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S5 Chemistry titration paper 2: titration data and purity calculations
S5 Chemistry titration paper 2: titration data and purity calculations
Ten items cover average titre, concentration from mass, dibasic acids, the two-stage carbonate titration and sample purity.
Direct answers Items 01 to 02 cover titre data and mass to concentration, 03 to 06 dibasic acid, g per dm3 and dilution, 07 to 09 an unknown acid, two-stage titration and purity, and 10 the unconverted volume unit.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Chemistry Acid-Base Titration |
| Level | S5 |
02 Questions (10 items)
- FoundationTitration of 25.0 cm3 of hydrochloric acid needs 23.50 cm3 of 0.100 M sodium hydroxide on average. Find the acid concentration.
- FoundationDissolve 4.0 g of sodium hydroxide, molar mass 40.0, to make 250 cm3 of solution. Find the molarity.
- IntermediateWhat volume of 0.150 M sodium hydroxide neutralises 20.0 cm3 of 0.100 M sulphuric acid?
- IntermediateExpress 0.250 M sodium hydroxide in g per dm3, taking the molar mass as 40.0.
- IntermediateDilute 25.0 cm3 of 2.00 M hydrochloric acid to 500 cm3. Find the new concentration.
- Foundation25.0 cm3 of 0.100 M sodium carbonate is titrated with hydrochloric acid to the phenolphthalein end point. What volume of 0.100 M acid is used?
- Challenge20.0 cm3 of a monobasic acid needs 18.75 cm3 of 0.120 M sodium hydroxide. Find the acid concentration.
- ChallengeThe same sodium carbonate solution is titrated first to the phenolphthalein end point and then to the methyl orange end point. How do the two acid volumes compare?
- ChallengeA 0.500 g impure sodium hydroxide sample is made up to 100 cm3. A 25.0 cm3 portion needs 22.50 cm3 of 0.100 M hydrochloric acid. Find the percentage purity.
- Error redoA student puts 25.0 cm3 straight into the volume and gets 1.0 times 10 to the minus 4 M. With 0.00250 mol of solute, what is the correct molarity?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.0940 M | The alkali supplies 0.002350 mol, equal to the acid moles, divided by 0.0250 dm3. |
| 2 | 0.400 M | 4.0 g is 0.100 mol dissolved in 0.250 dm3. |
| 3 | 26.7 cm3 | There are 0.00400 mol of hydrogen ions, divided by 0.150 M. |
| 4 | 10.0 g per dm3 | There are 0.250 mol in each dm3, multiplied by 40.0. |
| 5 | 0.100 M | The 0.0500 mol is unchanged and is divided by 0.500 dm3. |
| 6 | 25.0 cm3 | The carbonate ion first takes one hydrogen ion to become hydrogencarbonate, a one to one ratio. |
| 7 | 0.113 M | The alkali gives 0.002250 mol, equal to the acid moles, divided by 0.0200 dm3. |
| 8 | The two volumes are equal | The carbonate takes one hydrogen ion in each of two steps, so both steps use the same number of moles. |
| 9 | 72.0 percent | The whole 100 cm3 holds 0.00900 mol, which is 0.360 g and 72.0 percent of 0.500 g. |
| 10 | 0.100 M | Volume must be in dm3, and 25.0 cm3 is 0.0250 dm3. |
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.
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