IGCSE Chemistry 0620/52 October/November 2025 Past Paper with Answers

Question 1

In a titration, 25 cm³ of acidic solution B is titrated against aqueous sodium hydroxide solution C, recording the burette reading at the first end-point (methyl orange changing red to orange) and at the second end-point (thymolphthalein changing yellow to green). How should the burette readings be recorded, and what relationship is expected between the two end-points?

1. all readings are recorded to one decimal place using a consistent precision, and the volume to the second end-point is about twice the volume to the first end-point
2. only the final readings need recording, and the two end-point volumes should be exactly equal
3. readings are recorded as whole numbers of cm³, and the first end-point volume is about twice the second end-point volume
4. readings are estimated to the nearest 5 cm³, and the two end-point volumes are unrelated

Question 2

After Experiment 1 of a titration, the conical flask is rinsed with distilled water before Experiment 2 is carried out in the same flask. Why is the flask rinsed with distilled water?

1. to remove any solution left from Experiment 1 so it does not contaminate Experiment 2
2. to add extra water that increases the volume of base needed in Experiment 2
3. to warm the flask so the second reaction goes faster
4. to add more acid to the flask ready for Experiment 2

Question 3

A burette is rinsed with distilled water but not with sodium hydroxide solution C before being filled with solution C. Compared with a correctly rinsed burette, how is the volume of solution C needed to reach the first end-point affected, and why?

1. a greater volume is needed, because the water left in the burette dilutes solution C and makes it less concentrated
2. a smaller volume is needed, because the leftover water makes solution C more concentrated
3. the same volume is needed, because rinsing the burette has no effect on the result
4. a greater volume is needed, because the leftover water reacts with and uses up the acid

Question 4 · Answer A

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.

Question 5 · Answer A

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.

Question 6 · Answer A

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.

Question 7 · Answer A

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.

Question 8 · Answer A

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.

Question 9 · Answer A

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.

Question 10 · Answer A

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.

Question 11 · Answer A

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.

Question 12 · Answer A

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.

Question 13 · Answer A

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.

Question 14 · Answer A

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, $\text{(NH}_4\text{)}_2\text{SO}_4$.

Question 15 · Answer A

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)).

Question 16 · Answer A

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.

Question 17 · Answer A

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.

Question 18 · Answer A

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.

Question 19 · Answer A

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.

Question 20 · Answer A

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.