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) Candidate-dependent observations; expected: condensation, purple vapour, yellow solid pales on cooling (Test 1); effervescence, CO2 turns limewater milky, colourless/pale yellow solution (Test 2). · (a)(ii) Candidate-dependent observations; expected: brown solution then colourless (Test 1); white ppt soluble in excess NaOH and NH3 (Tests 2 & 3); ppt/solution with CuSO4 then white ppt or blue solution (Test 4); pale yellow ppt insoluble in NH3 (Test 5). · (a)(iii)…

Chemistry 9701/33 — May/June 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 EvaluationQualitative Analysis

When ammonium compounds are heated with sodium hydroxide or calcium hydroxide, ammonia is liberated. The ionic equation for the reaction is shown.

\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}

FA 1 is prepared by heating $10.0\text{ cm}^3$ of aqueous ammonium chloride, $\text{NH}_4\text{Cl}(\text{aq})$ , with $25.0\text{ cm}^3$ of $2.00\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ . The sodium hydroxide is in excess. The reaction mixture is cooled and diluted to $250\text{ cm}^3$ with distilled water.

You will determine the concentration of the aqueous solution of ammonium chloride by titrating the remaining sodium hydroxide from the preparation of FA 1 with a known concentration of sulfuric acid.

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

FA 2 is $0.0520\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .
FA 3 is bromophenol blue indicator.

(a)

Method

Fill the burette with FA 2.
Pipette $25.0\text{ cm}^3$ of FA 1 into a conical flask.
Add a few drops of FA 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 FA 2 added in each accurate titration.

Results

7M Similar

DifficultyEasy

Worked solution

Answer

Record the rough titre and carry out accurate titrations, repeating until at least two results are concordant.

Use a table whose headings include units, for example:

Titration	initial burette reading / $\text{cm}^3$	final burette reading / $\text{cm}^3$	titre / $\text{cm}^3$
rough	0.00	13.10	13.10
accurate 1	0.00	12.50	12.50
accurate 2	12.50	25.00	12.50

The actual readings are the candidate's own; all burette readings should be to the nearest $0.05\ \text{cm}^3$ , and the accurate titres should agree to within $0.10\ \text{cm}^3$ .

Final answer

A table of rough and accurate burette readings in cm^3, readings to nearest 0.05 cm^3 and at least two concordant accurate titres; candidate-dependent values.

Detailed explanation

Background Concept

In a titration the burette delivers a variable volume of FA 2 and the volume needed to just complete the reaction is the titre. Because the burette scale is marked so that the eye can read to $0.05\ \text{cm}^3$ , all readings should be recorded to the nearest $0.05\ \text{cm}^3$ . Reliable conclusions need repeated consistent results: two or more accurate titres that agree within $0.10\ \text{cm}^3$ are called concordant, and only concordant titres are used to calculate the mean.

Understanding the Question

This section asks you to perform the titration of FA 1 (the diluted reaction mixture containing excess NaOH) against FA 2 ( $0.0520\ \text{mol dm}^{-3}$ $\text{H}_2\text{SO}_4$ ) and to present the data properly. It is not asking for calculations yet. The examiner checks that the rough titre is recorded, that the accurate table has initial and final readings and a titre for each run, that headings have units, that readings show correct precision, and that you have repeated until results are consistent.

Approach

Carry out one rough titration to find the approximate end point, then repeat as many accurate titrations as needed. Use bromophenol blue indicator FA 3. While recording, keep a clear table with headings and units. After the titrations, check which accurate titres are concordant before averaging in part (b).

Step-by-Step Reasoning

Fill the burette with FA 2 and record the initial burette reading.
Pipette $25.0\ \text{cm}^3$ of FA 1 into a conical flask and add a few drops of FA 3.
Run in FA 2 fairly quickly for the rough titration, swirling the flask. When the indicator changes sharply, stop and record the final reading; the rough titre is final minus initial.
Repeat for accurate titrations, but near the end point add FA 2 dropwise.
For each accurate run record initial and final readings, then calculate the titre.
Aim for at least two accurate titres within $0.10\ \text{cm}^3$ of one another.

Key Takeaways

Correct recording is as important as the final answer: table headings must include quantities and units, readings must be to nearest $0.05\ \text{cm}^3$ , and at least two concordant accurate titres are needed to justify a mean.

Common Mistakes

Recording only the titre without the initial and final burette readings.
Leaving units out of the table headings.
Writing readings such as 12.53, which a burette cannot give; readings are to the nearest $0.05\ \text{cm}^3$ .
Averaging the rough titre with accurate titres.
Taking only one accurate titre; at least two concordant results are needed.

Things to Be Careful About

The rough titre is used only as a guide and should not be included in the mean. Accurate readings must have the subtraction checked (titre = final – initial). The unit $\text{cm}^3$ must appear for the readings, the titre on the answer line, and in each table heading; an unheaded column of numbers loses marks.

Techniques used

set up a burette and carry out a titrationrecord initial and final burette readingscalculate titres by subtractionrepeat accurate titrations until concordant results are obtainedconstruct a results table with headings and units

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

1M Similar

DifficultyEasy

Worked solution

Working

The two concordant accurate titres are $12.50$ and $12.50\ \text{cm}^3$ .

\text{mean titre} = \frac{12.50 + 12.50}{2} = 12.50\ \text{cm}^3

Answer

$25.0\ \text{cm}^3$ of FA 1 required $12.50\ \text{cm}^3$ of FA 2.

Final answer

12.50 cm^3 (example; candidate-dependent)

Detailed explanation

Background Concept

A mean titre is only meaningful if it is calculated from accurate, concordant results. The rough titre is ignored. The accepted mean is taken from at least two accurate titres that lie within a small spread (commonly $0.20\ \text{cm}^3$ total, and often $0.10\ \text{cm}^3$ ). The mean is quoted to two decimal places.

Understanding the Question

Part (b) asks you to take your own accurate titre values from part (a), choose the best concordant ones, and show the averaging that gives the value you will use in all later calculations. The line to fill in is simply the mean volume of FA 2 needed by $25.0\ \text{cm}^3$ of FA 1.

Approach

Look at your accurate titres, select two or more that agree best, add them, and divide by the number of readings. Show the working, then round the result to the nearest $0.01\ \text{cm}^3$ .

Step-by-Step Reasoning

Identify the accurate readings that are concordant. In the example, two titres are both $12.50\ \text{cm}^3$ .
Add them and divide by the number used:

\frac{12.50 + 12.50}{2} = 12.50\ \text{cm}^3

Quote the mean to two decimal places. Here it is already $12.50$ .

Key Takeaways

The mean used in calculations must come from the concordant accurate titres, with working shown. If your titres were $12.50$ and $12.55$ , the mean would be $12.525 \rightarrow 12.53$ ; the display of the calculation is part of the mark.

Common Mistakes

Including the rough titre in the mean.
Averaging all accurate titres even when some are wildly different.
Not showing which readings were averaged.
Quoting the mean to too many decimal places, e.g. $12.5250$ , when two decimal places are expected.

Things to Be Careful About

The mean must be to two decimal places. Check whether you have two or more titres within the required spread; if not, you should have repeated the titration in part (a). Any wrong titre chosen as the mean will be carried forward into the later calculations.

Techniques used

select concordant accurate titrescalculate a mean titrequote the mean to two decimal places

(c)

Calculations

(i)

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

1M Similar

(ii)

Calculate the amount, in mol, of sulfuric acid present in the volume of FA 2 in (b).

amount of $\text{H}_2\text{SO}_4$ = .............................. mol

1M Similar

DifficultyEasy

Worked solution

Working

Using the mean titre from (b), $12.50\ \text{cm}^3$ :

n(\text{H}_2\text{SO}_4) = \frac{0.0520 \times 12.50}{1000} = 6.50 \times 10^{-4}\ \text{mol}

Answer

$6.50 \times 10^{-4}$ mol

Final answer

6.50 × 10^-4 mol

Detailed explanation

Background Concept

The amount of a solute in a solution is given by $n = cV$ , where $c$ is concentration in $\text{mol dm}^{-3}$ and $V$ is volume in $\text{dm}^3$ . Since burette volumes are measured in $\text{cm}^3$ , they must be divided by 1000 before using the formula.

Understanding the Question

Part (c)(ii) asks for the amount of sulfuric acid in the mean titre of FA 2 found in part (b). This is the first calculation; all later steps depend on it.

Approach

Take the mean titre, convert it to $\text{dm}^3$ by dividing by 1000, and multiply by the concentration of FA 2, $0.0520\ \text{mol dm}^{-3}$ .

Step-by-Step Reasoning

$V = 12.50\ \text{cm}^3 = 12.50/1000 = 0.01250\ \text{dm}^3$ .
Use $n = cV$ :

n = 0.0520 \times 0.01250 = 6.50 \times 10^{-4}\ \text{mol}

Quote the answer to 3 significant figures: $6.50 \times 10^{-4}$ mol.

Key Takeaways

Always convert volumes to $\text{dm}^3$ in stoichiometric calculations. The relation between moles, concentration and volume is the foundation of every titration calculation.

Common Mistakes

Forgetting to divide the volume by 1000.
Quoting $0.00065$ or $6.5 \times 10^{-4}$ (2 s.f.) instead of $6.50 \times 10^{-4}$ (3 s.f.).
Using a volume other than the mean titre, such as the rough titre.

Things to Be Careful About

If your part (b) titre is different, your result will differ; the examiner allows error carried forward, so use your own value consistently. Write the final unit as mol.

Techniques used

calculate moles of H2SO4 from concentration and volumeconvert volume in cm^3 to dm^3apply significant figure rules

(iii)

Use your answer to (c)(ii) to calculate the amount, in mol, of sodium hydroxide in $25.0\text{ cm}^3$ of FA 1.

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The balanced equation is $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ , so 2 mol NaOH react with 1 mol $\text{H}_2\text{SO}_4$ .

n(\text{NaOH in }25.0\ \text{cm}^3\ \text{FA 1}) = 2 \times 6.50 \times 10^{-4} = 1.30 \times 10^{-3}\ \text{mol}

Answer

$1.30 \times 10^{-3}$ mol (in $25.0\ \text{cm}^3$ of FA 1)

Final answer

1.30 × 10^-3 mol

Detailed explanation

Background Concept

The titration reaction is

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

so the mole ratio of NaOH to H2SO4 is 2:1. For every mole of acid used in the titration, twice that number of moles of NaOH were present in the FA 1 sample titrated.

Understanding the Question

Part (c)(iii) asks how much NaOH remains in the $25.0\ \text{cm}^3$ sample of FA 1 that was titrated. The sulfuric acid titre measures exactly that remaining NaOH.

Approach

Multiply the amount of H2SO4 from (c)(ii) by 2, because the equation requires two moles of NaOH for every mole of acid.

Step-by-Step Reasoning

From (c)(ii), $n(\text{H}_2\text{SO}_4) = 6.50 \times 10^{-4}$ mol.
Ratio: $n(\text{NaOH}) = 2 \times n(\text{H}_2\text{SO}_4)$ .
$n(\text{NaOH}) = 2 \times 6.50 \times 10^{-4} = 1.30 \times 10^{-3}$ mol.
This is the NaOH remaining in the $25.0\ \text{cm}^3$ sample, not in the whole 250 $\text{cm}^3$ of FA 1.

Key Takeaways

The stoichiometric ratio comes directly from the balanced equation. It is useful to write the equation before doing the calculation.

Common Mistakes

Dividing by 2 instead of multiplying by 2.
Forgetting that this amount belongs to a $25.0\ \text{cm}^3$ sample only; the next part needs scaling to the full volume.
Using a different ratio because the equation was not written down.

Things to Be Careful About

Use your own value from (c)(ii) even if it differs from the example shown; the mark scheme will allow error carried forward. Keep the answer to 3 significant figures.

Techniques used

use the stoichiometric ratio from the balanced equationmultiply moles of H2SO4 by 2

(iv)

Use your answer to (c)(iii) and the information given to calculate the amount, in mol, of sodium hydroxide that reacted with the ammonium chloride when FA 1 was prepared.

amount of $\text{NaOH}$ that reacted with $\text{NH}_4\text{Cl}$ = .............................. mol

2M Similar

DifficultyMedium

Worked solution

Working

Initial amount of NaOH added before dilution:

n(\text{NaOH})_{\text{initial}} = 2.00 \times \frac{25.0}{1000} = 0.0500\ \text{mol}

FA 1 has a total volume of $250\ \text{cm}^3$ , so the NaOH remaining in the whole FA 1 is (c)(iii) $\times$ 10:

n(\text{NaOH})_{\text{remaining}} = 1.30 \times 10^{-3} \times 10 = 0.0130\ \text{mol}

Amount that reacted with ammonium chloride:

n(\text{NaOH reacted}) = 0.0500 - 0.0130 = 3.70 \times 10^{-2}\ \text{mol}

Answer

$3.70 \times 10^{-2}$ mol

Final answer

3.70 × 10^-2 mol

Detailed explanation

Background Concept

When FA 1 was prepared, $25.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ NaOH was added to the ammonium chloride and the mixture made up to $250\ \text{cm}^3$ . Dilution does not change the total amount of NaOH; it only spreads it through a bigger volume. A $25.0\ \text{cm}^3$ sample is therefore one tenth of the whole FA 1.

Understanding the Question

This part asks for the amount of NaOH that actually reacted with ammonium chloride during the preparation of FA 1. Since NaOH was in excess, the amount that reacted equals the initial amount added minus the amount that remained unreacted.

Approach

The value from (c)(iii) is the NaOH left in only $25.0\ \text{cm}^3$ of FA 1. Multiply it by 10 to find the total NaOH left in the full $250\ \text{cm}^3$ . Then subtract this total from the initial $0.0500$ mol of NaOH.

Step-by-Step Reasoning

Initial NaOH:

2.00 \times \frac{25.0}{1000} = 0.0500\ \text{mol}

Total NaOH remaining in FA 1:

1.30 \times 10^{-3} \times 10 = 0.0130\ \text{mol}

NaOH consumed by the ammonium chloride:

0.0500 - 0.0130 = 3.70 \times 10^{-2}\ \text{mol}

This is also the amount of ammonium ions that reacted, because the equation $\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}$ consumes hydroxide and ammonium ions in a 1:1 ratio.

Key Takeaways

The dilution factor is essential: $25.0\ \text{cm}^3$ is one tenth of $250\ \text{cm}^3$ , so multiply by 10. The excess-NaOH method works because reacted NaOH = initial NaOH – leftover NaOH.

Common Mistakes

Forgetting to multiply (c)(iii) by 10, and subtracting $1.30 \times 10^{-3}$ from $0.0500$ instead of $0.0130$ .
Using the titration amount directly as the amount that reacted.
Confusing the $10.0\ \text{cm}^3$ of ammonium chloride solution with the $250\ \text{cm}^3$ volume of FA 1.

Things to Be Careful About

Show both the initial amount and the scaling clearly; the examiner gives one mark for the initial amount and one for the subtraction. Keep working to 3 significant figures. Use your own (c)(iii) value if different.

Techniques used

scale the sample amount to the total diluted volumecalculate initial moles of NaOH from concentration and volumesubtract remaining NaOH from initial NaOH

(v)

Use your answer to (c)(iv) and the information given to calculate the concentration, in $\text{mol dm}^{-3}$ , of the ammonium chloride solution used to prepare FA 1.

concentration of $\text{NH}_4\text{Cl}$ = .......................................... $\text{mol dm}^{-3}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The reaction is 1:1, so the amount of ammonium chloride originally present in $10.0\ \text{cm}^3$ equals the amount of NaOH that reacted:
$n(\text{NH}_4\text{Cl}) = 3.70 \times 10^{-2}$ mol.

c(\text{NH}_4\text{Cl}) = \frac{3.70 \times 10^{-2}}{10.0/1000} = \frac{3.70 \times 10^{-2} \times 1000}{10.0} = 3.70\ \text{mol dm}^{-3}

Answer

$3.70\ \text{mol dm}^{-3}$

Final answer

3.70 mol dm^-3

Detailed explanation

Background Concept

Concentration is amount of solute per unit volume of solution: $c = n/V$ . The volume must be in $\text{dm}^3$ . Here the ammonium chloride was present in the original $10.0\ \text{cm}^3$ solution before the NaOH was added; diluting to 250 $\text{cm}^3$ does not change the amount, only the concentration.

Understanding the Question

Part (c)(v) asks for the concentration of the ammonium chloride solution that was originally used to prepare FA 1. The key is to use the amount from (c)(iv) (which equals the amount of NH4Cl) and the original volume, $10.0\ \text{cm}^3$ , not the diluted volume.

Approach

Set the amount of NH4Cl equal to the amount of NaOH that reacted. Then divide by $10.0\ \text{cm}^3$ converted to $\text{dm}^3$ to get the concentration.

Step-by-Step Reasoning

$\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}$ : 1 mol NH4+ needs 1 mol OH-, so $n(\text{NH}_4\text{Cl}) = n(\text{NaOH reacted}) = 3.70 \times 10^{-2}$ mol.
Convert the original volume: $10.0\ \text{cm}^3 = 0.0100\ \text{dm}^3$ .
Concentration:

\frac{3.70 \times 10^{-2}}{0.0100} = 3.70\ \text{mol dm}^{-3}

Key Takeaways

The dilution step creates a usable solution for titration, but the concentration being asked for refers to the original, undiluted ammonium chloride. Always identify which volume belongs to the solute of interest.

Common Mistakes

Dividing by $250\ \text{cm}^3$ instead of $10.0\ \text{cm}^3$ .
Forgetting that the amount of NH4Cl equals the amount of NaOH reacted due to the 1:1 equation.
Leaving the answer without a unit or with the wrong unit ( $\text{mol}$ instead of $\text{mol dm}^{-3}$ ).

Things to Be Careful About

Use the amount from (c)(iv) exactly, with the correct power of ten. Quote the final concentration to 3 significant figures and include $\text{mol dm}^{-3}$ .

Techniques used

relate reacted NaOH to original amount of NH4Clconvert amount and volume into concentration

(d)

When ammonia gas is prepared by heating one of its salts with calcium hydroxide, it can be dried by passing it through a drying agent.

Drying agents for gases include calcium oxide, calcium sulfate, concentrated sulfuric acid and phosphorus(V) oxide.

Select one of these drying agents and suggest why it is not suitable for drying ammonia gas.

...................................................... is not suitable because ......................................................

.............................................................................................................................................

1M Similar

DifficultyEasy

Worked solution

Answer

Concentrated sulfuric acid is not suitable because it is an acidic drying agent that neutralises/reacts with the basic ammonia gas:

2\text{NH}_3(\text{g}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow (\text{NH}_4)_2\text{SO}_4(\text{s})

Phosphorus(V) oxide is also unsuitable for the same reason.

Final answer

Concentrated sulfuric acid (or phosphorus(V) oxide); an acidic drying agent neutralises/reacts with basic ammonia.

Detailed explanation

Background Concept

A drying agent for a gas must remove water without reacting with the gas itself. Acidic drying agents such as concentrated sulfuric acid and phosphorus(V) oxide absorb water, but they also behave as acids. Ammonia is a basic gas because the nitrogen atom has a lone pair and can accept a proton.

Understanding the Question

Choose one of the four listed drying agents and state why it is not suitable for drying ammonia. The best choice is one of the acidic agents, because its acidity will destroy the basic ammonia.

Approach

Identify which of the four drying agents is acidic. Then explain that an acid neutralises or reacts with the basic gas, so ammonia would not be recovered.

Step-by-Step Reasoning

Concentrated $\text{H}_2\text{SO}_4$ is a strong acid and a dehydrating/acidic drying agent.
$\text{NH}_3$ is basic and can accept $\text{H}^+$ .
They react:

2\text{NH}_3(\text{g}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow (\text{NH}_4)_2\text{SO}_4(\text{s})

Therefore the ammonia is absorbed/consumed and cannot be dried this way. The same argument applies to acidic phosphorus(V) oxide.
Calcium oxide and calcium sulfate, which are not acidic, may be used to dry ammonia.

Key Takeaways

The suitability of a gas drying agent depends on its chemical nature relative to the gas: acidic drying agents fail for basic gases such as ammonia. The mark requires the idea of neutralisation or reaction with the base.

Common Mistakes

Choosing calcium oxide or calcium sulfate, which are suitable drying agents for ammonia.
Saying sulfuric acid is unsuitable merely because it is concentrated or absorbs water, without mentioning neutralisation/reaction with ammonia.
Saying ammonia is acidic; ammonia is basic.

Things to Be Careful About

You only need to select one unsuitable agent and give the reason. The credit comes from linking the agent's acidic nature to its reaction with basic ammonia. Including the equation is a clear way to show the reaction.

Techniques used

identify the acid-base reaction between a drying agent and ammoniaexplain why an acidic drying agent is unsuitable

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

A sample of ammonium bromide has been accidentally mixed with another salt.

A student suggests that the percentage purity of the ammonium bromide can be determined by measuring the change in temperature when a solution is prepared. The enthalpy change when the sample dissolves in water, $\Delta H$ , can then be calculated and compared with the literature value for ammonium bromide. You will carry out the student's experiment.

FA 4 is ammonium bromide, $\text{NH}_4\text{Br}$ , mixed with another salt.

(a)

Method

Support the cup in the $250\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ measuring cylinder to transfer $25.0\text{ cm}^3$ of distilled water into the cup.
Place the thermometer in the water and tilt the cup, if necessary, so that the bulb of the thermometer is fully covered. Record the temperature of the water in the space for results.
Weigh the container with FA 4. Record the mass.
Add all of FA 4 to the water in the cup.
Stir the mixture gently so that FA 4 dissolves.
Measure and record the minimum temperature reached.
Calculate and record the change in temperature.
Weigh the container with any residual FA 4. Record the mass.
Calculate and record the mass of FA 4 added to the cup.

Results

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record a results table with these six headings and units:

Heading	Unit
mass of container + FA 4	g
mass of container + residue	g
mass of FA 4	g
temperature of water / initial thermometer reading	°C
minimum / lowest thermometer reading	°C
change in temperature / ΔT	°C

Use balance readings to 2 or 3 decimal places and thermometer readings to 0.5 °C. Ensure the initial thermometer reading is between 10.0 °C and 40.0 °C.

Calculate:

mass of FA 4 = (mass of container + FA 4) − (mass of container + residue)

ΔT = initial temperature − minimum temperature

Example: container + FA 4 = 12.34 g, container + residue = 9.84 g, so mass FA 4 = 2.50 g; initial temperature = 24.0 °C, minimum = 20.5 °C, so ΔT = 3.5 °C.

Final answer

Results table with six headings; ΔT = initial − minimum; mass = (container + FA4) − (container + residue)

Detailed explanation

Background Concept

In this experiment the dissolving of ammonium bromide in water is endothermic: $\Delta H = +16.8\text{ kJ mol}^{-1}$ . Energy is taken from the water, so the temperature of the solution falls. The experiment uses the temperature fall to estimate how much ammonium bromide is present. To do this reliably you need to record the initial water temperature and the lowest temperature reached, together with the mass of solid added. The mass of solid is found by weighing the container before and after adding the solid, because some solid may stick to the container.

Understanding the Question

Part (a) is the results-recording stage. You are not asked to calculate the purity yet; you are asked to set up a clear results table, record the thermometer and balance readings, and then calculate two quantities: the mass of FA 4 added and the temperature change $\Delta T$ . The values are candidate-dependent, so there is no single correct numerical answer; the marks are for the headings, units, precision and correct calculations.

Approach

Make a table with six headings, each with its unit. Weigh the container with FA 4, then after adding the solid weigh the container with any residue. The mass of FA 4 added is the difference. Record the initial water temperature and the minimum temperature reached; the temperature change is the difference. Use balance readings to a consistent number of decimal places and thermometer readings to the nearest 0.5 °C.

Step-by-Step Reasoning

The mark scheme requires six unambiguous headings:

mass of container + FA 4 (g)
mass of container + residue (g)
mass of FA 4 (g)
temperature of water / initial thermometer reading (°C)
minimum / lowest thermometer reading (°C)
change in temperature / ΔT (°C)

Each heading must have a unit. The balance readings should be recorded to either 2 or 3 decimal places, and all readings in a column should use the same precision. Thermometer readings should be to 0.5 °C (e.g. 24.0 °C, 20.5 °C), and the initial reading should be between 10.0 °C and 40.0 °C.

Then calculate:

mass of FA 4 = (mass of container + FA 4) − (mass of container + residue)

ΔT = initial temperature − minimum temperature

For example, if the container + FA 4 has mass 12.34 g and the container + residue has mass 9.84 g, the mass of FA 4 is 2.50 g. If the initial temperature is 24.0 °C and the minimum is 20.5 °C, ΔT = 3.5 °C.

Key Takeaways

A results table must have headings and units for every column.
Mass by difference is a standard technique.
Temperature change is a difference of two readings.
Consistent precision is part of good practical recording.

Common Mistakes

Missing units on headings.
Using inconsistent decimal places in a column.
Recording the initial temperature outside the allowed range.
Writing ΔT as a negative number; the change in temperature is reported as a positive magnitude.
Forgetting to weigh the container after adding the solid, so the mass of FA 4 cannot be found.

Things to Be Careful About

The thermometer reading should be to 0.5 °C, not to 0.1 °C unless the thermometer allows it.
The balance readings should be to 2 or 3 decimal places, and all readings in the same column should have the same number of decimal places.
The mass of FA 4 is the difference between the two container readings, not the reading of the container + FA 4 alone.
Keep the initial temperature between 10.0 °C and 40.0 °C as required by the mark scheme.

Techniques used

record initial and minimum temperaturescalculate temperature changecalculate mass of solid by differencerecord results with units and consistent precision

(b)

(i)

Calculate the energy change, in J, for the reaction.

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Energy change $q = mc\Delta T$ , where $m = 25.0\text{ g}$ (from $25.0\text{ cm}^3$ water), $c = 4.18\text{ J g}^{-1}\text{ K}^{-1}$ .

q = 25.0 \times 4.18 \times \Delta T = 104.5\,\Delta T\text{ J}

Using the example $\Delta T = 3.5\text{ }^\circ\text{C}$ :

q = 25.0 \times 4.18 \times 3.5 = 365.75 \approx 366\text{ J}

Answer

$q = 25.0 \times 4.18 \times \Delta T$ J (example: 366 J for ΔT = 3.5 °C)

Final answer

q = 25.0 × 4.18 × ΔT J (example: 366 J for ΔT = 3.5 °C)

Detailed explanation

Background Concept

When a substance dissolves, the energy change can be estimated from the temperature change of the solution using $q = mc\Delta T$ , where $q$ is the energy change in joules, $m$ is the mass of the solution (usually taken as the mass of water) in grams, $c$ is the specific heat capacity of water ( $4.18\text{ J g}^{-1}\text{ K}^{-1}$ ), and $\Delta T$ is the temperature change in °C or K. Because the density of water is approximately $1\text{ g cm}^{-3}$ , $25.0\text{ cm}^3$ of water has a mass of about $25.0\text{ g}$ .

Understanding the Question

Part (b)(i) asks you to calculate the energy change, in J, for the reaction that occurred in your experiment. You need to use the temperature change you recorded in part (a). The value is candidate-dependent, so the working is the important part.

Approach

Substitute the mass of water, the specific heat capacity and your $\Delta T$ into $q = mc\Delta T$ . Keep the units consistent: mass in g, $c$ in J g^-1 K^-1, and $\Delta T$ in °C (the size of a degree Celsius is the same as a kelvin).

Step-by-Step Reasoning

The mass of water is $25.0\text{ g}$ because $25.0\text{ cm}^3$ of water has a mass of $25.0\text{ g}$ . The specific heat capacity of water is $4.18\text{ J g}^{-1}\text{ K}^{-1}$ . Therefore:

q = 25.0 \times 4.18 \times \Delta T = 104.5\,\Delta T\text{ J}

If, for example, $\Delta T = 3.5\text{ }^\circ\text{C}$ :

q = 25.0 \times 4.18 \times 3.5 = 365.75 \approx 366\text{ J}

The answer should be given to 2–4 significant figures.

Key Takeaways

$q = mc\Delta T$ is the key calorimetry equation.
Use the mass of water, not the mass of solid, in this calculation.
A temperature change in °C can be used directly because a kelvin and a degree Celsius are the same size.

Common Mistakes

Using the mass of FA 4 instead of the mass of water.
Forgetting to use $4.18$ as the specific heat capacity.
Using $\Delta T$ with the wrong sign; the magnitude is needed for the energy change.
Giving the answer to too many or too few significant figures.

Things to Be Careful About

The mark scheme expects $25 \times 4.18 \times \Delta T$ and an answer to 2–4 significant figures.
If you use a different $\Delta T$ from the example, your numerical answer will differ; the method is what matters.

Techniques used

apply q = mcΔTuse mass of water from volumeuse specific heat capacity of water

(ii)

When one mole of ammonium bromide dissolves in water the enthalpy change, $\Delta H$ , is $+16.8\text{ kJ mol}^{-1}$ .

Use this value to calculate the amount, in mol, of ammonium bromide in the $25.0\text{ cm}^3$ of solution FA 4 used in your experiment.

Assume that the enthalpy change when the other salt dissolves in water is $0\text{ kJ mol}^{-1}$ .

amount of $\text{NH}_4\text{Br}$ in $25.0\text{ cm}^3$ = .............................. mol

1M Similar

DifficultyMedium-Easy

Worked solution

Working

n = \frac{q}{16.8 \times 1000} = \frac{366}{16800} = 0.0218\text{ mol}

Answer

$2.18 \times 10^{-2}$ mol (example for ΔT = 3.5 °C)

Final answer

2.18 × 10^-2 mol (example for ΔT = 3.5 °C)

Detailed explanation

Background Concept

The enthalpy change of solution, $\Delta H$ , is the energy change when one mole of solute dissolves. Here $\Delta H = +16.8\text{ kJ mol}^{-1}$ . The energy change you calculated in (b)(i) is in joules, so you must convert $\Delta H$ to joules per mole before dividing. The amount of ammonium bromide that dissolved is then:

n = \frac{q}{\Delta H}

Understanding the Question

Part (b)(ii) asks for the amount, in mol, of ammonium bromide in the 25.0 cm^3 solution used in your experiment. You use your energy change from (b)(i) and the given enthalpy change. The assumption is that the other salt contributes no enthalpy change.

Approach

Convert $\Delta H$ from kJ mol^-1 to J mol^-1 by multiplying by 1000. Then divide the energy change from (b)(i) by this value.

Step-by-Step Reasoning

$\Delta H = +16.8\text{ kJ mol}^{-1} = 16800\text{ J mol}^{-1}$ .

Using the example $q = 366\text{ J}$ :

n = \frac{366}{16800} = 0.0218\text{ mol}

This is the amount of ammonium bromide in the 25.0 cm^3 of solution.

Key Takeaways

Always convert kJ to J when the energy change is in J.
Amount = energy change / enthalpy change per mole.
The sign of $\Delta H$ is positive for an endothermic process, but the magnitude is used for the calculation.

Common Mistakes

Dividing by 16.8 instead of 16800.
Using the energy change in kJ without converting.
Forgetting that the answer is in mol.

Things to Be Careful About

The mark scheme uses (b)(i) / (16.8 × 1000).
Give the answer to 2–4 significant figures.
If you could not calculate (b)(i), the question gives an assumed value of $2.06 \times 10^{-2}$ mol to allow you to continue.

Techniques used

convert kJ mol^-1 to J mol^-1calculate moles from energy change and enthalpy change

(iii)

Calculate the mass, in g, of ammonium bromide in your sample.

(If you were unable to calculate the amount of ammonium bromide in (b)(ii), then assume it is $2.06 \times 10^{-2}\text{ mol}$ . This may not be the correct answer.)

mass of $\text{NH}_4\text{Br}$ in sample = .............................. g

1M Similar

(iv)

Hence calculate the percentage by mass of ammonium bromide in your sample.

% by mass of $\text{NH}_4\text{Br}$ = .............................. %

1M Similar

(c)

Another student states that this method is not accurate as the enthalpy change when the other salt dissolves in water could be exothermic.

Suggest how the calculated value in (b)(iv) would change if the other salt dissolves in water exothermically. Explain your answer.

...................................................................................................................................................

.............................................................................................................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The other salt dissolving exothermically releases heat, so the observed temperature decrease is smaller. The calculated energy change $q = mc\Delta T$ is therefore smaller, giving a smaller apparent amount (and mass) of $\text{NH}_4\text{Br}$ . The calculated percentage by mass will be smaller.

Final answer

Percentage by mass will be smaller.

Detailed explanation

Background Concept

Ammonium bromide dissolves endothermically, so it takes in heat and lowers the temperature. If the other salt dissolves exothermically, it releases heat. The two effects oppose each other. The observed temperature change is the net result of both processes.

Understanding the Question

Part (c) asks how the calculated percentage by mass in (b)(iv) would change if the other salt dissolved exothermically, and why. This is a reasoning question, not a calculation.

Approach

Trace the effect step by step: exothermic salt releases heat → observed temperature decrease is smaller → calculated $q = mc\Delta T$ is smaller → calculated amount of ammonium bromide is smaller → calculated mass and percentage are smaller.

Step-by-Step Reasoning

If the other salt releases heat when it dissolves, some of the cooling caused by ammonium bromide is cancelled out. The thermometer therefore records a smaller temperature decrease than would occur if only ammonium bromide were present. Since the energy change is calculated from $q = mc\Delta T$ , a smaller $\Delta T$ gives a smaller $q$ . Dividing this smaller $q$ by the enthalpy change per mole gives a smaller apparent amount of ammonium bromide. Consequently the calculated mass and percentage by mass are smaller than the true values.

Key Takeaways

An exothermic impurity reduces the observed cooling.
The calculated amount of ammonium bromide is an apparent value based on the assumption that only ammonium bromide contributes to the enthalpy change.
The direction of the error can be predicted by following the sign of the enthalpy change.

Common Mistakes

Saying the percentage would increase.
Saying the temperature change would increase.
Not explaining that the calculated amount of ammonium bromide is affected.

Things to Be Careful About

The mark scheme requires three linked ideas: smaller temperature decrease, smaller apparent moles, smaller percentage.
The word "apparent" is useful because the true amount of ammonium bromide is unchanged; only the calculated value changes.

Techniques used

reason about sign of enthalpy changetrace effect on calculated moles and percentage

(d)

The maximum uncertainty in a single thermometer reading is $\pm 0.5\text{ }^\circ\text{C}$ .

Calculate the maximum percentage error in your temperature change recorded in (a). Show your working.

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

1M Similar

DifficultyMedium

Worked solution

Working

Uncertainty in $\Delta T = 2 \times 0.5 = 1.0\text{ }^\circ\text{C}$ .

\text{max percentage error} = \frac{1.0}{\Delta T} \times 100

Using example $\Delta T = 3.5\text{ }^\circ\text{C}$ :

\frac{1.0}{3.5} \times 100 = 28.6\%

Answer

28.6% (example for ΔT = 3.5 °C)

Final answer

28.6% (example for ΔT = 3.5 °C)

Detailed explanation

Background Concept

When a quantity is calculated as a difference of two readings, the uncertainties in the two readings add. The temperature change is:

\Delta T = T_{\text{initial}} - T_{\text{minimum}}

so the absolute uncertainty in $\Delta T$ is:

2 \times 0.5 = 1.0\text{ }^\circ\text{C}

The percentage error is then:

\text{percentage error} = \frac{\text{absolute uncertainty}}{\text{value}} \times 100

Understanding the Question

Part (d) asks for the maximum percentage error in the temperature change recorded in part (a). The maximum uncertainty in a single thermometer reading is $\pm 0.5\text{ }^\circ\text{C}$ , and two readings are used to find $\Delta T$ .

Approach

Double the single-reading uncertainty to get the uncertainty in $\Delta T$ . Then divide by your $\Delta T$ and multiply by 100.

Step-by-Step Reasoning

The uncertainty in each thermometer reading is $0.5\text{ }^\circ\text{C}$ . Because $\Delta T$ is the difference of two readings, the maximum absolute uncertainty is:

2 \times 0.5 = 1.0\text{ }^\circ\text{C}

Using the example $\Delta T = 3.5\text{ }^\circ\text{C}$ :

\text{max percentage error} = \frac{1.0}{3.5} \times 100 = 28.6\%

The answer should be given to 2–4 significant figures.

Key Takeaways

Uncertainties in a difference add.
Percentage error = (absolute uncertainty / value) × 100.
The uncertainty in $\Delta T$ is twice the uncertainty in a single reading.

Common Mistakes

Using $\pm 0.5$ instead of $1.0$ in the numerator.
Forgetting to multiply by 100.
Using the initial temperature instead of $\Delta T$ as the denominator.

Things to Be Careful About

The mark scheme expects $\{(2 \times 0.5)/\Delta T\} \times 100$ .
The percentage error depends on your $\Delta T$ ; a smaller $\Delta T$ gives a larger percentage error.
Give the answer to 2–4 significant figures.

Techniques used

propagate uncertainty in a differencecalculate 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.

(a)

FA 5 is a mixture containing two cations and two anions. Three of the ions are listed in the Qualitative analysis notes.

10M

(i)

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

Table 3.1

test	observations
Test 1 Heat a spatula measure of FA 5 in a hard-glass test-tube.
Test 2 Put a $2\text{ cm}$ depth of dilute nitric acid in a boiling tube. Add a spatula measure of FA 5. Dilute the solution from Test 2 with an equal volume of distilled water. Keep the resulting solution for use in (a)(ii). This solution is FA 6.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test 1 (heat FA 5 in a hard-glass test-tube):

condensation / droplets of moisture on the walls of the test-tube
purple gas / vapour given off
yellow solid forms on heating
yellow residue pales on cooling to white / cream / off-white

Test 2 (add dilute nitric acid to FA 5):

fizzing / bubbling / effervescence
gas given off which turns limewater milky (white precipitate) — $\text{CO}_2$
colourless / pale yellow solution formed

Final answer

Candidate-dependent observations; expected: condensation, purple vapour, yellow solid pales on cooling (Test 1); effervescence, CO2 turns limewater milky, colourless/pale yellow solution (Test 2).

Detailed explanation

Background Concept

FA 5 is a mixture of zinc carbonate, $\text{ZnCO}_3$ , and potassium iodide, KI. Heating a carbonate causes it to decompose into the metal oxide and carbon dioxide:

\text{ZnCO}_3(\text{s}) \rightarrow \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})

Zinc oxide is white when cold but turns yellow when hot — a characteristic property of this oxide. On cooling it reverts to white. The purple vapour seen on heating comes from iodine, $\text{I}_2$ , formed by oxidation of iodide ions (from KI) at high temperature in air.

Adding dilute nitric acid to a carbonate releases carbon dioxide:

\text{ZnCO}_3(\text{s}) + 2\text{HNO}_3(\text{aq}) \rightarrow \text{Zn(NO}_3)_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

The $\text{CO}_2$ is identified by bubbling it through limewater, which turns milky (white precipitate of $\text{CaCO}_3$ ).

Understanding the Question

This is a qualitative analysis practical. FA 5 is an unknown solid containing two cations and two anions. The candidate must carry out two tests — heating the solid, and adding dilute nitric acid — and record all observations in Table 3.1. These observations will later be used to identify the ions present.

The command is 'Carry out the following tests and record your observations.' This is a practical observation task — the candidate must actually perform the tests and write down what they see, at the stage each observation occurs.

Approach

Carry out each test carefully and watch for:

Colour changes (of the solid, the solution, and any gas).
Gas evolution — and test any gas with limewater.
Precipitate formation and its solubility in excess reagent.
Any residue changes on heating and cooling.

Record each observation at the stage it occurs.

Step-by-Step Reasoning

Test 1 — heating FA 5.

The carbonate decomposes: $\text{ZnCO}_3 \rightarrow \text{ZnO} + \text{CO}_2$ . Any moisture present condenses as droplets on the cooler upper walls of the test-tube.
The purple vapour is iodine, $\text{I}_2$ , from oxidation of iodide ions at high temperature.
The residue is ZnO, which is yellow while hot and turns white/cream/off-white on cooling.

Test 2 — adding dilute nitric acid.

The carbonate reacts with the acid: effervescence (bubbling) of $\text{CO}_2$ .
Test the gas with limewater: it turns milky, confirming $\text{CO}_2$ .
The zinc nitrate and potassium iodide dissolve, giving a colourless or pale yellow solution (FA 6).

Key Takeaways

Carbonates decompose on heating to the metal oxide and $\text{CO}_2$ .
$\text{CO}_2$ is identified by turning limewater milky.
Zinc oxide is yellow when hot and white when cold.
Iodine vapour is purple.
Record observations at the stage they occur, with enough detail.

Common Mistakes

Not testing the gas with limewater — the mark requires identification of $\text{CO}_2$ .
Not noting the colour change of ZnO on cooling (yellow → white).
Writing only 'gas evolved' without identifying it.
Using a boiling tube instead of a hard-glass test-tube for heating a solid (the instructions explicitly require a hard-glass test-tube).

Things to Be Careful About

The mark scheme requires two observation points per mark — record enough detail (6 points across both tests for 3 marks).
Distinguish the yellow colour of hot ZnO (which pales on cooling) from any other yellow colour.
State the name or correct formula of reagents used.
Write 'no change' where nothing is observed.

Techniques used

heat a solid and observe decomposition productstest a gas with limewaterrecord observations of colour changes and residuesobserve colour change of zinc oxide on heating and cooling

(ii)

Carry out the following tests using a $1\text{ cm}$ depth of FA 6 in a test-tube for each test. Record your observations in Table 3.2.

Table 3.2

test	observations
Test 1 Add a $1\text{ cm}$ depth of sodium chlorate(I), then add aqueous sodium thiosulfate dropwise.
Test 2 Add aqueous sodium hydroxide.
Test 3 Add aqueous ammonia.
Test 4 Add a $1\text{ cm}$ depth of aqueous copper(II) sulfate, then add aqueous sodium thiosulfate dropwise until no further change occurs.
Test 5 Add a few drops of aqueous silver nitrate, then add aqueous ammonia until no further change occurs.

4M Similar

DifficultyMedium

Worked solution

Answer

Test 1 (+ $\text{NaClO(aq)}$ , then $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ dropwise):

- $\text{NaClO(aq)}$ : brown / red-brown / orange-brown / yellow-brown solution formed
- $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ : colourless solution

Test 2 (+ $\text{NaOH(aq)}$ ):

white precipitate, soluble in excess $\text{NaOH(aq)}$ giving a colourless solution

Test 3 (+ $\text{NH}_3\text{(aq)}$ ):

white precipitate, soluble in excess $\text{NH}_3\text{(aq)}$ giving a colourless solution

Test 4 (+ $\text{CuSO}_4\text{(aq)}$ , then $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ dropwise):

- $\text{CuSO}_4\text{(aq)}$ : white / off-white / cream / brown / yellow-brown precipitate formed (or brown / yellow-brown solution)
- $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ : changes to white precipitate OR blue solution

Test 5 (+ $\text{AgNO}_3\text{(aq)}$ , then $\text{NH}_3\text{(aq)}$ ):

- $\text{AgNO}_3\text{(aq)}$ : pale yellow precipitate
- $\text{NH}_3\text{(aq)}$ : precipitate insoluble / turns paler

Final answer

Candidate-dependent observations; expected: brown solution then colourless (Test 1); white ppt soluble in excess NaOH and NH3 (Tests 2 & 3); ppt/solution with CuSO4 then white ppt or blue solution (Test 4); pale yellow ppt insoluble in NH3 (Test 5).

Detailed explanation

Background Concept

FA 6 is the solution obtained by dissolving FA 5 ( $\text{ZnCO}_3$ + KI) in dilute nitric acid and diluting with water. It therefore contains $\text{Zn}^{2+}$ , $\text{I}^-$ , and $\text{NO}_3^-$ ions (plus some $\text{H}^+$ from the acid). The five tests are designed to confirm the presence of zinc and iodide ions.

Key chemistry:

Iodide, $\text{I}^-$ : A reducing agent. Chlorate(I) ( $\text{NaClO}$ ) oxidises $\text{I}^-$ to iodine, $\text{I}_2$ , which is brown/red-brown in aqueous solution. Thiosulfate ( $\text{Na}_2\text{S}_2\text{O}_3$ ) reduces $\text{I}_2$ back to colourless $\text{I}^-$ .
Zinc, $\text{Zn}^{2+}$ : Forms a white gelatinous precipitate of $\text{Zn(OH)}_2$ with NaOH and with $\text{NH}_3$ . The hydroxide is amphoteric — it dissolves in excess NaOH (forming $[\text{Zn(OH)}_4]^{2-}$ ) and in excess $\text{NH}_3$ (forming $[\text{Zn(NH}_3)_4]^{2+}$ ).
Iodide with $\text{Cu}^{2+}$ : $\text{Cu}^{2+}$ oxidises $\text{I}^-$ to $\text{I}_2$ , forming copper(I) iodide, CuI (white/off-white precipitate) and iodine (brown). Adding thiosulfate removes the iodine, leaving the white CuI (or the blue $\text{Cu}^{2+}$ colour returns if $\text{Cu}^{2+}$ is in excess).
Iodide with $\text{Ag}^+$ : $\text{Ag}^+ + \text{I}^- \rightarrow \text{AgI}$ , a pale yellow precipitate. AgI is insoluble in $\text{NH}_3$ (unlike AgCl, which dissolves).

Understanding the Question

Five tests are carried out on FA 6 (a solution). The candidate must record observations for each test in Table 3.2. The tests are designed to confirm $\text{Zn}^{2+}$ (Tests 2 & 3) and $\text{I}^-$ (Tests 1, 4, 5). Each test involves adding reagents in a specific order, and observations must be recorded at each stage.

Approach

For each test, note:

The reagent added.
What happens immediately (colour change, precipitate, gas).
What happens in excess (for NaOH and $\text{NH}_3$ : does the precipitate dissolve?).

Step-by-Step Reasoning

Test 1: $\text{NaClO}$ oxidises $\text{I}^-$ to $\text{I}_2$ : $2\text{I}^- + \text{ClO}^- + 2\text{H}^+ \rightarrow \text{I}_2 + \text{Cl}^- + \text{H}_2\text{O}$ . The iodine gives a brown/red-brown solution. Adding $\text{Na}_2\text{S}_2\text{O}_3$ reduces $\text{I}_2$ back to $\text{I}^-$ : $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ . The solution becomes colourless.

Test 2: $\text{NaOH} + \text{Zn}^{2+} \rightarrow \text{Zn(OH)}_2$ (white ppt). Excess NaOH: $\text{Zn(OH)}_2 + 2\text{OH}^- \rightarrow [\text{Zn(OH)}_4]^{2-}$ (colourless, soluble).

Test 3: $\text{NH}_3 + \text{Zn}^{2+} \rightarrow \text{Zn(OH)}_2$ (white ppt). Excess $\text{NH}_3$ : $\text{Zn(OH)}_2 + 4\text{NH}_3 \rightarrow [\text{Zn(NH}_3)_4]^{2+} + 2\text{OH}^-$ (colourless, soluble).

Test 4: $\text{Cu}^{2+} + 2\text{I}^- \rightarrow \text{CuI}$ (white ppt) $+ \tfrac{1}{2}\text{I}_2$ (brown). The brown colour is from iodine. Adding $\text{Na}_2\text{S}_2\text{O}_3$ removes the iodine ( $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ ), leaving the white CuI precipitate (or a blue solution if $\text{Cu}^{2+}$ is in excess).

Test 5: $\text{Ag}^+ + \text{I}^- \rightarrow \text{AgI}$ (pale yellow ppt). $\text{NH}_3$ does not dissolve AgI (AgI is insoluble in $\text{NH}_3$ , unlike AgCl).

Key Takeaways

Iodide is oxidised to iodine (brown) by oxidising agents; thiosulfate reverses this.
$\text{Zn(OH)}_2$ is amphoteric — dissolves in excess NaOH and in excess $\text{NH}_3$ .
AgI is pale yellow and insoluble in $\text{NH}_3$ .
Record observations after each reagent addition, in order.

Common Mistakes

Confusing the brown colour of iodine with other colours.
Not noting that the precipitate dissolves in excess NaOH/ $\text{NH}_3$ — both parts are needed for the mark.
Saying AgI dissolves in $\text{NH}_3$ (it is AgCl that dissolves).
Mixing up the order of reagents in Test 4.

Things to Be Careful About

The mark scheme requires two observation points per mark (4 marks = 8 points across the 5 tests).
'White ppt AND soluble in excess' is one mark point — both parts must be stated.
In Test 4, the initial precipitate can be white/off-white/cream/brown/yellow-brown depending on the amount of iodine formed; after thiosulfate, it is white (or the solution is blue).

Techniques used

oxidise iodide to iodine with chlorate(I)reduce iodine back to iodide with thiosulfatetest for Zn2+ with NaOH and NH3 observing solubility in excesstest for iodide with silver nitrate and ammoniaobserve precipitate formation and colour changes in sequence

(iii)

FA 5 reacts with nitric acid to make FA 6 in Test 2 in Table 3.1.

State a change you would see in one of the observations in Table 3.2 if hydrochloric acid had been used to prepare FA 6.

...........................................................................................................................................

.....................................................................................................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

In Test 5 (with $\text{AgNO}_3$ ), a white precipitate would also form alongside the yellow precipitate, because $\text{AgCl}$ is white and insoluble.

Final answer

White precipitate (as well as yellow) in Test 5 with AgNO3.

Detailed explanation

Background Concept

Silver nitrate is used to test for halide ions. AgCl is white, AgBr is cream, AgI is pale yellow. AgCl dissolves in dilute $\text{NH}_3$ ; AgBr dissolves in concentrated $\text{NH}_3$ ; AgI is insoluble in $\text{NH}_3$ .

If hydrochloric acid (HCl) were used instead of nitric acid ( $\text{HNO}_3$ ) to prepare FA 6, the solution would contain $\text{Cl}^-$ ions as well as $\text{I}^-$ ions.

Understanding the Question

The question asks what change would be seen in the observations in Table 3.2 if HCl had been used to prepare FA 6. The key is that $\text{Cl}^-$ would be present, and in Test 5 ( $\text{AgNO}_3$ ), AgCl (white) would form alongside AgI (yellow).

Approach

Identify which test in Table 3.2 involves $\text{AgNO}_3$ (Test 5) and what chloride ions would do there.

Step-by-Step Reasoning

In Test 5, $\text{AgNO}_3$ is added to FA 6. With $\text{I}^-$ present, a pale yellow precipitate of AgI forms. If $\text{Cl}^-$ were also present (from HCl), a white precipitate of AgCl would also form. The precipitate would therefore be a mixture — white as well as yellow.

Key Takeaways

AgCl is white; AgI is pale yellow.
The choice of acid affects which anions are present in the solution.
AgCl is soluble in dilute $\text{NH}_3$ ; AgI is not.

Common Mistakes

Saying the precipitate would be 'white only' — the yellow AgI is still present.
Not specifying which test (Test 5 / with $\text{AgNO}_3$ ).

Things to Be Careful About

The mark requires 'white ppt (as well as yellow) AND in Test 5 / with $\text{AgNO}_3$ '.

Techniques used

predict the effect of chloride ions on the silver nitrate testcompare solubility of AgCl and AgI in ammonia

(iv)

Use your observations in (a)(i) and (a)(ii) to identify the formulae of three of the ions present in FA 5.

Formulae of ions present are ......................... and ......................... and ......................... .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Ions present: $\text{I}^-$ , $\text{Zn}^{2+}$ , $\text{CO}_3^{2-}$

Final answer

I⁻, Zn²⁺ and CO₃²⁻

Detailed explanation

Background Concept

The observations from (a)(i) and (a)(ii) are used to identify the ions. Each observation points to a specific ion:

Effervescence with acid + gas turns limewater milky → $\text{CO}_3^{2-}$ (carbonate).
Purple vapour on heating → $\text{I}_2$ (from $\text{I}^-$ ).
White ppt soluble in excess NaOH and $\text{NH}_3$ → $\text{Zn}^{2+}$ .
Pale yellow ppt with $\text{AgNO}_3$ , insoluble in $\text{NH}_3$ → $\text{I}^-$ (confirms).

Understanding the Question

Use the observations from (a)(i) and (a)(ii) to identify three of the four ions present in FA 5. The three ions asked for are those listed in the Qualitative analysis notes.

Approach

Match each key observation to the ion it identifies:

Acid + effervescence + limewater → carbonate.
Purple vapour on heating → iodide.
White ppt soluble in excess NaOH and $\text{NH}_3$ → zinc.

Step-by-Step Reasoning

From (a)(i) Test 2: effervescence with nitric acid, gas turns limewater milky → $\text{CO}_3^{2-}$ present.
From (a)(i) Test 1: purple vapour on heating → $\text{I}^-$ present (iodine formed).
From (a)(ii) Tests 2 & 3: white ppt soluble in excess NaOH and $\text{NH}_3$ → $\text{Zn}^{2+}$ present.
From (a)(ii) Test 5: pale yellow ppt with $\text{AgNO}_3$ , insoluble in $\text{NH}_3$ → confirms $\text{I}^-$ .

Key Takeaways

Cross-referencing observations from multiple tests identifies ions.
Carbonate: acid + effervescence + limewater test.
Zinc: white amphoteric hydroxide.
Iodide: purple iodine vapour on heating, pale yellow AgI.

Common Mistakes

Confusing $\text{Zn}^{2+}$ with $\text{Al}^{3+}$ (both give white amphoteric hydroxides) — but $\text{Al}^{3+}$ wouldn't give the other observations.
Forgetting $\text{CO}_3^{2-}$ even though it's obvious from the acid test.
Including $\text{K}^+$ (the fourth ion, not asked for).

Things to Be Careful About

2 marks for 3 correct ions; 1 mark for 2 correct.
The fourth ion ( $\text{K}^+$ ) is not required.

Techniques used

deduce ions from qualitative observationscross-reference observations from multiple testsmatch each observation to a specific ion

(b)

FA 7, FA 8 and FA 9 are each known to be propan-1-ol, propanal, propanoic acid or propanone. No two liquids are identical.

(i)

You will carry out tests to identify samples of FA 7, FA 8 and FA 9.

Test 1 has been carried out for you.

Carry out Test 2 and record your observations in Table 3.3.

Table 3.3

test	observations: FA 7	observations: FA 8	observations: FA 9
Test 1 Put a $1\text{ cm}$ depth in a test-tube. Add a few drops of 2,4-dinitrophenylhydrazine (2,4-DNPH reagent).	orange precipitate	orange precipitate	no change
Test 2 Put a $1\text{ cm}$ depth in a test-tube. Add a small spatula measure of solid sodium carbonate.

1M Similar

DifficultyEasy

Worked solution

Answer

FA 7: no reaction / no change / no bubbling
FA 8: no reaction / no change / no bubbling
FA 9: fizzing / effervescence / bubbling ( $\text{CO}_2$ given off)

Final answer

FA 7 and FA 8: no change; FA 9: effervescence/fizzing (CO2).

Detailed explanation

Background Concept

Sodium carbonate reacts with carboxylic acids to release $\text{CO}_2$ :

2\text{RCOOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{RCOONa} + \text{CO}_2 + \text{H}_2\text{O}

This produces effervescence. Aldehydes, ketones and alcohols do not react with $\text{Na}_2\text{CO}_3$ (no acidic proton).

Understanding the Question

Add solid $\text{Na}_2\text{CO}_3$ to each of the three liquids (FA 7, FA 8, FA 9) and record observations in Table 3.3. The three liquids are propan-1-ol, propanal, propanoic acid, or propanone.

Approach

Propanoic acid will fizz (effervescence); propanone and propanal will not.

Step-by-Step Reasoning

FA 9 is propanoic acid (from the marking scheme). It reacts with $\text{Na}_2\text{CO}_3$ :

2\text{CH}_3\text{CH}_2\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{CH}_2\text{COONa} + \text{CO}_2 + \text{H}_2\text{O}

Effervescence (bubbling) of $\text{CO}_2$ is observed.

FA 7 (propanone) and FA 8 (propanal) have no acidic hydrogen and do not react with $\text{Na}_2\text{CO}_3$ — no change.

Key Takeaways

Carboxylic acids react with carbonates to give $\text{CO}_2$ (effervescence).
This distinguishes carboxylic acids from aldehydes, ketones and alcohols.

Common Mistakes

Expecting propanal to react with $\text{Na}_2\text{CO}_3$ (it doesn't — no acidic H).
Not recording 'no change' for FA 7 and FA 8.

Things to Be Careful About

The mark requires 'no reaction / no change / no bubbling' for FA 7 and FA 8 AND 'fizzing / effervescence / bubbling' for FA 9.

Techniques used

test for a carboxylic acid with sodium carbonateobserve effervescence of carbon dioxiderecord no-change results for non-acidic compounds

(ii)

State what you can deduce from Tests 1 and 2 in Table 3.3 about the functional group present in each of FA 7, FA 8 and FA 9.

...........................................................................................................................................

.....................................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

FA 7 and FA 8: contain a carbonyl group — they are carbonyl compounds (either aldehydes or ketones), since they give an orange precipitate with 2,4-DNPH.
FA 9: contains a carboxyl group — it is a carboxylic acid (propanoic acid), since it fizzes with $\text{Na}_2\text{CO}_3$ (releasing $\text{CO}_2$ ).

Final answer

FA 7 & FA 8: carbonyl compounds (aldehyde or ketone); FA 9: carboxylic acid.

Detailed explanation

Background Concept

2,4-DNPH (Brady's reagent) reacts with aldehydes and ketones (carbonyl compounds) to give an orange/red precipitate. It does not react with alcohols or carboxylic acids. Sodium carbonate reacts with carboxylic acids (fizzing) but not with carbonyl compounds.

From Test 1: FA 7 and FA 8 give orange ppt (carbonyl); FA 9 does not.
From Test 2: FA 9 fizzes (carboxylic acid); FA 7 and FA 8 do not.

Understanding the Question

Deduce the functional group in each compound from the two tests. The question asks specifically about the functional group present in each of FA 7, FA 8 and FA 9.

Approach

Orange ppt with 2,4-DNPH → carbonyl (aldehyde or ketone).
Fizzing with $\text{Na}_2\text{CO}_3$ → carboxylic acid.

Step-by-Step Reasoning

FA 7 and FA 8: orange ppt with 2,4-DNPH → carbonyl compounds (aldehyde or ketone). At this stage we cannot distinguish aldehyde from ketone — both give the same result with 2,4-DNPH.

FA 9: no orange ppt, but fizzes with $\text{Na}_2\text{CO}_3$ → carboxylic acid. This is consistent with propanoic acid.

Key Takeaways

2,4-DNPH is a test for the carbonyl group.
$\text{Na}_2\text{CO}_3$ is a test for the carboxylic acid group.
Combining tests narrows down the functional group.

Common Mistakes

Saying FA 7/FA 8 are 'aldehydes' specifically (at this stage they could be ketones).
Not identifying FA 9 as a carboxylic acid.

Things to Be Careful About

The mark scheme: M1 = carbonyl compounds for FA 7 & FA 8; M2 = carboxyl group for FA 9.

Techniques used

interpret the 2,4-DNPH test result for carbonyl compoundsinterpret the sodium carbonate test for carboxylic acidsdeduce functional groups from combined test results

(iii)

Describe a test to identify FA 7 and FA 8 using the reagents provided.

Carry out your test. Record your observations and conclusions.

test ....................................................................................................................................

...........................................................................................................................................

observations ......................................................................................................................

...........................................................................................................................................

FA 7 is ...................................................... .

FA 8 is ...................................................... .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

test: Add a few drops of acidified aqueous potassium manganate(VII) ( $\text{KMnO}_4$ ) to a 1 cm depth of each liquid in a test-tube.

observations:

FA 7: purple colour remains / no change (propanone is not oxidised)
FA 8: purple decolourises / turns colourless (or pale brown) — propanal is oxidised by $\text{KMnO}_4$

FA 7 is propanone. FA 8 is propanal.

Final answer

Acidified KMnO4: FA 8 (propanal) decolourises KMnO4; FA 7 (propanone) does not.

Detailed explanation

Background Concept

Aldehydes are readily oxidised to carboxylic acids; ketones are not oxidised under mild conditions. Acidified potassium manganate(VII), $\text{KMnO}_4$ , is a strong oxidising agent. When it oxidises an aldehyde, the purple $\text{MnO}_4^-$ is reduced to colourless $\text{Mn}^{2+}$ (or pale brown $\text{MnO}_2$ ). Ketones do not react, so the purple colour persists.

The oxidation of propanal:

\text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH}

Understanding the Question

Design and carry out a test to distinguish FA 7 (propanone) from FA 8 (propanal) using the reagents provided. The test must be one that gives different results for an aldehyde and a ketone.

Approach

Use an oxidising agent that reacts with aldehydes but not ketones. Acidified $\text{KMnO}_4$ is the appropriate reagent — a colour change (purple → colourless) indicates oxidation (aldehyde), while no change indicates a ketone.

Step-by-Step Reasoning

Add a few drops of acidified $\text{KMnO}_4$ to a 1 cm depth of each liquid in a separate test-tube.
Observe the colour:
- FA 8 (propanal, an aldehyde) is oxidised to propanoic acid. The purple $\text{MnO}_4^-$ is reduced to colourless $\text{Mn}^{2+}$ — the purple colour disappears.
- FA 7 (propanone, a ketone) is not oxidised under these conditions — the purple colour persists.
Conclusion: FA 8 is propanal; FA 7 is propanone.

Key Takeaways

Aldehydes are easily oxidised; ketones are not.
Acidified $\text{KMnO}_4$ : purple → colourless indicates oxidation (aldehyde present).
This is a reliable way to distinguish aldehydes from ketones.

Common Mistakes

Using Fehling's or Tollens' reagent (not provided in this question).
Saying propanone is oxidised (it is not, under these conditions).
Not specifying 'acidified' $\text{KMnO}_4$ .

Things to Be Careful About

Both the observation (purple → colourless for FA 8) and the conclusion (FA 8 = propanal, FA 7 = propanone) are needed for the marks.
The reagent must be named with its correct formula or name.

Techniques used

distinguish aldehyde from ketone by oxidationuse acidified potassium manganate(VII) as an oxidising agentobserve decolourisation of KMnO4draw conclusions from the oxidation result