0620/62 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

20 questions · 40 marks · 60 minutes

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1. Question 1(a)(i)

   1 marksObserving a carbonate reacting with acid
   Step 1: A carbonate reacts with an acid to give a salt, water and carbon dioxide gas. Step 2: The carbon dioxide escapes as bubbles, which is seen as effervescence (fizzing). Step 3: There is no colour change and no precipitate, so what is seen is effervescence as bubbles of gas are given off.
   Method:

   Recall the products of an acid-carbonate reaction and pick the visible sign of the gas.
   Examiner tips

   • Carbonate + acid -> effervescence (bubbles of carbon dioxide).

2. Question 1(a)(ii)

   2 marksChecking that an acid is in excess
   Step 1: If the acid is in excess, all the carbonate is used up, so no more gas is produced when more acid is added. Step 2: Equally, leftover acid keeps the mixture acidic, so universal indicator turns (and stays) red or orange. Step 3: Temperature, colour of the residue and mass do not show excess acid, so the correct check is no more bubbles on adding acid, or a red/orange universal-indicator colour.
   Method:

   Link "acid in excess" to leftover acid that keeps the mixture acidic and stops further reaction.
   Examiner tips

   • Test for excess acid: no more bubbles on adding acid, or indicator stays acidic.

3. Question 1(b)(i)

   2 marksApparatus for filtration
   Step 1: Filtration separates an insoluble solid from a liquid using a filter paper held in a filter funnel. Step 2: The funnel is supported over a beaker (or flask) so the liquid (filtrate) runs through the paper while the solid (residue) stays on the paper. Step 3: A funnel with no paper, an evaporating basin or a measuring cylinder cannot trap the solid, so the correct apparatus is a filter paper in a funnel over a beaker.
   Method:

   Match the job (separate insoluble solid from liquid) to the standard filtration set-up.
   Examiner tips

   • Filtration = filter paper in a funnel over a beaker; residue on paper, filtrate through.

4. Question 1(b)(ii)

   2 marksSubstances removed when washing a residue
   Step 1: The acid is in excess, so leftover dilute hydrochloric acid clings to the residue. Step 2: The reaction makes soluble calcium chloride, which also clings to the residue. Step 3: Both dissolve in the wash water and are rinsed away, so the two substances removed by washing are hydrochloric acid and calcium chloride.
   Method:

   Identify the soluble substances present after the reaction that would cling to the solid.
   Examiner tips

   • Wash away the soluble products and leftover acid: calcium chloride and hydrochloric acid.

5. Question 1(b)(iii)

   2 marksCalculating a percentage by mass from experimental data
   Step 1: The calcium carbonate is the part of the 2.00 g sample that reacted, so its mass is 2.00 - 0.11 = 1.89 g. Step 2: The percentage by mass is (mass of calcium carbonate / mass of sample) x 100 = (1.89 / 2.00) x 100. Step 3: This gives 94.5%, so the eggshell is 94.5% calcium carbonate by mass.
   Method:

   Find the reacted mass by subtracting the residue, then express it as a percentage of the sample.
   Examiner tips

   • Reacted mass = total - residue; then divide by total and times 100.

6. Question 2(a)

   5 marksReading a burette and calculating titres
   Step 1: A titre is the final (end-point) burette reading minus the initial reading, recorded to one decimal place. Step 2: Experiment 1: first end-point 21.2 - 1.6 = 19.6 cm³, second end-point 41.0 - 1.6 = 39.4 cm³. Step 3: Experiment 2: first end-point 10.2 - 0.4 = 9.8 cm³, second end-point 20.1 - 0.4 = 19.7 cm³, giving titres of 19.6, 39.4, 9.8 and 19.7 cm³.
   Method:

   Subtract the initial reading from each end-point reading for both experiments.
   Examiner tips

   • Both end-points are measured from the one initial reading, not from each other.

7. Question 2(b)(i)

   1 marksWhy a flask is rinsed with distilled water
   Step 1: The flask was used in the previous experiment, so traces of the old solution could be left inside. Step 2: Rinsing with distilled water washes these contaminants away so they cannot affect the next result. Step 3: The rinse does not react with anything or change concentrations, so its purpose is simply to clean the flask and remove contaminants.
   Method:

   Recognise that a reused flask must be cleaned so old chemicals do not affect the new result.
   Examiner tips

   • Rinse used glassware with distilled water to remove contaminants.

8. Question 2(b)(ii)

   2 marksEffect of not rinsing a burette with the solution
   Step 1: Distilled water left clinging inside the burette mixes with solution C and dilutes it, so the solution C run out is less concentrated. Step 2: A less concentrated alkali contains fewer moles per cm³, so more of it is needed to neutralise the same acid. Step 3: Therefore the volume added to reach the first end-point would be greater, because the leftover water dilutes solution C.
   Method:

   Trace how leftover water changes the concentration delivered, then how that changes the volume needed.
   Examiner tips

   • Rinse a burette with the solution it will hold, or leftover water dilutes it.

9. Question 2(b)(iii)

   1 marksWhy a white tile is used in a titration
   Step 1: The end-point is judged from a colour change of the indicator in the flask. Step 2: A plain white background makes that colour stand out, so the change is easier and clearer to see. Step 3: The tile does not warm the flask or affect the burette reading, so its purpose is to make the colour change easier to see.
   Method:

   Link the white background to judging the indicator colour change at the end-point.
   Examiner tips

   • White tile = clear background for spotting the indicator colour change.

10. Question 2(c)

    3 marksComparing concentrations from titration volumes
    Step 1: The same amount of acid in solution B is neutralised in both experiments, so the amount of sodium hydroxide reacting is the same each time. Step 2: That same amount is supplied by 39.4 cm³ of C but only 19.7 cm³ of D, so D is more concentrated because it delivers the same amount in a smaller volume. Step 3: Since 39.4 is about twice 19.7, solution D is about twice the concentration of solution C.
    Method:

    Compare the volumes needed for the same acid, then use the ratio to compare concentrations.
    Examiner tips

    • Smaller titre for the same acid = more concentrated alkali; the volume ratio gives the concentration ratio.

11. Question 2(d)(i)

    2 marksScaling a titre with the volume of acid
    Step 1: Doubling the volume of solution B from 25 cm³ to 50 cm³ doubles the amount of acid to be neutralised. Step 2: Twice the acid needs twice the volume of solution C to reach the same first end-point. Step 3: So the volume of C is 19.6 x 2 = 39.2 cm³ (the unit is cm³).
    Method:

    Use the proportional link between amount of acid and volume of alkali needed.
    Examiner tips

    • Double the acid volume -> double the titre (19.6 -> 39.2 cm³).

12. Question 2(d)(ii)

    1 marksA titre that exceeds the burette capacity
    Step 1: Doubling solution B to 50 cm³ doubles the volume of C needed for the second end-point, from 39.4 cm³ to about 78.8 cm³. Step 2: The burette only holds 50 cm³, so it cannot deliver 78.8 cm³ in one filling. Step 3: The burette would have to be refilled part-way, which is awkward and less accurate, so the problem is that the volume needed is more than the burette can hold.
    Method:

    Predict the new titre and compare it with the burette capacity.
    Examiner tips

    • If a predicted titre exceeds 50 cm³, a single burette filling is not enough.

13. Question 2(e)

    2 marksEffect of warming the acid on the titre
    Step 1: The titre depends on the amount (moles) of acid in solution B, not on its temperature. Step 2: Warming the solution does not add or remove acid, so the number of moles of acid stays the same. Step 3: The same amount of acid needs the same amount of solution D to neutralise it, so there is no change in the volume of D required.
    Method:

    Decide what the titre depends on (amount of acid), then judge whether warming changes it.
    Examiner tips

    • The titre depends on moles of acid, which warming does not change.

14. Question 2(f)

    1 marksImproving the accuracy of measuring a volume
    Step 1: A measuring cylinder measures volume only roughly, so using it for the fixed 25 cm³ of solution B limits the accuracy. Step 2: A volumetric pipette delivers a fixed 25 cm³ much more accurately. Step 3: Swapping the measuring cylinder for a volumetric pipette therefore improves the accuracy of the results.
    Method:

    Find the least accurate measuring step and replace its apparatus with a more accurate one.
    Examiner tips

    • Use a volumetric pipette (not a measuring cylinder) to measure a fixed volume accurately.

15. Question 3(a)(i)

    1 marksTesting the gas from an ammonium salt with alkali
    Step 1: Warming an ammonium salt with sodium hydroxide releases ammonia gas. Step 2: Ammonia is an alkaline gas, and an alkaline gas turns damp red litmus paper blue. Step 3: Ammonia does not bleach litmus or pop with a splint, so the observation is that the damp red litmus paper turns blue.
    Method:

    Identify the gas released, then recall its effect on damp red litmus.
    Examiner tips

    • Damp red litmus turning blue = an alkaline gas (ammonia).

16. Question 3(a)(ii)

    1 marksIdentifying the gas from an ammonium salt
    Step 1: Warming an ammonium salt with sodium hydroxide produces ammonia gas. Step 2: Ammonia is the only common gas that is alkaline, which is why it turns damp red litmus paper blue. Step 3: Hydrogen, carbon dioxide and chlorine do not turn damp red litmus blue, so the gas is ammonia.
    Method:

    Use the alkaline-gas test result to name the gas as ammonia.
    Examiner tips

    • Ammonium salt + warm alkali -> ammonia (turns damp red litmus blue).

17. Question 3(b)

    1 marksBarium nitrate test for sulfate ions
    Step 1: Acidified barium nitrate is the test for sulfate ions. Step 2: Sulfate ions react with the barium ions to give a white precipitate of insoluble barium sulfate. Step 3: A cream or brown precipitate would mean other ions, so sulfate ions give a white precipitate.
    Method:

    Match the barium nitrate test result to the sulfate ion.
    Examiner tips

    • Acidified barium nitrate + sulfate -> white precipitate.

18. Question 3(c)

    1 marksResult of adding potassium manganate(VII) to a non-reducing solution
    Step 1: Acidified potassium manganate(VII) is decolourised (purple to colourless) only when a reducing agent is present. Step 2: Sulfate ions are not a reducing agent (unlike sulfite ions), so there is nothing to decolourise the manganate(VII). Step 3: The purple colour therefore remains, appearing pink/purple in the diluted few drops, so the solution is not decolourised.
    Method:

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

    • Manganate(VII) is decolourised only by a reducing agent; sulfate is not one.

19. Question 3(d)

    3 marksIdentifying three ions from combined test results
    Step 1: A yellow flame is the flame colour of sodium ions, so one ion is sodium, Na⁺. Step 2: A white precipitate that dissolves in excess of BOTH sodium hydroxide and ammonia (each to a colourless solution) is given by zinc, Zn²⁺; aluminium would not dissolve in excess ammonia. Step 3: A yellow precipitate with acidified silver nitrate is iodide, I⁻ (chloride is white, bromide cream), so the three ions are sodium, zinc and iodide.
    Method:

    Take each test result in turn and match it to one ion, ruling out the look-alikes.
    Examiner tips

    • Soluble in excess of BOTH NaOH and ammonia points to zinc (aluminium is not soluble in excess ammonia).

20. Question 4

    6 marksPlanning a fair test to compare fuels
    Step 1: For a fair test only the alcohol should change, so use the same fixed volume of water (measured with a measuring cylinder, burette or pipette) and record its starting temperature each time. Step 2: Heat the water with one alcohol burner and measure the heating in a comparable way: either time how long it takes to reach a set temperature, or heat for a fixed time and find the temperature rise. Step 3: Repeat under identical conditions for all three alcohols; the alcohol with the shortest time (or the largest temperature rise) heats the water fastest.
    Method:

    Control every variable except the alcohol, then choose a single comparable measurement of how fast the water heats.
    Examiner tips

    • Fair test: same volume of water, same conditions; compare time-to-set-temperature or temperature rise.