0620/52 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

20 questions · 40 marks · 75 minutes

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1. Question 1(a)

   6 marksRecording titration readings and titre relationships
   Step 1: Burette readings are recorded to one decimal place (to 0.05 cm³) and to a consistent number of decimal places. Step 2: With two indicators the first end-point marks the first stage of neutralisation and the second end-point the complete reaction, so the volume to the second end-point is about twice that to the first. Step 3: So all readings are recorded to one decimal place and the second end-point volume is about double the first.
   Method:

   Recall the rules for recording burette readings and the expected ratio of the two end-point volumes.
   Examiner tips

   • Always record burette readings to one decimal place and a consistent precision.

2. Question 1(b)(i)

   1 marksReason for rinsing the conical flask between titrations
   Step 1: Any solution left in the flask from Experiment 1 would mix with the new reactants and change the result. Step 2: Rinsing with distilled water washes out these contaminants, and pure water does not add any acid or base. Step 3: So the flask is rinsed to remove contamination left from Experiment 1.
   Method:

   Recall why apparatus is cleaned between repeat experiments.
   Examiner tips

   • Rinse glassware to remove contamination from a previous experiment.

3. Question 1(b)(ii)

   2 marksEffect of not rinsing the burette with the titrant
   Step 1: Distilled water left in the burette mixes with solution C and dilutes it. Step 2: The more dilute solution C contains fewer moles of hydroxide per cm³, so a larger volume is needed to neutralise the same amount of acid. Step 3: So a greater volume is needed, because the water dilutes solution C.
   Method:

   Link the leftover water to dilution of the titrant and then to a larger titre.
   Examiner tips

   • Rinse the burette with the titrant so it is not diluted by leftover water.

4. Question 1(b)(iii)

   1 marksReason for standing the flask on a white tile
   Step 1: The end-point is judged by a colour change of the indicator. Step 2: A white background makes the colour stand out, so the change is seen clearly and the end-point is judged accurately. Step 3: So a white tile is used to make the indicator colour change easier to see.
   Method:

   Link the white background to clearer detection of the indicator colour change.
   Examiner tips

   • Use a white tile so the indicator colour change is clear.

5. Question 1(c)

   3 marksComparing concentrations from titration volumes
   Step 1: The same 25 cm³ of acid always needs the same number of moles of hydroxide to neutralise it. Step 2: Only half the volume of solution D is needed compared with solution C, so each cm³ of D must contain twice as many moles, meaning D is twice as concentrated. Step 3: So solution D is twice as concentrated as solution C.
   Method:

   Relate the volume needed to the concentration for a fixed amount of acid.
   Examiner tips

   • For the same amount of acid: smaller titre = more concentrated base.

6. Question 1(d)(i)

   2 marksScaling the titre volume with the amount of acid
   Step 1: Doubling the volume of acid from 25 to 50 cm³ doubles the moles of acid to be neutralised. Step 2: Twice as many moles of acid need twice the volume of solution C: 2 × 12.0 = 24.0. Step 3: So 24.0 cm³ of solution C is needed, with the unit cm³.
   Method:

   Scale the first end-point volume in proportion to the volume of acid and attach the unit.
   Examiner tips

   • Volume of base scales in proportion to the amount of acid.

7. Question 1(d)(ii)

   1 marksApparatus limitation: burette capacity
   Step 1: Doubling the acid to 50 cm³ roughly doubles every titre, and the second end-point already needs about twice the first end-point volume. Step 2: This combined volume would exceed the 50.0 cm³ capacity of the burette. Step 3: So the problem is that the volume of solution C needed would be more than the burette can hold.
   Method:

   Compare the expected titre with the fixed capacity of the burette.
   Examiner tips

   • Check that the expected titre fits within the 50 cm³ burette.

8. Question 1(e)

   2 marksEffect of warming the acid on the titre
   Step 1: The number of moles of acid in 25 cm³ of solution B is fixed and does not change when the solution is warmed. Step 2: Neutralisation needs a number of moles of hydroxide that depends only on the moles of acid, not on the temperature. Step 3: So warming has no effect and the same volume of solution D is needed.
   Method:

   Recall that the titre depends on the amount (moles) of acid, which warming does not change.
   Examiner tips

   • The titre depends on moles of reactant, not on temperature.

9. Question 1(f)

   1 marksImproving accuracy with a volumetric pipette
   Step 1: A measuring cylinder measures a fixed volume only roughly. Step 2: A volumetric pipette delivers a precise, fixed volume far more accurately than a measuring cylinder. Step 3: So accuracy is improved by using a volumetric pipette instead of a measuring cylinder to measure the 25 cm³ of solution B.
   Method:

   Choose the apparatus that measures a fixed volume most accurately.
   Examiner tips

   • Use a pipette, not a measuring cylinder, for an accurate fixed volume.

10. Question 2(a)(i)

    1 marksEffect of sodium hydroxide on an ammonium salt without warming
    Step 1: Ammonium ions do not form a precipitate with aqueous sodium hydroxide. Step 2: Ammonia gas is only released when the mixture is warmed, so in the cold nothing visible happens. Step 3: So at room temperature, without warming, no change is seen.
    Method:

    Recall that ammonium salts give no change with cold sodium hydroxide.
    Examiner tips

    • Ammonium ions give no precipitate with sodium hydroxide; warming is needed for the gas.

11. Question 2(a)(ii)

    1 marksTesting the gas released from an ammonium salt on warming
    Step 1: Warming an ammonium salt with sodium hydroxide releases ammonia gas. Step 2: Ammonia is alkaline, so it turns damp red litmus paper blue. Step 3: So the damp red litmus paper turns blue.
    Method:

    Match the alkaline gas ammonia to its damp red litmus test result.
    Examiner tips

    • Ammonia turns damp red litmus paper blue.

12. Question 2(b)

    1 marksTest for sulfate ions with barium nitrate
    Step 1: Acidified barium nitrate reacts with sulfate ions to form barium sulfate, which is insoluble. Step 2: Barium sulfate appears as a white precipitate. Step 3: So a white precipitate forms, showing that sulfate ions are present.
    Method:

    Apply the acidified barium nitrate test for sulfate ions.
    Examiner tips

    • Acidified barium nitrate + sulfate = white precipitate.

13. Question 2(c)

    1 marksResult of acidified potassium manganate(VII) with a non-reducing salt
    Step 1: Acidified potassium manganate(VII) is decolourised from purple to colourless only by a reducing agent, such as a sulfite. Step 2: A sulfate is not a reducing agent, so it cannot decolourise the manganate(VII). Step 3: So the few drops stay pink/purple, showing no reducing agent is present.
    Method:

    Decide whether a reducing agent is present, then predict if the manganate(VII) is decolourised.
    Examiner tips

    • Manganate(VII) only loses its colour with a reducing agent.

14. Question 2(d)

    2 marksIdentifying a salt from its ion tests
    Step 1: A gas that turns damp red litmus paper blue on warming with sodium hydroxide shows ammonium ions. Step 2: A white precipitate with acidified barium nitrate shows sulfate ions. Step 3: Combining ammonium ions with sulfate ions gives ammonium sulfate, $\text{(NH}_4\text{)}_2\text{SO}_4$.
    Method:

    Deduce the cation and anion separately, then name the compound.
    Examiner tips

    • Identify the cation and the anion from their tests, then combine them.

15. Question 2(e)

    1 marksFlame colour for sodium
    A flame test on a sodium compound gives a characteristic yellow flame (in contrast to lilac for potassium, orange-red for calcium and blue-green for copper(II)).
    Method:

    Recall the flame colour produced by sodium ions.
    Examiner tips

    • Flame colours: Na yellow, K lilac, Ca orange-red, Cu blue-green.

16. Question 2(f)(i)

    1 marksSilver nitrate test for iodide ions
    Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide and yellow for iodide. Step 2: Iodide ions give a yellow precipitate of silver iodide. Step 3: So a yellow precipitate forms.
    Method:

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

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

17. Question 2(f)(ii)

    2 marksEffect of sodium hydroxide on a zinc salt
    Step 1: Sodium hydroxide added to a zinc salt gives a white precipitate of zinc hydroxide. Step 2: Zinc hydroxide is amphoteric, so it dissolves in excess sodium hydroxide to give a colourless solution. Step 3: So a white precipitate forms dropwise and it dissolves in excess.
    Method:

    Match the white-precipitate-soluble-in-excess result to a zinc ion.
    Examiner tips

    • Zinc: white precipitate with sodium hydroxide, soluble in excess.

18. Question 2(f)(iii)

    2 marksEffect of aqueous ammonia on a zinc salt
    Step 1: Aqueous ammonia added to a zinc salt gives a white precipitate of zinc hydroxide. Step 2: Zinc hydroxide dissolves in excess ammonia to give a colourless solution; this also distinguishes zinc from aluminium, whose precipitate stays insoluble in excess ammonia. Step 3: So a white precipitate forms dropwise and it dissolves in excess.
    Method:

    Use the dissolving in excess ammonia to identify zinc and rule out aluminium.
    Examiner tips

    • Zinc dissolves in excess ammonia; aluminium does not.

19. Question 2(g)

    3 marksIdentifying the ions in a salt from combined tests
    Step 1: A yellow flame test shows sodium ions. Step 2: A white precipitate with both sodium hydroxide and ammonia that dissolves in excess of each shows zinc ions (aluminium would not dissolve in excess ammonia). Step 3: A yellow precipitate with silver nitrate shows iodide ions, so solid F contains sodium, zinc and iodide ions.
    Method:

    Read each test result, identify the ion it indicates, and list all three ions.
    Examiner tips

    • Combine the flame, hydroxide/ammonia and silver nitrate results to find all three ions.

20. Question 3

    6 marksPlanning a fair comparison of fuels heating water
    Step 1: To compare fairly, use the same measured volume of water (measured with a measuring cylinder) at the same starting temperature for each alcohol, with the burner the same distance from the boiling tube. Step 2: Heat the water with each alcohol and time how long it takes to reach a chosen higher temperature (or, alternatively, heat for a fixed time and measure the temperature rise). Step 3: The alcohol that reaches the set temperature in the shortest time (or gives the greatest temperature rise in a fixed time) heats the water fastest.
    Method:

    Design a fair test that controls the water and measures either the time to a set temperature or the temperature rise in a set time.
    Examiner tips

    • Control the volume of water and start temperature; measure the time to a set temperature rise.