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S4 Chemistry: Neutralisation and Titration (Paper 2: intermediate)
S4 Chemistry: Neutralisation and Titration (Paper 2: intermediate)
Ten intermediate items on finding concentration from titration data, the 1 to 2 volume ratio of sulphuric acid, mean titre, end point against equivalence point, indicator choice and the error of adding volumes.
Direct answers Answers in order: 0.125 mol/dm3, 20.0 cm3, 40.0 cm3, 20.0 cm3, 10.0 cm3, end point and equivalence point need not coincide, 100 cm3, 0.0900 mol/dm3, methyl orange, and volumes may not be added.
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)
- FoundationTitrating 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide needs 20.0 cm3 of hydrochloric acid. Find the acid concentration.
- FoundationWhat volume of 0.150 mol/dm3 sodium hydroxide titrates 20.0 cm3 of 0.150 mol/dm3 hydrochloric acid?
- FoundationWhat volume of 0.100 mol/dm3 sodium hydroxide neutralises 20.0 cm3 of 0.100 mol/dm3 sulphuric acid?
- IntermediateThree titres read 19.8 cm3, 20.0 cm3 and 20.2 cm3. What is the mean titre?
- IntermediateWhat volume of 0.100 mol/dm3 sulphuric acid neutralises 25.0 cm3 of 0.0800 mol/dm3 sodium hydroxide?
- IntermediateWhat is the difference between the end point and the equivalence point of a titration?
- ChallengeWhat volume of 0.125 mol/dm3 sulphuric acid neutralises 0.0250 mol of sodium hydroxide?
- Challenge25.0 cm3 of sodium hydroxide solution needs 22.5 cm3 of 0.100 mol/dm3 hydrochloric acid. Find the alkali concentration.
- ChallengeWhich indicator is normally used for a titration of a strong acid with a weak base?
- Error redoA student adds 20.0 cm3 and 25.0 cm3 and uses the sum as the total volume in the titration formula. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.125 mol/dm3 | The alkali holds 2.50 x 10-3 mol, and dividing by 0.0200 dm3 gives 0.125 mol/dm3. |
| 2 | 20.0 cm3 | The ratio is 1 to 1 and the concentrations are equal, so the volumes are equal. |
| 3 | 40.0 cm3 | The acid gives 2.00 x 10-3 mol, the ratio is 1 to 2, so 4.00 x 10-3 mol of alkali is needed, which is 40.0 cm3. |
| 4 | 20.0 cm3 | Adding the three readings and dividing by three gives 20.0 cm3. |
| 5 | 10.0 cm3 | The alkali gives 2.00 x 10-3 mol, which needs 1.00 x 10-3 mol of acid, and dividing by 0.100 gives 0.0100 dm3. |
| 6 | The end point is when the indicator changes colour, while the equivalence point is where the acid and alkali have reacted exactly | The two do not always coincide, and the colour range of the indicator decides the size of the error. |
| 7 | 100 cm3 | The acid needed is 0.0125 mol, and dividing by 0.125 gives 0.100 dm3. |
| 8 | 0.0900 mol/dm3 | The acid gives 2.25 x 10-3 mol, which equals the moles of alkali, and dividing by 0.0250 dm3 gives 0.0900 mol/dm3. |
| 9 | Methyl orange | The equivalence point of such a titration lies in the acidic range, and methyl orange changes between pH 3.1 and 4.4. |
| 10 | Each side must be turned into moles separately, the two volumes may not be added as if diluted | A titration compares moles on each side, so adding the two volumes has no meaning. |
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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