S4 Chemistry: Neutralisation And Titration (Paper 4: Misconception Drill)

S4 Chemistry: Neutralisation and Titration (Paper 4: misconception drill)

2026-11-12

S4 Chemistry: Neutralisation and Titration (Paper 4: misconception drill)

Ten misconception items on cm3 to dm3 conversion, the 1 to 2 ratio of sulphuric acid, the titration formula, the wrong indicator, subtracting burette readings, rejecting outlying titres and common errors in purity work.

Direct answers　Answers in order: 2.50 x 10-3 mol, 0.020 mol, 12.5 cm3, the end point is too late so the reading is too large, 20.80 cm3, reject the outlying reading, 500 cm3, 40.0 cm3, 6.5 degrees Celsius, and the unconverted volume makes the answer 1000 times too large.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	20
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Chemistry: Neutralisation and Titration
Level	S4

02　Questions (10 items)

FoundationAfter converting 25.0 cm3 into dm3, how many moles are in a 0.100 mol/dm3 solution?
FoundationThe mole ratio of sulphuric acid to sodium hydroxide is 1 to 2. How many moles of alkali does 0.010 mol of acid need?
FoundationUsing c1V1 over n1 equals c2V2 over n2, what volume of 0.100 mol/dm3 sulphuric acid matches 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide?
IntermediateWhat happens if phenolphthalein is wrongly used for a strong acid with a weak base?
IntermediateA burette reads 0.50 cm3 at the start and 21.30 cm3 at the end. What volume was delivered?
IntermediateA titre of 25.6 cm3 that is clearly off is included in the mean. What should be done?
ChallengeWhat volume of 0.200 mol/dm3 hydrochloric acid neutralises 4.00 g of sodium hydroxide, relative molecular mass 40?
ChallengeTreating sulphuric acid as monobasic halves the volume of alkali calculated. How much 0.100 mol/dm3 sodium hydroxide does 20.0 cm3 of 0.100 mol/dm3 acid really need?
ChallengeNeutralisation warms the solution from 25.0 degrees Celsius to 31.5 degrees Celsius. What is the temperature rise?
Error redoA student writes 2.50 mol instead of 2.50 x 10-3 mol. What is the root cause?

03　Answers and one-line notes

#	Answer	One-line note
1	2.50 x 10-3 mol	25.0 cm3 is 0.0250 dm3, and 0.100 times 0.0250 gives 2.50 x 10-3 mol.
2	0.020 mol	Each mole of acid needs two moles of alkali, so the value is doubled.
3	12.5 cm3	The alkali gives 2.50 x 10-3 mol, which matches 1.25 x 10-3 mol of acid, and dividing by 0.100 gives 0.0125 dm3.
4	The end point comes too late and the acid volume recorded is too large	Phenolphthalein only changes in the alkaline range, long after the equivalence point, so the reading is too large.
5	20.80 cm3	Subtracting the start reading from the final one gives 20.80 cm3.
6	Reject the reading that is clearly off and average only the close values	A reading far from the rest comes from an operational error, and including it distorts the mean.
7	500 cm3	4.00 divided by 40 gives 0.100 mol, and dividing by 0.200 gives 0.500 dm3.
8	40.0 cm3	The ratio is 1 to 2, so the volume of alkali is twice that of the acid.
9	6.5 degrees Celsius	Subtracting 25.0 from 31.5 gives 6.5 degrees, which does not depend on the time taken.
10	The volume was not converted from cm3 to dm3, so the answer is 1000 times too large	25.0 cm3 is 0.0250 dm3, and multiplying by 0.100 gives only 2.50 x 10-3 mol.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

For every item write concentration times volume to get moles, compare the mole ratio, then give the answer with its unit.Tutor Registration →

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