Chemistry 5070/31 — May/June 2025
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Experimental Contexts · Use of Techniques, Apparatus and Materials · Analysis, Conclusions and Evaluation · Observations and Measurements · Qualitative Analysis · Planning Experiments and Investigations
Solid A is impure calcium carbonate.
You are going to determine the number of moles of calcium carbonate in A by reacting it with excess hydrochloric acid.
The unreacted acid in this mixture is then titrated with sodium hydroxide.
You are provided with a sample of A.
Read all the instructions carefully before starting the experiments.
Instructions
Preparation of mixture B
- Place the sample of A in a beaker.
- Use a measuring cylinder to add of hydrochloric acid, , to the beaker.
- Stir the mixture until no further effervescence is observed.
- Label this mixture B.
Calculate the number of moles of added to the beaker.
number of moles = _____
Working
Answer
0.025
0.025
Walkthrough
To find the number of moles of hydrochloric acid added, convert the volume from to by dividing by 1000, and multiply by the concentration in :
Key Takeaways
- Always convert volumes from to before multiplying by concentration.
Common Mistakes
- Forgetting to divide by 1000, giving an answer of .
Things to Be Careful About
- Write the answer clearly as a decimal value.
Titration of B with sodium hydroxide
- Rinse a burette with water and then with sodium hydroxide.
- Fill the burette with sodium hydroxide.
- Record in Table 1.1 the initial burette reading.
- Use a volumetric pipette to add of B to a conical flask.
- Add five drops of methyl orange indicator to the conical flask.
- Add aqueous sodium hydroxide from the burette while swirling the flask, adding drop by drop near the end-point, until the solution just changes colour.
- Record in Table 1.1 the final burette reading.
- Repeat this titration two more times.
Record in Table 1.1 the burette readings from your titrations and complete the table with the volume used in each titration.
Tick (✓) the best titration results.
Table 1.1
| titration 1 | titration 2 | titration 3 | |
|---|---|---|---|
| final burette reading / | |||
| initial burette reading / | |||
| volume used / | |||
| best titration results (✓) |
Working
- Complete all initial and final burette readings for all three titrations.
- Ensure final readings are larger than initial readings.
- Record all burette readings to 1 decimal place (e.g. , ).
- Calculate each volume used correctly:
- Tick (✓) the best titration results that are concordant (within of each other, or the two closest values).
Answer
Three complete sets of burette readings recorded to 1 decimal place, volumes used calculated correctly, and concordant titres (within ) ticked.
Candidate's own practical readings recorded to 1 d.p. with concordant titres (within 0.2 cm3) ticked
Walkthrough
- For each titration, record the initial and final burette readings to one decimal place (e.g., or depending on standard Cambridge practical instructions).
- Calculate the titre (volume used) by subtracting the initial reading from the final reading: .
- Identify concordant titres, which are those within of each other, and place a tick (✓) in the bottom row for those titrations.
Key Takeaways
- All burette readings in Cambridge 5070 practicals must be recorded to at least 1 decimal place.
- Concordant titres agree within .
Common Mistakes
- Omitting the trailing zero (e.g., writing instead of , or instead of ).
- Ticking all three results when one is an outlier.
Things to Be Careful About
- Ensure subtraction is arithmetic-error free.
Use the best titration results (✓) to calculate the average volume of sodium hydroxide, , used.
average volume = _____
Working
Answer
Average volume calculated from ticked results.
Average volume of ticked titres in cm3
Walkthrough
Take only the values from the columns ticked in (b)(i), add them together, and divide by the number of ticked columns to calculate the mean titre.
Key Takeaways
- Never include non-concordant or unticked rough titres in the average calculation.
Common Mistakes
- Averaging all three titrations even if one was not ticked.
Things to Be Careful About
- Report the average clearly to 1 or 2 decimal places.
The acid used in (a) to prepare mixture B is in excess.
Use your answer to (b)(ii) to calculate the number of moles of that react with of B.
number of moles = _____
Working
Answer
(average volume / 1000) * 0.100
Walkthrough
Using the average titre calculated in (b)(ii), convert the volume to and multiply by the concentration of the sodium hydroxide solution ():
Key Takeaways
- .
Common Mistakes
- Forgetting to convert volume from to .
Things to Be Careful About
- Maintain at least two significant figures in the calculated value.
Working
Answer
answer to (c) * 4
Walkthrough
The titration in (b) used a sample (aliquot) of mixture B. The total volume of mixture B prepared was . Therefore, the total number of moles in is four times the number of moles in :
Key Takeaways
- In back-titrations, always scale the moles found in the pipetted aliquot up to the full volume of the original solution.
Common Mistakes
- Multiplying by 100 instead of 4.
Things to Be Careful About
- Give the evaluated answer to at least two significant figures.
The answer to (d) is equal to the number of moles of that remain in the beaker after the acid reacts with the calcium carbonate in the sample of A.
Use your answers to (a) and (d) to calculate the number of moles of that react with the calcium carbonate in the sample of A.
number of moles = _____
Working
Answer
0.025 - answer to (d)
Walkthrough
Since reacts with unreacted in a 1:1 mole ratio (), the moles of in (d) equals the moles of unreacted remaining in the solution.
To find how many moles of reacted with :
Key Takeaways
- Back-titration relationship: .
Common Mistakes
- Adding the values instead of subtracting.
Things to Be Careful About
- Correctly evaluate to at least 2 significant figures.
The equation for the reaction between hydrochloric acid and calcium carbonate is shown.
Calculate the number of moles of calcium carbonate in the sample of A.
number of moles = _____
Working
From the equation, react with .
Answer
answer to (e) / 2
Walkthrough
The balanced equation is:
The stoichiometric mole ratio is . Therefore, the number of moles of calcium carbonate is half the number of moles of hydrochloric acid that reacted:
Key Takeaways
- Use the coefficients from the balanced equation to relate moles of reactants.
Common Mistakes
- Multiplying by 2 instead of dividing by 2.
Things to Be Careful About
- Ensure the final value is given to at least two significant figures.
In (a) the mixture of A and acid is stirred until effervescence stops.
Answer
To make the reaction happen faster / to dissolve the solid quicker.
to make it react faster
Walkthrough
Stirring mixes the solid particles with the acid, increasing the frequency of collisions between acid molecules and solid calcium carbonate particles. This increases the rate of the reaction so that it finishes in a reasonable time.
Key Takeaways
- Stirring increases the rate of reaction by ensuring efficient contact/mixing between reactants.
Common Mistakes
- Writing vague statements such as "to mix them" without explaining that it speeds up the reaction.
Things to Be Careful About
- Relate the answer to speed/rate of reaction or dissolution.
Answer
To ensure that all the calcium carbonate has reacted / to ensure the reaction is complete.
to ensure the reaction is complete
Walkthrough
Effervescence (bubbling) is due to the production of carbon dioxide gas (). When bubbling stops, no more carbon dioxide is being produced, indicating that all the calcium carbonate (the limiting reactant) has reacted completely.
Key Takeaways
- In reactions producing a gas, the cessation of effervescence indicates that the reaction has reached completion.
Common Mistakes
- Stating that the acid has run out (acid is in excess, so calcium carbonate is limiting).
Things to Be Careful About
- Specify that the calcium carbonate has completely reacted or the reaction is complete.
In (a) a measuring cylinder is used to add of to the beaker.
Explain why using the measuring cylinder makes the volume of used inaccurate. Suggest an improvement.
explanation _____
improvement _____
Answer
explanation: A measuring cylinder has low resolution / is not as accurate as volumetric glassware.
improvement: Use a volumetric pipette / burette / volumetric flask.
explanation: measuring cylinder has low resolution; improvement: use a burette / pipette
Walkthrough
- Explanation: Measuring cylinders have a relatively wide diameter and low precision/resolution (typically reading only to or ). This introduces a large percentage error in the measurement of volume.
- Improvement: Using high-precision volumetric apparatus such as a burette, a volumetric pipette, or a volumetric flask gives a far more accurate volume.
Key Takeaways
- Measuring cylinders are used for approximate volumes; burettes and pipettes are used for precise, quantitative measurements.
Common Mistakes
- Suggesting a beaker as an improvement (beakers are even less accurate).
Things to Be Careful About
- Ensure both the explanation (relating to resolution/accuracy) and the named replacement apparatus are provided.
In (b) the burette is rinsed with water and then with .
Suggest and explain the effect on the titration results if the burette is not rinsed with after rinsing with water.
effect _____
explanation _____
Answer
effect: The titration volume / titre increases.
explanation: Residual water in the burette dilutes the sodium hydroxide solution, lowering its concentration, so a larger volume is needed to neutralise the acid.
effect: titration result increases; explanation: residual water dilutes the NaOH solution
Walkthrough
- Effect: The volume of sodium hydroxide required to reach the end-point (the titre) will be larger (increases).
- Explanation: Water droplets remaining inside the burette will mix with the added sodium hydroxide solution. This dilutes the solution and decreases its concentration. Since the concentration of ions per unit volume is lower, a greater volume of solution must be delivered from the burette to provide the same number of moles of required to neutralize the acid.
Key Takeaways
- A burette must always be rinsed with the solution it is to contain after washing with water to avoid dilution.
- Diluting the titrant increases the required titre volume (, so lower means higher ).
Common Mistakes
- Claiming that the titre decreases.
- Failing to state that residual water dilutes the / reduces its concentration.
Things to Be Careful About
- Clearly separate the effect (increase) from the scientific explanation (dilution/lower concentration).
You are provided with solutions W and X.
Do the following tests on W and X.
Record your observations and conclusions for these tests.
Do a flame test on solution W. Describe the method you use.
method _____
observations _____
conclusion _____
Answer
Method: Dip a clean nichrome wire (or wooden splint) into solution W, then place it in a blue (roaring / non-luminous) Bunsen flame. Observe the initial colour.
Observations: Lilac / purple flame colour.
Conclusion: Potassium ions are present.
Method: dip clean wire in W, blue flame; observation: lilac/purple; conclusion: potassium ions present
Walkthrough
A flame test is used to identify metal cations by the colour they give to a Bunsen flame. First, dip a clean nichrome wire (or a wooden splint) into solution W so that some of the solution sticks to it. A clean wire is important because impurities, especially sodium compounds, give a strong yellow flame that can hide the colour you are looking for. Hold the wire in the blue (roaring, non-luminous) part of the flame and look at the initial colour. The blue flame is used because it is hot and does not itself add much colour. A lilac/purple flame is characteristic of potassium ions, so W contains potassium ions.
Key Takeaways
- Flame tests identify metal cations by their flame colour.
- Potassium ions give a lilac/purple flame.
- The wire must be clean and the flame must be blue/non-luminous.
Common Mistakes
- Using a yellow/luminous flame, which can mask the colour.
- Not cleaning the wire, so sodium contamination gives a yellow flame.
- Writing 'potassium' instead of 'potassium ions'.
- Saying 'purple' without 'lilac' (the mark scheme accepts either, but 'lilac' is the standard word).
Things to Be Careful About
- The mark scheme wants the method (dipping and placing in a blue flame) and the observation and conclusion as separate points.
- 'Initial colour' matters because a sodium impurity may appear later.
- The conclusion must be 'potassium ions present', not 'potassium metal'.
To depth of W in a test-tube, add depth of aqueous chlorine. Keep this mixture for use in (b)(ii).
observations _____
conclusion _____
Answer
Observations: Red/brown solution formed (or black solid formed).
Conclusion: W contains iodide ions; chlorine displaces iodine.
Red/brown solution formed; W contains iodide ions (iodine displaced by chlorine)
Walkthrough
Chlorine is a more reactive halogen than iodine, so when chlorine water is added to a solution containing iodide ions, chlorine displaces iodine:
Iodine dissolved in water gives a red/brown solution; if the iodine concentration is high, a black solid may form. This colour change tells you that iodide ions were present in W and that iodine has been displaced.
Key Takeaways
- More reactive halogens displace less reactive halogens from their salts.
- Iodine in solution is red/brown (or black solid if concentrated).
- This is evidence for iodide ions in W.
Common Mistakes
- Saying chlorine has been displaced (chlorine is the displacing halogen).
- Calling the colour 'orange' or 'brown' without 'red/brown'.
- Writing 'iodine ions' instead of 'iodide ions'.
Things to Be Careful About
- The mixture is kept for part (b)(ii), so do not discard it.
- The conclusion should be 'iodide ions present' or 'iodine displaced', not just 'iodine formed'.
Add depth of starch solution to the contents of the solution from (b)(i).
observations _____
Answer
Observations: Blue-black colour forms.
Blue-black colour
Walkthrough
Starch solution is a test for iodine. Iodine forms a blue-black complex with starch. Since part (b)(i) produced iodine by displacement, adding starch gives a blue-black colour. This confirms that iodine was produced.
Key Takeaways
- Starch turns blue-black in the presence of iodine.
- This is a confirmatory test for iodine.
Common Mistakes
- Writing 'blue' or 'black' instead of 'blue-black'.
- Saying the starch itself is blue.
Things to Be Careful About
- The observation is 'blue-black', the exact term used in the mark scheme.
Test W for the presence of sulfate ions. Describe how you do the test and record your observations.
test _____
observations _____
Answer
Test: Add dilute nitric acid (or dilute hydrochloric acid) to solution W, then add aqueous barium nitrate (or barium chloride).
Observations: Colourless solution; no white precipitate formed.
Conclusion: Sulfate ions are not present.
No white precipitate; sulfate ions absent
Walkthrough
The test for sulfate ions uses barium ions. Barium sulfate is insoluble, so a white precipitate forms if sulfate is present. The solution is first acidified with dilute nitric acid to remove carbonate ions, which would also give a white precipitate with barium ions. Here, after adding barium nitrate, the solution stays colourless and no white precipitate appears, so W does not contain sulfate ions.
Key Takeaways
- Sulfate test: acidify with dilute nitric acid, add barium nitrate; white precipitate = sulfate present.
- Acidification removes carbonate interference.
- A negative result is shown by no white precipitate.
Common Mistakes
- Using sulfuric acid (it already contains sulfate ions).
- Not acidifying, so carbonate could give a false positive.
- Saying 'white precipitate formed' when the observation is no precipitate.
Things to Be Careful About
- The mark scheme accepts nitric acid or hydrochloric acid, and barium nitrate or barium chloride.
- The observation must be 'no white precipitate' or 'colourless solution'.
To depth of W in a test-tube, add depth of dilute nitric acid and depth of aqueous silver nitrate.
observations _____
conclusion _____
Answer
Observations: Pale yellow precipitate forms.
Conclusion: Iodide ions are present.
Pale yellow precipitate; iodide ions present
Walkthrough
Silver nitrate is used to test for halide ions. The solution is acidified with dilute nitric acid to remove carbonate ions, which would also give a precipitate with silver ions. Silver iodide is pale yellow and insoluble, so a pale yellow precipitate shows iodide ions are present. Chloride would give a white precipitate and bromide a cream precipitate.
Key Takeaways
- Silver halide precipitates: chloride white, bromide cream, iodide yellow.
- Acidification with nitric acid removes carbonate interference.
- A pale yellow precipitate identifies iodide ions.
Common Mistakes
- Confusing the colours: cream is bromide, white is chloride.
- Saying 'iodine' instead of 'iodide ions'.
- Not acidifying, which could give a false result from carbonate.
Things to Be Careful About
- The mark scheme says 'pale yellow precipitate'.
- The conclusion must be 'iodide ions present'.
To depth of W in a test-tube, add depth of aqueous iron(III) nitrate.
observations _____
Answer
Observations: Red/brown solution formed (or black solid formed).
Red/brown solution formed (or black solid formed)
Walkthrough
Iron(III) ions are oxidising agents. When iron(III) nitrate is added to a solution containing iodide ions, iron(III) oxidises iodide to iodine:
The iodine formed gives a red/brown solution (or a black solid if concentrated). This observation is consistent with the iodide ions already indicated by earlier tests.
Key Takeaways
- Iron(III) can oxidise iodide ions to iodine.
- Iodine is red/brown in solution.
- This is a test for a reducing agent (iodide).
Common Mistakes
- Saying a brown precipitate forms (iodine is a solution colour, not a precipitate).
- Writing 'iron(II)' instead of 'iron(III)' if an equation is attempted.
Things to Be Careful About
- The mark scheme only requires the observation: red/brown solution formed or black solid formed.
- Do not confuse this with the silver nitrate test.
Put depth of X into a boiling tube.
Add aqueous sodium hydroxide until no further change is seen.
Keep the mixture for use in (f)(ii).
observation _____
conclusion about solution X _____
Answer
Observation: White precipitate forms; it dissolves in excess sodium hydroxide giving a colourless solution.
Conclusion: Solution X contains zinc ions or aluminium ions.
White precipitate, soluble in excess NaOH; X contains Zn2+ or Al3+
Walkthrough
Aqueous sodium hydroxide is used to test for cations. Many metal ions form insoluble hydroxides. A white precipitate with sodium hydroxide could be from several cations, but the key clue is what happens in excess. Zinc hydroxide and aluminium hydroxide are amphoteric: they dissolve in excess sodium hydroxide to give a colourless solution. So a white precipitate that dissolves in excess NaOH suggests zinc ions or aluminium ions. Other white hydroxides, such as magnesium hydroxide, stay insoluble in excess.
Key Takeaways
- NaOH gives a white precipitate with and .
- Amphoteric hydroxides dissolve in excess NaOH.
- The conclusion is limited to zinc ions or aluminium ions.
Common Mistakes
- Saying the precipitate is insoluble in excess.
- Concluding only 'zinc' or only 'aluminium' when both are possible.
- Confusing with magnesium or calcium (white ppt insoluble in excess).
Things to Be Careful About
- The mark scheme accepts 'soluble in excess sodium hydroxide' or 'colourless solution with excess sodium hydroxide'.
- The conclusion must include both zinc ions and aluminium ions.
Put a depth of the mixture from (f)(i) into a clean boiling tube.
Add a small piece of aluminium foil to the mixture and warm gently.
Test any gas evolved.
observations _____
conclusion about solution X _____
Answer
Observations: A gas is evolved that turns damp red litmus paper blue (ammonia).
Conclusion: Solution X contains nitrate ions.
Gas turns red litmus blue; nitrate ions present
Walkthrough
The mixture from (f)(i) is alkaline because excess sodium hydroxide is present. Adding aluminium foil and warming causes aluminium to reduce nitrate ions to ammonia gas. Ammonia is an alkaline gas, so it turns damp red litmus paper blue. This is the confirmatory test for nitrate ions.
Key Takeaways
- Nitrate ions are reduced to ammonia by aluminium in alkaline solution.
- Ammonia turns damp red litmus blue.
- This test confirms nitrate ions in X.
Common Mistakes
- Saying the gas is hydrogen (hydrogen would pop with a lighted splint, not turn litmus blue).
- Saying the gas turns blue litmus red (that would be an acidic gas).
- Not mentioning warming.
Things to Be Careful About
- The observation must link the gas to ammonia: 'gas turns damp red litmus blue'.
- The conclusion is 'nitrate ions present'.
You are not expected to do any experimental work for this question.
Q is a mixture of solid magnesium oxide and solid barium sulfate.
Magnesium oxide is insoluble in water. It reacts with dilute hydrochloric acid to make a solution of magnesium chloride.
Barium sulfate is insoluble in water and does not react with dilute hydrochloric acid.
Plan an investigation to obtain pure magnesium chloride crystals and pure barium sulfate solid from Q.
Your plan should describe the use of common laboratory apparatus, dilute hydrochloric acid and Q. No other chemicals should be used.
Your plan should include:
- the apparatus needed
- the preparation of magnesium chloride solution
- the method to obtain pure magnesium chloride crystals
- the method to obtain pure barium sulfate solid
- how to test that the barium sulfate is pure.
You may draw a diagram to help answer the question.
Answer
1. Apparatus needed
- Beaker, glass rod, filter funnel, filter paper, conical flask, evaporating basin, Bunsen burner (or water bath / tripod and gauze), melting point apparatus.
2. Preparation of magnesium chloride solution
- Place mixture Q in a beaker and add dilute hydrochloric acid.
- Add excess dilute hydrochloric acid and stir until no more magnesium oxide reacts (all magnesium oxide dissolves).
3. Obtaining pure barium sulfate solid
- Filter the mixture using a filter funnel and filter paper to separate the insoluble barium sulfate residue from the magnesium chloride solution (filtrate).
- Wash the barium sulfate residue on the filter paper with distilled water to remove any remaining magnesium chloride solution.
- Dry the residue between sheets of filter paper (or in a warm oven / desiccator).
4. Obtaining pure magnesium chloride crystals
- Transfer the filtrate (magnesium chloride solution) into an evaporating basin.
- Heat with a Bunsen burner to evaporate water until the crystallisation point is reached (solution is saturated / crystals start to form around the edge).
- Allow the saturated solution to cool slowly to form crystals.
- Filter off the crystals and dry them between sheets of filter paper.
5. Testing the purity of barium sulfate
- Measure the melting point of the dry barium sulfate using melting point apparatus and compare it with the known (book) value for pure barium sulfate (a pure substance melts sharply at its exact melting point).
Add excess dilute hydrochloric acid to Q, filter to separate insoluble barium sulfate residue from magnesium chloride filtrate. Wash barium sulfate residue with distilled water and dry. Heat filtrate to saturation point, allow to cool and crystallise, then filter and dry magnesium chloride crystals. Test barium sulfate purity by measuring its melting point and comparing to literature value.
Walkthrough
This is a standard practical planning question worth 6 marks, requiring one scoring point from each of five logical sections plus one additional detail:
- Apparatus (M1 & M2): Specify standard laboratory apparatus: a beaker and stirring rod for reacting, a filter funnel and filter paper with a conical flask/beaker for filtration, an evaporating basin and heat source (Bunsen burner or water bath) for crystallisation, and melting point apparatus.
- Preparation of solution (M3 & M4): is a basic oxide that reacts with to give soluble and :
Adding excess dilute hydrochloric acid ensures that all completely reacts and dissolves, leaving only pure insoluble in the solid residue.
3. Obtaining pure solid (M7 & M8): Filter the resulting mixture. The insoluble remains on the filter paper as the residue. Wash the residue with distilled water to remove adhering solution/acid, and dry the solid.
4. Obtaining pure crystals (M5 & M6): Pour the filtrate (containing dissolved ) into an evaporating basin. Heat to evaporate water until the solution is saturated (crystallisation point). Allow the solution to cool slowly so that crystals of form. Finally, filter and dry the crystals.
5. Testing purity of (M9): Measure the melting point of the dry solid using melting point apparatus. A sharp, distinct melting point matching the accepted literature value confirms purity.
Key Takeaways
- For a 6-mark practical planning question, structure your answer using subheadings directly matching the bullet points in the prompt.
- Excess acid is crucial when reacting an insoluble oxide with an acid if another insoluble inert substance is present, ensuring total conversion.
- To form hydrated crystals, never evaporate to dryness; always heat to saturation / crystallisation point and allow to cool.
- Purity of an insoluble solid is determined by measuring its melting point (sharp and at the reference value).
Common Mistakes
- Heating the salt solution to dryness rather than to the crystallisation point.
- Forgetting to specify washing the residue with distilled water and drying it.
- Omitting the specific purity test for a solid (melting point determination) or suggesting chemical tests instead.
Things to Be Careful About
- Ensure you specify that excess hydrochloric acid is used so that no unreacted contaminates the residue.
- Ensure each section asked for in the question prompt is clearly addressed to avoid penalties for missing sections.