2.7.6Inorganic Chemistry
Prepare a Soluble Salt II — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 20 of 23 Unit quiz
Part 1 of 6Why Use a Titration?
This method is used when both reactants are soluble.
Soluble base
- Acids react with alkalis, which are soluble, so no excess solid can be filtered off to remove leftover reactant.
Exact volumes
- A titration finds the exact volume of acid that neutralises a fixed volume of alkali.
Pure salt
- Repeating without indicator gives a salt solution that contains no indicator or leftover reactant.
Example
- Sodium hydroxide and sulfuric acid give sodium sulfate: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l).
Part 2 of 6Method
The titration fixes the volumes, and the second run gives a pure product.
1
Use a pipette to place a measured volume of alkali into a conical flask and add a few drops of indicator such as phenolphthalein or methyl orange.
2
Fill a burette with acid and note the starting volume.
3
Add the acid slowly until the indicator just changes colour, then record the final volume and find the volume added.
4
Repeat the process in a clean flask with the same volumes of acid and alkali but with no indicator.
5
Heat the solution gently to evaporate some water, leaving a saturated solution.
6
Leave to crystallise, decant the remaining liquid and dry the crystals.
Part 3 of 6Key Practical Points
Key Practical Points
Partial evaporation
- Some water is left behind so the salt does not overheat or decompose and can keep its water of crystallisation.
Saturation test
- Dip a cold glass rod in the solution;
- crystals forming on it show the solution is saturated enough to crystallise.
Indicator
- Indicator is left out of the final preparation because it would contaminate the crystals.
Drying
- Crystals are filtered or decanted off and dried between filter paper or in a warm place.
| Indicator | In acid | In alkali |
|---|---|---|
| Methyl orange | Red | Yellow |
| Phenolphthalein | Colourless | Pink |
Part 4 of 6Comparing Methods
Comparing Methods
Acid plus alkali
- Titration is needed because neither reactant can be added in excess and filtered off.
Acid plus insoluble base
- Add excess solid, filter it off, then crystallise the filtrate.
Precipitation
- For insoluble salts, mix two soluble salt solutions and filter off the solid.
Part 5 of 6Worked Example
Question
- 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is exactly neutralised by sulfuric acid.
- Calculate the maximum mass of anhydrous sodium sulfate that can be obtained.
- (Ar: Na = 23, S = 32, O = 16)
- Moles of NaOH = 0.100 × 25.0 ÷ 1000 = 0.00250 mol.
- From 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, moles of Na₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol.
- Mr of Na₂SO₄ = (2 × 23) + 32 + (4 × 16) = 142.
- Mass = 0.00125 × 142 = 0.1775 g.
Answer Maximum mass of Na₂SO₄ = 0.178 g (3 s.f.)
Part 6 of 6Exam Checklist & Tips
2.7.6 – Key Points for the Exam
- Titration A method to find the exact volumes of two solutions that react together.
- Saturated solution A solution that cannot dissolve any more solute at that temperature.
- Crystallisation Formation of solid crystals as a hot saturated solution cools.
- Burette Apparatus used to deliver accurately measured volumes of liquid.
Common Mistakes
- Leaving indicator in the final mixture, which contaminates the salt.
- Evaporating to dryness, which can decompose the salt and lose water of crystallisation.
- Using different volumes of acid and alkali in the second run.
- Choosing a titration for an insoluble base; excess solid and filtration is used instead.
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