Chemistry 5070/32 — May/June 2025
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Experimental Contexts · Observations and Measurements · Analysis, Conclusions and Evaluation · Use of Techniques, Apparatus and Materials · Qualitative Analysis · Planning Experiments and Investigations
Solid W is an impure organic acid, .
You are going to determine the mass of in a sample of W by titration with sodium hydroxide, .
You are provided with a sample of W.
Read all the instructions carefully before starting the experiments.
Instructions
Preparation of mixture X
- Place the sample of W in a beaker.
- Use a measuring cylinder to add of distilled water to the beaker.
- Stir the mixture until the solid is fully dissolved.
- Label this mixture X.
Explain why it is important to use distilled water and not tap water for the experiment.
______
Answer
Tap water is impure / contains dissolved substances / minerals that would react with the acid (H₃A) and affect the titration results.
Tap water is impure / contains dissolved substances
Walkthrough
This question asks why distilled water is used instead of tap water. The key idea is that tap water is not pure — it contains dissolved minerals and other substances. If tap water were used, these impurities could react with the acid H₃A, changing the amount of acid available to react with the sodium hydroxide. This would give an inaccurate titration result.
Key Takeaways
- Distilled water is pure water with no dissolved substances.
- Tap water contains dissolved minerals and ions.
- Impurities in tap water could react with the acid and alter the titration results.
Common Mistakes
- Saying "tap water is dirty" — this is not the reason; it is about dissolved substances.
- Saying "distilled water is cleaner" without linking to the effect on the experiment.
Things to Be Careful About
- The mark is for the idea that tap water contains dissolved substances/impurities — not for saying tap water is 'dirty'.
- Do not say the impurities would just 'dilute' the acid — the point is they could react with it.
Answer
Colourless solution / colourless liquid
Colourless solution
Walkthrough
The solid W is an organic acid that dissolves fully in water. Since the acid and its solution are both colourless, the resulting mixture X is a colourless solution. The word 'solution' is important here because the solid has fully dissolved.
Key Takeaways
- A solid that dissolves completely in water forms a solution.
- The appearance must describe both colour and state (solution/liquid).
Common Mistakes
- Writing 'clear' instead of 'colourless' — 'clear' is not a colour and would not score.
- Writing 'white' — this would describe the solid, not the solution.
Things to Be Careful About
- The answer must be 'colourless solution' — both words are needed for full credit.
Titration of with X.
- Rinse a burette with distilled water and then with X.
- Fill the burette with X.
- Record in Table 1.1 the initial burette reading.
- Use a volumetric pipette to add of the to a conical flask.
- Add five drops of thymolphthalein indicator to the conical flask.
- Add X 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 (✓) |
Answer
M1 Three sets of initial and final burette readings with the final reading greater than the initial reading.
M2 All readings given to one decimal place (e.g. 25.0, 12.5, 0.0).
M3 Volume used calculated correctly for all three titrations: volume used = final reading − initial reading.
M4 At least two values within 0.2 cm³ of each other.
M5 Tick (✓) the values that are within 0.2 cm³ of each other (or the two closest values if none are within 0.2 cm³).
See working — three sets of readings to 1 dp with volumes calculated and the best results ticked
Walkthrough
This part is about recording titration data correctly. The candidate must:
- Record initial and final burette readings for each of the three titrations. The final reading must be greater than the initial reading (since X is being added from the burette).
- Give all readings to one decimal place — burettes read to 0.05 cm³, so readings like 0.00, 12.50, 25.00 are appropriate.
- Calculate the volume used in each titration: volume = final reading − initial reading.
- Identify the best (most concordant) results — those within 0.2 cm³ of each other.
- Tick the best results.
Key Takeaways
- Burette readings must be recorded to one decimal place (e.g. 25.0, not 25).
- Volume used = final reading − initial reading.
- Concordant results are those within 0.2 cm³ of each other.
- Ticking the best results allows an average to be calculated in the next part.
Common Mistakes
- Recording readings without a decimal place (e.g. 25 instead of 25.0).
- Final reading less than initial reading — impossible for a titration.
- Ticking all three results even when only two are concordant.
- Forgetting to calculate the volume used in each titration.
Things to Be Careful About
- The mark scheme requires readings to 1 decimal place — a zero must be written even if the reading is exactly 25.0 cm³.
- If no two values are within 0.2 cm³, tick the two closest values — the mark scheme explicitly allows this.
Use the best titration results (✓) to calculate the average volume of X used.
average volume = ______
Working
Add the ticked values from (b)(i) and divide by the number of values ticked.
For example, if the ticked values are 24.8 cm³ and 25.0 cm³:
Answer
Candidate-dependent — average of the ticked volumes from (b)(i).
Candidate-dependent — average of the ticked volumes from (b)(i)
Walkthrough
This part requires the candidate to calculate the average of the best (ticked) titration results from (b)(i). The average is found by adding the ticked volumes together and dividing by the number of values. Only the ticked values are used — not all three titrations.
Key Takeaways
- The average is calculated from the ticked (best) results only.
- Average = sum of ticked values ÷ number of ticked values.
Common Mistakes
- Including all three titrations in the average instead of just the ticked ones.
- Arithmetic errors in the division.
Things to Be Careful About
- The average must be quoted to a sensible number of decimal places (usually 1 dp to match the readings).
Working
Answer
0.00250 mol
0.00250 mol
Walkthrough
This is a straightforward application of the mole formula. The concentration is given as 0.100 mol/dm³ and the volume is 25.0 cm³. Since the concentration is in mol/dm³, the volume must be converted from cm³ to dm³ by dividing by 1000. Then multiply concentration by volume to find the number of moles.
Key Takeaways
- The formula is moles = concentration × volume (in dm³).
- To convert cm³ to dm³, divide by 1000.
Common Mistakes
- Forgetting to convert cm³ to dm³ (would give a huge wrong answer).
- Using the wrong formula (e.g. dividing by volume instead of multiplying).
Things to Be Careful About
- The volume is 25.0 cm³ (from the pipette), not 100 cm³ (the total volume of X).
- The answer should be given to 3 significant figures (0.00250 mol) to match the data.
The equation for the reaction between and is shown.
Calculate the number of moles of that react with of .
number of moles = ______
Working
From the equation: 1 mol H₃A reacts with 3 mol NaOH.
Answer
8.33 × 10⁻⁴ mol
8.33 × 10⁻⁴ mol
Walkthrough
The balanced equation shows that 1 mole of H₃A reacts with 3 moles of NaOH. From part (c), the number of moles of NaOH used is 0.00250 mol. To find the moles of H₃A that react, divide the moles of NaOH by the stoichiometric ratio (3).
Key Takeaways
- The balanced equation gives the mole ratio: H₃A : NaOH = 1 : 3.
- Moles of H₃A = moles of NaOH ÷ 3.
Common Mistakes
- Using a 1:1 ratio instead of 1:3.
- Multiplying by 3 instead of dividing.
Things to Be Careful About
- The answer must be given to a minimum of two significant figures (8.33 × 10⁻⁴ has 3 significant figures, which is fine).
Working
Let V be the average volume of X from (b)(ii) in cm³.
Moles of H₃A in V cm³ of X = 8.33 × 10⁻⁴ mol
Total volume of X = 100 cm³
Answer
(8.33 × 10⁻⁴ / V) × 100 mol, where V is the average volume from (b)(ii) in cm³
(8.33 × 10⁻⁴ / V) × 100 mol, where V is the average titre
Walkthrough
The titration used only a portion of the 100 cm³ of solution X. The volume V (from part b(ii)) is the average volume of X needed to neutralise the NaOH. The moles of H₃A in that volume V is 8.33 × 10⁻⁴ mol (from part d). To find the moles in the whole 100 cm³ sample, scale up proportionally: multiply the moles in V cm³ by (100 / V).
Key Takeaways
- The sample is not fully used in one titration — only a portion is.
- Scaling factor = total volume ÷ volume used in titration.
Common Mistakes
- Forgetting to scale up — giving the answer to (d) as the answer to (e).
- Using the wrong total volume (e.g. 250 cm³ for the beaker instead of 100 cm³ for the solution).
Things to Be Careful About
- The total volume of X is 100 cm³, not the 250 cm³ beaker volume.
The relative molecular mass of is 210.
Calculate the mass of in the sample of W.
mass = ______
Working
Answer
(answer to (e)) × 210 g
(answer to (e)) × 210 g
Walkthrough
This is a direct application of the formula mass = moles × Mᵣ. The moles of H₃A in the sample were calculated in part (e), and the relative molecular mass is given as 210. Multiply them together to find the mass in grams.
Key Takeaways
- The formula mass = moles × Mᵣ links mass, amount and molar mass.
- Units: grams = mol × g/mol.
Common Mistakes
- Using the wrong value for Mᵣ (e.g. 210 g/mol is correct here).
- Forgetting to carry over the answer from (e) correctly.
Things to Be Careful About
- The answer must be given to a minimum of two significant figures, matching the precision of the data.
In (a) a measuring cylinder is used to add of distilled water to the beaker.
Explain why using the measuring cylinder makes the volume of distilled water used inaccurate. Suggest an improvement.
explanation ______
improvement ______
Answer
Explanation: The measuring cylinder does not have high resolution / does not read to 1 decimal place / only reads to 0.5 cm³.
Improvement: Use a burette or pipette.
Measuring cylinder has low resolution; use a burette or pipette
Walkthrough
A measuring cylinder is designed for approximate measurements — it typically has a resolution of 0.5 cm³ or worse. This means the volume of distilled water added is not known precisely. A burette or pipette measures to 0.05 cm³, giving a much more accurate volume.
Key Takeaways
- Measuring cylinders are for approximate volumes only.
- Burettes and pipettes are used when precise volumes are needed.
- The improvement must address the lack of precision in the original method.
Common Mistakes
- Saying 'use a beaker' — beakers are even less precise.
- Not explaining why the measuring cylinder is inaccurate (just saying 'it is inaccurate').
Things to Be Careful About
- The mark scheme accepts 'does not read to 1 dp' as well as 'only reads to 0.5 cm³'.
- Both the explanation and the improvement are needed for both marks.
In (b) the burette is rinsed with distilled water and then with X.
Answer
To remove impurities from the burette.
To remove impurities from the burette
Walkthrough
The burette may contain dust or other impurities from storage. Rinsing with distilled water removes these impurities so they do not contaminate the solution X and affect the titration results.
Key Takeaways
- Distilled water is used for rinsing because it leaves no dissolved solids behind.
- The purpose is to clean the apparatus without introducing new contaminants.
Common Mistakes
- Saying 'to clean the burette' without mentioning removing impurities.
- Confusing this with the reason for rinsing with the solution itself (which is to avoid dilution).
Things to Be Careful About
- The mark is for 'remove impurities' — a simple 'to clean' may not score.
Suggest and explain the effect on the titration results if the burette is not rinsed with X after rinsing with distilled water.
effect ______
explanation ______
Answer
Effect: The titration result (volume of X used) increases.
Explanation: The water left in the burette will dilute the X, so the concentration of X decreases. A larger volume of the diluted X is therefore needed to neutralise the same amount of NaOH.
Result increases; water dilutes X so concentration decreases
Walkthrough
If the burette is not rinsed with X after rinsing with distilled water, a thin film of water remains on the inside walls of the burette. When X is added, this water mixes with it, diluting the solution. A more dilute solution of X contains fewer moles of H₃A per cm³, so a larger volume is needed to neutralise the fixed amount of NaOH in the conical flask. Hence the titre volume increases.
Key Takeaways
- Rinsing with the solution itself (X) removes the water film and prevents dilution.
- Dilution lowers concentration, which increases the volume needed for neutralisation.
Common Mistakes
- Saying the result 'decreases' — the opposite is true.
- Confusing the effect on concentration with the effect on volume.
Things to Be Careful About
- Both the effect (increases) and the explanation (dilution) are needed for both marks.
You are provided with solid A and solution B.
Do the following tests on A and B.
Record your observations and conclusions for these tests.
Test and name any gases evolved.
Do a flame test on one sample of A. Describe the method you use.
method ______
observations ______
conclusion ______
Answer
method: Dip a clean nichrome wire (or wooden splint) into the solid and place it into a non-luminous / roaring / blue Bunsen flame.
observations: Blue-green flame
conclusion: Copper(II) ions / present
Method: dip wire/splint into solid and place into blue/roaring flame; Observation: blue-green flame; Conclusion: copper(II) ions / Cu2+ present
Walkthrough
To carry out a flame test:
- Method: A clean nichrome/platinum wire or soaked wooden splint is dipped into the solid sample (often moistened with concentrated/dilute acid). The sample on the wire/splint is then introduced into the hot, non-luminous (blue/roaring) flame of a Bunsen burner with the airhole open.
- Observation: A distinct blue-green colour is observed in the flame.
- Conclusion: A blue-green flame test confirms the presence of copper(II) ions, .
Key Takeaways
- Flame tests must be conducted in a non-luminous (roaring/blue) Bunsen flame so the colour of the flame is clearly visible against a colourless background.
- Copper(II) produces a characteristic blue-green flame.
Common Mistakes
- Stating a yellow/luminous flame for the flame test method (which masks the true flame colour).
- Writing just "green" or "blue" instead of the expected "blue-green" for .
Things to Be Careful About
- Ensure the cation is specified with its correct charge/oxidation state, or copper(II).
Place the other sample of A in a test-tube and add depth of dilute hydrochloric acid.
When the reaction has finished, keep the contents of the test-tube for use in (c).
observations ______
conclusion ______
Answer
observations:
- Effervescence / fizzing / bubbling
- Gas turns limewater milky / cloudy
- Blue-green solution formed
conclusion:
- Carbon dioxide () formed / evolved
- Carbonate () ions present
Observations: effervescence, gas turns limewater milky, blue-green solution forms; Conclusion: carbon dioxide evolved, carbonate (CO3^2-) present
Walkthrough
When solid reacts with dilute hydrochloric acid:
- An acid-carbonate reaction occurs, producing effervescence (bubbles of gas).
- Testing the evolved gas with limewater gives a white precipitate / turns limewater milky or cloudy, confirming the gas is carbon dioxide ().
- The solid dissolves to form a blue-green solution of aqueous copper(II) chloride ().
- Thus, the presence of carbonate ions () is confirmed.
Key Takeaways
- Metal carbonates react with dilute acids to produce a salt, water, and carbon dioxide gas.
- The definitive test for is bubbling through limewater (aqueous calcium hydroxide), which turns milky/cloudy.
Common Mistakes
- Omitting the solution colour formed after the reaction finishes.
- Forgetting to name the gas as carbon dioxide when describing the limewater test result.
Things to Be Careful About
- Distinguish clearly between an observation (what is seen: "bubbles", "limewater turns cloudy", "blue-green solution") and a conclusion ("carbon dioxide formed", "carbonate ion present").
Answer
Effervescence / bubbling stops (or no more solid remains undissolved).
Effervescence stops / no more bubbling
Walkthrough
The reaction produces carbon dioxide gas as long as reactants are actively reacting. Once all the solid carbonate (or acid) has reacted completely, gas production ceases, meaning no more bubbles/effervescence are observed.
Key Takeaways
- A reaction producing a gas is complete when effervescence/fizzing stops completely.
Common Mistakes
- Stating that the temperature stops rising, which is harder to detect visually without a thermometer in a test-tube.
Things to Be Careful About
- Use clear terminology: "bubbling / fizzing / effervescence stops".
Decant depth of the solution from (b)(i) into each of two test-tubes.
To one of these test-tubes, add aqueous ammonia until no further change is observed.
observations ______
Answer
Blue precipitate forms, which dissolves in excess aqueous ammonia to form a deep blue solution.
Blue precipitate, dissolves in excess to form a deep blue solution
Walkthrough
When aqueous ammonia is added to a solution containing ions:
- On initial addition, hydroxide ions from the weak base react to form an insoluble light blue precipitate of copper(II) hydroxide, .
- On adding excess aqueous ammonia, the precipitate dissolves because a soluble tetraamminecopper(II) complex ion is formed, giving a characteristic intense deep blue / royal blue solution.
Key Takeaways
- with aqueous : blue ppt., soluble in excess to give a deep blue solution.
Common Mistakes
- Forgetting to state both stages (the initial precipitate AND what happens in excess).
- Confusing the colour in excess with the initial pale blue precipitate.
Things to Be Careful About
- Always state both the precipitate colour ("blue ppt.") and the effect of excess reagent ("dissolves to give a deep blue solution").
To the second test-tube, add aqueous sodium hydroxide until no further change is observed.
observations ______
Answer
Blue precipitate (which is insoluble in excess).
Blue precipitate (insoluble in excess)
Walkthrough
Adding aqueous sodium hydroxide to a solution containing copper(II) ions forms a light blue precipitate of copper(II) hydroxide:
Unlike with ammonia, this precipitate does not dissolve in excess sodium hydroxide and remains as an insoluble blue precipitate.
Key Takeaways
- with aqueous : light blue precipitate, insoluble in excess.
Common Mistakes
- Stating that the precipitate dissolves in excess sodium hydroxide (it only dissolves in excess ammonia).
Things to Be Careful About
- Clearly identify the colour and state of the product: "blue precipitate".
Answer
Copper(II) carbonate /
copper(II) carbonate
Walkthrough
Combining the deductions from parts (a), (b), and (c):
- Flame test and precipitation reactions confirmed the presence of copper(II) ions ().
- Acid reaction and limewater test confirmed the presence of carbonate ions ().
Therefore, solid A is copper(II) carbonate, .
Key Takeaways
- A binary ionic compound is identified by combining the deduced cation and anion.
Common Mistakes
- Writing only the formula with incorrect stoichiometry (e.g. ).
Things to Be Careful About
- Either the correct chemical name ("copper(II) carbonate" / "copper carbonate") or the correct formula ("") is accepted.
Place depth of aqueous silver nitrate into each of four test-tubes.
Add depth of dilute nitric acid to each of these test-tubes.
Keep the contents of all four test-tubes until you have completed all of part (e).
To the first test-tube add depth of aqueous sodium chloride.
To the second test-tube add depth of aqueous potassium bromide.
To the third test-tube add depth of aqueous potassium iodide.
To the fourth test-tube add depth of B.
Leave the test-tubes for at least 1 minute before recording your observations in Table 2.1.
Table 2.1
| solution | observation |
|---|---|
| sodium chloride | |
| potassium bromide | |
| potassium iodide | |
| B |
State the conclusion about B you can make from these observations.
conclusion ______
Answer
Table 2.1
| solution | observation |
|---|---|
| sodium chloride | white precipitate |
| potassium bromide | cream / off-white precipitate |
| potassium iodide | yellow / pale yellow precipitate |
| B | cream / off-white precipitate |
conclusion:
Bromide ions / present
Table: chloride = white ppt, bromide = cream ppt, iodide = yellow ppt, B = cream ppt; Conclusion: bromide / Br- ions present
Walkthrough
When dilute nitric acid followed by aqueous silver nitrate is added to halide solutions:
- Chloride (): forms a white precipitate of silver chloride ().
- Bromide (): forms a cream / off-white precipitate of silver bromide ().
- Iodide (): forms a pale yellow / yellow precipitate of silver iodide ().
- Solution B: produces a cream / off-white precipitate identical to that of potassium bromide.
Conclusion: Solution B contains bromide ions ().
Key Takeaways
- Halide ions are distinguished using acidified silver nitrate solution:
- gives a white precipitate.
- gives a cream precipitate.
- gives a yellow precipitate.
- Nitric acid is added first to remove any interfering carbonate or sulfite ions.
Common Mistakes
- Confusing the cream precipitate of bromide with the yellow precipitate of iodide.
- Forgetting to write the word "precipitate" (just writing the colour alone is often penalised).
Things to Be Careful About
- Clearly specify both the colour and the state ("precipitate") for each table entry.
- Conclude with the specific halide ion: bromide / .
You are not expected to do any experimental work for this question.
Q is a mixture of solid copper(II) carbonate and solid lead sulfate.
Lead sulfate is insoluble in water and does not react with dilute sulfuric acid.
Copper(II) carbonate is insoluble in water. It reacts with dilute sulfuric acid to form copper(II) sulfate solution.
Plan an investigation to obtain pure copper(II) sulfate crystals and pure lead sulfate solid from Q.
Your plan should describe the use of common laboratory apparatus, dilute sulfuric acid and Q. No other chemicals should be used.
Your plan should include:
- the apparatus needed
- the preparation of copper(II) sulfate solution
- the method to obtain pure copper(II) sulfate crystals
- the method to obtain pure lead sulfate solid
- how to test that the lead sulfate is pure.
You may draw a diagram to help answer the question.
Answer
1. Apparatus needed
- Beaker, glass stirring rod, filter funnel, filter paper, conical flask, evaporating dish, Bunsen burner (or water bath/tripod/gauze), and melting point apparatus.
2. Preparation of solution
- Place mixture into a beaker and add dilute sulfuric acid.
- Continue adding dilute sulfuric acid in excess (until effervescence/fizzing stops) and stir to ensure all the has reacted.
3. Method to obtain pure solid
- Filter the reaction mixture using a filter funnel and filter paper to separate the insoluble as the residue.
- Wash the residue (lead sulfate) with a small amount of distilled water to remove any remaining acid/solution.
- Dry the solid between sheets of filter paper or in a warm oven.
4. Method to obtain pure crystals
- Collect the filtrate (copper(II) sulfate solution) in an evaporating dish.
- Heat the solution to evaporate water until the crystallisation point is reached (saturated solution / crystals appear on a glass rod).
- Leave the hot saturated solution to cool and crystallise.
- Filter off the crystals and dry them between filter papers.
5. Test for the purity of
- Measure the melting point of the dried solid using melting point apparatus and compare it to the known melting point in data tables (a sharp melting point at the literature value indicates purity).
Add excess dilute sulfuric acid to Q until fizzing stops. Filter to collect insoluble lead sulfate residue; wash with distilled water and dry. Heat the copper(II) sulfate filtrate to the point of crystallisation, leave to cool and crystallise, then filter and dry crystals. Test the purity of lead sulfate by determining its melting point and comparing it with data book values.
Walkthrough
This is a structured 6-mark experimental planning question. To obtain full marks, each required aspect of the brief must be addressed logically:
-
Chemistry of the Mixture:
- is an insoluble carbonate that reacts with to form soluble , water, and carbon dioxide gas:
- is insoluble in water and unreactive towards dilute , so it remains as a solid throughout.
-
Reacting the Mixture:
- To ensure that all copper(II) carbonate reacts and none remains contaminating the lead sulfate, dilute sulfuric acid must be added in excess (until effervescence ceases).
-
Isolating and Purifying the Insoluble Salt ():
- Separate the solid lead sulfate residue from the copper(II) sulfate filtrate by filtration using a funnel and filter paper.
- Wash the residue with distilled water to remove any adhering copper(II) sulfate and sulfuric acid solution, then dry it.
-
Obtaining Crystals:
- The filtrate contains aqueous . To obtain pure crystals, evaporate water by heating gently until the saturation point is reached (or crystals begin to form on the edge of the dish / on a cold glass rod).
- Allow it to cool slowly to form large crystals, then filter and pat dry with filter paper.
-
Testing Purity:
- Pure solids have a sharp, fixed melting point. By measuring the melting point of the dried lead sulfate and checking if it matches the literature/data book value, its purity is confirmed.
Key Takeaways
- When preparing a soluble salt from an insoluble reactant and an acid, adding an excess of acid is critical here because the unreacted solid would contaminate the desired insoluble product ().
- Pure insoluble residues obtained via filtration must always be washed with distilled water and dried.
- Crystallisation of a soluble salt involves heating to the saturation point followed by cooling, rather than evaporating to dryness (which yields anhydrous powder or decomposes the salt).
- Purity of a solid is routinely verified by checking that it has a sharp melting point matching standard data book values.
Common Mistakes
- Evaporating to dryness: Boiling away all water gives anhydrous powder or decomposed copper(II) oxide rather than hydrated copper(II) sulfate crystals.
- Omitting the washing/drying steps for the residue: Simply filtering is insufficient to obtain a pure solid; washing removes traces of dissolved solute.
- Suggesting a chemical test for purity instead of physical: Qualitative tests (e.g., adding sodium hydroxide) show the presence of an ion, not the purity of the bulk solid. Melting point determination is the required test for solid purity.
Things to Be Careful About
- Ensure every bullet point in the question prompt is explicitly covered in a distinct section of your answer to avoid losing marks for missing sections.
- Note the constraint: "No other chemicals should be used." Therefore, do not suggest adding other reagents (such as sodium hydroxide or barium nitrate) in the preparation.