中文（繁體）
中文（简体）
English
S5 Chemistry Titration and pH, Paper 4: concentrations, dilution and back titration
S5 Chemistry Titration and pH, Paper 4: concentrations, dilution and back titration
Ten graded items on acid-alkali volumes, strong acid and base pH, diprotic and diluted pH, a mixture, a back titration and a mean titre, each with an answer.
Focus Find the moles first, double them for a diprotic acid, take the excess in a mixture, subtract in a back titration and discard the odd reading.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 Chemistry: Acid-Base Titration and pH |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- FoundationWhat volume of 0.100 mol dm^-3 sodium hydroxide neutralises 25.0 cm^3 of 0.200 mol dm^-3 hydrochloric acid?
- FoundationWhat is the pH of 0.0100 mol dm^-3 hydrochloric acid?
- FoundationWhat is the pH of 0.00050 mol dm^-3 sodium hydroxide at 25 C?
- FoundationWhat volume of 0.100 mol dm^-3 hydrochloric acid fully reacts with 25.0 cm^3 of 0.100 mol dm^-3 sodium carbonate?
- IntermediateWhat is the pH of 0.0050 mol dm^-3 sulphuric acid, fully ionised in both stages?
- Intermediate20.0 cm^3 of 0.200 mol dm^-3 hydrochloric acid is diluted to 500 cm^3. What is the new pH?
- Intermediate40.0 cm^3 of 0.100 mol dm^-3 sodium hydroxide is mixed with 10.0 cm^3 of 0.100 mol dm^-3 hydrochloric acid. What is the pH?
- Challenge0.800 g of a sample containing calcium carbonate is treated with 25.0 cm^3 of 0.500 mol dm^-3 hydrochloric acid and the excess needs 10.0 cm^3 of 0.250 mol dm^-3 sodium hydroxide. Find the percentage by mass of calcium carbonate.
- ChallengeTitres of 26.60, 26.50 and 24.10 cm^3 are obtained against 20.0 cm^3 of 0.100 mol dm^-3 sodium carbonate. Find the acid concentration.
- Error redoError repair: 20.0 cm^3 of distilled water is put into the conical flask before 25.0 cm^3 of the alkali is pipetted in. Does the calculated acid concentration come out high, low or unchanged?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 50.0 cm^3 | The acid is 0.00500 mol and neutralises one to one, so the same number of moles divided by 0.100 gives 0.0500 dm^3. |
| 2 | pH 2.00 | A strong acid ionises completely, so the hydrogen ion concentration is 0.0100 and its negative logarithm is 2.00. |
| 3 | pH 10.70 | The hydroxide concentration is 0.00050, so the pOH is 3.30 and the pH is 14 minus 3.30. |
| 4 | 50.0 cm^3 | The carbonate is 0.00250 mol and reacts one to two with the acid, so 0.00500 mol is needed, which doubles the volume at equal concentration. |
| 5 | pH 2.00 | Each mole of sulphuric acid releases two moles of hydrogen ions, giving 0.0100 whose negative logarithm is 2.00. |
| 6 | pH 2.10 | Diluting 25 times takes the hydrogen ion concentration from 0.200 down to 0.0080, whose negative logarithm is 2.10. |
| 7 | pH 12.78 | The alkali is in excess by 0.00300 mol, which over the 0.0500 dm^3 total gives 0.0600 mol dm^-3 of hydroxide and a pOH of 1.22. |
| 8 | 62.5% | The acid totals 0.01250 mol and the back titration uses 0.00250 mol, so 0.01000 mol reacted, which is 0.00500 mol or 0.500 g of calcium carbonate. |
| 9 | 0.151 mol dm^-3 | The 24.10 cm^3 reading is far off and discarded, so the mean of 26.60 and 26.50 is 26.55 cm^3, and 0.00200 mol of carbonate takes 0.00400 mol of acid. |
| 10 | Unchanged, because the moles of alkali are the same | Adding water only makes the end point easier to see; the moles of alkali are unchanged, so the acid needed and the calculated concentration stay the same. |
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.
Book an S5 Chemistry tutorTutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
