ChemistryICSE Class 10 · Study of Acids, Bases and Salts

Methods of Preparing Normal Salts with Equations (ICSE Class 10)

Updated 17 October 2026 · Fact-checked

Normal salts are prepared by six methods: direct combination of elements (FeS, FeCl₃), displacement of hydrogen from a dilute acid by a metal (ZnSO₄), precipitation by double decomposition (BaSO₄), neutralisation of an insoluble base by an acid (CuSO₄ from CuO), neutralisation of an alkali by an acid in a titration (NaCl) and action of a dilute acid on a carbonate or hydrogen carbonate. The method chosen depends on whether the salt is soluble and on the reactivity of the metal.

Understand Methods of Preparing Normal Salts

1. Direct combination (synthesis of a binary salt from its elements). A metal burns or is heated in a non-metal.

$F e + S Δ F e S$ (iron(II) sulphide)
$2 F e + 3 C l_{2} Δ 2 F e C l_{3}$ (iron(III) chloride, from dry chlorine over heated iron)

Note that iron forms the chloride FeCl₃ (not FeCl₂) when it burns in chlorine.

2. Displacement. A metal more reactive than hydrogen displaces hydrogen from a dilute acid. Metals such as zinc and iron are used (sodium and potassium are too violent; copper does not react).

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
$F e + H_{2} S O_{4} \to F e S O_{4} + H_{2}$

Here iron gives iron(II) salts. A more reactive metal can also displace a less reactive metal from its salt: $F e + C u S O_{4} \to F e S O_{4} + C u$ .

3. Precipitation (double decomposition). Two soluble salt solutions are mixed so that the ions exchange partners and an insoluble salt is formed. It is filtered, washed and dried.

$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ (white precipitate of barium sulphate)
$P b (N O_{3})_{2} + 2 N a C l \to P b C l_{2} + 2 N a N O_{3}$
$A g N O_{3} + N a C l \to A g C l + N a N O_{3}$

Use this method only for insoluble salts.

4. Neutralisation of an insoluble base (metal oxide or hydroxide) by an acid. The solid base is added in excess to the warm dilute acid; when the reaction stops, the excess is filtered off and the salt solution is crystallised.

$C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$
$Z n O + 2 H C l \to Z n C l_{2} + H_{2} O$

This is the method for salts of metals that do not displace hydrogen, such as copper.

5. Neutralisation of an alkali (titration). For sodium and potassium salts, an alkali is neutralised by an acid. Equal volumes cannot be assumed, so the exact amount is found by a titration with an indicator (phenolphthalein, methyl orange), then the titration is repeated without the indicator and the solution is crystallised.

$N a O H + H C l \to N a C l + H_{2} O$
$2 K O H + H_{2} S O_{4} \to K_{2} S O_{4} + 2 H_{2} O$

6. Action of a dilute acid on a carbonate or hydrogen carbonate. The carbonate is added to the acid until effervescence stops.

$C u C O_{3} + H_{2} S O_{4} \to C u S O_{4} + H_{2} O + C O_{2}$
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$

Use this method when the carbonate of the metal is available (for example calcium chloride from marble and HCl). Do not use dilute sulphuric acid with calcium carbonate for a pure sample, as it coats the lump.

7. Choosing a method.

Salt required	Method
Sodium or potassium salt	alkali + acid (titration)
Salt of a reactive metal (Zn, Fe)	displacement, or base or carbonate + acid
Salt of a less reactive metal (Cu)	insoluble base or carbonate + acid
Insoluble salt	precipitation
Binary salt of a metal and a non-metal	direct combination

(Apparatus and procedure details are not asked in the syllabus.)

Methods of Preparing Normal Salts: formulas and key facts

Direct combination: $F e + S Δ F e S$ ; $2 F e + 3 C l_{2} Δ 2 F e C l_{3}$
Displacement: $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
Precipitation: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Insoluble base and acid: $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$
Alkali and acid: $N a O H + H C l \to N a C l + H_{2} O$
Carbonate and acid: $C u C O_{3} + H_{2} S O_{4} \to C u S O_{4} + H_{2} O + C O_{2}$

How to solve Methods of Preparing Normal Salts questions

Choose the method from the nature of the salt, then write the equation.

1Decide whether the salt is soluble or insoluble in water: insoluble means precipitation.
2For a soluble salt, look at the metal: sodium or potassium means titration; zinc or iron means displacement (or a base/carbonate); copper means the oxide or carbonate method.
3Write the reactants (acid, base, carbonate, metal or salt solutions) and balance the equation.
4Mention the separation: filtration for a precipitate or excess solid; crystallisation for a soluble salt.
5For masses, convert to moles, use the equation ratio and multiply by the molar mass of the salt.

Methods of Preparing Normal Salts in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Making blue copper sulphate crystals for a school science fair

Suppose a student wants to prepare blue crystals of copper(II) sulphate. She finds that copper metal does not react with dilute sulphuric acid, but she has a jar of black copper(II) oxide.

Which method will she use and why?

Try it first, then show the solutionSolution

Copper lies below hydrogen in the reactivity series, so displacement will not work.
She can add copper(II) oxide to warm dilute sulphuric acid in excess, filter and crystallise: $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$ .

Answer: Neutralisation of an insoluble base (CuO) with dilute sulphuric acid.

Memory hook: No displacement for copper: use the oxide.

Tricks to remember Methods of Preparing Normal Salts

Short memory aids. Say each one aloud once, then use it on a question the same day.

D-D-P-I-A-C
Direct combination, Displacement, Precipitation, Insoluble base, Alkali (titration), Carbonate with acid: the six methods in order.
Use it when: Listing all methods.
Solid in excess for insoluble base and carbonate
For base or carbonate with an acid, keep the solid in excess and filter it off. For an alkali, use a titration.
Use it when: Explaining the procedure.
Precipitation = insoluble salt
If the salt you want is insoluble, precipitate it; if it is soluble, do not.
Use it when: Choosing the method.

Methods of Preparing Normal Salts: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Name the method · 3 marks · Short answerBoard level

Name the method used to prepare (a) FeS, (b) BaSO₄, (c) NaCl and write one equation each.

Try it first, then show the solutionSolution

(a) Direct combination: $F e + S Δ F e S$ .
(b) Precipitation: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ .
(c) Neutralisation of an alkali (titration): $N a O H + H C l \to N a C l + H_{2} O$ .

Answer: (a) Direct combination (b) Precipitation (c) Titration with alkali.

Example 2: Mass of copper sulphate · 3 marks · Short answerHarder board level

Calculate the mass of copper(II) sulphate formed when 4.0 g of copper(II) oxide completely reacts with dilute sulphuric acid. (Cu = 64, O = 16, S = 32)

Try it first, then show the solutionSolution

$C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$ ; molar mass of CuO = 80 g/mol; CuSO₄ = 64 + 32 + 64 = 160 g/mol.
Moles of CuO = 4.0 / 80 = 0.05 mol, giving 0.05 mol CuSO₄.
Mass = 0.05 x 160 = 8.0 g.

Answer: 8.0 g of copper(II) sulphate.

Example 3: Precipitate mass · 3 marks · Short answerHarder board level

Excess sulphuric acid is added to a solution containing 20.8 g of barium chloride. Find the mass of barium sulphate precipitated. (Ba = 137, Cl = 35.5, S = 32, O = 16)

Try it first, then show the solutionSolution

$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ .
Molar mass of BaCl₂ = 137 + 71 = 208 g/mol; moles = 20.8 / 208 = 0.1 mol.
BaSO₄ = 0.1 mol; molar mass = 137 + 32 + 64 = 233 g/mol; mass = 23.3 g.

Answer: 23.3 g.

Example 4: Displacement with gas volume · 4 marks · Long answerHarder board level

Zinc (6.5 g) is added to excess dilute sulphuric acid. Find (a) the mass of zinc sulphate formed, (b) the volume of hydrogen at S.T.P. (Zn = 65, S = 32, O = 16)

Try it first, then show the solutionSolution

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ ; moles of Zn = 6.5 / 65 = 0.1 mol.
(a) ZnSO₄ = 0.1 mol; molar mass = 65 + 32 + 64 = 161 g/mol; mass = 16.1 g.
(b) H₂ = 0.1 mol = 0.1 x 22.4 = 2.24 dm³.

Answer: (a) 16.1 g (b) 2.24 dm³.

Example 5: Select methods and justify · 4 marks · Long answerHOTS / competency

Suggest the most suitable method to prepare each salt and justify: (a) iron(II) sulphate, (b) lead(II) sulphate, (c) copper(II) sulphate, (d) sodium sulphate.

Try it first, then show the solutionSolution

(a) Displacement: iron filings with dilute sulphuric acid, $F e + H_{2} S O_{4} \to F e S O_{4} + H_{2}$ (iron is above hydrogen and gives iron(II) salts).
(b) Precipitation, since lead sulphate is insoluble: mix lead nitrate and sodium sulphate solutions.
(c) Copper oxide (or carbonate) with dilute sulphuric acid, as copper does not displace hydrogen.
(d) Titration of sodium hydroxide with dilute sulphuric acid, as sodium is too reactive to displace hydrogen safely.

Answer: (a) Displacement (b) Precipitation (c) CuO and acid (d) Titration.

Where students lose marks in Methods of Preparing Normal Salts

The slips examiners see most often, and what to write instead.

Mistake: Using displacement for copper salts
Copper does not displace hydrogen from dilute acids.
Do this: Use copper(II) oxide or carbonate with dilute sulphuric acid.
Usually costs: 1 to 2 marks.
Mistake: Writing FeCl₂ for iron burning in chlorine
Chlorine oxidises iron to iron(III).
Wrong working
Fe + Cl₂ -> FeCl₂
Right working
2Fe + 3Cl2 -> 2FeCl3
Do this: Remember: chlorine gives iron(III) chloride; HCl gives iron(II) chloride.
Usually costs: 1 mark.
Mistake: Using precipitation for a soluble salt
No solid forms when the product is soluble.
Do this: Precipitation is only for insoluble salts.
Usually costs: 1 mark.
Mistake: Leaving out the separation step
The question asks how to obtain the salt, but filtration/crystallisation is not stated.
Do this: Mention filtering the excess solid and crystallising the filtrate.
Usually costs: 1 mark.

Exam tips

State the type of reaction and the equation together; the marking scheme gives separate credit.
Remember the colour of common precipitates (BaSO₄ white, PbCl₂ white, AgCl white).
Balance equations and include the state (precipitate) where the question asks for observations.
Only the listed methods are asked; details of apparatus are not required.

Practice questions on Methods of Preparing Normal Salts

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which method is used to prepare lead(II) chloride from lead(II) nitrate?
(a) Displacement
(b) Precipitation
(c) Titration
(d) Direct combination
Hint
Lead chloride is only sparingly soluble in cold water, so it separates as a solid.
Answer
(b) Precipitation.
Q2 · 1 mark · MCQBoard level
To prepare zinc sulphate by displacement, the correct pair of reactants is:
(a) zinc and dilute sulphuric acid
(b) zinc and sodium sulphate solution
(c) copper and dilute sulphuric acid
(d) zinc oxide and sodium hydroxide
Hint
The metal must be above hydrogen.
Answer
(a) zinc and dilute sulphuric acid.
Q3 · 2 marks · Short answerBoard level
Write the equation for the preparation of zinc chloride from zinc oxide and state what is done to remove the unreacted solid.
Hint
Excess base.
Answer
$Z n O + 2 H C l \to Z n C l_{2} + H_{2} O$ ; the excess zinc oxide is removed by filtration.
Q4 · 3 marks · Short answerHarder board level
Why is iron(III) chloride, and not iron(II) chloride, formed when iron is heated in dry chlorine? How would you get iron(II) chloride?
Hint
Chlorine is a strong oxidising agent.
Answer
Chlorine oxidises iron to the higher state: $2 F e + 3 C l_{2} \to 2 F e C l_{3}$ . Iron(II) chloride is obtained by displacement with dilute HCl: $F e + 2 H C l \to F e C l_{2} + H_{2}$ .
Q5 · 3 marks · Short answerHarder board level
Calculate the mass of sodium chloride obtained by neutralising 4.0 g of sodium hydroxide with dilute hydrochloric acid. (Na = 23, O = 16, H = 1, Cl = 35.5)
Hint
NaOH : NaCl = 1 : 1.
Answer
Moles of NaOH = 4.0 / 40 = 0.1; NaCl = 0.1 mol x 58.5 = 5.85 g.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): Sodium sulphate is usually made by titrating sodium hydroxide with dilute sulphuric acid and not by adding sodium to the acid.
Reason (R): Sodium reacts violently with the acid and the reaction is unsafe.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Why is displacement not chosen for very reactive metals?
Answer
(a).
Q7 · 4 marks · Long answerHarder board level
A student prepares copper(II) sulphate by adding copper(II) carbonate in excess to warm dilute sulphuric acid. (a) Write the equation. (b) Why is the carbonate added in excess? (c) What is seen during the reaction? (d) How is the excess removed?
Hint
Effervescence.
Answer
(a) $C u C O_{3} + H_{2} S O_{4} \to C u S O_{4} + H_{2} O + C O_{2}$ . (b) To use up all the acid. (c) Effervescence of CO₂ and a blue solution. (d) By filtration.
Q8 · 5 marks · Long answerHOTS / competency
Compound A is a white solid insoluble in water, formed when two clear solutions of salts P and Q are mixed. P is barium chloride. (a) Name Q (a sulphate of sodium). (b) Write the equation and name the type of reaction. (c) How is A separated and purified? (d) Calculate the mass of A formed from 41.6 g of P. (Ba = 137, Cl = 35.5, S = 32, O = 16)
Hint
The reaction is precipitation.
Answer
(a) Sodium sulphate. (b) $B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l$ ; precipitation (double decomposition). (c) Filter, wash and dry. (d) Moles of BaCl₂ = 41.6 / 208 = 0.2; mass of BaSO₄ = 0.2 x 233 = 46.6 g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Name the six methods of preparing normal salts.Show answer
Direct combination, displacement, precipitation, neutralisation of an insoluble base, neutralisation of an alkali by titration, action of dilute acid on a carbonate.
2.Method for sodium chloride?Show answer
Neutralisation of an alkali (titration).
3.Method for insoluble salts?Show answer
Precipitation.
4.Product of iron with dry chlorine?Show answer
Iron(III) chloride.
5.How is excess insoluble base removed?Show answer
By filtration.

Methods of Preparing Normal Salts: frequently asked questions

Why is a titration needed to make sodium chloride?

Because both reactants are soluble, there is no excess solid to filter, so the exact amount of acid needed is found with an indicator.

What is double decomposition?

A reaction in which two compounds in solution exchange ions to give an insoluble product.

Can copper sulphate be made from copper and dilute sulphuric acid?

No. Copper is less reactive than hydrogen.

What is a normal salt?

A salt with no replaceable hydrogen or hydroxyl, such as NaCl or CuSO₄.