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S4 Chemistry: Neutralisation and Titration (Paper 3: challenge)
S4 Chemistry: Neutralisation and Titration (Paper 3: challenge)
Ten challenge items on the 1 to 2 titration of sodium carbonate, volume from mass, neutralising a diluted acid, the leftover of the limiting reagent, heat of neutralisation, percentage purity and the pH of the end point.
Direct answers Answers in order: 25.0 cm3, 50.0 cm3, 0.100 mol, 0.100 mol/dm3, 5.0 x 10-4 mol of acid left, 2.85 kJ, 96.0 percent, 0.0800 mol/dm3, 10.0 cm3, and the end point may lie in the acidic range.
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)
- Foundation25.0 cm3 of 0.100 mol/dm3 sodium carbonate reacts fully with 0.200 mol/dm3 hydrochloric acid. What volume of acid is needed?
- FoundationWhat volume of 0.100 mol/dm3 hydrochloric acid neutralises 0.200 g of sodium hydroxide, relative molecular mass 40?
- FoundationWhen 0.0500 mol of sulphuric acid reacts fully with excess sodium hydroxide, how many moles of water are formed?
- Intermediate25.0 cm3 of hydrochloric acid of unknown concentration needs 12.5 cm3 of 0.100 mol/dm3 sodium carbonate. Find the acid concentration.
- Intermediate20.0 cm3 of 0.100 mol/dm3 sulphuric acid is mixed with 30.0 cm3 of 0.100 mol/dm3 sodium hydroxide. Which reactant is left and in what amount?
- IntermediateThe molar enthalpy of neutralisation is -57 kJ/mol. How much heat is released when 0.0500 mol of hydrogen ions is neutralised?
- Challenge1.00 g of impure sodium hydroxide needs 96.0 cm3 of 0.250 mol/dm3 hydrochloric acid. Find the percentage purity, relative molecular mass 40.
- Challenge25.0 cm3 of sulphuric acid is titrated with 0.100 mol/dm3 sodium hydroxide and 40.0 cm3 is used. Find the acid concentration.
- Challenge100 cm3 of 0.100 mol/dm3 hydrochloric acid is diluted to 500 cm3. What volume of 0.0500 mol/dm3 sodium hydroxide neutralises 25.0 cm3 of the diluted acid?
- Error redoA student claims that the end point of a titration is always neutral with pH 7. Why is this wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 25.0 cm3 | The carbonate gives 2.50 x 10-3 mol, which needs 5.00 x 10-3 mol of acid, and dividing by 0.200 gives 0.0250 dm3. |
| 2 | 50.0 cm3 | 0.200 divided by 40 gives 5.00 x 10-3 mol, the ratio is 1 to 1, and dividing by 0.100 gives 0.0500 dm3. |
| 3 | 0.100 mol | The equation gives two moles of water per mole of acid, so 0.0500 mol gives 0.100 mol. |
| 4 | 0.100 mol/dm3 | The carbonate gives 1.25 x 10-3 mol, which needs 2.50 x 10-3 mol of acid, and dividing by 0.0250 dm3 gives 0.100 mol/dm3. |
| 5 | Sulphuric acid is left, 5.0 x 10-4 mol | The acid needs 4.00 x 10-3 mol of alkali but only 3.00 x 10-3 mol is present, so 5.0 x 10-4 mol of acid remains. |
| 6 | 2.85 kJ | Multiplying 57 by 0.0500 gives 2.85 kJ. |
| 7 | 96.0 percent | The acid gives 0.0240 mol, which corresponds to 0.960 g of sodium hydroxide, and 0.960 over 1.00 gives 96.0 percent. |
| 8 | 0.0800 mol/dm3 | The alkali gives 4.00 x 10-3 mol, which corresponds to 2.00 x 10-3 mol of acid, and dividing by 0.0250 dm3 gives 0.0800 mol/dm3. |
| 9 | 10.0 cm3 | The diluted acid is 0.0200 mol/dm3, so 25.0 cm3 holds 5.00 x 10-4 mol, and dividing by 0.0500 gives 0.0100 dm3. |
| 10 | The end point is where the indicator changes colour, and for a strong acid with a weak base it lies in the acidic range | Only a strong acid with a strong base gives an equivalence point near pH 7, other combinations are not neutral. |
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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