Preparation of salts
Section: Chemistry | Syllabus: Cambridge IGCSE Combined Science (0653)
Naming Salts
A salt is formed when the H⁺ ions of an acid are replaced by metal or ammonium ions.
| Acid | Salt ending | Example |
|---|---|---|
| Hydrochloric acid | -chloride | sodium chloride, NaCl |
| Sulfuric acid | -sulfate | copper sulfate, CuSO₄ |
| Nitric acid | -nitrate | potassium nitrate, KNO₃ |
There are four methods for preparing a soluble salt, depending on what's available: excess metal, excess insoluble base (oxide/hydroxide), excess insoluble carbonate, or titration with an alkali.
Method 1: Acid + Excess Metal
Suitable for moderately reactive metals (Mg, Zn, Fe) - too reactive (Na, K, Ca) is dangerous; too unreactive (Cu, Ag) won't react at all.
acid + metal → salt + hydrogen
- Add excess metal to the acid; it reacts and bubbles of H₂ are given off
- When bubbling stops and metal remains, all the acid has reacted
- Filter to remove unreacted excess metal
- Evaporate the filtrate to concentrate it, then cool to crystallise
- Filter and dry the crystals
Method 2: Acid + Excess Insoluble Base
Used for unreactive metals, via their oxide or hydroxide (a base).
acid + base → salt + water
Worked Example: Copper(II) Chloride from Copper(II) Oxide
- Question: copper(II) chloride crystals are prepared by reacting dilute hydrochloric acid with excess solid copper(II) oxide. Name the method used to remove unreacted copper(II) oxide. [1]
- Answer: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l); the excess unreacted solid is removed by filtration
After filtering, the blue filtrate is evaporated and cooled to crystallise, exactly as in Method 1.
Method 3: Acid + Excess Insoluble Carbonate
This is listed separately from Method 2 because it produces an extra product - a gas.
acid + carbonate → salt + water + carbon dioxide
e.g. CuCO₃(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) + CO₂(g)
Procedure is otherwise identical to Method 2: add excess insoluble carbonate (fizzing confirms reaction), filter off the excess, then evaporate/crystallise/dry the filtrate.
Worked Example: Full Practical Method
- Question: hydrated chromium(III) nitrate is made by reacting the insoluble base chromium(III) oxide with dilute nitric acid, which reacts very slowly at room temperature. Describe how to make pure, dry crystals. [6]
- Answer: add chromium(III) oxide to dilute nitric acid in a beaker, in excess, so all the acid is used up [1]; stir, and gently heat/warm the mixture, since the reaction is slow at room temperature [1]; filter to remove the excess unreacted oxide [1]; gently heat/evaporate the filtrate to reduce the volume of water, without evaporating to dryness [1]; cool to let crystals form, then filter to collect them [1]; dry the crystals, e.g. between filter papers or in a warm oven [1]
The same excess-then-filter-then-crystallise sequence works whether the insoluble reactant is a base or a carbonate
Method 4: Acid + Alkali (Titration)
Used when the base is soluble (an alkali) - since both reactants dissolve, there is no excess solid to filter off, so exact volumes are needed instead.
- Titrate: add acid from a burette to a measured volume of alkali with a few drops of indicator, until the colour just changes - this finds the exact volume of acid needed
- Repeat using the same volumes, but with no indicator (it would contaminate the salt)
- Evaporate, cool to crystallise, filter, and dry, as before
A suitable indicator for the titration is methyl orange, which gives a sharp colour change at the neutralisation point (litmus's colour change is too gradual to pinpoint the endpoint).
Worked Example: Titration Technique
- Question: dilute hydrochloric acid is added to a known volume of aqueous sodium hydroxide, in the presence of an indicator, until neutral. (i) Write the ionic equation. (ii) Name a suitable indicator. (iii) Name the technique. [3]
- Answer (i): H⁺ + OH⁻ → H₂O
- Answer (ii): methyl orange
- Answer (iii): titration
Common Mistakes
- ❌ Adding excess alkali in Method 4, as in Methods 1-3
- ✓ both reactants in Method 4 are soluble, so excess alkali would remain dissolved in the salt solution - exact volumes (found by titration) are essential instead
- ❌ Evaporating a salt solution to complete dryness
- ✓ some salts decompose on strong heating - evaporate only until crystals start to form, then cool and filter
Preparing Insoluble Salts by Precipitation
An insoluble salt cannot be made by the acid-based methods (it wouldn't dissolve to be crystallised). Instead, mix two soluble solutions that each supply one of the required ions - the insoluble salt forms immediately as a precipitate. (You will be told, or can work out from the question, which starting compounds are soluble - you do not need to memorise general solubility rules.)
soluble salt A(aq) + soluble salt B(aq) → insoluble salt(s) + soluble salt(aq)
- Mix the two soluble solutions - the precipitate forms at once
- Filter to collect the precipitate
- Wash with distilled water to remove soluble impurities
- Dry between filter paper or in a warm oven (no crystallisation needed)
Worked Example: Lead(II) Chloride
- Question: which two compounds react together to form the insoluble salt lead(II) chloride, given: 1 - lead(II) nitrate (soluble), 2 - lead(II) sulfate (insoluble), 3 - silver chloride (insoluble), 4 - sodium chloride (soluble)? [1]
- Answer: 1 and 4. Both reactants must be soluble so they can be mixed as solutions: Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
Ionic equation: Pb²⁺(aq) + 2Cl⁻(aq) → PbCl₂(s) - the Na⁺ and NO₃⁻ are spectator ions and don't appear.
Mix two soluble solutions, filter out the precipitate, wash, then dry - no crystallisation step is needed
Common Mistakes
- ❌ Evaporating/crystallising an insoluble salt like a soluble one
- ✓ insoluble salts are already solid the moment they form - just filter, wash, and dry (no heating/evaporating solution needed)
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