Chemistry 5070/31 — October/November 2024
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Experimental Contexts · Analysis, Conclusions and Evaluation · Observations and Measurements · Use of Techniques, Apparatus and Materials · Qualitative Analysis · Planning Experiments and Investigations
You are provided with:
- an aqueous solution of ethanedioic acid, X
- aqueous sodium hydroxide, , Y.
You are going to investigate the reaction between X and Y.
Read all the instructions carefully before starting the experiments.
Instructions
You are going to do four titration experiments.
Rinse and fill a burette with X.
Experiment 1
- Use a volumetric pipette to add of Y to a conical flask.
- Add five drops of thymolphthalein indicator to the conical flask.
- Place the conical flask on a white tile.
- Record the initial burette reading in Table 1.1.
- Add X from the burette while swirling the flask, adding drop by drop near the end-point, until the solution just changes colour.
- Record the final burette reading in Table 1.1.
Experiments 2, 3 and 4
- Empty the conical flask and rinse it with distilled water.
- Refill the burette if necessary.
- Repeat Experiment 1.
Calculate the volume used in each experiment and record your values in Table 1.1.
Table 1.1
| experiment number 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| final burette reading / | ||||
| initial burette reading / | ||||
| volume of X used / | ||||
| best titration results (✓) |
Answer
All initial and final burette readings for the four experiments, each recorded to one decimal place (including 0.0). The volume of X used in each experiment is calculated as final reading − initial reading, also to one decimal place. The results are comparable to the supervisor's values.
Candidate's own readings; all readings and volumes to 1 decimal place
Walkthrough
In this part you perform a titration and record your results. The mark scheme rewards complete and precise recording: all initial and final burette readings must be entered, each to one decimal place (including 0.0). The volume used is the difference between final and initial readings, also to one decimal place. Your results are then compared with the supervisor's values, so they should be sensible for the reaction.
Key Takeaways
- Titration readings must be recorded to the nearest 0.05 cm3, but here to one decimal place.
- Always include the trailing zero (e.g., 0.0) to show precision.
- The volume used is always final − initial.
Common Mistakes
- Forgetting to record the initial reading as 0.0.
- Recording volumes as whole numbers.
- Arithmetic errors when subtracting.
- Not showing units in the table.
Things to Be Careful About
- Use the same number of decimal places for all readings.
- Rinse the pipette and burette appropriately, though not asked here.
- The white tile is used to help see the colour change.
Answer
Tick the two results that are closest together (concordant). The two results are within of each other.
The two results within 0.2 cm3 of each other
Walkthrough
After four titrations, you need to choose the two best results. These are the two that are most concordant, i.e., closest together, ideally within 0.2 cm3 of each other. Tick these two in the table and explain that they are concordant.
Key Takeaways
- The best titration results are those that agree closely with each other.
- Concordant results are usually within 0.1–0.2 cm3.
Common Mistakes
- Ticking any two results without checking closeness.
- Saying 'the two highest' or 'the two lowest' without reference to concordance.
Things to Be Careful About
- The explanation must mention the volume difference (within 0.2 cm3).
Use the ticked (✓) titration results in Table 1.1 to calculate the average volume of X needed to neutralise of Y.
volume = ______
Working
Let the two ticked volumes be and (in ).
Answer
Average of the two ticked volumes in cm3
Walkthrough
The average volume is the mean of the two ticked results. Add them and divide by two. This gives the average titre of X.
Key Takeaways
- Mean = sum of values / number of values.
- Use only the ticked results.
Common Mistakes
- Using all four results.
- Forgetting to divide by two.
Things to Be Careful About
- Keep the unit cm3.
Working
Answer
0.02 mol
Walkthrough
Moles of NaOH = concentration × volume in dm3. Convert 25.0 cm3 to 0.0250 dm3. Multiply by 0.800 to get 0.02 mol.
Key Takeaways
- Moles = concentration (mol/dm3) × volume (dm3).
- Volume must be in dm3.
Common Mistakes
- Forgetting to divide by 1000.
- Units error.
Things to Be Careful About
- The answer is 0.02 mol, not 20 mol.
One mole of ethanedioic acid is neutralised by two moles of sodium hydroxide.
Use your answers from (c) and (d) to calculate the concentration, in , of ethanedioic acid in X.
Give your answer to three significant figures.
concentration = ______
Working
From part (d), moles of .
From the equation, 1 mol acid reacts with 2 mol , so:
Let the average titre from part (c) be . Then:
Answer
(to 3 s.f.)
10/V mol/dm3 (to 3 s.f.)
Walkthrough
Use the mole ratio from the neutralisation: 1 mol acid : 2 mol NaOH. So moles of acid = moles NaOH ÷ 2. Then concentration = moles ÷ volume in dm3. The volume is the average titre from (c) in cm3, so divide by 1000. Give answer to 3 s.f.
Key Takeaways
- Stoichiometric ratio is essential.
- Concentration = moles / volume (dm3).
Common Mistakes
- Using the wrong mole ratio.
- Forgetting to convert cm3 to dm3.
- Not giving answer to 3 s.f.
Things to Be Careful About
- The final answer depends on your titre; ensure you use your own value.
The formula of ethanedioic acid is .
X contains of in of solution.
Use your answer from (e) to calculate the relative formula mass, , of .
Working
Mass of acid in = , so in :
Let the concentration from part (e) be . Then:
Using (from the expected titre), .
Answer
126
Walkthrough
The solution has 6.3 g in 100 cm3, so 63 g/dm3. Since concentration in mol/dm3 = (mass concentration in g/dm3) / Mr, we have Mr = 63 / c, where c is your answer from (e). With the expected titre, c = 0.5 mol/dm3, giving Mr = 126.
Key Takeaways
- Mass concentration (g/dm3) = Mr × molar concentration (mol/dm3).
- Convert volume to dm3.
Common Mistakes
- Using 6.3 g directly without scaling to 1 dm3.
- Confusing mass and molar concentration.
Things to Be Careful About
- Use the value from (e) consistently.
Use your answer from (f)(i) to deduce the value for in .
Give your answer to the nearest whole number.
[: H, 1; C, 12; O, 16]
Working
Answer
2
Walkthrough
The Mr of the hydrate is the Mr of the anhydrous acid plus n times the Mr of water. Calculate Mr(C2H2O4) = 90, Mr(H2O) = 18. Then n = (Mr(hydrate) − 90) / 18 = (126 − 90)/18 = 2.
Key Takeaways
- Water of crystallisation can be found from Mr difference.
- Mr(H2O) = 18.
Common Mistakes
- Using Mr of H2O as 17.
- Arithmetic error.
Things to Be Careful About
- n must be a whole number.
State why the conical flask is placed on a white tile before X is added from the burette.
______
Answer
To make the colour change at the end-point easier to see clearly.
To see the colour change more easily/clearly
Walkthrough
The white tile provides a white background against which the colour change at the end-point is easier to see.
Key Takeaways
- White tile improves contrast for colour change.
Common Mistakes
- Saying it is to hold the flask.
Things to Be Careful About
- Mention 'colour change' and 'easily/clearly'.
State why a measuring cylinder is not used to measure of aqueous in this experiment.
______
Answer
A measuring cylinder is not accurate/precise enough for measuring ; a volumetric pipette is used for accurate measurement.
It is not accurate/precise enough
Walkthrough
A measuring cylinder is not calibrated accurately enough for precise volume measurement. A volumetric pipette is used to measure 25.0 cm3 accurately.
Key Takeaways
- Volumetric pipette is more accurate than measuring cylinder.
- Accuracy matters in titrations.
Common Mistakes
- Saying it is too small or too large.
Things to Be Careful About
- Use 'accurate' or 'precise'.
You are provided with solid P and solution Q.
You will do a series of tests on P and Q.
Tests on solid P
You should:
- record your observations for each of these tests
- test and identify any gases evolved
- describe the gas test used that identifies any gas evolved.
Put the sample of P into a boiling tube. Add depth of dilute hydrochloric acid.
Keep the mixture for use in (b).
______
Answer
- Solid disappears.
- Effervescence (bubbles of gas).
- Gas test: bubble the gas through limewater; limewater turns milky/cloudy.
- Gas is carbon dioxide, .
Solid disappears; effervescence; gas turns limewater milky, so carbon dioxide
Walkthrough
P is a carbonate. When dilute hydrochloric acid is added, the carbonate reacts to give a salt, water and carbon dioxide. The solid disappears because it is used up, and bubbles of carbon dioxide cause effervescence. To identify the gas, bubble it through limewater; carbon dioxide turns limewater milky/cloudy. The equation is:
Keep the mixture for part (b): it now contains zinc chloride solution.
Key Takeaways
Acid + carbonate gives salt + water + carbon dioxide. The limewater test is the standard test for carbon dioxide.
Common Mistakes
- Writing only 'bubbles' without saying the gas was tested.
- Saying limewater turns 'white' instead of 'milky/cloudy'.
- Forgetting to state that the solid disappears.
Things to Be Careful About
Use the word 'effervescence' for vigorous bubbling. The gas test must name both the reagent (limewater) and the result (milky/cloudy).
Divide the mixture from (a) equally into two boiling tubes.
To the first portion of the mixture from (a), add aqueous sodium hydroxide drop by drop until a change is seen.
Then add excess aqueous sodium hydroxide.
______
Answer
- White precipitate forms.
- On adding excess sodium hydroxide, the precipitate dissolves.
- A colourless solution is formed.
White precipitate; dissolves in excess NaOH giving a colourless solution
Walkthrough
The mixture from (a) contains zinc ions. Adding sodium hydroxide supplies hydroxide ions, which react with zinc ions to form a white precipitate of zinc hydroxide. When more sodium hydroxide is added, this precipitate dissolves because zinc hydroxide is soluble in excess sodium hydroxide, giving a colourless solution. This behaviour is the key test for zinc ions.
Key Takeaways
Zinc hydroxide is a white precipitate that dissolves in excess sodium hydroxide to give a colourless solution. This identifies Zn2+.
Common Mistakes
- Saying the precipitate is insoluble in excess.
- Missing the 'colourless solution' mark.
- Describing the precipitate as 'white solid' instead of 'white precipitate'.
Things to Be Careful About
Add the sodium hydroxide drop by drop first; the precipitate appears before excess is added. The mark scheme wants three points: white ppt, soluble in excess, colourless solution.
To the second portion of the mixture from (a), add aqueous ammonia drop by drop until a change is seen.
Then add excess aqueous ammonia.
______
Answer
- White precipitate forms.
- On adding excess aqueous ammonia, the precipitate dissolves.
White precipitate; dissolves in excess ammonia
Walkthrough
This is the same zinc ion test but using aqueous ammonia. Ammonia provides hydroxide ions in solution, so a white precipitate of zinc hydroxide forms. Zinc hydroxide also dissolves in excess ammonia, so the precipitate disappears. This confirms the cation is zinc.
Key Takeaways
Zinc hydroxide dissolves in excess ammonia as well as excess sodium hydroxide; both tests give a white precipitate that is soluble in excess.
Common Mistakes
- Writing 'blue precipitate' (that would be copper).
- Saying the precipitate is insoluble in excess ammonia.
- Not mentioning 'white'.
Things to Be Careful About
The mark scheme gives two marks: white precipitate and soluble in excess. No colourless solution is required here, but it is true.
Answer
cation = zinc
anion = carbonate
zinc; carbonate
Walkthrough
The acid test produced carbon dioxide, so the anion in P must be carbonate. The hydroxide and ammonia tests gave a white precipitate soluble in excess, which is the standard test for zinc ions. Therefore P is zinc carbonate.
Key Takeaways
Observations from several tests are combined to identify an unknown compound: gas test gives the anion, precipitate behaviour gives the cation.
Common Mistakes
- Writing the formula instead of the name if the question asks for name.
- Confusing carbonate with hydrogencarbonate.
- Identifying the cation from only one test when both tests agree.
Things to Be Careful About
Use the names 'zinc' and 'carbonate' exactly as the blanks ask. The mark scheme awards one mark for each.
Tests on solution Q
You should record your observations for each of these tests.
Do a flame test on Q.
flame colour = ______
Answer
flame colour = lilac
lilac
Walkthrough
A flame test is done by dipping a clean wire into the solution and holding it in a blue Bunsen flame. Potassium ions give a lilac flame. The colour is the observation needed.
Key Takeaways
Flame colours: potassium = lilac, sodium = yellow, calcium = brick red, copper = blue-green.
Common Mistakes
- Writing 'purple' instead of 'lilac'.
- Confusing potassium with sodium (yellow).
- Not cleaning the wire before the test.
Things to Be Careful About
The question asks for the flame colour only; one mark for 'lilac'.
Answer
No observable change.
No observable change
Walkthrough
Nitric acid is added to acidify the solution before testing for anions. It removes carbonate ions that would otherwise interfere with the silver nitrate test. With potassium iodide solution, adding nitric acid produces no visible reaction.
Key Takeaways
Acidification with nitric acid is part of the test for halide ions; it prevents false results from carbonates.
Common Mistakes
- Inventing bubbles or a precipitate.
- Using hydrochloric acid instead of nitric acid (would add chloride ions and confuse the test).
Things to Be Careful About
The mark is for 'no observable change'. Do not write 'no reaction' if the question asks for observation; either is usually accepted, but 'no observable change' is safest.
Answer
- Yellow precipitate forms.
yellow precipitate
Walkthrough
Silver nitrate is added to the acidified solution. Silver ions react with iodide ions to form silver iodide, which is insoluble and appears as a yellow precipitate. The equation is:
Key Takeaways
Silver nitrate tests for halides: chloride gives white, bromide gives cream, iodide gives yellow. The precipitate colour identifies the halide.
Common Mistakes
- Writing 'white precipitate' (chloride).
- Writing 'cream precipitate' (bromide).
- Forgetting to say 'precipitate' and only writing 'yellow'.
Things to Be Careful About
The mark scheme gives one mark for 'yellow' and one for 'precipitate'. Acidification with nitric acid must have been done first.
Answer
cation = potassium
anion = iodide
potassium; iodide
Walkthrough
The lilac flame colour shows the cation is potassium. The yellow precipitate with acidified silver nitrate shows the anion is iodide. Therefore Q is potassium iodide, KI.
Key Takeaways
A flame test identifies the cation; the silver nitrate test identifies the halide anion. Together they identify the compound.
Common Mistakes
- Writing 'K' and 'I' instead of names.
- Confusing iodide with chloride or bromide.
- Not linking the flame colour to potassium.
Things to Be Careful About
The blanks ask for names, so write 'potassium' and 'iodide'. One mark for each.
You are not expected to do any practical work for this question.
Argentan is an alloy containing only zinc, nickel and copper.
Zinc and nickel both react with dilute hydrochloric acid. Copper does not react with dilute hydrochloric acid.
Plan an investigation to find the percentage by mass of copper in a powdered sample of argentan.
Your plan must include the use of common laboratory apparatus, argentan and hydrochloric acid. No other chemicals should be used.
Your plan must include:
- the apparatus needed
- the method to use and the measurements to take
- procedures to ensure that the percentage determined is as accurate as possible
- how the measurements are used to determine the percentage by mass of copper in the sample.
You may draw a diagram to help answer the question.
Answer
- Apparatus: balance, beaker or conical flask, measuring cylinder, filter funnel and filter paper, evaporating basin or watch glass.
- Weigh a fixed mass of the powdered argentan sample on the balance and record it.
- Place the sample in the beaker and add an excess of dilute hydrochloric acid. Zinc and nickel react (effervescence of hydrogen); copper does not react.
- Wait until effervescence stops, showing all the zinc and nickel have reacted.
- Filter the mixture. The unreacted copper is collected on the filter paper; the solution contains zinc and nickel chlorides.
- Wash the copper with distilled water and dry it thoroughly.
- Weigh the dry copper and record its mass.
- Repeat the whole procedure with fresh samples and calculate the average percentage.
- Calculate the percentage by mass of copper:
Weigh the argentan sample, add excess HCl, filter and dry the unreacted copper, weigh it, repeat and average, then %Cu = (mass Cu / mass sample) × 100.
Walkthrough
This question asks for a plan, not actual results. The key chemistry is that zinc and nickel are more reactive than hydrogen, so they displace hydrogen from dilute hydrochloric acid and dissolve, forming their chlorides. Copper is below hydrogen in the reactivity series, so it does not react with the acid and remains as a solid. Therefore, adding excess acid to the alloy dissolves the zinc and nickel completely, leaving only the copper behind. The copper can then be separated by filtration, dried and weighed. Comparing the mass of copper with the original mass of the alloy gives the percentage by mass of copper.
Step by step: weigh the alloy sample; add excess acid and wait until effervescence stops so all reactive metal has reacted; filter to collect the copper; wash and dry the copper to remove acid and solution; weigh the copper; repeat with fresh samples and average. The calculation is simply the mass of copper divided by the mass of the alloy, multiplied by 100.
Key Takeaways
- Metals above hydrogen in the reactivity series react with dilute acids; copper does not.
- Excess acid and waiting for effervescence to stop ensure complete reaction.
- Filtration separates an insoluble solid (copper) from a solution.
- Percentage by mass = (mass of component / total mass) × 100%.
- Repeating and averaging improves reliability.
Common Mistakes
- Not using excess acid, so some zinc or nickel remains unreacted and the copper mass is overestimated.
- Weighing the copper while it is still wet, which adds water to the mass.
- Not washing the copper, so dried acid salts remain and increase the mass.
- Forgetting to filter, so the copper cannot be separated.
- Using the wrong denominator in the calculation (e.g. mass of solution instead of mass of alloy).
- Not repeating the experiment, so the result is less reliable.
Things to Be Careful About
- Use the same units for both masses (e.g. grams) in the calculation.
- Make sure the copper is completely dry before the final weighing.
- If the copper is weighed with the filter paper, subtract the mass of the dry filter paper.
- Record masses to the precision of the balance (e.g. two decimal places).
- Hydrochloric acid is corrosive/irritant; wear eye protection and avoid skin contact.
- The mark scheme awards marks in four areas: apparatus, method and measurements, accuracy, and calculation; make sure all four are covered.