May/June 2025 Paper 61 Worked Answers (IGCSE Chemistry 0620 Extended)

22 questions · 40 marks · 60 minutes

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1. Question 1(a)

   1 marksCalculating the volume added from burette readings
   Step 1: The volume delivered from a burette is the final reading minus the initial reading. Step 2: That is 18.8 cm³ − 4.2 cm³ = 14.6 cm³. Step 3: So the volume of dilute sulfuric acid added is 14.6 cm³.
   Method:

   Subtract the initial burette reading from the final burette reading.
   Examiner tips

   • Always do final burette reading minus initial burette reading.

2. Question 1(b)

   1 marksComparing concentrations from reacting volumes
   Step 1: The reaction is 1:1, so equal numbers of moles of potassium carbonate and sulfuric acid react. Step 2: Those equal moles of potassium carbonate fill the larger volume (25.0 cm³) while the same moles of acid fill only 14.6 cm³. Step 3: Concentration is moles per volume, so the larger volume holding the same moles is the more dilute, making the potassium carbonate the least concentrated.
   Method:

   Use the 1:1 ratio to see equal moles react, then compare moles per volume.
   Examiner tips

   • In a 1:1 reaction, the solution needing the larger volume is the more dilute.

3. Question 1(c)

   1 marksChoosing apparatus to measure a fixed volume accurately
   Step 1: Measuring a fixed volume accurately needs apparatus calibrated for that exact volume. Step 2: A volumetric pipette delivers a precise fixed volume such as 25.0 cm³, far more accurately than a beaker or flask. Step 3: So the apparatus is a volumetric pipette.
   Method:

   Match the need for an accurate fixed volume to the correct measuring apparatus.
   Examiner tips

   • Use a volumetric pipette for an accurate fixed volume.

4. Question 1(d)

   2 marksIndicator and end-point colour change in a titration
   Step 1: Methyl orange is a suitable indicator for an acid being added to an alkaline carbonate. Step 2: In the alkaline solution it is yellow, and at the end-point the solution becomes acidic, turning the methyl orange orange. Step 3: So a suitable indicator is methyl orange, changing from yellow to orange at the end-point.
   Method:

   Recall the colour of methyl orange in alkali and how it changes as the acid neutralises it.
   Examiner tips

   • Methyl orange: yellow in alkali, orange/red in acid.

5. Question 1(e)

   1 marksMixing the contents during a titration
   Step 1: As the acid is added it must be mixed evenly with the carbonate so the reaction reaches the end-point fairly. Step 2: Swirling the conical flask mixes the two solutions thoroughly. Step 3: So the student should swirl (mix) the contents of the conical flask.
   Method:

   Recall the standard titration action of swirling to mix the reagents.
   Examiner tips

   • Swirl continually so the acid mixes evenly with the carbonate.

6. Question 1(f)

   3 marksObtaining pure crystals of a soluble salt
   Step 1: Repeat the reaction with the known volumes (25.0 cm³ carbonate and 14.6 cm³ acid) but without indicator, so the salt is not contaminated by the indicator. Step 2: Warm the solution in an evaporating basin until it reaches the crystallisation point, when about half the water has gone and crystals begin to form. Step 3: Leave the solution to cool so that pure crystals of potassium sulfate form, then filter them off and dry them.
   Method:

   Apply the soluble-salt preparation method: make the salt cleanly, concentrate by evaporation, then crystallise on cooling.
   Examiner tips

   • Make the salt without indicator, evaporate part of the water, then cool to crystallise.

7. Question 2(a)

   2 marksRecording a reaction time from a stop-watch to the nearest second
   Step 1: The stop-watch shows 0 minutes and 20 seconds, so no whole minute has passed. Step 2: The total time is therefore just 20 seconds, which is already a whole number of seconds. Step 3: So the time recorded for Experiment 1 is 20 s.
   Method:

   Convert the stop-watch reading to a whole number of seconds.
   Examiner tips

   • Read minutes and seconds separately; with 0 minutes the time is just the seconds.

8. Question 2(b)

   4 marksPlotting a graph of reaction time against concentration
   Step 1: A good graph uses a linear scale (equal steps) chosen so the plotted points fill more than half of the available grid. Step 2: Every result is plotted accurately, then a single smooth line of best fit is drawn through the trend. Step 3: So the correct approach is a linear y-axis scale filling over half the grid, all points plotted accurately and one line of best fit.
   Method:

   Recall the rules for a good line graph: linear scale, points spread out, single best-fit line.
   Examiner tips

   • Choose a scale so the points use more than half the grid; draw one best-fit line.

9. Question 2(c)

   2 marksReading a value by interpolation from a best-fit line
   Step 1: To read an intermediate value, find 1.3 mol/dm³ on the concentration axis. Step 2: Draw a vertical construction line up to the line of best fit, then a horizontal line across to the time axis. Step 3: The value where the horizontal line meets the time axis is the answer, so you draw a vertical line up from 1.3 then read horizontally across to the time axis.
   Method:

   Use construction lines to interpolate the time at the stated concentration.
   Examiner tips

   • Interpolation: up to the line, then across to the axis.

10. Question 2(d)(i)

    2 marksCalculating mean rate of reaction with units
    Step 1: Mean rate = length ÷ time = 5 cm ÷ 20 s. Step 2: 5 ÷ 20 = 0.25, and the unit is length over time, so cm/s. Step 3: So the mean rate of reaction is 0.25 cm/s.
    Method:

    Substitute the 5 cm length and the 20 s time into rate = length ÷ time and attach the unit cm/s.
    Examiner tips

    • rate = length ÷ time; keep the unit as cm/s.

11. Question 2(d)(ii)

    1 marksIdentifying the slowest reaction
    Step 1: A lower concentration of acid gives fewer collisions per second and so a slower reaction. Step 2: Experiment 5 uses the lowest concentration, 0.5 mol/dm³. Step 3: So the mean rate of reaction is slowest in Experiment 5.
    Method:

    Match the slowest rate to the experiment with the lowest acid concentration.
    Examiner tips

    • Lowest concentration = slowest mean rate.

12. Question 2(e)

    1 marksReason for repeating an experiment
    Step 1: Repeating an experiment lets you compare the results from each run. Step 2: Close (concordant) results show the data is repeatable, and any value that differs a lot can be spotted as an anomaly and left out. Step 3: So repeating each experiment lets you check the results are repeatable and identify anomalous values.
    Method:

    Recall why repeat readings improve the reliability of experimental data.
    Examiner tips

    • Repeats improve reliability and reveal anomalies.

13. Question 2(f)(i)

    2 marksCalculating a temperature change
    Step 1: The temperature change is the final temperature minus the initial temperature. Step 2: That is 46.5 °C − 24.0 °C = 22.5 °C, written to the same one decimal place as the readings. Step 3: So the temperature change is 22.5 °C.
    Method:

    Subtract the initial temperature from the final temperature, keeping consistent decimal places.
    Examiner tips

    • Always do final minus initial, keeping the same decimal places.

14. Question 2(f)(ii)

    1 marksWhy temperature must be controlled in a rate experiment
    Step 1: Temperature is a factor that affects the rate of reaction: a higher temperature speeds the reaction up. Step 2: If the temperature varied between experiments, the rate would change for that reason as well as the concentration, so the comparison would be unfair. Step 3: So keeping the temperature constant is an improvement because a change in temperature would change the rate of reaction.
    Method:

    Recall that temperature is a rate factor that must be controlled to keep the test fair.
    Examiner tips

    • Control temperature so only the concentration is varied.

15. Question 2(f)(iii)

    1 marksWhy an insulated cup does not keep temperature constant
    Step 1: The reaction of magnesium with acid is exothermic, so it releases heat. Step 2: Polystyrene is a good insulator, so it prevents that heat escaping to the surroundings. Step 3: With the heat trapped, the temperature still rises, so a polystyrene cup does not keep the temperature constant because it prevents heat escaping.
    Method:

    Link the insulating property of polystyrene to trapped heat and a rising temperature.
    Examiner tips

    • Insulation traps heat, so the temperature is not controlled.

16. Question 2(f)(iv)

    1 marksUsing a water bath to control temperature
    Step 1: To keep a temperature constant you surround the reaction vessel with something held at that temperature. Step 2: A water bath kept at a fixed temperature absorbs or supplies heat to hold the acid at that temperature. Step 3: So you stand the conical flask in a water bath kept at a constant temperature.
    Method:

    Recall that a water bath at a fixed temperature is used to control temperature.
    Examiner tips

    • Use a thermostatically controlled water bath to fix the temperature.

17. Question 3(a)

    2 marksObservations when acid reacts with a carbonate
    Step 1: A carbonate reacts with dilute acid to give a salt, water and carbon dioxide, so you see effervescence as the gas bubbles off. Step 2: Carbon dioxide turns limewater milky. Step 3: So the observations are effervescence and the gas turning limewater milky.
    Method:

    Apply the acid + carbonate reaction and the limewater test for the gas.
    Examiner tips

    • Carbonate + acid → effervescence + CO₂ (limewater milky).

18. Question 3(b)(i)

    1 marksTesting for sulfate ions with barium nitrate
    Step 1: Acidified barium nitrate gives a white precipitate of barium sulfate if sulfate ions are present. Step 2: Here no precipitate forms, so there are no sulfate ions in solution H. Step 3: So the observation is no change, showing that sulfate ions are absent.
    Method:

    Apply the barium nitrate test for sulfate and interpret the absence of a precipitate.
    Examiner tips

    • Barium nitrate + sulfate = white ppt; no ppt = no sulfate.

19. Question 3(b)(ii)

    2 marksEffect of sodium hydroxide on a calcium salt solution
    Step 1: Adding sodium hydroxide to a calcium salt gives a white precipitate of calcium hydroxide. Step 2: Calcium hydroxide is insoluble in excess sodium hydroxide, so the precipitate stays when more is added. Step 3: So a white precipitate forms dropwise and it remains insoluble in excess.
    Method:

    Match the white-precipitate-insoluble-in-excess result to a calcium ion.
    Examiner tips

    • White ppt, insoluble in excess NaOH = calcium (or magnesium).

20. Question 3(b)(iii)

    1 marksSilver nitrate test giving a white precipitate for chloride
    Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide, yellow for iodide. Step 2: Solution H was made using hydrochloric acid, so it contains chloride ions, which give a white precipitate. Step 3: So the observation is a white precipitate, showing chloride ions are present.
    Method:

    Match the silver nitrate precipitate colour to the halide ion present.
    Examiner tips

    • Silver nitrate: chloride white, bromide cream, iodide yellow.

21. Question 3(c)

    2 marksIdentifying a solid from flame and halide tests
    Step 1: A lilac flame shows a potassium ion. Step 2: A cream precipitate with silver nitrate, and an orange colour when chlorine displaces it, both show a bromide ion. Step 3: Combining a potassium ion with a bromide ion gives potassium bromide, KBr.
    Method:

    Deduce the cation from the flame test and the anion from the silver nitrate test, then name the salt.
    Examiner tips

    • Identify the metal from the flame test and the halide from the silver nitrate colour.

22. Question 4

    6 marksPlanning a comparison of two endothermic reactions
    Step 1: To compare fairly, use the same known mass of sodium hydrogencarbonate and the same measured volume of each acid, measured with suitable apparatus in a suitable container. Step 2: Measure the temperature of the acid before mixing, add and stir, then measure the temperature after the reaction. Step 3: Endothermic reactions take in heat, so the temperature falls; the reaction with the larger temperature decrease is the more endothermic.
    Method:

    Design a fair test that measures the temperature change and links the larger decrease to the more endothermic reaction.
    Examiner tips

    • Fair test: same mass, same volume; the bigger temperature decrease is more endothermic.