Chemistry 9701/33 — 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

When a metal carbonate reacts with a suitable acid, carbon dioxide is produced. You will determine the relative formula mass, $M_r$ , of a metal carbonate by reacting it with excess hydrochloric acid and measuring the mass of gas produced.

FA 1 is the metal carbonate.
FA 2 is $2.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .

(a)

Method

● Use the measuring cylinder to transfer $30.0\text{ cm}^3$ of FA 2 into a conical flask.
● Weigh the conical flask containing FA 2. Record the mass.
● Weigh the container with FA 1. Record the mass.
● Tip all the FA 1 slowly into the conical flask. When the reaction slows, swirl the flask gently.
● Weigh the container with any residual FA 1. Record the mass.
● Calculate the mass of FA 1 added. Record the mass.
● Leave the conical flask and its contents for 15 minutes. Swirl the flask occasionally during this time.

During this period begin work on Question 2.

● After 15 minutes weigh the flask and its contents. Record the mass.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the following readings in a table, all in g:

Mass of conical flask + FA 2 = ...
Mass of container + FA 1 = ...
Mass of container + residual FA 1 = ...
Mass of FA 1 added = (mass of container + FA 1) − (mass of container + residual FA 1) = ...
Mass of conical flask + contents after 15 min = ...

Record all balance readings to the same number of decimal places (2 or 3 d.p.).

Final answer

Candidate-dependent: results table with these five headings and units, all readings to consistent 2/3 d.p., and mass FA1 by subtraction.

Detailed explanation

Background Concept

This is a gas-evolution experiment. Excess hydrochloric acid reacts with the metal carbonate, releasing CO2 gas. Because the flask is open, the CO2 escapes and the mass of the flask and contents falls. The loss in mass equals the mass of CO2 produced.

Understanding the Question

Part (a) does not ask for a calculation yet: it asks you to carry out the procedure and record the relevant masses carefully. The examiner will look for a properly headed results table, consistent precision, and the calculation of the mass of FA 1 by difference.

Approach

Identify every mass needed to work out the mass of CO2 later: the mass before reaction (flask + acid and FA1) and the mass after reaction (flask + contents after 15 min). Weigh FA1 before and after transfer so you can find exactly how much was added.

Step-by-Step Reasoning

Use the measuring cylinder to put 30.0 cm3 of FA 2 in the conical flask and weigh the flask + acid; record it.
Weigh the container holding FA 1; record it.
Tip the FA 1 into the acid, swirl when the reaction slows, and reweigh the container with any residual FA 1.
Mass of FA 1 added = initial container mass − residual container mass. This is weighing by difference, which avoids errors from incomplete transfer.
Leave for 15 min (swirling occasionally) to ensure all the carbonate reacts and the CO2 escapes, then weigh the flask + contents; record the final mass.
Enter all values in a table with proper headings and units; use the same number of decimal places for all balance readings.

Key Takeaways

Weighing by difference gives the exact mass transferred. A clear, unit-labelled table and consistent precision are essential for full marks.

Common Mistakes

Mixing 2 d.p. and 3 d.p. readings in the same table.
Forgetting to state units (g).
Not weighing the residual FA 1, so the mass added cannot be found accurately.
Recording only the mass after reaction instead of all the required masses.

Things to Be Careful About

Keep the flask open to the atmosphere so CO2 can escape. Swirl gently to avoid splashing. Wait the full 15 minutes before the final weighing so the mass loss is complete.

Techniques used

weigh by differencerecord readings to consistent precisionconstruct a results table with headings and unitscalculate mass of FA1 by subtraction

(b)

(i)

Use your readings to calculate the mass of carbon dioxide produced.

1M Similar

DifficultyEasy

Worked solution

Working

Let $x$ = mass of conical flask + FA 2, $y$ = mass of FA 1 added, $z$ = final mass of conical flask + contents.

The mass loss is the mass of CO2 produced.

\text{mass of CO}_2 = x + y - z

Example only: $x = 148.52\text{ g}$ , $y = 0.50\text{ g}$ , $z = 148.80\text{ g}$ :

\text{mass of CO}_2 = 148.52 + 0.50 - 148.80 = 0.22\text{ g}

Answer

Mass of CO2 = $x + y - z$ using your readings; e.g. $0.22\text{ g}$ .

Final answer

m(CO2) = x + y − z (g), e.g. 0.22 g

Detailed explanation

Background Concept

Carbon dioxide is a gas and escapes from the open flask. Every other reactant and product stays in the flask: the acid and the metal salt remain in solution, and water is part of the solution. So the decrease in total mass equals the mass of CO2 that has left.

Understanding the Question

This part asks you to process your readings from part (a). Before reaction the total mass is (mass of flask + acid) + (mass of FA1). After reaction, the only thing lost is CO2, so total initial mass − final mass gives CO2.

Approach

Recognise that the final mass already includes the acid, the metal salt solution and any unreacted material, but not the CO2. Therefore you should subtract the final mass from the sum of all masses present before mixing.

Step-by-Step Reasoning

Initial total mass = mass(flask + FA2) + mass(FA1 added) = $x + y$ .
Final total mass = $z$ .
Mass loss = initial total mass − final total mass = $x + y - z$ .
This mass loss equals the mass of CO2 produced.
Give the result to 2–4 significant figures, matching the precision of your readings.

Key Takeaways

In an open-system gas-evolution experiment, mass loss directly gives the mass of gas evolved.

Common Mistakes

Taking the final flask mass alone as the mass loss.
Forgetting to add the mass of FA1 to the initial flask + acid mass.
Using the mass of FA1 as if it were the CO2 mass.

Things to Be Careful About

Make sure the reaction has finished and the flask is at room temperature before the final weighing. Give units (g) and an appropriate number of significant figures.

Techniques used

calculate mass loss by differenceapply conservation of mass

(ii)

The ionic equation for the reaction of FA 1 with FA 2 is shown.

\text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

Calculate the relative formula mass, $M_r$ , of the metal carbonate FA 1. Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From (b)(i), mass of CO2 = $0.22\text{ g}$ (example).

n(\text{CO}_2) = \frac{\text{mass of CO}_2}{44} = \frac{0.22}{44} = 0.0050\text{ mol}

The ionic equation shows 1 mol FA 1 gives 1 mol CO2, so $n(\text{FA 1}) = n(\text{CO}_2)$ .

M_r(\text{FA 1}) = \frac{\text{mass of FA 1}}{n(\text{FA 1})} = \frac{0.50}{0.0050} = 100.0

Equivalently, $M_r = \frac{44 \times \text{mass of FA 1}}{\text{mass of CO}_2}$ .

Answer

Using your readings: $M_r = \frac{44 \times \text{mass of FA 1}}{\text{mass of CO}_2}$ ; e.g. $100.0$ (2–4 sig figs).

Final answer

Mr = 44 × mass(FA1) / mass(CO2), e.g. 100.0

Detailed explanation

Background Concept

The relative formula mass is the mass of one mole of a substance in grams: $M_r = \frac{\text{mass}}{\text{number of moles}}$ . Carbon dioxide has $M_r = 12.0 + 2(16.0) = 44.0$ . The reaction stoichiometry is one carbonate ion to one CO2, so moles of metal carbonate equal moles of CO2.

Understanding the Question

You already know the mass of CO2 and the mass of FA1 used. The question wants you to turn the CO2 mass into moles, remember the mole ratio, then divide the FA1 mass by those moles to get Mr.

Approach

Work in moles: mass loss → n(CO2) → n(FA1) → Mr = mass(FA1)/n(FA1). Because one mole carbonate produces one mole CO2, the stoichiometric factor is 1.

Step-by-Step Reasoning

Calculate n(CO2) = mass(CO2)/44.
Use the 1:1 ratio from the ionic equation: n(FA1) = n(CO2).
Calculate Mr = mass(FA1)/n(FA1). This is equivalent to Mr = 44 × mass(FA1)/mass(CO2).
Give the answer to 2–4 significant figures (e.g. 100.0, not a value with too many decimals from a 2 d.p. balance).

Key Takeaways

The trick is to go through moles. The ratio in the balanced equation determines the conversion, not the ratio of molar masses.

Common Mistakes

Dividing mass of CO2 by 44 but then using the result as the moles of FA1 without accounting for the ratio; here it is 1:1, but you must state it.
Using mass of FA1 as if it were mass of CO2 in the Mr equation.
Leaving the answer with too many significant figures.

Things to Be Careful About

The units cancel in Mr, so Mr has no unit. If you use your own readings, carry the error forward correctly: a wrong mass in (b)(i) should still be used consistently in (b)(ii).

Techniques used

convert mass of CO2 to molesuse stoichiometric ratio 1:1calculate molar mass

(c)

(i)

A student carries out the experiment as described in (a), except that the acid used is $15^\circ\text{C}$ colder than the acid you used. The student calculates the $M_r$ correctly from the readings obtained.

State whether the value of the $M_r$ calculated by the student is higher or lower than the value you calculated in (b)(ii). Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

The calculated $M_r$ is higher.

At the lower temperature, CO2 is more soluble in the acid (or the reaction is slower), so less CO2 escapes and the measured mass loss is reduced. The calculated moles of CO2 (and hence moles of FA 1) are therefore too low. Since $M_r = \frac{\text{mass of FA 1}}{n}$ , a smaller denominator gives a higher $M_r$ .

Final answer

Higher, because less CO2 escapes, so calculated moles of CO2 (and carbonate) are too low.

Detailed explanation

Background Concept

Gas solubility increases as temperature falls. In an open flask, the measured mass loss represents only the CO2 that has actually escaped. If some CO2 stays dissolved, the apparent mass loss is smaller than the true mass of CO2 produced.

Understanding the Question

This is an error-analysis question. The student uses colder acid but does exactly the same calculation. You must decide whether the calculated Mr will come out too high or too low and justify it through the chain: temperature → solubility/rate → mass loss → moles → Mr.

Approach

Trace the effect step by step rather than jumping to a conclusion. Identify the intermediate quantity that is too small, then see which direction that pushes Mr.

Step-by-Step Reasoning

Colder acid: CO2 is more soluble and/or the reaction is slower, so less CO2 is lost as gas.
Therefore the recorded mass loss (mass of CO2) is smaller than the true mass produced.
In the calculation, n(CO2) = mass loss/44 is too small.
Since n(FA1) = n(CO2), the calculated moles of FA1 are too small.
Mr = mass(FA1)/n(FA1): denominator too small → Mr too high.

Key Takeaways

Error questions ask you to follow the effect through the calculation. Identify whether the measured quantity is too high or too low before deciding the direction of the final error.

Common Mistakes

Saying Mr would be lower because less CO2 is produced; you must link smaller moles to a larger Mr.
Confusing 'mass loss reduced' with 'mass of FA1 reduced'.
Not mentioning that the mass of FA1 stays correct; only the moles are mis-estimated.

Things to Be Careful About

The mark scheme allows either reason (greater solubility OR slower reaction), but both must end with 'less CO2 escaped / mass loss reduced'. Then the second mark requires 'lower moles of CO2/carbonate and higher Mr'.

Techniques used

predict effect of temperature on gas solubilitylink reduced mass loss to calculated Mr

(ii)

Another student suggests that the experiment will be more accurate if the conical flask is tilted carefully to an almost horizontal position while the reaction is taking place.

Explain why the student is correct.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Tilting the flask almost horizontal allows air to enter and displace/replace the CO2, or tips out the denser CO2, so no CO2 remains trapped in the flask. The measured mass loss then equals the total mass of CO2 produced. It also helps the acid reach solid FA1 stuck to the flask walls.

Final answer

Allows air to displace CO2, so all evolved gas is lost and the mass loss is complete.

Detailed explanation

Background Concept

CO2 is denser than air. In an upright flask, CO2 can accumulate and some may remain above the liquid instead of leaving through the neck. The measured mass loss would then be too small.

Understanding the Question

The student proposes a practical modification: tilting the flask. You must explain why it improves accuracy.

Approach

Think about where the CO2 goes after it bubbles out of the solution. The flask has a wide base and narrow neck; tilting it makes it easier for the gas to escape and for air to fill the space.

Step-by-Step Reasoning

CO2 produced in the reaction is denser than the air in the flask.
In an upright flask, CO2 may pool and not leave through the neck completely.
Tilting the flask lets air enter and displaces/replaces the CO2, or tips the dense gas out.
As a result, all the CO2 produced is removed, so the mass after reaction reflects the true mass loss. Additionally, tilting helps acid wash solid carbonate from the flask walls into solution.

Key Takeaways

In mass-loss experiments, any gas that remains in the flask causes the mass loss to be underestimated. Practical modifications that promote complete escape of the gas improve accuracy.

Common Mistakes

Saying tilting increases the rate of reaction — the mark is about gas escape, not rate.
Giving only 'it mixes the contents' without saying how it affects CO2 removal.

Things to Be Careful About

Keep the answer tied to the measured mass loss. The mark scheme accepts three equivalent ideas: air replaces CO2, denser CO2 is tipped out, or acid reaches solid stuck on the walls.

Techniques used

ensure all evolved gas is expelleddisplace dense gas with air

(d)

State the uncertainty in a single balance reading.

\text{uncertainty} = \pm \text{.............................. g}

Give an expression that would enable you to calculate the percentage error in your weighing of FA 1.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Uncertainty in a single balance reading = $\pm 0.01\text{ g}$ (or $\pm 0.005\text{ g}$ ).

Percentage error in weighing FA 1:

\% \text{ error} = \frac{2 \times U}{\text{mass of FA 1}} \times 100

where $U$ is the uncertainty in one balance reading.

Final answer

U = ±0.01 g (or ±0.005 g); % error = (2U / mass FA1) × 100

Detailed explanation

Background Concept

Every measuring instrument has an uncertainty. For a balance that reads to two decimal places, the uncertainty in a single reading is usually taken as ±0.01 g (some conventions use ±0.005 g, half the smallest division). When a mass is found by difference, two separate readings contribute, so the uncertainty is doubled.

Understanding the Question

Part (d) has two parts: state the uncertainty of one balance reading, then write the percentage error in the mass of FA1. The mass of FA1 is obtained by weighing the container before and after transfer, so two readings matter.

Approach

Recall the uncertainty in a single reading, then write the general formula for percentage error of a weighing by difference: (2 × U / measured mass) × 100.

Step-by-Step Reasoning

A 2 d.p. balance has a single-reading uncertainty U = ±0.01 g (or ±0.005 g).
The mass of FA1 = mass(container + FA1) − mass(container + residual FA1), so two balance readings are involved.
Total absolute uncertainty = 2U.
Percentage error = (2U / mass of FA1) × 100.

Key Takeaways

For any measurement obtained by difference, the uncertainty is the sum of the uncertainties of the two readings. Here the two readings each have the same uncertainty, so it is 2U.

Common Mistakes

Using U instead of 2U, forgetting that two readings are involved.
Writing the percentage error without ×100.
Stating ±0.02 instead of ±0.01 for a single reading; the question asks for a single reading, not the difference.

Things to Be Careful About

If you choose U = ±0.01 g, the expression uses 2 × 0.01 = 0.02 g in the numerator. The mark scheme allows U = 0.01 or 0.005, so both conventions are acceptable as long as the expression is correct.

Techniques used

state balance uncertaintypropagate uncertainty in weighing by difference

Q2Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

The relative formula mass, $M_r$ , of metal carbonate FA 1 can also be determined by titration of a solution of FA 1 with an acid such as hydrochloric acid.

\text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

FA 3 is an aqueous solution of FA 1 containing $15.50\text{ g dm}^{-3}$ of the metal carbonate.
FA 4 is a solution of hydrochloric acid containing $4.02\text{ g dm}^{-3}$ of $\text{HCl}$ .
FA 5 is bromophenol blue indicator.

(a)

Method

● Fill the burette with FA 4.
● Pipette $25.0\text{ cm}^3$ of FA 3 into a conical flask.
● Add a few drops of FA 5 into the same 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 all your burette readings and the volume of FA 4 added in each accurate titration.

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the rough titre.
For the accurate titrations, record in a table:
- initial burette reading / cm $^3$
- final burette reading / cm $^3$
- titre / cm $^3$ (or volume of FA 4 added)
Read every burette reading to the nearest 0.05 cm $^3$ .
Carry out at least two accurate titrations; two titres should agree to within 0.10 cm $^3$ (no more than 0.20 cm $^3$ apart if only two are used).
Give the unit cm $^3$ in each heading and in each recorded reading.

Final answer

Candidate-dependent: record a rough titre and accurate burette readings to 0.05 cm3, with units; use two concordant titres.

Detailed explanation

Background Concept

In a titration, a solution of known concentration (the titrant) is added from a burette to a measured volume of analyte in a conical flask. The burette reading before and after addition gives the volume added. The endpoint is judged by an indicator colour change. Because the endpoint judgement is subjective and burette readings are limited by the scale divisions, good practical work requires repeat titrations and concordant results.

Understanding the Question

The question asks you to carry out the titration of FA 3 (metal carbonate solution) with FA 4 (hydrochloric acid) and to record the raw data properly. The marks are awarded for the method of recording: rough titre, accurate titres, correct headings and units, and precision of readings.

Approach

Set up the burette and fill it with FA 4. Pipette 25.0 cm $^3$ of FA 3 into a conical flask, add a few drops of indicator, then do a rough titration to find the approximate end point. Use that to judge how much to add during accurate titrations. Record initial and final readings, calculate titres, and repeat until results agree closely.

Step-by-Step Reasoning

The rough titre gives an approximate volume and tells you whether the burette has enough solution for subsequent titrations. Fill the burette and remove air bubbles below the tap.
For each accurate titration, record both the initial and final burette readings. The titre is the difference. Always read the burette at eye level to avoid parallax error and estimate to 0.05 cm $^3$ .
Repeat until two accurate titres agree within 0.10 cm $^3$ . This is called concordance and shows the end point was judged consistently.
Use a table with clear columns: initial reading, final reading, titre. Every heading must include the unit cm $^3$ . This is a mark-scheme requirement.

Key Takeaways

A good titration result depends on careful technique: correct burette use, precise readings, and repeating until results are concordant. Accurate recording with units is just as important as the numbers themselves.

Common Mistakes

Recording burette readings to only 0.1 or 0.2 cm $^3$ instead of 0.05 cm $^3$ .
Forgetting to quote units in headings.
Using only one accurate titre, or using non-concordant titres.
Not recording initial and final readings, only the titre.

Things to Be Careful About

Read the burette scale at eye level, read the bottom of the meniscus, and estimate between the scale divisions to 0.05 cm $^3$ . For a rough titre a single reading is enough, but accurate titrations must be repeated and only concordant values should be used for the mean.

Techniques used

fill a burette and read it correctlyrecord readings to the nearest 0.05 cm3perform a rough and accurate titrationsselect concordant titrestabulate readings with headings and units

(b)

From your accurate titration results calculate a suitable mean value to be used in your calculations. Show clearly how you obtained the mean value.

$25.0\text{ cm}^3$ of FA 3 required .............................. $\text{cm}^3$ of FA 4.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Choose two or more accurate titres that agree to within 0.20 cm $^3$ .

Example using titres 24.55 cm $^3$ , 24.65 cm $^3$ and 24.60 cm $^3$ :

\text{mean titre} = \frac{24.55 + 24.65 + 24.60}{3} = 24.60\text{ cm}^3

Round the mean to 2 decimal places.

Answer

Mean titre = 24.60 cm $^3$ (candidate-dependent, quoted to 2 d.p.)

Final answer

24.60 cm3 (example using concordant titres 24.55, 24.65 and 24.60 cm3)

Detailed explanation

Background Concept

The mean titre is the average of accurate, concordant titration results. It reduces random errors associated with judging the end point. In Cambridge practical papers the mean must be quoted to 2 decimal places, even if the raw readings were recorded to 0.05 cm $^3$ .

Understanding the Question

Part (b) asks for a mean value to be used in the later calculations. You must show how you obtained it, usually by writing the sum of the chosen titres divided by their number, or by ticking the selected readings.

Approach

Select only the accurate titres that agree closely, average them, and report the mean to 2 decimal places. Do not include the rough titre or outlying results.

Step-by-Step Reasoning

Identify two or more accurate titres that are within 0.20 cm $^3$ of each other, ideally within 0.10 cm $^3$ .
Add these values and divide by how many there are.
Round the result to the nearest 0.01 cm $^3$ . For example, 26.675 cm $^3$ becomes 26.68 cm $^3$ .
Clearly show the calculation so the examiner can see which readings were used.

Key Takeaways

The mean is only useful if it is based on concordant results. Rounding it correctly to 2 decimal places is an explicit mark in this paper.

Common Mistakes

Including the rough titre in the mean.
Averaging non-concordant values.
Quoting the mean to 3 decimal places.
Forgetting to show the working or which readings were selected.

Things to Be Careful About

The mark is only awarded if the candidate shows the working or indicates which accurate titres were used. Round half to nearest even? In this context, round 26.675 to 26.68 as normal rounding.

Techniques used

select concordant accurate titrescalculate the mean titreround to 2 decimal places

(c)

(i)

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

1M Similar

(ii)

Calculate the amount, in mol, of hydrochloric acid present in the volume of FA 4 calculated in (b).

amount of HCl = .............................. mol

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Concentration of HCl:

[\text{HCl}] = \frac{4.02}{36.5}\,\text{mol dm}^{-3} = 0.110\,\text{mol dm}^{-3}

Using the mean titre from (b), $V$ cm $^3$ :

\text{amount of HCl} = \frac{4.02}{36.5} \times \frac{V}{1000}\,\text{mol}

Example using $V = 24.60$ cm $^3$ :

\text{amount of HCl} = 0.110\, \times \, 0.02460 = 0.00271\,\text{mol}

Answer

amount of HCl = 0.00271 mol (example using $V = 24.60$ cm $^3$ )

Final answer

0.00271 mol (example using V = 24.60 cm3)

Detailed explanation

Background Concept

The amount of a solute in moles is obtained by multiplying its concentration in mol dm $^{-3}$ by its volume in dm $^3$ :

\text{amount} = \text{concentration} \times \text{volume}

Here the concentration of FA 4 is given as a mass concentration in g dm $^{-3}$ , so it must first be converted to mol dm $^{-3}$ by dividing by the molar mass of HCl, which is 36.5 g mol $^{-1}$ .

Understanding the Question

You need the amount of HCl that reacted with the 25.0 cm $^3$ of FA 3. This amount depends on the mean titre, $V$ , obtained in part (b).

Approach

Convert the concentration of HCl from g dm $^{-3}$ to mol dm $^{-3}$ .
Convert the titre volume from cm $^3$ to dm $^3$ by dividing by 1000.
Multiply the two.

Step-by-Step Reasoning

$M_r(\text{HCl}) = 1 + 35.5 = 36.5$ .
$[\text{HCl}] = 4.02 / 36.5 = 0.110\,\text{mol dm}^{-3}$ (to 3 s.f.).
If the mean titre is $V$ cm $^3$ , then volume in dm $^3$ is $V/1000$ .
Amount of HCl $= 0.110 \times V/1000$ mol.
For $V = 24.60$ cm $^3$ : amount $= 0.110 \times 0.02460 = 0.00271$ mol.

Key Takeaways

Always convert mass concentration to molar concentration before using the amount-concentration-volume relationship. Volume must be in dm $^3$ .

Common Mistakes

Using volume in cm $^3$ without dividing by 1000.
Using 36.5 as $M_r$ of HCl but forgetting unit conversion.
Quoting too many significant figures.

Things to Be Careful About

Use the unrounded mean titre from part (b) in this calculation mirroring the mark scheme and carry the unrounded value forward to part (c)(iii).

Techniques used

convert mass concentration to amount concentrationconvert volume in cm3 to dm3calculate moles from concentration and volume

(iii)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of metal carbonate in FA 3.

concentration of metal carbonate in FA 3 = .............................. mol dm $^{-3}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation:

\text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

1 mol carbonate reacts with 2 mol HCl, so:

\text{amount of carbonate in 25.0 cm}^3 = \frac{\text{amount of HCl in (c)(ii)}}{2}

Converting to concentration in mol dm $^{-3}$ :

[\text{carbonate}] = \frac{\text{(c)(ii)}}{2} \times \frac{1000}{25}\,\text{mol dm}^{-3}

Example using (c)(ii) = 0.00271 mol:

[\text{carbonate}] = \frac{0.00271}{2} \times 40 = 0.0542\,\text{mol dm}^{-3}

Answer

concentration of metal carbonate in FA 3 = 0.0542 mol dm $^{-3}$ (example using mean titre 24.60 cm $^3$ )

Final answer

0.0542 mol dm-3 (example using mean titre 24.60 cm3)

Detailed explanation

Background Concept

The titration reaction is:

\text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O}

This tells us the stoichiometric ratio: one carbonate ion reacts with two hydrogen ions. Therefore the amount of carbonate in the sample is half the amount of HCl used.

Understanding the Question

You know the amount of HCl that reacted with the 25.0 cm $^3$ sample. You are asked for the concentration of the metal carbonate in mol dm $^{-3}$ , not just in the 25.0 cm $^3$ sample.

Approach

Divide the amount of HCl by 2 to obtain the amount of carbonate in 25.0 cm $^3$ .
Multiply by 1000/25 to scale from 25.0 cm $^3$ to 1 dm $^3$ .

Step-by-Step Reasoning

From part (c)(ii), amount of HCl = 0.00271 mol (example).
Moles carbonate in 25.0 cm $^3$ = 0.00271 / 2 = 0.001355 mol.
Concentration = moles / volume in dm $^3$ = 0.001355 / 0.0250 = 0.0542 mol dm $^{-3}$ .
This is the same as multiplying by 1000/25.

Key Takeaways

When a titration sample volume is not 1 dm $^3$ , always convert the amount found in the sample to a concentration by dividing by the sample volume in dm $^3$ .

Common Mistakes

Forgetting to halve the moles of HCl.
Using 25.0 instead of 0.0250 dm $^3$ when dividing.
Confusing amount (mol) with concentration (mol dm $^{-3}$ ).

Things to Be Careful About

Carry forward the unrounded amount of HCl from (c)(ii) to avoid rounding errors. Quote the final concentration to 3 or 4 significant figures.

Techniques used

apply the 1:2 stoichiometry between carbonate and HClcalculate moles of carbonate in the 25.0 cm3 sampleconvert moles to concentration in mol dm-3

(iv)

Calculate the relative formula mass, $M_r$ , of the metal carbonate in FA 1.

$M_r$ of metal carbonate = ..............................

1M Similar

DifficultyEasy

Worked solution

Working

FA 3 contains 15.50 g dm $^{-3}$ of the metal carbonate. Therefore:

M_r = \frac{15.50}{(\text{c)(iii)}}

Example using (c)(iii) = 0.0542 mol dm $^{-3}$ :

M_r = \frac{15.50}{0.0542} = 286\,\text{g mol}^{-1}

Answer

$M_r$ of metal carbonate = 286 (example using mean titre 24.60 cm $^3$ )

Final answer

286 (example using mean titre 24.60 cm3)

Detailed explanation

Background Concept

The concentration of a solution in mol dm $^{-3}$ is related to the mass concentration in g dm $^{-3}$ by the molar mass:

\text{concentration in mol dm}^{-3} = \frac{\text{mass concentration in g dm}^{-3}}{M_r}

Rearranged:

M_r = \frac{\text{mass concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}}

Understanding the Question

FA 3 is an aqueous solution of FA 1 containing 15.50 g dm $^{-3}$ of the metal carbonate. Using the molar concentration found in (c)(iii), you can find the relative formula mass.

Approach

Divide 15.50 g dm $^{-3}$ by the concentration in mol dm $^{-3}$ from part (c)(iii). The units cancel to give g mol $^{-1}$ .

Step-by-Step Reasoning

Mass concentration = 15.50 g dm $^{-3}$ .
Molar concentration = 0.0542 mol dm $^{-3}$ (example).
$M_r = 15.50 / 0.0542 = 286$ .

Key Takeaways

The same relationship can be used in reverse: if you know the concentration and the Mr, you can find the mass concentration.

Common Mistakes

Using the volume 25.0 cm $^3$ somewhere in this calculation. It is already included in the concentration in (c)(iii).
Forgetting that relative formula mass has no unit, although the calculation gives g mol $^{-1}$ .

Things to Be Careful About

Use the unrounded concentration from (c)(iii). State the result to 3 or 4 significant figures.

Techniques used

use mass concentration and molar concentration to find Mrdivide 15.50 g dm-3 by concentration in mol dm-3

(v)

The metal carbonate in FA 1 is hydrated sodium carbonate, $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ .

Calculate the value of $x$ to the nearest whole number. Show your working.

$x$ = ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Working

M_r(\text{Na}_2\text{CO}_3) = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106

M_r(\text{H}_2\text{O}) = 2(1) + 16 = 18

For $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ :

x = \frac{M_r(\text{FA 1}) - 106}{18}

Example using $M_r = 286$ :

x = \frac{286 - 106}{18} = \frac{180}{18} = 10

Answer

$x$ = 10 (example using mean titre 24.60 cm $^3$ ; to nearest whole number)

Final answer

10 (example using mean titre 24.60 cm3)

Detailed explanation

Background Concept

A hydrated salt has a formula such as $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ . Its relative formula mass is the sum of the $M_r$ of the anhydrous salt and $x$ times the $M_r$ of water. Knowing the total $M_r$ allows $x$ to be found.

Understanding the Question

You have found the $M_r$ of the metal carbonate in part (c)(iv), and you are told the metal carbonate is hydrated sodium carbonate. You now calculate the number of water molecules, $x$ .

Approach

Calculate the $M_r$ of anhydrous sodium carbonate)Skip? Calculate: $\text{Na}_2\text{CO}_3$ = 106. Then subtract this from the total $M_r$ and divide the remainder by 18.

Step-by-Step Reasoning

$M_r(\text{Na}_2\text{CO}_3) = 2 \times 23 + 12 + 3 \times 16 = 106$ .
$M_r(\text{H}_2\text{O}) = 2 \times 1 + 16 = 18$ .
Total $M_r = 106 + 18x$ .
Rearranged: $x = (M_r - 106)/18$ .
For example, if $M_r = 286$ , then $x = (286 - 106)/18 = 10$ .

Key Takeaways

The water of crystallisation can be found from the difference between the hydrated and anhydrous formula masses.

Common Mistakes

Using 23 for sodium and forgetting there are two sodium atoms.
Dividing by 18 but not subtracting the anhydrous mass first.
Rounding $x$ incorrectly: if the calculation gives 9.7, round to 10, not 9.

Things to Be Careful About

The answer must be the nearest whole number)Skip. If the value is close to 10 or another integer, use the nearest whole number.

Techniques used

calculate Mr of anhydrous Na2CO3subtract Mr of Na2CO3 from Mr of hydrated saltdivide by Mr of water to find x

(d)

Describe a different method to determine the value of $x$ in FA 1.

This method should not involve the reaction of an acid. Explain how the method will ensure that the value of $x$ is as accurate as possible.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Weigh a sample of FA 1 in a crucible.
Heat gently at first, with the crucible lid on, to drive off the water of crystallisation and convert the salt to the anhydrous form.
Allow the crucible to cool (in a desiccator) and reweigh.
Repeat the heating and cooling until constant mass is obtained.
Mass loss = mass of water lost. Calculate:

x = \frac{\text{moles of water}}{\text{moles of anhydrous Na}_2\text{CO}_3} = \frac{(\text{mass loss})/18}{(\text{mass of anhydrous salt})/106}

Accuracy is ensured by heating to constant mass so all water is removed, and cooling in a desiccator to prevent the anhydrous salt from reabsorbing moisture.

Final answer

Heat to constant mass, cool in desiccator, reweigh; calculate x from mass loss divided by 18 and anhydrous mass divided by 106.

Detailed explanation

Background Concept

Water of crystallisation can be determined gravimetrically. When a hydrated salt is heated, water is driven off and the mass decreases. The loss in mass is the mass of water originally present. By comparing the moles of water with the moles of anhydrous salt, the value of $x$ in $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ can be calculated.

Understanding the Question

The question asks for a different method, one that does not involve reaction with an acid. A gravimetric heating method is appropriate. You also need to explain how the method ensures accuracy.

Approach

Weigh the hydrated salt, heat it until all water is removed, cool it so it does not regain moisture, reweigh, and repeat until constant mass. The mass difference gives moles of water; the final mass gives moles of anhydrous salt.

Step-by-Step Reasoning

Weigh an empty clean, dry crucible, then a known mass of FA 1.
Heat the crucible gently at first, with the lid on, so the salt does not spit out.
Heat more strongly but avoid decomposing the anhydrous carbonate.
Cool in a desiccator to prevent absorption of atmospheric moisture, then weigh.
Repeat heating, cooling and weighing until two successive masses agree. This is constant mass and shows all water has been removed.
Mass of water lost = initial mass of hydrated salt - final mass of anhydrous salt.
Moles of water = mass loss / 18. Moles of anhydrous sodium carbonate = final mass / 106.
$x$ = moles of water / moles of anhydrous salt.

Key Takeaways

Heating to constant mass is the key to a reliable gravimetric determination. Cooling in a desiccator avoids reabsorption of water, which would make the mass loss too small.

Common Mistakes

Heating so strongly that sodium carbonate decomposes, losing CO $_2$ as well as water, giving a false mass loss.
Weighing while hot; convection currents and reabsorption of moisture cause inaccurate results.
Not repeating until constant mass.
Confusing mass loss with moles; always convert masses to moles using molar masses.

Things to Be Careful About

The mark scheme awards one mark for heating FA 1 to remove water and a second mark for heating to constant mass. Mention the lid, gentle heating and desiccator in your accuracy explanation.

Techniques used

heat a sample to drive off water of crystallisationheat to constant massweigh before and after heatingcalculate mole ratio of water to anhydrous salt

Q3MediumQualitative AnalysisManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

(a)

(i)

FA 6 contains one cation and one anion both of which are listed in the Qualitative analysis notes.

Heat a small spatula measure of FA 6 in a hard-glass test-tube, until no further change occurs.

Record all your observations. Identify any gas produced.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

FA 6 is a (light) green / blue-green solid / powder.
On heating: condensation / steam / water vapour produced; the solid fluidises / jumps about in the test-tube; a black residue / solid remains.
The gas turned limewater milky (white precipitate), so the gas is carbon dioxide, CO2.

Final answer

Gas is carbon dioxide (CO2) — turns limewater milky; black residue of CuO forms; condensation produced.

Detailed explanation

Background Concept

Copper(II) carbonate, $\text{CuCO}_3$ , and its common basic form, basic copper(II) carbonate (malachite), $\text{Cu}_2\text{(OH)}_2\text{CO}_3$ , are green or blue-green solids. On strong heating both decompose to give black copper(II) oxide, CuO. The simple carbonate gives carbon dioxide:

\text{CuCO}_3\text{(s)} \rightarrow \text{CuO(s)} + \text{CO}_2\text{(g)}

The basic carbonate also loses water:

\text{Cu}_2\text{(OH)}_2\text{CO}_3\text{(s)} \rightarrow 2\text{CuO(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

Carbon dioxide is confirmed by bubbling the gas into limewater (aqueous $\text{Ca(OH)}_2$ ): $\text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O}$ , producing a white / milky precipitate.

Understanding the Question

FA 6 is an unknown solid. You heat it in a hard-glass test-tube and record everything you see — the starting appearance, what happens during heating, the colour of the final residue — and then identify any gas produced using limewater. The question carries 3 marks, so three or more distinct observations plus the gas identification are expected.

Approach

Split the task into observation of (i) the starting solid, (ii) the changes on heating (steam, movement of the solid, colour change), and (iii) the gas test. Work from what is visible to the confirmatory limewater test.

Step-by-Step Reasoning

Record the original appearance: FA 6 is a (light) green / blue-green solid powder. This colour is characteristic of copper carbonates.
Heat strongly. Condensation / steam collects on the cooler upper part of the tube, signalling that water is being driven off — which is why a basic (hydroxide-containing) carbonate is present.
The solid swells / jumps about / fluidises as gas is released from within it.
The residue turns black: this is solid CuO, the decomposition product.
Collect the gas and bubble it into limewater. A white precipitate / milky appearance confirms CO2.
Conclusion: the gas is carbon dioxide.

Key Takeaways

A decomposition practical follows a fixed script: note the starting colour, watch the physical changes during heating, describe the final residue, and confirm any gas with the appropriate test. For a carbonate the test is limewater.

Common Mistakes

Forgetting to record the starting colour of the solid — it is a credited observation.
Writing that the gas 'turns limewater cloudy' without specifying the white precipitate / milky colour the mark scheme requires.
Naming the gas by guesswork instead of carrying out (or describing) the limewater test.
Missing the condensation; if you watch only the bottom of the tube, the steam at the top is easily lost.

Things to Be Careful About

Use a hard-glass test-tube: ordinary test-tubes can crack under strong heating.
Condensation, the fluidising of the solid and the black residue are separate credited observations — record each.
The decomposition equation ( $\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2$ ) is worth being able to write, even though this part only asks you to identify the gas.

Techniques used

heat a solid carbonate in a hard-glass test-tubetest a gas with limewaterrecord observations of colour change and residue

(ii)

Allow the residue to cool for 2 minutes. Then transfer a small quantity of the residue from (a)(i) into a test-tube containing a $2\text{ cm}$ depth of dilute sulfuric acid. Shake the test-tube.

Record your observations.

1M Similar

DifficultyEasy

Worked solution

Answer

A (light) blue solution formed as the black solid dissolved.

Final answer

Light blue solution formed.

Detailed explanation

Background Concept

The black residue from part (a)(i) is copper(II) oxide, CuO — a typical insoluble basic (metal) oxide. Insoluble basic oxides react with dilute acids to produce a salt and water:

\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}

The aqueous copper(II) sulfate that forms is strongly blue because of the hydrated $\text{Cu}^{2+}\text{(aq)}$ ion.

Understanding the Question

You take a small amount of the cooled residue, add it to dilute sulfuric acid, shake, and record what you see. The single mark is for recognising that a (light) blue solution forms.

Approach

Identify the black residue as a basic oxide, then recall that a basic oxide + acid gives a soluble salt plus water. Look for the solid dissolving and the colour of the resulting solution.

Step-by-Step Reasoning

The black solid is CuO.
CuO is insoluble in water but reacts with dilute H2SO4.
The solid particles disappear (dissolve/react) and a light blue solution forms — this is CuSO4(aq).
No gas is produced in this reaction.

Key Takeaways

Basic (metallic) oxides are neutralised by acids to give salt + water; the colour of the aqueous transition-metal salt (here blue for Cu2+) is the key observation.

Common Mistakes

Describing a 'blue precipitate'. No precipitate forms here — the solid dissolves to give a blue solution. (A blue precipitate would appear, for example, on adding NaOH, which is a different test.)
Writing that a gas is given off; this is a simple neutralisation with no gas.

Things to Be Careful About

The observation is about the solution colour, not the solid.
Allow the residue to cool for 2 minutes before adding acid, exactly as instructed — adding acid to a very hot tube can cause a violent reaction.
Only a small quantity of residue is needed.

Techniques used

add dilute sulfuric acid to the black residueobserve the colour change of the solution

(iii)

Give the equation for the reaction in (a)(ii).

Include state symbols.

1M Similar

DifficultyEasy

Worked solution

Answer

\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}

Final answer

CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)

Detailed explanation

Background Concept

This is a neutralisation reaction between a basic oxide and an acid. The two products are the salt and water. Here the basic oxide is copper(II) oxide and the acid is dilute sulfuric acid, giving copper(II) sulfate and water.

Understanding the Question

Write the balanced symbol equation, with state symbols, for the reaction that produced the blue solution in part (a)(ii). This is a cao (correct answer only) mark — both the balancing and the state symbols must be correct.

Approach

Identify the reactants (CuO and H2SO4) and products (CuSO4 and H2O), then balance and add state symbols based on the phases seen in the experiment.

Step-by-Step Reasoning

Reactants: CuO is the black solid, so CuO(s); dilute H2SO4 is an aqueous solution, H2SO4(aq).
Products: the blue solution is aqueous copper(II) sulfate, CuSO4(aq); water is a liquid, H2O(l).
Check balance: 1 Cu on each side; 1 S on each side; 5 O on the left (1 in CuO + 4 in H2SO4) and 5 on the right (4 in CuSO4 + 1 in H2O); 2 H on each side. Balanced.

\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}

Key Takeaways

For oxide + acid reactions, salt + water is the product pattern; remember every species needs its state symbol, chosen from the phases in the experiment.

Common Mistakes

Missing state symbols, or giving CuO(aq) / CuSO4(s) — the phases must match what was actually present.
Writing an unbalanced equation, or adding an extra water molecule.

Things to Be Careful About

The equation must balance in atoms (no ions here, so no charge check needed).
There is exactly one water molecule — a common slip is to add more.

Techniques used

write a balanced symbol equationinclude state symbols

(b)

FA 7 is a sample of the solution produced by the experiment described in (a)(ii).
FA 8 is a solution of a salt containing one anion.

10M

(i)

Use a $1\text{ cm}$ depth of FA 7 in a test-tube for each of the following tests.
Record your observations in Table 3.1.

Table 3.1

test	observations
Test 1 Add aqueous sodium hydroxide.
Test 2 Add aqueous ammonia, then add dilute hydrochloric acid.
Test 3 Add aqueous sodium carbonate.
Test 4 Add several drops of FA 8, then add a few drops of aqueous starch.
Test 5 Add a small spatula measure of iron powder. Leave the test-tube to stand.
Test 6 Add a few drops of aqueous barium nitrate or aqueous barium chloride.

6M Similar

DifficultyMedium

Worked solution

Answer

Test	Observations
1 (NaOH)	(Pale/light) blue precipitate forms; insoluble in excess NaOH.
2 (NH3 then dilute HCl)	(Pale) blue precipitate; soluble / disappears in excess NH3 giving a deep / dark blue solution; adding dilute HCl makes the solution paler blue or re-forms a blue precipitate.
3 (Na2CO3)	Fizzing / effervescence; blue precipitate forms.
4 (FA 8, then starch)	(Off-)white precipitate / solid forms; with starch goes dark blue / blue-black / black.
5 (Fe powder, leave to stand)	Brown solid / precipitate forms; fizzing or solution goes paler (blue) or colourless; the gas pops with a lighted splint (H2).
6 (Ba(NO3)2 or BaCl2)	White precipitate forms.

Final answer

Test-by-test observations as in the table: blue ppt insoluble in excess NaOH; blue ppt soluble in excess NH3 giving deep blue; effervescence + blue ppt with carbonate; off-white ppt then blue-black with starch; brown solid + fizzing/pop with Fe; white ppt with barium.

Detailed explanation

Background Concept

FA 7 is the solution from part (a)(ii) — acidified copper(II) sulfate, i.e. $\text{CuSO}_4\text{(aq)}$ in dilute $\text{H}_2\text{SO}_4$ . FA 8 is a salt solution containing a single anion — here potassium iodide, so the reagent supplies $\text{I}^-\text{(aq)}$ . The six tests explore two areas of qualitative chemistry: the reactions of the $\text{Cu}^{2+}\text{(aq)}$ cation with alkalis and ammonia, and confirmations of anions (sulfate, iodide) and gases ( $\text{H}_2$ ).

The key reactions are:

Test 1 (NaOH): $\text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Cu(OH)}_2\text{(s)}$ , a (pale) blue precipitate that is insoluble in excess NaOH.
Test 2 (NH3 then HCl): $\text{Cu}^{2+} + 2\text{NH}_3 + 2\text{H}_2\text{O} \rightarrow \text{Cu(OH)}_2\text{(s)} + 2\text{NH}_4^+$ ; in excess NH3 the precipitate dissolves to give the deep / dark blue complex $[\text{Cu(NH}_3)_4]^{2+}$ . Adding dilute HCl removes ammonia and converts the complex back, so the solution becomes paler blue or a blue precipitate reappears.
Test 3 (Na2CO3): the H2SO4 already in FA 7 reacts with carbonate to give CO2 (fizzing / effervescence), and $\text{Cu}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \rightarrow \text{CuCO}_3\text{(s)}$ , a blue precipitate.
Test 4 (FA 8 = KI, then starch): $\text{Cu}^{2+}$ oxidises iodide: $2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$ . CuI is an (off-)white precipitate; the iodine produced turns starch dark blue / blue-black.
Test 5 (Fe powder): Fe displaces copper: $\text{Fe(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Cu(s)}$ , depositing a brown solid; Fe also reacts with the acid $\text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ , giving fizzing and a gas that pops with a lighted splint.
Test 6 (Ba(NO3)2 or BaCl2): $\text{SO}_4^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}$ , a white precipitate — the classic sulfate test.

Understanding the Question

You are given six tests to carry out on the acidified copper(II) sulfate solution and asked to record the observation for each in Table 3.1, worth 6 marks total. The observations are the whole point — no interpretation is asked for here, just precise recording. Remember that FA 7 already contains dilute sulfuric acid, which changes some results (the fizzing in Test 3 and the H2 in Test 5).

Approach

Work test by test. For each, think about which ion reacts (Cu2+, SO4^2-, H+) and what the product looks like and whether it dissolves in excess reagent.

Step-by-Step Reasoning

Test 1: Any solution of Cu2+ gives a blue precipitate with NaOH. Since Cu(OH)2 is insoluble in excess NaOH, the precipitate stays. Give both observations: blue ppt, insoluble in excess.
Test 2: Same blue precipitate with NH3, but Cu(OH)2 dissolves in excess NH3 to give the deep blue complex. Then HCl reverses this — paler blue or re-formed blue ppt. State both stages.
Test 3: The carbonate reacts with the acid already present — effervescence — and gives a blue CuCO3 precipitate.
Test 4: KI provides I^-, which Cu2+ oxidises, giving white CuI and iodine. Iodine + starch gives dark blue / blue-black. Record: off-white ppt; with starch gives blue-black.
Test 5: Fe is more reactive than Cu, so a brown solid (Cu) deposits; the acid gives H2 — fizzing and a pop with a lighted splint.
Test 6: Barium ions precipitate sulfate as white BaSO4.

Key Takeaways

Cu2+ is distinguished from most cations by giving a blue hydroxide precipitate that is insoluble in excess NaOH but soluble in excess NH3 (deep blue solution).
The sulfate test is a white precipitate with barium ions.
Cu2+ with iodide is a redox reaction producing white CuI and iodine (starch gives blue-black).
When the solution already contains acid, carbonate tests fizz (CO2) and reactive metals give H2.

Common Mistakes

For Test 2, writing only 'blue ppt' and missing the deep blue complex in excess NH3 and the effect of HCl.
For Test 4, calling the precipitate 'blue' — CuI is off-white; the blue-black colour appears only after adding starch, from the iodine.
For Test 5, mentioning the brown solid but not the fizzing / pop (the H2 from the acid).
For Test 3, missing the effervescence — it is credited separately from the precipitate.

Things to Be Careful About

FA 7 is acidified copper(II) sulfate — that acid is responsible for both the fizzing in Test 3 and the H2 in Test 5.
Record each observation as it is seen; several tests yield two credited observations.
The starch only gives its colour if iodine is present — an off-white solid alone is not the blue-black sign.

Techniques used

test cations with aqueous sodium hydroxidetest cations with aqueous ammonia then dilute hydrochloric acidtest for sulfate with barium nitrate or barium chloridetest for iodide with starchdetect hydrogen with a lighted splintrecord observations in a results table

(ii)

Carry out a different test from those you have already carried out in (b)(i) to confirm the identity of the anion in FA 8.

Record the reagent(s) used and your observations and give the formula of the anion.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Add (aqueous) silver nitrate, AgNO3(aq). A (pale) yellow precipitate formed that is insoluble in (excess) aqueous ammonia. Therefore FA 8 contains I⁻.

(Alternatively: add acidified aqueous potassium manganate(VII), KMnO4 — a brown / yellow solution forms, giving a dark blue / blue-black colour with starch.)

Final answer

Add AgNO3(aq) → pale yellow ppt, insoluble in excess NH3(aq) → FA 8 contains I⁻.

Detailed explanation

Background Concept

The standard confirmatory test for halide ions uses aqueous silver nitrate. Silver halides are insoluble and each has a distinguishing colour: AgCl white, AgBr cream and AgI pale yellow. They also differ in solubility in ammonia: AgCl dissolves in dilute NH3, AgBr in concentrated NH3, and AgI is insoluble even in excess (concentrated) ammonia. Iodide can also be identified by oxidation to iodine — acidified KMnO4 oxidises I^- to I2, which colours starch blue-black.

Understanding the Question

FA 8 is a solution of a salt containing one anion. In part (b)(i) you have already used NaOH, NH3/HCl, Na2CO3, KI/starch, iron and barium reagents. You must now choose a different test to confirm the anion, record the reagent(s) and observations, and give the formula of the anion. The mark scheme credits the reagent, the key observation(s) and the conclusion that the anion is I^-.

Approach

Pick the AgNO3 test (not used in (b)(i)) — it directly gives the pale yellow AgI precipitate that does not dissolve in excess ammonia — or the acidified KMnO4 + starch route. Both arrive at iodide via iodine chemistry.

Step-by-Step Reasoning

Add a few drops of aqueous silver nitrate, $\text{AgNO}_3\text{(aq)}$ .
A (pale) yellow precipitate forms: $\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}$ .
Add excess aqueous ammonia: AgI does not dissolve, confirming it is iodide rather than chloride (white, soluble) or bromide (cream, sparingly soluble).
Conclusion: FA 8 contains I^-.

(Alternative route: add acidified aqueous KMnO4 — a brown / yellow solution forms as I^- is oxidised to I2; adding starch then gives a dark blue / blue-black colour, the positive test for iodine.)

Key Takeaways

Silver halide precipitates are distinguished by colour AND their solubility in ammonia; AgI is the pale yellow one that stays insoluble. Oxidising iodide to iodine and testing with starch is an equally valid route.

Common Mistakes

Using a reagent already used in (b)(i) — the question asks for a different test.
Confusing the colours of the silver halides: AgCl white, AgBr cream, AgI pale yellow.
Reporting the precipitate as dissolving in ammonia (that is AgCl behaviour) when the observation should be 'insoluble'.

Things to Be Careful About

State the reagent clearly (AgNO3(aq), or acidified KMnO4).
The 'insoluble in excess ammonia' observation is the decisive point that separates iodide from chloride.
The formula of the anion is I^-, not KI.

Techniques used

confirm iodide with aqueous silver nitratetest solubility of the precipitate in excess ammoniaoxidise iodide with acidified potassium manganate(VII) and test with starch

(iii)

List all the numbered tests you carried out in (b)(i) which involved redox reactions.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Tests 4 and 5.

Final answer

Tests 4 and 5

Detailed explanation

Background Concept

A redox reaction is one in which electrons are transferred, i.e. the oxidation numbers (oxidation states) of some atoms change. Precipitation, complexation and acid-carbonate reactions do not change oxidation numbers, so they are not redox. Redox here happens in Tests 4 and 5.

Understanding the Question

List all the numbered tests in (b)(i) that involved redox reactions. This requires applying the definition of redox to each of the six tests performed. The answer is a list: '4 and 5'.

Approach

Go through each test and check whether any atoms change oxidation number. If no oxidation state changes, the reaction is not redox.

Step-by-Step Reasoning

Test 1 (NaOH): Cu2+ → Cu(OH)2. No change in oxidation number; precipitation only. Not redox.
Test 2 (NH3/HCl): forms [Cu(NH3)4]2+ which retains Cu2+; complexation and neutralisation. Not redox.
Test 3 (Na2CO3): Cu2+ → CuCO3 and H+ + CO3^2- → CO2 + H2O; no oxidation number change. Not redox.
Test 4 (KI + starch): Cu2+ is reduced to Cu+ (in CuI), from +2 to +1, while I^- is oxidised to I2, from -1 to 0. Redox.
Test 5 (Fe powder): Fe (0) → Fe2+ (+2) while Cu2+ (+2) → Cu (0); also Fe + H+ → Fe2+ + H2. Redox.
Test 6 (Ba salt): SO4^2- + Ba2+ → BaSO4; precipitation only. Not redox.

So the redox tests are 4 and 5.

Key Takeaways

Use oxidation-number change as the criterion for redox, not whether a solid forms or a colour changes. Precipitation and acid-carbonate reactions are usually not redox.

Common Mistakes

Thinking precipitation or colour change implies redox.
Missing Test 4, whose Cu2+/I^- electron transfer is the giveaway.
Including Test 6 (barium sulfate precipitation), which is not redox.

Things to Be Careful About

The answer must be a list of the test numbers exactly as labelled in (b)(i): '4 and 5'.
Only the numbers are needed.

Techniques used

identify redox reactions by change in oxidation numberclassify the six qualitative tests

(iv)

Give the ionic equation for the reaction taking place in Test 5 in (b)(i).

Include state symbols.

1M Similar

DifficultyEasy

Worked solution

Answer

\text{Cu}^{2+}\text{(aq)} + \text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Cu(s)}

(or $\text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ )

Final answer

Cu2+(aq) + Fe(s) → Fe2+(aq) + Cu(s)

Detailed explanation

Background Concept

Test 5 adds iron powder to acidified copper(II) sulfate. Two redox reactions can occur:

Displacement of copper: $\text{Cu}^{2+}\text{(aq)} + \text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Cu(s)}$ . Iron, being more reactive, donates two electrons and becomes Fe2+ while Cu2+ is reduced to copper metal.
Reaction with the acid: $\text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ , producing the hydrogen that fizzes and pops.

An ionic equation shows only the species actually undergoing change — spectator ions (here SO4^2-) are omitted. Charge and atoms must both balance, and state symbols are required.

Understanding the Question

Write the ionic equation, with state symbols, for the reaction in Test 5. Either of the two reactions above is accepted (1 mark).

Approach

Choose the displacement reaction as the main one; write the half-equations, combine them, then add state symbols and check balance.

Step-by-Step Reasoning

Reduction: $\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}$ .
Oxidation: $\text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2\text{e}^-$ .
Combine (electrons cancel): $\text{Cu}^{2+}\text{(aq)} + \text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Cu(s)}$ .
Check: atoms (1 Cu and 1 Fe each side) and charge (+2 + 0 = +2 on both sides) balance.

(Alternative accepted: $\text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ .)

Key Takeaways

To write an ionic equation: identify the reacting species, cancel spectator ions, balance atoms and charge, and add state symbols based on what is actually present as solid, aqueous, liquid or gas.

Common Mistakes

Leaving out state symbols — the mark specifically requires them.
Writing a full equation with SO4^2- included as a spectator ion.
Not balancing the charges when writing the combined equation.

Things to Be Careful About

The solids Fe and Cu are (s); the ions are (aq); H2 is (g).
Both the metal-displacement and the acid reaction are credited — give only one balanced ionic equation.
The equation must match the reaction you claim is happening in Test 5.

Techniques used

write an ionic equationbalance atoms and chargeinclude state symbols