correct use of 106, and M2 for the comparison of moles/masses of Na2CO3 used versus needed, or a statement that moles Na2CO3 > moles CO2.
The calculation must be complete enough to earn both marks; do not just write a conclusion.

Techniques used

calculate moles of sodium carbonate using molar mass 106compare stoichiometric amounts in the 1:1 reactionidentify the excess reagent

(iii)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of sulfuric acid in FA 2.

concentration of $\text{H}_2\text{SO}_4$ = .............................. $\text{mol dm}^{-3}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation, 1 mol $\text{H}_2\text{SO}_4$ produces 1 mol $\text{CO}_2$ , so:

\text{amount of H}_2\text{SO}_4 \text{ in } 25.0 \text{ cm}^3 = \text{amount of CO}_2 = n \text{ mol}

\text{concentration} = \frac{n}{25.0/1000} = \frac{n \times 1000}{25.0} = 40n \text{ mol dm}^{-3}

Substitute the value of $n$ from part (b)(i) and give the final answer to 2 – 4 significant figures.

Answer

\frac{n \times 1000}{25} \text{ mol dm}^{-3}

Final answer

n × 1000/25 mol dm^-3

Detailed explanation

Background Concept

Concentration is amount of solute per unit volume of solution: $c = n/V$ , with volume in dm^3. Since 1 dm^3 = 1000 cm^3, a volume in cm^3 is converted to dm^3 by dividing by 1000.

Understanding the Question

This part wants the concentration of sulfuric acid in FA 2. The only data you have are the amount of CO2 from part (b)(i) and the volume of acid used, 25.0 cm^3. The balanced equation shows H2SO4 and CO2 react 1:1, so the amount of acid in the 25.0 cm^3 sample equals the amount of CO2.

Approach

First recognise that moles H2SO4 = moles CO2. Then divide this amount by the volume of acid in dm^3. Alternatively, multiply the amount in moles by 1000/25, because 25.0 cm^3 is 0.0250 dm^3.

Step-by-Step Reasoning

Use $n = n(\text{CO}_2)$ from part (b)(i).
Convert $25.0 \text{ cm}^3$ to $25.0/1000 = 0.0250 \text{ dm}^3$ .
Apply $c = n/V$ : $c = n/0.0250 = 40n$ .
Quote the result in $\text{mol dm}^{-3}$ to 2 – 4 sf.

Key Takeaways

Acid-carbonate reactions commonly have a 1:1 stoichiometry between acid and CO2.
Concentration requires amount in mol and volume in dm3.
The factor 1000/25 = 40 arises from converting 25 cm3 into dm3.

Common Mistakes

Forgetting to divide by 1000 and leaving the volume in cm3.
Using a 2:1 or 1:2 ratio when the equation gives 1:1.
Giving the concentration without a unit or with a unit of g dm^-3.
Rounding to too many significant figures.

Things to Be Careful About

The mark scheme expects the calculation to use the amount of H2SO4 in 25 cm3 equal to the amount of CO2 from part (b)(i).
Answer must be in mol dm^-3.
If you made an earlier arithmetic error in part (b)(i), later work here is normally awarded ecf as long as the method is correct.

Techniques used

use the 1:1 stoichiometric ratioconvert 25.0 cm3 to dm3calculate concentration in mol dm^-3

(c)

(i)

A student does not have a conical flask and uses a small beaker for the reaction. Explain why a conical flask is better.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

When carbonate reacts with acid the mixture froths / effervesces rapidly. A conical flask has a narrow neck, so it is much less likely that the reaction mixture is splashed or lost from the container. This ensures the only significant mass lost is carbon dioxide, not acid or reaction mixture, so the mass-loss measurement remains valid.

A beaker has a wide open mouth, so the effervescing mixture could overflow or splash out, causing a loss of material and making the calculated mass of CO2 too large.

Final answer

Narrow neck of conical flask prevents loss of frothing/effervescing mixture, so only CO2 is lost

Detailed explanation

Background Concept

The experiment relies on measuring the mass lost as CO2 gas. For this to work, the only matter that should leave the apparatus is carbon dioxide. Any liquid or solid lost by splashing would be recorded as if it were CO2, making the apparent mass of CO2 too large.

Understanding the Question

The question asks why a conical flask is better than a small beaker. The relevant feature is the shape: the conical flask has a relatively narrow neck, the beaker has a wide open top. You need to connect this shape to the vigorous release of gas.

Approach

Think about what happens during the reaction: bubbles of CO2 form rapidly rathe reaction mixture. A narrow opening confines the bubbles and any spray; a wide opening lets the froth escape more easily.

Step-by-Step Reasoning

Acid + carbonate produces CO2 rapidly, causing frothing/effervescence.
In a beaker, the wide mouth makes it easy for the froth to overflow or for drops to be spat out.
Any lost liquid/solid counts as apparent CO2 mass loss, giving an inaccurate result.
A conical flask's narrow neck keeps the mixture inside, so the only mass loss is due to gas escaping.

Key Takeaways

The success of a mass-loss experiment depends on ensuring that only the gas is lost.
Apparatus shape can be as important as the reagents in controlling systematic error.

Common Mistakes

Saying only that a conical flask is 'safer' without linking it to mass loss.
Saying the conical flask 'collects the gas' since here the gas is allowed to escape and is measured by mass loss.
Missing the idea that splashing/loss of reaction mixture would affect the final mass.

Things to Be Careful About

The mark scheme explicitly rewards the idea that the mixture froths/effervesces and that without a narrow neck some mixture is lost/overflows.
Mention both the cause (effervescence) and the consequence (loss of mixture) for full marks.

Techniques used

relate apparatus shape to loss of massidentify the risk of frothing or overflow

(ii)

Two students made suggestions of how they thought the experiment in (a) could be adapted to determine the concentration of sulfuric acid, FA 2, by using other reactions. In each case their teacher told them that this method was not suitable.

Explain, in each case, why the method is not suitable. Do not consider factors based on quantities of any reagent.

Student 1 suggested using magnesium in place of sodium carbonate.

Student 2 suggested using calcium carbonate in place of sodium carbonate.

3M Similar

DifficultyMedium

Worked solution

Answer

Student 1 – magnesium

$\text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2$

The gas produced is hydrogen, not carbon dioxide. Hydrogen has a very low density, so the mass of gas lost is very small. The uncertainty in measuring such a small mass loss gives a large percentage error, so the method is not reliable.

Student 2 – calcium carbonate

$\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O}$

Calcium sulfate is insoluble / sparingly soluble. It forms a coating on the calcium carbonate solid and prevents the acid from reaching the remaining carbonate. The reaction therefore stops before completion, so the mass loss is not a true measure of the amount of acid present.

Final answer

Mg gives very light H2 causing a large % error; CaSO4 coats CaCO3 and stops the reaction

Detailed explanation

Background Concept

The whole method treats the mass lost by the flask as the mass of carbon dioxide evolved. For a gas-evolution mass-loss method to be usable, two conditions are needed: the reaction should go to completion, and the evolved gas should have a reasonably large molar mass so that the mass loss is not too small to measure accurately.

Understanding the Question

Two alternative reactions are suggested. For each, you must explain why the method is not suitable, without referring to quantities of reagents. The reasons come from the properties of the gas produced and the properties of the solid product.

Approach

For Student 1, focus on the gas: magnesium with acid gives H2, which has a very low density. Since the mass-loss method relies on measuring a change in mass, a very light gas gives a tiny mass loss and hence a large percentage error.

For Student 2, focus on the solid product: CaCO3 with H2SO4 gives CaSO4, which is insoluble. An insoluble solid can coat the remaining CaCO3, preventing further reaction and making the reaction incomplete.

Step-by-Step Reasoning

Student 1: Write the reaction with magnesium. The gas is H2, not CO2.
Hydrogen has a very low density / very small mass per mole.
Therefore the measured mass loss is small and the experimental percentage error is large.
Student 2: Write the reaction with calcium carbonate. The product is CaSO4.
CaSO4 is insoluble or sparingly soluble in water.
It coats the unreacted CaCO3, stopping the acid from reaching it, so the reaction may not go to completion.

Key Takeaways

A gas-evolution method works best when the gas has a reasonable molar mass so its mass loss is measurable.
Insoluble products can coat reactants and stop a reaction reaching completion.
When evaluating a method, connect the chemical property directly to the measurement being made.

Common Mistakes

For Student 1, saying only that H2 is produced without linking the low density to the large percentage error.
For Student 2, saying calcium sulfate 'is insoluble' but not explaining the effect on the reaction.
Confusing 'sulfate is insoluble' with 'sulfuric acid stops the reaction' or mentioning quantities when the question forbids that.

Things to Be Careful About

The mark scheme gives M1 for hydrogen having a very low density, M2 for the link between small mass loss and large percentage error, and M3 for calcium sulfate being insoluble/sparingly soluble or coating the solid and preventing further reaction.
Do not discuss quantities of reagents; the reasons must be based on the chemistry of the alternatives.

Techniques used

evaluate alternative gas-evolution methodsrelate low gas density to a large percentage errorexplain how an insoluble product stops the reaction

(d)

State the uncertainty in a single reading of your balance.

uncertainty = $\pm$ .............................. g

Calculate the maximum percentage error in the mass of FA 3 that you weighed out in (a).

maximum percentage error = .............................. %

1M Similar

DifficultyMedium-Easy

Worked solution

Working

If balance readings are recorded to 2 decimal places, the uncertainty in a single reading is $U = \pm 0.01$ g (or $\pm 0.005$ g). If they are recorded to 3 decimal places, $U = \pm 0.001$ g (or $\pm 0.0005$ g).

The mass of FA 3 is a difference of two balance readings, so its uncertainty is $2U$ .

\text{maximum percentage error} = \frac{2U}{\text{mass of FA 3}} \times 100\%

Substitute your value of the mass of FA 3 from part (a).

Answer

\frac{2U}{\text{mass FA 3}} \times 100\%

where $U$ is the single-reading balance uncertainty.

Final answer

2U / mass of FA3 × 100%, with U = ±0.01 g (or ±0.005 g) for a 2 dp balance

Detailed explanation

Background Concept

Every measuring instrument has an uncertainty. For a digital balance, the uncertainty of a single reading is often either the smallest division shown (0.01 g for a 2 dp balance) or half that division (0.005 g). When a quantity is obtained by subtracting two readings, the uncertainties of both readings add, so the uncertainty in the difference is twice the uncertainty of one reading.

Understanding the Question

You are asked two things: state the uncertainty in one balance reading, and calculate the maximum percentage error in the mass of FA 3 weighed out. The mass of FA 3 comes from the difference between two readings: (mass of container + FA 3) – (mass of container after), so its uncertainty is double the single-reading uncertainty.

Approach

Choose the uncertainty appropriate to the precision shown by the balance. Then apply the formula for percentage error: absolute uncertainty divided by the measured value, multiplied by 100. Remember to use the absolute uncertainty of the difference, which is $2U$ .

Step-by-Step Reasoning

Decide the precision of the balance from the readings recorded in part (a), e.g. 2 dp means readings such as 12.34 g.
State the single-reading uncertainty: $U = 0.01$ g or $0.005$ g for 2 dp; $0.001$ g or $0.0005$ g for 3 dp.
Recognise that the mass of FA 3 is found from two balance readings, so its absolute uncertainty is $2U$ .
Use percentage error = (absolute uncertainty / measured mass) × 100.
If, for example, $U = 0.01$ g and the mass of FA 3 is 2.50 g, the maximum percentage error is $(0.02/2.50) × 100 = 0.8\%$ .

Key Takeaways

A digital balance's uncertainty is either its smallest division or half that division.
A quantity obtained by a difference of two readings carries twice the single-reading uncertainty.
Percentage error is absolute uncertainty divided by the measured value, then multiplied by 100.

Common Mistakes

Forgetting the factor of 2, i.e. using $U$ instead of $2U$ for a mass obtained by difference.
Giving the uncertainty of the balance as a volume or as a concentration.
Omitting the ×100 when calculating a percentage.
Using the wrong value of mass, e.g. using the mass of acid instead of the mass of FA 3.

Things to Be Careful About

The mark scheme allows either 0.01 g or 0.005 g for a 2 dp balance and either 0.001 g or 0.0005 g for a 3 dp balance.
The percentage error must be expressed as a percentage and should be based on the mass of FA 3 weighed out in part (a).
If you carry an incorrect mass from part (a), apply ecf so the method still scores.

Techniques used

state the uncertainty of a single balance readingdouble the uncertainty for a mass found by differencecalculate a maximum percentage error

Q3MediumQualitative AnalysisManipulation, Measurement and ObservationAnalysis, 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.

FA 4, FA 5 and FA 6 are compounds of the same metal in different oxidation states.

(a)

(i)

Place a small spatula measure of FA 4 in a hard-glass test-tube. Heat the tube gently at first and then more strongly.

Record all your observations.

Leave the tube and contents to cool and keep for use in (a)(iii).

2M Similar

DifficultyMedium

Worked solution

Answer

Observations on heating FA 4 (potassium manganate(VII)):

Dark purple crystals/solid initially; the solid crackles/pops and jumps about on heating.
On stronger heating the purple solid turns to a black powder/residue (black smoke seen).
A gas is evolved which relights a glowing splint (oxygen).

2\text{KMnO}_4(\text{s}) \rightarrow \text{K}_2\text{MnO}_4(\text{s}) + \text{MnO}_2(\text{s}) + \text{O}_2(\text{g})

Final answer

Purple crystals crackle on heating, forming a black powder/residue; gas evolved relights a glowing splint (oxygen).

Detailed explanation

Background Concept

Potassium manganate(VII), KMnO4, is a dark purple crystalline solid containing manganese in the +7 oxidation state. On heating it decomposes, giving oxygen gas and manganese in lower oxidation states (manganate(VI), K2MnO4, which is dark green, and MnO2, which is black). Oxygen is identified by its ability to relight a glowing (not burning) splint — the classic test for O2.

Understanding the Question

This is a Paper 3 qualitative analysis task. You physically heat a small spatula measure of FA 4, gently at first then strongly, in a hard-glass test-tube (required because a solid is being heated), and record every observation at the stage it occurs. The mark scheme awards marks in pairs (2 observations = 1 mark, rounded down), so quantity and precision of observations matter.

Approach

Watch the solid throughout: its initial appearance, any physical behaviour (crackling), any colour change on heating, and any gas evolved. Test any gas with a glowing splint. Record each observation at the correct stage — 'initially', 'on stronger heating', 'gas evolved'.

Step-by-Step Reasoning

FA 4 is dark purple crystals. As heating begins, the crystals often crackle/pop and jump about (this is a creditable observation).
On stronger heating the purple colour is lost and a black powder/residue forms (MnO2); black smoke may be seen.
Oxygen is released; a glowing splint inserted into the tube relights. This is the gas identification mark.
The underlying chemistry: 2KMnO4 -> K2MnO4 + MnO2 + O2, a redox decomposition in which Mn goes from +7 to +6 and +4.

Key Takeaways

Thermal decomposition of KMnO4 produces O2 and black MnO2-containing residue.
Observations must be recorded at the correct stage of the test.
Marks are awarded per pair of correct observations (2* = 1 mark, round down), so list all of them.

Common Mistakes

Writing only 'black solid forms' with no initial purple observation — you lose the pairing mark.
Saying 'burns brightly' instead of 'relights a glowing splint' — the splint must be glowing, not lit.
Using a boiling tube instead of a hard-glass test-tube for heating a solid (instruction breach).
Writing 'smoke' without colour, or omitting the crackling/jumping observation.

Things to Be Careful About

Keep the residue for part (a)(iii) — do not discard it.
Use 'relights a glowing splint', not 'ignites' or 'burns'.
Record 'no change' only if genuinely nothing happens; here changes do occur.

Techniques used

heat a solid in a hard-glass test-tube and record stage-by-stage observationstest for oxygen with a glowing splintdescribe colour changes of a transition-metal compound on heating

(ii)

Dissolve a small spatula measure of FA 4 in approximately $5\text{ cm}$ depth of distilled water in a boiling tube and add approximately $1\text{ cm}$ depth of dilute sulfuric acid.

Carry out the following tests and record your observations in Table 3.1.

Table 3.1

test	observations
Test 1 To a $1\text{ cm}$ depth of aqueous FA 4 in a test-tube add aqueous iron(II) sulfate.
Test 2 To a $1\text{ cm}$ depth of aqueous FA 4 in a test-tube add hydrogen peroxide.

2M Similar

DifficultyMedium

Worked solution

Answer

Test 1 — FA 4 solution + aqueous iron(II) sulfate:

The purple solution reacts and the mixture becomes a yellow solution (Fe^3+ formed as the purple MnO4^- is reduced).

Test 2 — FA 4 solution + hydrogen peroxide:

The purple solution becomes colourless (decolourises).
Effervescence/fizzing/bubbling; the gas evolved relights a glowing splint (oxygen).

Final answer

Test 1: purple solution turns yellow. Test 2: purple decolourises with effervescence; gas relights a glowing splint (O2).

Detailed explanation

Background Concept

Aqueous manganate(VII), MnO4^-, is an intense purple colour and a strong oxidising agent. Iron(II) ions, Fe^2+, are pale green and are oxidised to iron(III), Fe^3+, which gives yellow/brown solutions. Hydrogen peroxide can act as a reducing agent towards very strong oxidants like MnO4^- in neutral/alkaline conditions, reducing Mn(VII) to colourless Mn(II) (or MnO2) while itself being oxidised to O2, which is seen as effervescence.

Understanding the Question

FA 4 has been dissolved in water with a little dilute sulfuric acid. Two tests are performed on 1 cm depths of this purple solution, and all observations (colour changes, effervescence, gas tests) must be recorded in Table 3.1. Marks come in pairs (2 observations = 1 mark).

Approach

For each test, note the initial colour, the final colour of the solution, any bubbling, and test any gas with a glowing splint.

Step-by-Step Reasoning

Test 1: MnO4^- oxidises Fe^2+ to Fe^3+. The purple colour is consumed and the Fe^3+ formed gives a yellow solution. So: purple -> yellow solution.
Test 2: H2O2 reduces the MnO4^-; the purple solution decolourises (goes colourless). O2 gas is released, seen as effervescence, and the gas relights a glowing splint.

Key Takeaways

Purple MnO4^- losing its colour indicates reduction of manganese.
Yellow solution after adding Fe^2+ indicates Fe^3+ formation.
Effervescence + relighting a glowing splint = oxygen.

Common Mistakes

Writing 'clear' instead of 'colourless' — 'colourless solution' is the accepted term.
Omitting the gas test observation; 'fizzing' alone and 'relights glowing splint' are separate creditable points.
Saying the yellow is a precipitate — it is a solution colour.
Confusing which test gives which colour change.

Things to Be Careful About

Record observations, not conclusions: write 'purple solution becomes yellow', not 'Fe^2+ is oxidised'.
Note the stage: initial purple, then the change.

Techniques used

observe the oxidising action of manganate(VII) on iron(II) ionsobserve the reaction of manganate(VII) with hydrogen peroxidetest for oxygen with a glowing splint

(iii)

To the cooled test-tube in (a)(i) add a $5\text{ cm}$ depth of distilled water. Observe and record the colour formed.

1M Similar

(b)

Dissolve a small spatula measure of FA 5 in a boiling tube half-filled with distilled water. Warming may be needed to dissolve the FA 5.

Carry out the following tests and record your observations in Table 3.2.

For each of the tests use a $1\text{ cm}$ depth of this FA 5 solution in a test-tube.

Table 3.2

test	observations
Test 1 Add dilute nitric acid, then
add aqueous silver nitrate.
Test 2 Add aqueous barium chloride or barium nitrate, then
add dilute nitric acid.
Test 3 Add aqueous sodium hydroxide, then
add hydrogen peroxide.

3M Similar

DifficultyMedium

Worked solution

Answer

Test 1 — add dilute nitric acid, then aqueous silver nitrate:

With HNO3: no change.
With AgNO3(aq): no change (no precipitate — no halide present).

Test 2 — aqueous barium chloride/nitrate, then dilute nitric acid:

White precipitate forms with Ba^2+ (aq).
On adding dilute nitric acid: no change — the precipitate does not dissolve (sulfate confirmed).

Test 3 — aqueous sodium hydroxide, then hydrogen peroxide:

With NaOH(aq): an off-white/cream/fawn/buff/pale brown precipitate forms (Mn(OH)2), insoluble in excess; the precipitate darkens on standing.
On adding H2O2(aq): the precipitate turns black/dark brown (MnO2).
Effervescence/fizzing OR the gas relights a glowing splint (oxygen).

Final answer

Test 1: no change with HNO3 and with AgNO3. Test 2: white ppt insoluble in HNO3 (sulfate). Test 3: pale brown ppt darkening, turning black with H2O2, with effervescence (O2).

Detailed explanation

Background Concept

FA 5 is manganese(II) sulfate, MnSO4. Standard anion tests: sulfate ions give a white precipitate of BaSO4 with Ba^2+ ions, insoluble in dilute nitric acid; halides would give silver halide precipitates with AgNO3 (none here). Mn^2+ ions react with OH^- to give Mn(OH)2, an off-white/cream/pale brown precipitate that is readily oxidised by air or by H2O2 to black/dark brown MnO2. H2O2 decomposes (catalysed by manganese species) releasing O2, which relights a glowing splint.

Understanding the Question

Three tests are carried out on the FA 5 solution, each in two stages ('add X, then add Y'). Both stages of each test must be recorded — the mark scheme pairs observations, and 2 correct observations = 1 mark (round down), so recording every stage matters.

Approach

For each test, record the observation after the first reagent, then after the second. Where nothing happens, write 'no change' explicitly — this is itself creditable.

Step-by-Step Reasoning

Test 1: HNO3 is added first to acidify (ruling out carbonate/interference); no change. AgNO3 then gives no precipitate, showing no chloride/bromide/iodide — consistent with the anion being sulfate, not a halide.
Test 2: Ba^2+ gives a white precipitate, BaSO4. Adding dilute nitric acid does not dissolve it, confirming sulfate (a carbonate precipitate would effervesce and dissolve).
Test 3: OH^- precipitates Mn(OH)2, described as off-white/cream/fawn/buff/pale brown; it darkens as it is air-oxidised. Adding H2O2 oxidises the precipitate to black/dark brown MnO2, and O2 is evolved (effervescence; relights a glowing splint).

Key Takeaways

Sulfate test: Ba^2+ then white ppt insoluble in dilute acid.
'No change' is a valid, markable observation.
Mn(OH)2 is pale/off-white to pale brown and darkens; oxidation by H2O2 gives black MnO2 plus O2.

Common Mistakes

Omitting the 'no change' observations in Test 1 — both stages must be recorded.
Describing the Mn(OH)2 precipitate as simply 'white' or 'brown' — the accepted descriptors are off-white/cream/fawn/buff/pale brown.
Saying the white ppt 'dissolves' in acid — BaSO4 is insoluble in dilute nitric acid.
Forgetting to test the gas in Test 3 or to record the effervescence.

Things to Be Careful About

Record observations at the correct stage (before/after the second reagent).
Use 'precipitate (ppt)' and state its colour precisely; note insolubility in excess NaOH.
Use a 1 cm depth of solution for each test as instructed.

Techniques used

test for sulfate ions with barium ions and nitric acidtest for halide ions with silver nitrate and nitric acidobserve precipitation of manganese(II) hydroxide and its oxidation by hydrogen peroxide

(c)

Carry out the following tests and record your observations in Table 3.3. Identify any gases produced.

For each of the tests use a small spatula measure of FA 6 in a test-tube.

Table 3.3

test	observations and gases produced
Test 1 Add a $1\text{ cm}$ depth of dilute nitric acid.
Test 2 Add a few drops of concentrated hydrochloric acid. CARE Hydrochloric acid is corrosive. Fill the test-tube with water as soon as you have made your observation.
Test 3 Add a $1\text{ cm}$ depth of hydrogen peroxide.

3M Similar

DifficultyMedium

Worked solution

Answer

Test 1 — FA 6 (MnO2) + dilute nitric acid:

No change — the solid does not dissolve.

Test 2 — FA 6 + concentrated hydrochloric acid:

Bubbles/effervescence; the gas bleaches (moist) litmus — chlorine is produced.

Test 3 — FA 6 + hydrogen peroxide:

Effervescence; the gas relights a glowing splint — oxygen is produced.

Final answer

Test 1: no change. Test 2: bubbles, gas bleaches litmus (chlorine). Test 3: effervescence, gas relights glowing splint (oxygen).

Detailed explanation

Background Concept

FA 6 is manganese(IV) oxide, MnO2, a black insoluble solid. MnO2 is a strong oxidising agent: with concentrated hydrochloric acid it oxidises Cl^- to Cl2 (the laboratory preparation of chlorine), the gas bleaching moist litmus. MnO2 also catalyses the decomposition of hydrogen peroxide, 2H2O2 -> 2H2O + O2, giving vigorous effervescence of oxygen, which relights a glowing splint. MnO2 does not dissolve in dilute nitric acid, so Test 1 gives no change.

Understanding the Question

Three tests on solid FA 6, each with observations AND identification of any gas. Negative results ('no change') must still be recorded. The safety note warns that concentrated HCl is corrosive and that chlorine is toxic — fill the test-tube with water after Test 2.

Approach

For each test record: does the solid dissolve/React? Is gas evolved? Test the gas (bleaching of litmus for chlorine; glowing splint for oxygen) and name the gas.

Step-by-Step Reasoning

Test 1: MnO2 is insoluble and unreactive towards dilute HNO3, so the correct observation is 'no change / solid does not dissolve'.
Test 2: MnO2 + 4HCl(conc) -> MnCl2 + Cl2 + 2H2O. Bubbles form and the chlorine bleaches litmus — both the bubbling and the bleaching are creditable, and naming 'chlorine' is a separate mark.
Test 3: H2O2 decomposes on the MnO2 surface giving O2 — effervescence or a glowing splint relighting, plus the identification 'oxygen'.

Key Takeaways

MnO2 + conc. HCl is the classic laboratory preparation of chlorine.
MnO2 catalyses H2O2 decomposition — a classic oxygen-generation reaction.
Distinguish gas tests: chlorine bleaches damp litmus; oxygen relights a glowing splint.

Common Mistakes

Saying the solid 'dissolves' in dilute nitric acid — it does not.
Confusing the gas tests: oxygen relights a glowing splint; chlorine bleaches litmus (and smells of bleach).
Omitting the gas identification when the question explicitly asks for it.
Describing chlorine as 'green' gas observation only without the litmus test.

Things to Be Careful About

Concentrated HCl is corrosive and chlorine is toxic — work carefully and fill the tube with water immediately after the observation, as instructed.
Record 'no change' for Test 1; it earns a mark.

Techniques used

test gases evolved from reactions of a solid with acids and hydrogen peroxideidentify chlorine with litmus and oxygen with a glowing splint

(d)

(i)

Use your observations from the tests on FA 4, FA 5 and FA 6 to suggest the identity of the metal present in all 3 compounds.

Metal identity ..............................

1M Similar

DifficultyMedium

Worked solution

Answer

The metal is manganese (Mn).

Final answer

Manganese (Mn)

Detailed explanation

Background Concept

Manganese shows several oxidation states with characteristic colours: +7 purple (MnO4^-), +6 dark green (MnO4^2-), +4 black (MnO2), +2 very pale pink/off-white (Mn^2+). A compound that is purple and releases O2 on heating, gives a dark green solution after heating, forms a pale precipitate with NaOH that blackens with H2O2, and gives a black solid that generates Cl2 with conc. HCl and O2 with H2O2 — all point to manganese.

Understanding the Question

Using all observations from FA 4, FA 5 and FA 6 (stated to be compounds of the same metal in different oxidation states), name the metal. One mark.

Approach

Match each observation to known manganese chemistry: purple KMnO4, dark green manganate(VI), black MnO2, pale Mn(OH)2 darkening to black — the only common metal with this set of coloured states is manganese.

Step-by-Step Reasoning

FA 4 is a purple solid releasing O2 on heating and decolourising with H2O2: potassium manganate(VII), Mn in +7.
The heated residue gives a dark green solution: manganate(VI).
FA 5 gives a white ppt with Ba^2+ insoluble in acid (sulfate) and a pale ppt with NaOH that blackens with H2O2: MnSO4, Mn in +2.
FA 6 is a black solid producing Cl2 with conc. HCl and O2 with H2O2: MnO2, Mn in +4.
All three are manganese compounds.

Key Takeaways

Manganese's oxidation-state colour set (+7 purple, +6 green, +4 black, +2 pale) is a diagnostic fingerprint.

Common Mistakes

Guessing chromium (also coloured states) — chromium does not give the purple MnO4^- colour or the black MnO2 behaviour with HCl/H2O2 in this pattern.
Suggesting iron — Fe^2+/Fe^3+ colours (pale green/yellow) do not match the purple and black observations.

Things to Be Careful About

The answer must be the metal (manganese), not a compound name.

Techniques used

deduce the identity of a metal from its characteristic oxidation-state colours and reactions

(ii)

Identify the oxidation state of the metal in each compound.

FA 4 ..............................

FA 5 ..............................

FA 6 ..............................

2M Similar

DifficultyMedium

Worked solution

Answer

FA 4 (KMnO4): oxidation state of Mn = +7 (VII)
FA 5 (MnSO4): oxidation state of Mn = +2 (II)
FA 6 (MnO2): oxidation state of Mn = +4 (IV)

Final answer

FA 4: +7; FA 5: +2; FA 6: +4

Detailed explanation

Background Concept

Oxidation state rules: K is always +1, O is -2 (except peroxides), SO4^2- carries charge -2. The sum of oxidation states in a neutral compound is zero; in a polyatomic ion it equals the ion charge.

Understanding the Question

Assign the oxidation state of manganese in each of the three compounds, using both the formulae and the observed chemistry/colours. All three correct = 2 marks; any two correct = 1 mark.

Approach

For each compound, apply oxidation-state bookkeeping, then confirm against the observed colours (purple = +7, pale/off-white = +2, black = +4).

Step-by-Step Reasoning

FA 4, KMnO4: K is +1, each O is -2 (4 x -2 = -8), so Mn = +7. This is the purple manganate(VII) ion, consistent with the purple solid and its strong oxidising behaviour.
FA 5, MnSO4: SO4 is 2-, so Mn = +2. Consistent with the pale/off-white Mn(OH)2 precipitate in Test 3 of (b).
FA 6, MnO2: each O is -2 (total -4), so Mn = +4. Consistent with the black solid that oxidises Cl^- to Cl2 and catalyses H2O2 decomposition.

Key Takeaways

Oxidation states are deduced from formulae using standard rules and confirmed by characteristic colours.
Manganese commonly shows +2, +4, +6 and +7.

Common Mistakes

Writing 'VII' without the sign or confusing oxidation state with charge notation — either +7 or VII is accepted, but be consistent.
Assigning FA 5 as +6 by confusing MnSO4 with manganate(VI), K2MnO4.
Forgetting that in MnO2 the two oxygens give -4, not -2.

Things to Be Careful About

All three must be correct for 2 marks — check each one.
Use the observations (purple, dark green, pale ppt, black solid) as a cross-check on your arithmetic.

Techniques used

deduce oxidation states from compound formulae and colourslink observed colours to manganese oxidation states