Last updated: 6 October 2026
When you are asked to prepare a salt in the laboratory, the hardest part is usually not remembering the apparatus. It is deciding what should react, which substance should be added in excess, whether filtration is needed and whether the product will be found in the residue or the filtrate.
These decisions are all connected. A salt is made from positive and negative ions, and its solubility determines what happens to those ions when the reactants are mixed. If the product is insoluble, it appears as a solid precipitate. If it is soluble, it remains dissolved and must later be crystallised from the solution.
So instead of memorising three unrelated experimental procedures, we will build one logical decision-making system that works for almost every O-Level salt-preparation question.
Try these five questions before reading the chapter. They test the ideas about acids, solubility and filtration that you will need when choosing a salt-preparation method.
- 5 quick questions
- Instant feedback for every option
- Find misconceptions before you start
What Is a Salt?
A salt is an ionic compound formed when the hydrogen ions in an acid are replaced by metal ions or ammonium ions, NH₄⁺.
For example, when hydrochloric acid reacts with sodium hydroxide, the H⁺ ions from the acid are replaced by Na⁺ ions:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Sodium chloride contains Na⁺ and Cl⁻ ions. The Na⁺ ion comes from sodium hydroxide, while the Cl⁻ ion comes from hydrochloric acid.
The acid determines the negative ion in the salt
Before choosing the other reactant, look at the name of the salt you need. The ending of the salt name usually tells you which acid is required.
- Hydrochloric acid, HCl supplies Cl⁻ ions and forms chlorides.
- Sulfuric acid, H₂SO₄ supplies SO₄²⁻ ions and forms sulfates.
- Nitric acid, HNO₃ supplies NO₃⁻ ions and forms nitrates.
Reading the name of a salt
The first part of a salt name tells you the positive ion. The second part tells you the negative ion.
| Salt | Positive ion | Negative ion | Acid needed |
|---|---|---|---|
| Sodium chloride, NaCl | Na⁺ | Cl⁻ | Hydrochloric acid |
| Copper(II) sulfate, CuSO₄ | Cu²⁺ | SO₄²⁻ | Sulfuric acid |
| Magnesium nitrate, Mg(NO₃)₂ | Mg²⁺ | NO₃⁻ | Nitric acid |
| Ammonium sulfate, (NH₄)₂SO₄ | NH₄⁺ | SO₄²⁻ | Sulfuric acid |
Solubility Rules for Salts
Before deciding how to prepare a salt, you must know whether that salt is soluble in water.
A soluble salt dissolves in water to form an aqueous solution. An insoluble salt does not dissolve to any significant extent and remains as a solid.
At O-Level, you do not need to test every salt individually. You use solubility rules.
Sodium, potassium and ammonium salts are soluble
A useful memory cue is SPA:
- Sodium salts
- Potassium salts
- Ammonium salts
If a compound is a sodium, potassium or ammonium salt, treat it as soluble in water.
The main solubility rules
| Type of compound | General rule | Important exceptions |
|---|---|---|
| Sodium, potassium and ammonium salts | Soluble | No stated exceptions at this level |
| Nitrates | Soluble | No stated exceptions |
| Chlorides | Usually soluble | AgCl and PbCl₂ are insoluble or only slightly soluble |
| Sulfates | Usually soluble | BaSO₄ and PbSO₄ are insoluble; CaSO₄ is only slightly soluble |
| Carbonates | Usually insoluble | Na⁺, K⁺ and NH₄⁺ carbonates are soluble |
| Hydroxides | Usually insoluble | NaOH and KOH are soluble; Ca(OH)₂ is slightly soluble |
Apply the general rule, then check the exceptions
Consider silver nitrate, AgNO₃. Silver compounds are not automatically insoluble. AgNO₃ is a nitrate, and all nitrates are soluble. Therefore, silver nitrate is soluble.
Silver chloride, AgCl, is different. Most chlorides are soluble, but AgCl is one of the exceptions, so it is insoluble.
Do not judge solubility from the metal alone
AgNO₃ is soluble because all nitrates are soluble.
AgCl is insoluble because silver chloride is an exception to the general chloride rule.
Always identify the type of salt first, apply the general rule and then check for an exception.
How to Choose a Salt-Preparation Method
Every preparation question can be approached through the same sequence.
- Identify the salt that must be prepared.
- Identify the positive and negative ions required.
- Choose the correct acid to supply the negative ion.
- Use the solubility rules to decide whether the product is soluble.
- Choose the preparation method from its solubility.
Three preparation routes cover the main O-Level cases
- Required salt is insoluble → Precipitation
- Required salt is soluble and is a sodium, potassium or ammonium salt → Titration
- Required salt is another soluble salt → Acid + excess insoluble reactant
Preparing an Insoluble Salt by Precipitation
An insoluble salt cannot first be made as an aqueous solution and then crystallised because the salt does not remain dissolved. Instead, the required ions are brought together in two separate aqueous solutions.
When the solutions are mixed, the required positive and negative ions combine to form an insoluble solid. This solid is called a precipitate.
Both starting compounds must be soluble
Suppose you want to prepare an insoluble salt containing X²⁺ and Y²⁻. You need one soluble compound that provides X²⁺ ions and another soluble compound that provides Y²⁻ ions.
This is why nitrates are often useful sources of metal ions: all nitrates are soluble. Sodium and potassium salts are also useful because their salts are soluble.
Worked example: preparing barium sulfate
Barium sulfate, BaSO₄, is insoluble. It must therefore be made by precipitation.
We need:
- a soluble compound supplying Ba²⁺ ions; and
- a soluble compound supplying SO₄²⁻ ions.
Barium nitrate and sodium sulfate are suitable:
Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaNO₃(aq)
The essential change can also be represented by the ionic equation:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
BaSO₄ is the insoluble product. The Na⁺ and NO₃⁻ ions remain dissolved in the solution.
Procedure for precipitation
- Mix two suitable aqueous solutions containing the ions required in the salt.
- Stir the mixture so that precipitation is complete.
- Filter the mixture.
- Collect the insoluble salt as the residue on the filter paper.
- Wash the residue with distilled water.
- Dry the solid between sheets of filter paper or in a suitable warm place.
Where is the required product after filtration?
In a precipitation experiment, the salt you want is insoluble.
Therefore, the required product is the residue on the filter paper, not the filtrate.
Why must the precipitate be washed?
After filtration, some solution remains trapped between and around the particles of precipitate. That solution contains soluble ions from the reaction mixture.
Washing the residue with distilled water removes these soluble impurities. Distilled water is used because tap water contains dissolved ions that could contaminate the product.
Preparing a Soluble Salt Using an Excess Insoluble Reactant
Now suppose the required salt is soluble. If it is not a sodium, potassium or ammonium salt, a common method is to react the appropriate acid with an insoluble substance containing the required positive ion.
Suitable substances may include:
- a suitable metal;
- a metal oxide;
- a metal hydroxide; or
- a metal carbonate.
Why is the insoluble reactant added in excess?
The aim is to make sure that all the acid has reacted. If acid remained in the mixture, it would contaminate the salt solution.
The insoluble substance is therefore added a little at a time until some remains unreacted.
That remaining solid tells you that there is no longer enough acid to react with it. Since the solid is insoluble, it can then be removed by filtration.
The excess substance must be removable
Adding excess reactant is useful here because the reactant is insoluble. Any amount left over can be trapped by the filter paper.
This would not work with an excess soluble alkali because the excess alkali would remain dissolved together with the salt.
General procedure
- Place the appropriate dilute acid in a beaker.
- Warm the acid gently if appropriate.
- Add the insoluble reactant in small portions while stirring.
- Continue adding it until no further reaction occurs and some solid remains.
- Filter the mixture to remove the excess unreacted solid.
- Collect the filtrate, which contains the soluble salt.
- Heat the filtrate gently to evaporate some of the water.
- Stop heating when a hot saturated solution has formed.
- Allow the solution to cool so that crystals form.
- Filter the crystals.
- Wash them with a small amount of cold distilled water.
- Dry the crystals between sheets of filter paper.
Worked example: copper(II) sulfate from copper(II) oxide
Copper(II) sulfate, CuSO₄, is soluble. It is not a sodium, potassium or ammonium salt, so the excess-insoluble-reactant method is suitable.
Sulfuric acid is required because sulfate ions are needed. Copper(II) oxide supplies Cu²⁺ ions.
CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
Black copper(II) oxide is added to warm dilute sulfuric acid until some black solid remains. The excess CuO is then removed by filtration.
The blue filtrate contains copper(II) sulfate solution. Water is partly evaporated and the concentrated solution is allowed to cool, producing blue copper(II) sulfate crystals.
Using a metal carbonate
A carbonate can also react with an acid to form a salt. Carbon dioxide and water are formed at the same time.
For example:
CuCO₃(s) + 2HNO₃(aq) → Cu(NO₃)₂(aq) + CO₂(g) + H₂O(l)
Effervescence is observed because CO₂ is produced. Copper(II) carbonate is added until the bubbling stops and some unreacted solid remains.
Using a metal
Some suitable metals can react with dilute hydrochloric acid or dilute sulfuric acid to form a salt and hydrogen.
For example:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
However, not every metal is suitable.
- Very reactive metals such as sodium and potassium react too violently with dilute acids.
- Metals below hydrogen in the reactivity series, such as copper, do not react with dilute hydrochloric acid or dilute sulfuric acid.
- Metals such as magnesium and zinc are more suitable in typical school experiments.
If the metal itself is unsuitable, its oxide, hydroxide or carbonate may provide a safer and more convenient route.
How a Soluble Salt Is Crystallised
After the excess solid has been filtered off, the required salt is still dissolved in water. Filtration cannot separate dissolved salt from water, so another separation step is needed.
The solution is first concentrated by evaporation and then allowed to cool.
From salt solution to crystals
- Heat the salt solution gently.
- Allow some water to evaporate.
- Stop heating when a hot saturated solution is obtained.
- Leave the solution to cool.
- Crystals form as the solution cools.
- Filter the crystals from the remaining solution.
- Wash the crystals with a little cold distilled water.
- Dry them between sheets of filter paper.
Saturated does not mean dry
A saturated solution still contains water. It contains as much dissolved solute as it can hold under those conditions.
Do not continue heating until all the water has disappeared.
Why do crystals form on cooling?
Many salts are less soluble in cooler water than in hot water. When a hot saturated solution cools, it may no longer be able to keep all the dissolved salt in solution.
The excess dissolved salt then separates from the solution as crystals.
Why should the crystals be washed with cold water?
Some concentrated solution remains on the surface of freshly filtered crystals. Washing removes this solution and its dissolved impurities.
Only a small amount of cold distilled water should be used. The crystals are soluble, so using too much water or warm water would dissolve more product and reduce the yield.
Preparing Sodium, Potassium and Ammonium Salts by Titration
The excess-solid method depends on adding an insoluble reactant that can later be filtered away. That is difficult for sodium, potassium and ammonium salts because their salts are soluble and their commonly used bases are soluble too.
If excess sodium hydroxide, for example, were added to an acid, the excess NaOH would remain dissolved in the salt solution. Filtration would not remove it.
Instead, the acid and alkali must be reacted in the correct quantities. This is why titration is used.
What does the first titration achieve?
A pipette is used to measure a fixed volume of alkali into a conical flask. A few drops of indicator are added. Acid is then delivered accurately from a burette until the indicator changes colour at the end-point.
The titre tells us the volume of acid needed to react with the measured volume of alkali.
Why is the reaction repeated without indicator?
The first titration contains indicator. If that solution were crystallised, the indicator would remain as an impurity and could contaminate or colour the salt crystals.
Once the correct volumes are known, the acid and alkali are mixed again in exactly those volumes, but without indicator.
This produces a pure salt solution suitable for crystallisation.
Why titration instead of adding excess alkali?
An alkali is a soluble base. If it were added in excess, the extra alkali would remain dissolved together with the salt.
Since dissolved alkali cannot be removed by ordinary filtration, the correct reacting volumes must be found by titration instead.
General titration procedure
- Use a pipette to transfer a fixed volume of alkali into a conical flask.
- Add a few drops of a suitable indicator.
- Fill a burette with the acid and record the burette reading.
- Add acid gradually while swirling the flask.
- Near the end-point, add the acid drop by drop until the indicator shows the required colour change.
- Record the volume of acid used.
- Repeat as necessary to obtain reliable titres.
- Mix the correct measured volumes of acid and alkali again without indicator.
- Evaporate some water from the pure salt solution.
- Allow the concentrated solution to cool and crystallise.
- Filter, wash and dry the crystals.
Worked example: preparing potassium nitrate
Potassium nitrate, KNO₃, is soluble because it is a potassium salt and also because all nitrates are soluble.
Nitric acid supplies NO₃⁻ ions, while potassium hydroxide supplies K⁺ ions.
KOH(aq) + HNO₃(aq) → KNO₃(aq) + H₂O(l)
Both reactants are soluble, so any excess would remain in solution. Titration is therefore used to determine the correct reacting volumes.
Comparing the Three Salt-Preparation Methods
| Method | Used for | Key idea | Where is the product? | Final treatment |
|---|---|---|---|---|
| Precipitation | Insoluble salt | Mix two soluble solutions | Residue | Filter, wash and dry |
| Excess insoluble reactant | Most soluble salts other than sodium, potassium and ammonium salts | Acid reacts completely with an excess insoluble solid | Salt solution in the filtrate | Evaporate, cool, crystallise, filter, wash and dry |
| Titration | Soluble sodium, potassium and ammonium salts | Determine exact reacting amounts of soluble reactants | Salt solution | Evaporate, cool, crystallise, filter, wash and dry |
Residue or filtrate?
Precipitation: the required salt is the residue.
Excess insoluble reactant method: the residue is the unwanted excess reactant, while the required salt is dissolved in the filtrate.
This is one of the most important distinctions to remember in experimental salt-preparation questions.
Worked Method-Selection Examples
Example 1: silver chloride, AgCl
AgCl is insoluble because silver chloride is an exception to the general solubility of chlorides.
Therefore, use precipitation.
Silver nitrate supplies Ag⁺ ions and sodium chloride supplies Cl⁻ ions:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Example 2: zinc sulfate, ZnSO₄
ZnSO₄ is soluble and is not a sodium, potassium or ammonium salt.
Use sulfuric acid with an excess suitable zinc compound, such as zinc oxide or zinc carbonate. After reaction, filter off the excess solid and crystallise ZnSO₄ from the filtrate.
Example 3: sodium nitrate, NaNO₃
NaNO₃ is soluble and is a sodium salt.
Use titration between nitric acid and sodium hydroxide. After determining the correct reacting volumes, repeat the neutralisation without indicator and crystallise the sodium nitrate.
Example 4: lead(II) sulfate, PbSO₄
PbSO₄ is insoluble, so use precipitation.
Lead(II) nitrate and sodium sulfate are suitable soluble reactants:
Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) + 2NaNO₃(aq)
Common Mistakes and Why They Are Wrong
| Common mistake | Why it is wrong | Better reasoning |
|---|---|---|
| Choosing a preparation method before checking solubility | The method depends on whether the product is soluble or insoluble. | Apply the solubility rules first, then choose the method. |
| Using precipitation to prepare a soluble salt | A soluble product remains dissolved and passes through the filter paper. | Use precipitation only when the required product is insoluble. |
| Using an insoluble starting compound in a precipitation reaction | The required ions must be present in aqueous solution so that they can mix and form the precipitate. | Choose two soluble starting compounds. |
| Adding excess soluble alkali when preparing a sodium or potassium salt | Excess alkali remains dissolved and cannot be removed by filtration. | Use titration to determine exact reacting volumes. |
| Stopping the excess-solid method while all the solid has dissolved | Acid may still remain unreacted. | Continue adding solid until a little remains unreacted. |
| Keeping the residue after preparing a soluble salt with excess solid | That residue is the unwanted excess reactant. | Keep the filtrate because it contains the dissolved salt. |
| Forgetting to wash a precipitate | Soluble impurities may remain on the surface of the solid. | Wash the residue with distilled water before drying it. |
| Evaporating a salt solution to complete dryness | Strong heating may decompose some salts and does not produce good crystals. | Concentrate the solution to saturation, then cool it to crystallise. |
| Washing soluble crystals with a large amount of warm water | More of the product dissolves, reducing the yield. | Use a small amount of cold distilled water. |
| Crystallising the solution from the indicator titration | The indicator contaminates the salt crystals. | Repeat the neutralisation using the correct volumes without indicator. |
| Choosing the wrong acid | The acid supplies the negative ion in the salt. | Use HCl for chlorides, H₂SO₄ for sulfates and HNO₃ for nitrates. |
| Using any metal with dilute acid | Very reactive metals may react too violently, while metals such as copper do not react with dilute HCl or dilute H₂SO₄. | Choose a suitable metal, or use its oxide, hydroxide or carbonate. |
Salt-Preparation Checklist
1. Identify the ions
- What positive ion must the salt contain?
- What negative ion must it contain?
- Which acid supplies that negative ion?
2. Check solubility
- Apply the general solubility rule.
- Check for stated exceptions.
- Decide whether the required salt is soluble or insoluble.
3. Choose the preparation method
- Insoluble salt → precipitation.
- Na⁺, K⁺ or NH₄⁺ soluble salt → titration.
- Other soluble salt → excess insoluble reactant.
4. Separate and purify the product
- Know whether the product is in the residue or filtrate.
- Wash precipitates before drying.
- Crystallise soluble salts rather than evaporating to dryness.
- Wash crystals with a little cold distilled water and dry them.
These ten questions go beyond simple recall. Use solubility rules, experimental observations and separation logic to decide what a chemist should actually do.
- 10 application questions
- Exam-style method selection
- Explanation for every answer option
Frequently asked questions
Questions Students Often Ask About Salts
Once the three preparation methods are clear, the remaining questions usually come down to understanding why each practical step is necessary.
Why are nitrate solutions so useful in precipitation reactions?
All nitrates are soluble. This means a metal nitrate can often be used to introduce the required metal ion into an aqueous solution. For example, Ba(NO₃)₂ supplies Ba²⁺ ions when preparing BaSO₄, while AgNO₃ supplies Ag⁺ ions when preparing AgCl.
Why are sodium salts often used as the second reactant in precipitation?
Sodium salts are soluble, so they can provide the required negative ion in aqueous solution. Sodium chloride can supply Cl⁻ ions, while sodium sulfate can supply SO₄²⁻ ions. The Na⁺ ions usually remain dissolved as spectator ions.
Can the same soluble salt sometimes be prepared using different solid reactants?
Yes. Depending on the salt and the substances available, a suitable metal, metal oxide, metal hydroxide or metal carbonate may sometimes be used. For example, a zinc salt may be prepared using zinc, zinc oxide or zinc carbonate with an appropriate acid.
The important point is that the chosen reaction must be suitable and any reactant deliberately added in excess must be removable.
Why can copper(II) sulfate be prepared from CuO but not from copper metal and dilute H₂SO₄?
Copper is below hydrogen in the reactivity series and does not react with dilute sulfuric acid in the usual metal–acid reaction. Copper(II) oxide is a basic oxide, however, so it reacts with sulfuric acid to form CuSO₄ and H₂O.
Why does filtration collect the product in one method but remove an impurity in another?
Filtration always separates an insoluble solid from a liquid. What changes is the identity of that solid. During precipitation, the required salt is insoluble, so it becomes the residue. In the excess-solid method, the required salt is soluble, so the residue is merely the unwanted excess reactant and the product remains in the filtrate.
What is the difference between a concentrated solution and a saturated solution?
A concentrated solution contains a relatively large amount of dissolved solute, but it may still be able to dissolve more. A saturated solution contains the maximum amount of solute that can remain dissolved under the particular conditions.
In salt preparation, the solution is concentrated until it is sufficiently close to saturation and then allowed to cool so that crystals can form.
Why are crystals washed with cold water instead of warm water?
Washing removes concentrated solution and soluble impurities from the surface of the crystals. Cold water is used because the product is itself soluble. Warm water would generally dissolve more of the crystals and reduce the amount recovered.
Why does the indicator have to be left out of the final titration mixture?
The indicator is needed only to determine the end-point during the trial titration. If that mixture were crystallised, the indicator would remain as an impurity. The reaction is therefore repeated using the known accurate volumes but without indicator.
Does every salt solution have pH 7?
No. The word “salt” describes the ionic compound, not the pH of its aqueous solution. Some salt solutions may be neutral, while others may be acidic or alkaline. For O-Level salt-preparation questions, the more important issue is usually the salt's solubility and the method used to obtain it.
What is the fastest way to choose the correct preparation method in an examination?
Ask the questions in the same order every time. Is the required salt insoluble? If yes, use precipitation. If it is soluble, ask whether it is a sodium, potassium or ammonium salt. If yes, use titration. If not, an excess suitable insoluble reactant with the correct acid is usually the appropriate route.
Salt preparation is ultimately an exercise in chemical reasoning. Identify the ions you need, choose the acid that supplies the correct negative ion, use the solubility rules to predict where the product will be, and then choose a separation method that removes impurities without losing the salt. Once you can follow that chain of reasoning, precipitation, excess-reactant preparation and titration stop feeling like three procedures to memorise and become three logical solutions to three different solubility situations.