Chemistry 9701/52 — February/March 2020

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

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Topics Planning · Analysis, Conclusions and Evaluation

Q1PlanningAnalysis, Conclusions and EvaluationFree sample

Brass is an alloy of copper and zinc. Typical copper concentrations vary from 50% to 85%, depending upon the properties needed in the alloy. There may be small amounts of other metals present.

A student found a method to determine the percentage of copper in a sample of brass.

A known mass of brass powder is reacted with excess concentrated nitric acid. Both the copper and the zinc and any other metals present are oxidised into aqueous ions by the nitric acid. The amount of $\text{Cu}^{2+}\text{(aq)}$ ions present can be determined by a titration technique.

step 1 Use a weighing boat to accurately weigh by difference approximately 2g of brass powder and place the brass into a small glass beaker.

step 2 In a fume cupboard add approximately $20\text{ cm}^3$ of concentrated nitric acid to the brass in the beaker. Allow the brass to completely react to form solution A.

The equation for the reaction is shown.

\text{Cu(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3\text{)}_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

step 3 Dilute all of solution A to form exactly $250.0\text{ cm}^3$ of solution B.

step 4 Place $25.00\text{ cm}^3$ of solution B into a conical flask.

step 5 Use a dropping pipette to add aqueous sodium carbonate, $\text{Na}_2\text{CO}_3\text{(aq)}$ , to solution B in the conical flask until there is no more acid present.

step 6 Add approximately $20\text{ cm}^3$ of aqueous potassium iodide, KI(aq), to the conical flask. A white precipitate forms as well as a brown solution of aqueous iodine, $\text{I}_2\text{(aq)}$ .

step 7 Fill a burette with $0.100\text{ mol dm}^{-3}$ sodium thiosulfate solution, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ , so it is ready for the titration in step 8.

step 8 Carry out a titration of the aqueous iodine produced in the conical flask against the $0.100\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ .

(a)

Outline how the student should accurately weigh by difference in step 1 in order that the exact mass of brass transferred into the small glass beaker is known. Include a results table, with appropriate headings, ready for the student to fill in.

DifficultyEasy

Worked solution

Answer

Weigh the weighing boat with the brass powder.
Transfer the brass powder into the small glass beaker.
Reweigh the empty weighing boat.
Mass of brass transferred = mass of boat + brass before transfer − mass of empty boat after transfer.

	/ g
Mass of boat + brass before transfer
Mass of boat after transfer
Mass of brass transferred

Final answer

Weigh boat + brass, transfer brass, reweigh empty boat; table with before/after/difference headings in g.

Detailed explanation

Background Concept

Weighing by difference is a standard quantitative technique. It is used when a solid is transferred from one vessel to another and the exact amount transferred must be known. The balance reading before transfer minus the balance reading after transfer gives the mass transferred, even if a small residue remains in the original container.

Understanding the Question

Step 1 says “accurately weigh by difference approximately 2 g of brass powder”. The question asks for the procedure and a results table with headings. Two marks are available: one for the order of weighing, one for the table with units.

Approach

Weigh the boat + brass, transfer the brass to the beaker, then reweigh the empty boat. Subtract the two readings. Present the values in a table with rows for before, after and difference, with the unit g.

Step-by-Step Reasoning

Place the brass powder in a weighing boat and record its mass.
Tip the powder into the small beaker.
Reweigh the now-empty boat.
Mass transferred = (mass of boat + brass before) − (mass of empty boat after).

The table should have three rows and a /g heading, ready for the student to fill in.

Key Takeaways

Weighing by difference avoids the need for a perfect transfer and gives an accurate mass.

Common Mistakes

Not reweighing the empty boat.
Recording only one mass.
Omitting units or using inconsistent precision.

Things to Be Careful About

Use the same balance and record both masses to the same number of decimal places. The table heading should include the unit g.

Techniques used

weigh by differenceconstruct a results table with headings and units

(b)

Suggest why it is necessary to do step 2 in a fume cupboard.

DifficultyEasy

Worked solution

Answer

Concentrated nitric acid oxidises the metals and produces nitrogen dioxide, NO₂, which is a toxic/poisonous gas; the fume cupboard removes it.

Final answer

Toxic/poisonous nitrogen dioxide gas is produced.

Detailed explanation

Background Concept

Concentrated nitric acid is a strong oxidising agent. When it reacts with metals such as copper and zinc, the metals are oxidised to aqueous ions and the nitrate is reduced, producing nitrogen dioxide, NO₂, a brown and toxic gas.

Understanding the Question

Step 2 is done in a fume cupboard. The question asks why, for one mark. The key is to identify the hazardous gas produced.

Approach

Recognise that the reaction produces a toxic gas, so it must be carried out in a fume cupboard to prevent inhalation.

Step-by-Step Reasoning

The equation given shows NO₂(g) as a product. NO₂ is toxic/poisonous. A fume cupboard removes the gas and protects the student.

Key Takeaways

Reactions that produce toxic gases must be done in a fume cupboard.

Common Mistakes

Saying “acid fumes” without naming NO₂ or toxic gas.
Saying “to prevent explosion”, which is not the reason here.

Things to Be Careful About

The mark requires “toxic/poisonous gas given off”; naming NO₂ is a good addition.

Techniques used

identify the toxic gas producedrelate fume cupboard use to gas hazard

(c)

Outline how the student should carry out step 3. Include the name and capacity of the suitable piece of apparatus in which solution B should be prepared.

DifficultyMedium-Easy

Worked solution

Answer

Transfer all of solution A into a 250 cm³ volumetric flask.
Rinse the beaker and stirring rod with distilled water and add the rinsings to the flask.
Make up to the mark with distilled water, then mix thoroughly.

Final answer

Transfer to 250 cm³ volumetric flask, rinse beaker, make up to mark with distilled water.

Detailed explanation

Background Concept

A volumetric flask is used to prepare a solution of exactly known volume. To transfer a solution quantitatively, all material must be rinsed into the flask and the flask made up to the mark with distilled water.

Understanding the Question

Step 3 requires diluting all of solution A to exactly 250.0 cm³ of solution B. The question asks for the name and capacity of the apparatus and an outline of the procedure.

Approach

Use a 250 cm³ volumetric flask. Transfer solution A, rinse the beaker, then make up to the mark with distilled water.

Step-by-Step Reasoning

Pour solution A into a 250 cm³ volumetric flask.
Rinse the beaker and any stirring rod with distilled water and add the rinsings to the flask. This ensures all brass ions are transferred.
Add distilled water until the bottom of the meniscus is on the graduation mark.
Stopper and invert to mix.

The apparatus is a 250 cm³ volumetric flask.

Key Takeaways

Quantitative dilution requires rinsing and making up to the mark in a volumetric flask.

Common Mistakes

Using a measuring cylinder instead of a volumetric flask.
Not rinsing the beaker.
Filling above the mark.

Things to Be Careful About

Read the mark at eye level at the bottom of the meniscus.

Techniques used

prepare a solution in a volumetric flaskrinse apparatus quantitativelymake up to the mark

(d)

Name the apparatus needed to transfer solution B into the conical flask in step 4.

DifficultyEasy

Worked solution

Answer

25.00 cm³ pipette.

Final answer

25 cm³ pipette.

Detailed explanation

Background Concept

A pipette delivers a fixed, accurate volume, for example 25.00 cm³. It is the correct apparatus for transferring a precise aliquot.

Understanding the Question

Step 4 transfers 25.00 cm³ of solution B into a conical flask. The question asks for the name of the apparatus.

Approach

Recall that a 25 cm³ pipette is used for accurate fixed volumes.

Step-by-Step Reasoning

A measuring cylinder is not accurate enough; a burette is for variable volumes; a pipette delivers exactly 25.00 cm³.

Key Takeaways

Pipettes are used for accurate fixed-volume transfers.

Common Mistakes

Saying “measuring cylinder” or “burette”.

Things to Be Careful About

Include “25 cm³” in the answer.

Techniques used

select the correct volumetric apparatus

(e)

State how the student would know there was no more acid present in the mixture in step 5.

DifficultyEasy

Worked solution

Answer

No more effervescence/bubbles of carbon dioxide are seen when more Na₂CO₃(aq) is added.

Final answer

No more effervescence is seen.

Detailed explanation

Background Concept

Sodium carbonate reacts with acid to produce carbon dioxide gas, causing effervescence. When all acid has been neutralised, no more CO₂ is produced.

Understanding the Question

Step 5 adds Na₂CO₃ until no more acid is present. The question asks how the student would know when this point is reached.

Approach

Observe the mixture for effervescence; when it stops, the acid has been neutralised.

Step-by-Step Reasoning

Acid + carbonate → salt + water + CO₂. While acid remains, adding carbonate produces bubbles. When no more bubbles appear, no acid remains.

Key Takeaways

Effervescence indicates the acid–carbonate reaction; its cessation indicates neutralisation.

Common Mistakes

Saying “use pH paper”, which is not part of the described method.
Saying “the solution becomes colourless”, which is not relevant.

Things to Be Careful About

The answer should be “no more effervescence is seen”.

Techniques used

observe effervescence as an end-point indicator

(f)

The student is given $200\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ .

Outline how the student should use this solution to fill the burette in step 7 so it is ready for titration. Include any relevant procedures the student should follow to ensure the burette is correctly filled before any readings are taken.

DifficultyMedium-Easy

Worked solution

Answer

Rinse the burette with a little Na₂S₂O₃(aq).
Fill the burette with the thiosulfate solution.
Run some solution through the tap into a waste beaker to remove any air bubble below the tap.
Ensure the jet is filled and record the initial reading.

Final answer

Rinse burette with Na₂S₂O₃(aq), fill, run solution through tap to remove air.

Detailed explanation

Background Concept

A burette must be rinsed with the solution it will contain, not water, to avoid diluting it. Air bubbles below the tap must be removed because they would be delivered as part of the titre, giving an incorrect reading.

Understanding the Question

Step 7 requires filling a burette with 0.100 mol dm⁻³ Na₂S₂O₃(aq). The question asks for the procedure, including any relevant steps before readings are taken.

Approach

Rinse with thiosulfate, fill, run solution through the tap to remove air, then record the initial reading.

Step-by-Step Reasoning

Rinse the burette with a little Na₂S₂O₃(aq) to remove water and avoid dilution.
Fill the burette with thiosulfate using a funnel.
Open the tap briefly to fill the jet below the tap and remove air bubbles.
Ensure no air bubble remains, then record the initial reading.

The two marks are for rinsing with thiosulfate and running solution through the tap.

Key Takeaways

Burette preparation: rinse with the solution, fill, remove air from the jet.

Common Mistakes

Rinsing with distilled water only.
Not removing the air bubble.

Things to Be Careful About

Do not rinse with water; use the thiosulfate solution.

Techniques used

rinse a burette with the titrantremove air from the burette jet

(g)

The titration table the student used is shown.

titration number	rough	1	2	3
final burette reading/ $\text{cm}^3$	20.50	40.25	19.90	39.65
initial burette reading/ $\text{cm}^3$	0.00	20.60	0.00	19.90
titre/ $\text{cm}^3$

(i)

Complete the table and calculate the mean titre to be used in calculating the percentage of copper in brass.
Show your working.

DifficultyMedium-Easy

Worked solution

Working

Titres:

rough: 20.50 − 0.00 = 20.50 cm³
1: 40.25 − 20.60 = 19.65 cm³
2: 19.90 − 0.00 = 19.90 cm³
3: 39.65 − 19.90 = 19.75 cm³

Concordant titres are 1 and 3:
mean = (19.65 + 19.75)/2 = 19.70 cm³

Answer

19.70 cm³

Final answer

19.70 cm³

Detailed explanation

Background Concept

Titre = final burette reading − initial burette reading. Concordant titres are those within 0.10 cm³ of each other; the mean of concordant titres is used in the calculation.

Understanding the Question

The table gives burette readings for a rough titration and three further titrations. The question asks you to complete the titre column and calculate the mean titre to use.

Approach

Calculate each titre by subtraction. Identify the concordant set. Average only those concordant values.

Step-by-Step Reasoning

Titres:

rough: 20.50 − 0.00 = 20.50 cm³
1: 40.25 − 20.60 = 19.65 cm³
2: 19.90 − 0.00 = 19.90 cm³
3: 39.65 − 19.90 = 19.75 cm³

Titres 1 and 3 are within 0.10 cm³ of each other (19.65 and 19.75). Titre 2 at 19.90 is outside this range, so it is discarded. Mean = (19.65 + 19.75)/2 = 19.70 cm³.

Key Takeaways

Always calculate titres and average only concordant values.

Common Mistakes

Averaging all three titres including the outlier.
Arithmetic errors in subtraction.

Things to Be Careful About

Record titres to 2 decimal places; use concordant values.

Techniques used

calculate titres from burette readingsidentify concordant titrescalculate a mean titre

(ii)

The burette used by the student has graduations of $0.10\text{ cm}^3$ .

Determine the percentage error in the titre measured in titration number 2.

Show your working.

DifficultyMedium-Easy

Worked solution

Working

Each burette reading has an uncertainty of ±0.05 cm³. A titre uses two readings, so absolute error = 2 × 0.05 = 0.10 cm³.

Percentage error = (0.10/19.90) × 100 = 0.503%

Answer

0.503%

Final answer

0.503%

Detailed explanation

Background Concept

A burette with graduations of 0.10 cm³ can be read to ±0.05 cm³. A titre is the difference of two readings, so the absolute error is 2 × 0.05 = 0.10 cm³. Percentage error = (absolute error / measured value) × 100.

Understanding the Question

Titre 2 is 19.90 cm³. The burette has graduations of 0.10 cm³. Determine the percentage error in this titre, showing working.

Approach

Use absolute error 0.10 cm³, divide by 19.90 cm³, multiply by 100.

Step-by-Step Reasoning

Absolute error = 2 × 0.05 = 0.10 cm³.
Percentage error = (0.10/19.90) × 100 = 0.503%.

Key Takeaways

Percentage error in a titre uses two readings, so the reading uncertainty is doubled.

Common Mistakes

Using only 0.05 cm³ instead of 0.10 cm³.
Forgetting to multiply by 100.

Things to Be Careful About

Show working; answer 0.503% (or 0.50%).

Techniques used

calculate absolute error in a titrecalculate percentage error

(iii)

Other than a change in apparatus, suggest one change to the experiment which would lead to a reduction in the percentage error in a measured titre.

DifficultyEasy

Worked solution

Answer

Increase the mass of brass used, so a larger titre is obtained.

Decrease the concentration of Na₂S₂O₃(aq).

Final answer

Increase mass of brass (or decrease concentration of thiosulfate).

Detailed explanation

Background Concept

Percentage error = absolute error / titre × 100. Increasing the titre reduces the percentage error. This can be done by increasing the amount of analyte or decreasing the concentration of titrant.

Understanding the Question

Suggest one change to the experiment, other than a change in apparatus, that would reduce the percentage error in a measured titre.

Approach

Think of variables that increase the titre volume.

Step-by-Step Reasoning

Increase the mass of brass → more Cu²⁺ → more I₂ → larger titre. Or decrease the concentration of Na₂S₂O₃ → more volume needed for the same moles of iodine. Both reduce percentage error.

Key Takeaways

Larger titres have smaller percentage errors.

Common Mistakes

Suggesting “use a more accurate burette”, which is a change of apparatus.
Suggesting “repeat the titration”, which reduces random error but not the percentage error in a single titre.

Things to Be Careful About

The answer must be a change to the experiment, not to the apparatus.

Techniques used

suggest a change to reduce percentage error

(h)

Steps 1–8 were repeated, this time using $1.88\text{ g}$ of brass. The end-point of the titration was found to be $16.50\text{ cm}^3$ .

The equations for the reactions occurring are shown.

\begin{aligned} \text{equation 1 (step 6)} && 2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} &\rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)} \\ \text{equation 2 (step 8)} && \text{I}_2\text{(aq)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} &\rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)} \end{aligned}

(i)

Determine the number of moles of $\text{I}_2$ formed when excess KI(aq) was added to $25.00\text{ cm}^3$ of solution B in step 6.

Use the data from the repeated experiment in your calculations.

DifficultyMedium-Easy

Worked solution

Working

Moles of thiosulfate = 0.100 × 16.50/1000 = 1.65 × 10⁻³ mol

From I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, moles I₂ = 1.65 × 10⁻³ / 2 = 8.25 × 10⁻⁴ mol

Answer

8.25 × 10⁻⁴ mol

Final answer

8.25 × 10^-4 mol

Detailed explanation

Background Concept

In the titration, thiosulfate reacts with iodine: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. So moles of I₂ = half the moles of thiosulfate.

Understanding the Question

Given the end-point is 16.50 cm³ of 0.100 mol dm⁻³ thiosulfate for 25.00 cm³ of solution B, determine the moles of I₂ formed in step 6.

Approach

Calculate moles of thiosulfate used, then use the balanced equation to convert to moles of iodine.

Step-by-Step Reasoning

Moles thiosulfate = concentration × volume = 0.100 × 16.50/1000 = 1.65 × 10⁻³ mol.
From the equation, 1 mol I₂ reacts with 2 mol S₂O₃²⁻, so moles I₂ = 1.65 × 10⁻³ / 2 = 8.25 × 10⁻⁴ mol.

Key Takeaways

Use the balanced equation to convert moles of titrant to moles of analyte.

Common Mistakes

Forgetting to divide by 2.
Not converting cm³ to dm³.

Things to Be Careful About

Volume must be in dm³ when using concentration in mol dm⁻³.

Techniques used

calculate moles from concentration and volumeuse stoichiometry of the iodine-thiosulfate reaction

(ii)

Use your answer to (h)(i) to determine the mass of $\text{Cu}^{2+}$ ions in solution A and therefore the percentage by mass of copper in this sample of brass.
If you were unable to obtain an answer to (h)(i), assume the number of moles of $\text{I}_2$ to be $8.85 \times 10^{-4}\text{ mol}$ . This is not the correct value.
[ $A_r$ : Cu, 63.5]

DifficultyMedium

Worked solution

Working

From equation 1, 2Cu²⁺ → I₂, so moles Cu²⁺ in 25.00 cm³ = 2 × 8.25 × 10⁻⁴ = 1.65 × 10⁻³ mol.

Moles Cu²⁺ in 250.0 cm³ = 1.65 × 10⁻³ × 250/25 = 1.65 × 10⁻² mol.

Mass Cu = 1.65 × 10⁻² × 63.5 = 1.04775 g.

Percentage Cu = (1.04775 / 1.88) × 100 = 55.7%

Answer

55.7%

Final answer

55.7%

Detailed explanation

Background Concept

Equation 1 is 2Cu²⁺ + 4I⁻ → 2CuI + I₂, so moles Cu²⁺ = 2 × moles I₂. Solution B is 250.0 cm³, but only 25.00 cm³ was titrated, so the total amount in solution A is 10 times the amount in the aliquot. Mass = moles × A_r.

Understanding the Question

Use the answer to (h)(i) to find the mass of Cu²⁺ in solution A and hence the percentage by mass of copper in the 1.88 g brass sample.

Approach

Convert moles I₂ to moles Cu²⁺ in the aliquot, scale up to 250.0 cm³, convert to mass, then divide by the brass mass.

Step-by-Step Reasoning

Moles Cu²⁺ in 25.00 cm³ = 2 × 8.25 × 10⁻⁴ = 1.65 × 10⁻³ mol.
In 250.0 cm³ = 1.65 × 10⁻³ × 10 = 1.65 × 10⁻² mol.
Mass Cu = 1.65 × 10⁻² × 63.5 = 1.04775 g.
Percentage = (1.04775 / 1.88) × 100 = 55.7%.
If the fallback value 8.85 × 10⁻⁴ mol were used, the result would be 59.8%, but that is not the correct value.

Key Takeaways

Both the stoichiometric ratio and the dilution factor must be applied correctly.

Common Mistakes

Forgetting to multiply by 250/25.
Using moles I₂ directly as moles Cu²⁺.
Not converting to a percentage.

Things to Be Careful About

Use A_r(Cu) = 63.5; give the final percentage to 3 significant figures.

Techniques used

apply stoichiometry from Cu²⁺ to I₂apply dilution factorcalculate percentage by mass

(i)

A small percentage of silver is sometimes found in some brass alloys.

In step 2, when concentrated nitric acid is added, silver metal is oxidised to silver ions, $\text{Ag}^+\text{(aq)}$ .

At the end of step 6 the $\text{Ag}^+\text{(aq)}$ ions no longer remain in solution.

Explain why.

DifficultyEasy

Worked solution

Answer

Ag⁺(aq) reacts with I⁻(aq) from the potassium iodide to form a precipitate of silver iodide, AgI(s), so Ag⁺ is removed from solution.

Final answer

Ag+ reacts with I- to form AgI precipitate.

Detailed explanation

Background Concept

Silver iodide is very insoluble. In step 6, iodide ions are added in excess. Ag⁺ reacts with I⁻ to form AgI(s), a precipitate, removing Ag⁺ from solution.

Understanding the Question

The question asks why Ag⁺(aq) ions no longer remain in solution after step 6.

Approach

Identify that Ag⁺ forms an insoluble salt with iodide ions.

Step-by-Step Reasoning

Equation: Ag⁺(aq) + I⁻(aq) → AgI(s). Since KI is added in excess, all Ag⁺ is precipitated as AgI.

Key Takeaways

Insoluble salt formation removes ions from solution.

Common Mistakes

Saying Ag⁺ is reduced or oxidised.
Not mentioning the precipitate.

Things to Be Careful About

The answer must mention reaction with iodide to form a precipitate.

Techniques used

explain removal of Ag⁺ by precipitation

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