A salt is an ionic compound formed when the hydrogen in an acid is replaced by a metal ion (or ammonium ion). The method you choose to prepare a specific salt depends on whether the salt is soluble or insoluble in water and which reactants are available — each route gives a different practical procedure to follow.
What is a salt and how are salts named?
Salts are produced when an acid reacts with a base, a metal, or a carbonate. The name of a salt has two parts:
- The first part comes from the metal (or ammonium) — it matches the base or metal used.
- The second part comes from the acid: hydrochloric acid → chloride; sulfuric acid → sulfate; nitric acid → nitrate.
Examples: sodium chloride (NaCl) from NaOH + HCl; copper sulfate (CuSO₄) from CuO + H₂SO₄; zinc nitrate from Zn + HNO₃.
Method 1 — Reaction of an acid with an excess insoluble base or metal
This method is used when the salt is soluble in water and the base is an insoluble metal oxide, hydroxide, or carbonate (or when using a metal directly with a dilute acid).
Why use excess solid? If you add just enough base to neutralise the acid, you cannot tell exactly when neutralisation is complete without an indicator. Adding excess solid guarantees all the acid reacts. The unreacted solid is simply filtered off.
Steps:
- Add the acid to a beaker. Add the insoluble base (or metal) in small portions, stirring continuously.
- Continue adding until no more solid dissolves and some excess remains at the bottom.
- Filter the mixture to remove excess solid. The filtrate is a solution of the salt.
- Transfer the filtrate to an evaporating basin. Heat gently to evaporate most of the water until a small volume of saturated solution remains.
- Leave to cool — salt crystals grow as the solution becomes supersaturated.
- Filter to collect the crystals. Pat dry with filter paper and leave to dry fully.
Example: Copper(II) sulfate — add excess copper(II) oxide (CuO) to warm dilute sulfuric acid (H₂SO₄), filter off excess CuO, evaporate and crystallise.
Word equation: copper(II) oxide + sulfuric acid → copper(II) sulfate + water
Method 2 — Neutralisation of an acid with a soluble alkali (titration method)
Use this when both the acid and the base (alkali) are soluble in water — you cannot filter off excess alkali if you add too much. An indicator is used to find the exact volume of alkali needed for complete neutralisation.
Steps:
- Pipette a measured volume of alkali into a conical flask. Add a few drops of a suitable indicator (e.g. phenolphthalein).
- Fill a burette with the acid. Slowly add the acid from the burette, swirling constantly.
- Record the volume added at the end point (when the indicator just changes colour permanently).
- Repeat the titration without indicator, using exactly the same volumes of acid and alkali.
- Evaporate and crystallise the salt solution as in Method 1.
Example: Sodium chloride (NaCl) — titrate hydrochloric acid against sodium hydroxide solution.
Why repeat without indicator? Many indicators are organic dyes that would contaminate the product.
Method 3 — Precipitation of an insoluble salt
Use this when the desired salt is insoluble in water. Mixing two solutions whose ions combine to form the insoluble salt gives an immediate precipitate.
Steps:
- Prepare two solutions, each containing one of the required ions.
- Mix the solutions in a beaker — the insoluble salt precipitates immediately.
- Filter to collect the precipitate.
- Wash the residue on the filter paper with distilled water to remove soluble impurities.
- Dry in a low oven or leave at room temperature.
Example: Barium sulfate (BaSO₄) — mix barium chloride solution with sulfuric acid (or sodium sulfate solution). The barium ions and sulfate ions combine to form the insoluble barium sulfate precipitate.
Ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Which method should I choose for a given salt?
| Salt required | Salt soluble? | Base soluble? | Method |
|---|---|---|---|
| Copper(II) sulfate | Yes | CuO — insoluble | Method 1 (excess insoluble base) |
| Zinc chloride | Yes | Zn — insoluble metal | Method 1 (excess metal) |
| Sodium nitrate | Yes | NaOH — soluble | Method 2 (titration) |
| Potassium sulfate | Yes | KOH — soluble | Method 2 (titration) |
| Barium sulfate | No | n/a | Method 3 (precipitation) |
| Lead iodide | No | n/a | Method 3 (precipitation) |
What is crystallisation and why is it used?
Crystallisation is the process of obtaining solid, pure crystals from a solution. It works because the solubility of most salts decreases as temperature decreases — so a hot saturated solution, when cooled, deposits pure crystals as the dissolved ions arrange into a regular lattice.
Slow cooling produces larger, purer crystals; rapid cooling produces fine crystals that may trap impurities. For GCSE practical work, evaporating to about half the original volume and then leaving to cool slowly is the standard approach.
Frequently asked questions
What are the three methods for preparing a salt at GCSE?
The three main methods are: (1) reacting an acid with an excess insoluble base or metal, filtering off the excess, then crystallising the salt; (2) titrating an acid against a soluble alkali to find the exact volumes needed, then repeating without indicator and crystallising; (3) mixing two solutions whose ions form an insoluble salt as a precipitate, then filtering, washing, and drying. The choice depends on whether the salt and the base are soluble.
Why do you use excess solid base in Method 1?
Using excess solid base ensures all of the acid is neutralised — no acid remains in the product. Since the base is insoluble, any unreacted excess simply forms a solid that can be removed by filtration. If you used exactly the right amount of base without excess, it would be very difficult to confirm that neutralisation was complete, and you risk having unreacted acid in your salt solution.
What is an ionic equation and why is it useful for precipitation reactions?
An ionic equation shows only the ions that actually change during a reaction — the spectator ions (which do not change) are omitted. For precipitation reactions, the ionic equation is particularly useful because it shows exactly which ions combine to form the precipitate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). This highlights the essential chemistry regardless of which soluble starting compounds were used to supply those ions.
How do you obtain pure, dry crystals from a salt solution?
Heat the salt solution gently in an evaporating basin to drive off most of the water, until a small volume of concentrated solution remains (or until crystals just begin to form at the edges — this is the "point of crystallisation"). Remove from heat and allow to cool slowly. As the solution cools, crystals grow. Filter to collect the crystals, wash briefly with a small amount of cold distilled water, then press dry with filter paper and leave in a warm place or low oven to dry completely.
For Socratic GCSE chemistry with Professor Curie — reasoning particle-first about why each preparation method works before writing a single equation — visit aitutors.me.