Form 4 Acid bases and salts

Chapter 6: Acids, Bases and Salts

6.1 Role of Water in Showing Acidic and Alkaline
Properties

Definition of Acid
1. Based on Arrhenius theory, an acid is a chemical compound that dissolves

in water to produce hydrogen ions, H+ or hydroxonium ions, H3O+.
2. The formation of freely moving hydrogen or hydroxonium ions in water

resulting in a substance with acidic properties.
3. Hydrogen ions and anions are produced during the dissociation of acids in

water. For examples:
HCl(g) → H+(aq) + Cl-(aq)

HNO3(l) → H+(aq) + NO3-(aq)
H2SO4(l) → 2H+(aq) + SO42-(aq)
CH3COOH(l) → H+(aq) + CH3COO-(aq)

4. A substance will only shows its acidic property when freely moving
hydrogen ions are present. For example, glacial ethanoic acid or dry
hydrogen chloride gas will not show any acidic property if dissolved in
organic solvent.

Basicity of Acids
1. Basicity of an acid is defined as the number of moles of OH- ions required

to react with one mole of acid.
2. Basicity of an acid can also be defined as the number of moles of H+ ions

produced by one mole of acid when it dissolves in water during
neutralisation.
3. There are three types of acids
 Monoprotic acid (monobasic acid)

 Acid that produced one mole of H+ ion when one mole of acid
dissolves in water.

 Example: Ethanoic acid

CH3COOH(aq) → H+(aq) + CH3COO-(aq)

 Diprotic acid (dibasic acid)
 Acid that produced two moles of H+ ions when one mole of acid
dissolves in water.
 Example: Sulphuric acid

H2SO4(aq) → 2H+(aq) + SO42-(aq)

 Triprotic acid (tribasic acid)
 Acid that produced three moles of H+ ions when one mole of acid
dissolves in water.
 Example: Phosphoric acid

H3PO4(aq) → 3H+(aq) + PO43-(aq)

4. Uses of acidic substances in our daily lives

In food In industries

 Sour milk (lactic acid)

 Vinegar (ethanoic acid)  Electrolyte in lead-acid
 Tea leaves (tannic acid)  accumulator (sulphuric acid)
 Fruits (ascorbic acid) Coagulate latex in rubber industry
 Fizzy soft drink (carbonic acid) (methanoic acid)
 Food preservatives (benzoic acid)

 Aspirin (salicylic acid)

Definition of Base and Alkali
1. A base is a chemical substance that can neutralise an acid to produce salt

and water.

HCl + NaOH → NaCl + H2O
(acid) (base) (salt) (water)

2. Examples of bases: Copper(II) oxide (contains oxide ions, O2-) and
magnesium hydroxide (contains hydroxide ions, OH-)

3. Neutralisation is the reaction between acid and base. The O2- ions or OH-
ions in base react with H+ ions in acid to from water, H2O.
O2- + 2H+ → H2O
OH- + H+ → H2O

4. Most bases are not soluble in water. Bases that can dissolve in water to
form hydroxide ions, OH- are called alkalis.

5. The formation of freely moving hydroxide ions in water results in a
substance with alkaline properties.

6. Hydroxide ions and cations are produced during the dissociation of alkali in
water. For examples:
KOH(s) → K+(aq) + OH-(aq)
NaOH(s) → Na+(aq) + OH-(aq)
NH3(g) → NH4+(aq) + OH-(aq)

7. A substance will only show its alkaline property when freely moving
hydroxide ions are present. For example, dry ammonia gas will not show
any alkaline property if they are dissolved in organic solvent.

8. Uses of alkali in our daily lives

In household products In agriculture

 Cleaning agent for grease  Fertilisers (ammonia)
(ammonia)  Soften latex (ammonia)
 Neutralise acidic soil (calcium
 Cleaning of ovens and drains
(sodium hydroxide) oxide and calcium hydroxide)

 Bleaching powder (calcium Other industries
hydroxide)

In pharmaceutical industries

 Antacids to neutralise excess acid  Soaps, detergents, paper and

in stomach (magnesium hydroxide rayon (sodium hydroxide)

and aluminium hydroxide)  Electrolyte in alkaline batteries

 Smelling salts (ammonia) (potassium hydroxide)

6.2 pH Value

1. To measure the acidity or alkalinity of solution, pH value is used.
2. pH value is based on the concentration of hydrogen ions H+ in a solution.
3. To calculate the acidity or alkalinity of a solution, formula below is used.

pH = -log[H+], where [H+] is the concentration of H+ in mol dm-3
pOH = -log[OH-], where [OH-] is the concentration of OH- in mol dm-3

pH + pOH = 14

4. In the dissociation of water, H+ and OH- ions are formed.
H2O ↔ H+ + OH-

‘↔’ indicates partial dissociation

5. Figure below shows the pH scale

 pH < 7, acidic solution. The higher the concentration of H+ ions, the
lower the pH, and the more acidic the solution is.

 pH = 7, neutral solution.
 pH >7, alkaline solution. The higher the concentration of OH- ions, the

higher the pH, and the more alkaline the solution is.

Measurements of pH Value in a Solution
1. The pH value of a solution can be measured by using universal indicator

(pH paper) or pH meter.
2. Universal indicator is a mixture of indicator that changes its colour

according to the pH value of the solution being tested. It is always used in
the form of solution or paper strips. The colour changed of the indicator
during testing is compared to the standard colour chart to determine the
exact pH value.

3. pH meter is an electric meter that measures the pH value of a solution

accurately. The probe of pH meter is immersed in the solution to be tested
and the exact pH value will be shown.

6.3 Strength of Acids and Alkalis

Relationship between pH Values and Strength of Acids or Alkalis
1. The pH value of acid and alkali depends on their concentration (molarity)

and degree of dissociation (strength).
2. For acid or alkali with the same concentration, their pH value depends on

the degree of dissociation (or degree of ionisation).
 The higher the degree of dissociation of acid, the lower the pH value
 The higher the degree of dissociation of alkali, the higher the pH value
3. The degree of dissociation is a measurement of the percentage or fraction
of molecules that will dissociate when dissolved in water.

Strong Acids and Weak Acids
1. Based on the degree of dissociation, acids can be divided into strong and

weak acids.
2. A strong acid is an acid that dissociates completely in water (degree of

dissociation is 100%) to produce high concentration of hydrogen ions, H+.
It has low pH value. Examples of strong acids are hydrochloric acid, nitric
acid and sulphuric acid, they are all mineral acids.

HCl → H+ + Cl-
HNO3 → H+ + NO3-
H2SO4 → 2H+ + SO42-

‘→’ indicates complete dissociation

3. A weak acid is an acid that dissociates partially in water (incomplete
dissociation) to produce low concentration of hydrogen ions, H+. It has
higher pH value than strong acid. Examples of weak acids are ethanoic
acid, carbonic acid, ethanedioic acid, methanoic acid, citric acid and
tartaric acid, they are all organic acids.

CH3COOH ↔ CH3COO- + H+
H2CO3 ↔ 2H+ + CO32-

‘↔’ indicates partial dissociation

4. Weak acids exist as molecules in its solution and only a small portion of it

dissociates into ions.

5. At the same concentration, the stronger acid has lower pH value than the

weaker acid.

6. Both strong and weak acids have the same chemical properties. However,

the rate of reaction and electrical conductivity of weak acid is lower.

7. Table below shows the comparison between HCl and CH3COOH of the

same concentration.

Test 0.1 mol dm-3 HCl 0.1 mol dm-3 CH3COOH
Universal indicator Red, pH~1 Orange-red, pH~4
Reaction with metal Gas evolves slowly
Gas evolves vigorously

Electrical conductivity Light bulb lights up brightly Light bulb lights up dimly

Conclusion Has high concentration of H+ Has low concentration of
ions, it is a strong acid H+ ions, it is a weak acid

Strong Alkalis and Weak Alkalis
1. Based on the degree of dissociation, alkalis can be divided into strong and

weak alkalis.
2. A strong alkali is an alkali that dissociates completely in water (degree of

dissociation is 100%) to produce high concentration of hydroxide ions, OH-.

It has high pH value. Examples of strong alkalis are sodium hydroxide and
potassium hydroxide.

NaOH → Na+ + OH-
KOH → K+ + OH-

3. A weak alkali is an alkali that dissociates partially in water (incomplete
dissociation) to produce low concentration of hydroxide ions, OH-. It has
lower pH value than a strong alkali. Examples of weak alkalis are aqueous
ammonia, calcium hydroxide and magnesium hydroxide.

NH3 + H2O ↔ NH4+ + OH-
Ca(OH)2 ↔ Ca2+ + 2OH-

4. Table below shows the comparison between NaOH and NH3 of the same
concentration.

Test 0.1 mol dm-3 NaOH 0.1 mol dm-3 NH3
Universal indicator
Electrical conductivity Purple, pH~13 Blue, pH~10

Conclusion Light bulb lights up brightly Light bulb lights up dimly

Has high concentration of Has low concentration of
OH- ions, it is a strong alkali OH- ions, it is a weak alkali

6.4 Chemical Properties of Acids and Alkalis

Chemical Properties of Acids
1. Reaction with bases to produce salt and water.

Acid + base → salt + water

For example:

HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
CuO(s) + 2CH3COOH(aq) → Cu(CH3COO)2(aq) + 2H2O(l)

2. Reaction with reactive metals to produce salt and hydrogen gas.
Acid + metal → salt + hydrogen

For example:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Mg(s) + 2CH3COOH(aq) → Mg(CH3COO)2(aq) + H2(g)

3. Reaction with metal carbonates to produce salt, carbon dioxide gas and
water.
Acid + metal carbonate → salt + carbon dioxide + water
For example:
2HNO3(aq) + Na2CO3(s) → 2NaNO3(aq) + CO2(g) + H2O(l)
2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)

Chemical Properties of Alkalis
1. Reaction with acids to produce salt and water.

Base + acid → salt + water

For example:

KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)

2. Heating with ammonium salt to produce ammonia gas and water.
Base + ammonium salt → ammonia gas + water

For example:
NH4+(aq) + OH-(aq) → NH3(g) + H2O(l)

3. Reaction with metal ions solution to produce metal hydroxide precipitate.
Base + metal ions solution → metal hydroxide precipitate

For example:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

6.5 Concentration of Aqueous Solution

Meanings of Concentration and Molarity and Their Relationship

1. Solution is the mixture between solute and solvent.

2. Concentration is the measurement of the mass of solutes dissolved in a

given volume of solvent when 1.00 dm3 solution is formed.

3. Molarity is the number of moles of solute dissolves in a given volume of

solvent when 1.00 dm3 solution is formed.

4. Unit of solute: ‘gram’ or ‘mole’ Unit of concentration: ‘g dm-3’

Unit of solution: ‘dm3’ Unit of molarity: ‘mol dm-3’ or ‘M’

5. Formula for concentration and molarity:

where 1dm3 = 1000cm3

6. Concentration (g dm-3) can be converted to molarity (mol dm-3) using the
formula below.

7. Molarity is more useful in calculating the concentration of solution in
chemical experiment as the unit used in chemical equation are mol.

Examples

a) 5.00g of potassium hydroxide, KOH is dissolved in distilled water to form
400cm3 of solution. Calculate the concentration of the solution in g dm-3.

Solution:
Mass of KOH = 5.00g
Volume of solution = 400cm3
Concentration of solution = 5.00

0.40
=12.5g dm-3

b) Calculate the mass of nitric acid in 500cm3 of 3.0mol dm-3 nitric acid.

Solution:
Concentration of solution = 3.0mol dm-3
Volume of solution = 500cm3
Number of moles = 3.0mol dm-3 × 0.50dm3

= 1.5mol
Mass of HNO3 = 1.5mol × (1+14+(16×3))g

= 94.5g

c) Calculate the number of moles of hydrogen ions in 200cm3 of 2.0mol dm-3
sulphuric acid.

Solution:
Volume of solution = 200cm3
Molarity of solution = 2mol dm-3
Number of moles = 2mol dm-3 × 0.20dm3

= 0.40mol
Equation of sulphuric acid = H2SO4(aq) → 2H+(aq) +SO42-(aq)
From the equation, one mole of H2SO4 contains 2 hydrogen ions.
Thus, 0.40mol × 2 = 0.80mol

6.6 Standard Solution

1. A standard solution is a solution that has a known concentration. It can be

prepared by using a volumetric flask (or standardised flask) with a known

volume.

Preparation of a Standard Solution by Dilution Method
1. Dilution is a process that adds solvent such as water in a concentrated

standard solution to obtain a diluted solution.
2. In a diluted solution, the volume of solvent increases but the number of

moles of solute remains constant. Hence, the concentration of the solution
decreases.
3. Formula applied to calculate the new concentration of the diluted solution:

Where M1 = Molarity before dilution
M2 = Molarity after dilution
V1 = Volume of original solution (cm3)
V2 = Volume of diluted solution (cm3) = V1 + volume of water added
Example

200cm3 of 2mol dm-3 NaOH is diluted by 50cm3 of water. Calculate
the molarity of the diluted solution.

Solution:
2mol dm-3 × 200cm3 = M × 250cm3
M = 1.6mol dm-3

6.7 Neutralisation

The Definition of Neutralisation
1. Neutralisation is the reaction between an acid and a base (or alkali) to

produce the only products which are salt and water.
2. During neutralisation, alkali neutralised the acidity of acid while acid

neutralised the alkalinity of alkali to produce salt and water.
3. In the process, acids, bases and salts are dissociated to form ions

whereas water exists as molecules. For example,

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
Simplified ionic equation: H+(aq) + OH-(aq) → H2O(l)

Acid-base Titration
1. Titration helps in determining the exact quantity of acid needed to

neutralise a certain quantity of alkali.
2. Acid-base titration is a qualitative analysis that involves the addition of an

acid of known concentration from a burette to an alkali of unknown
concentration in a conical flask until neutralisation occurs.
3. Acid-base indicator (methyl orange or phenolphthalein) is added during the
titration process to show a colour change at the end point of neutralisation.

Calculations Involving Neutralisation
1. The general balanced equation of neutralisation is

Where a = number of moles of acid
b = number of moles of base

2. Formula applied to calculate the value of a and b:

Where MA = Molarity of acid VA = Volume of acid (cm3)
MB = Molarity of base VB = Volume of base (cm3)

3. Since a moles of acid reacts completely with b moles of base, the ratio of a

and b can be calculated from the balanced equation by

Example

a) Calculate the volume of 0.2mol dm-3 ammonia gas required to neutralise
30cm3 of 1.2mol dm-3 sulphuric acid.

Solution:

2NH3 + H2SO4 → (NH4)2SO4

M AVA  1

MBVB 2

1.2 30  1
0.2VB 2

VB 21.2 30  360cm3
0.2

b) Calculate the volume of 0.5mol dm-3 nitric acid, which is required to
neutralise 2.50g of sodium hydroxide.
[Relative atomic mass: Na = 23, O = 16, H = 1]

Solution:
Relative molecular mass of NaOH = 23+16+1 = 40g
Number of mole of NaOH = 2.50 ÷40 = 0.0625mol

HNO3 + NaOH → NaNO3 + H2O
From the equation, one mole of HNO3 reacts with one mole of NaOH
Thus, 0.0625 mole of HNO3 reacts with 0.0625 mole of NaOH
0.0625  0.5V

1000
V  125cm3

Neutralisation in the Manufacture of Fertilisers
1. Neutralisation reaction between acid and alkali are used in manufacturing

chemical fertilisers. These fertilisers are used to replace the elements
used up in the soil of plants.
2. Examples of fertilisers manufactured by neutralisation reaction:

Ammonium sulphate: H2SO4 + 2NH3 → (NH4)2SO4
Potassium sulphate: H2SO4 + 2KOH → K2SO4 + H2O

Ammonium nitrate: HNO3 + NH3 → NH4NO3
Urea: 2NH3 + CO2 → (NH2)2CO + H2O

3. Fertilisers with higher percentage of nitrogen is more effective. Formula
below is applied to determine the effectiveness of nitrogenous fertilisers.

6.8 Salts, Crystals and Their Uses in Daily Life

1. Salt is an ionic compound formed when the hydrogen ion in an acid is
replaced by a metal ion or ammonium ion (NH4+). Salt consists of a cation
(other than H+ ion) and an anion (other than OH- ion) bonded by ionic
bonds.

2. Table below shows the salts formed from their corresponding acids.

Acid General names Examples of salt
Hydrochloric acid of salts
NaCl, KCl, CuCl2, ZnCl2, NH4Cl
Nitric acid Chloride salts
Sulphuric acid NaNO3, KNO3, Mg(NO3)2, Pb(NO3)2,
Carbonic acid Nitrate salts NH4NO3

Sulphate salts Na2SO4, K2SO4, FeSO4, CaSO4, (NH4)SO4
Carbonate salts Na2CO3, CaCO3, MgCO3, ZnCO3, PbCO3

3. If an acid has more than one H+ ions to be replaced, it could be diprotic
acid or triprotic acid. These acids can form more than one type of salt.

Types of acid Example of acid Types of salt Examples of salt
Diprotic acid H2SO4 formed
Triprotic acid H3PO4 2 NaHSO4, Na2SO4
NaH2PO4, Na2HPO4,
3
Na3PO4,

Physical Properties of Salt Crystals
1. Salts are made up of anions and cations. These ions are packed closely

with a regular and repeated arrangement in an orderly manner. A lattice,
which is a solid with definite geometry is formed.
2. Unit cell is the repeating basic unit in the crystal structure. All crystals have
specific geometrical shapes, flat surfaces, straight edges, sharp vertices,
and fixed angles between two adjacent surfaces.

3. Crystals of the same salt have the same shape but different sizes. The size of
the salt crystals formed depends on the rate of crystallisation. Faster
crystallisation forms a smaller crystal .

4. Due to the regular arrangement of salt crystals, it can be cut easily into
different shapes and is hard and brittle.

5. Table below shows the uses of salts in various fields.

Fields Uses
Medicine
Antacid medicine: CaCO3, CaHCO3
Agricultural Smelling salts: NH4Cl
Food preservation Plaster of Paris: CaSO4
Food preparation Clearing intestine: Epsom salt and Glauber salt
Other uses Antiseptic: KMnO4
Iron pills: iron(II) sulphate

Nitrogenous fertilisers: KNO3, NaNO3, (NH4)2SO4,
NH4NO3, (NH4)3PO4,
Pesticides: CuSO4, FeSO4, HgCl

Salted fish and salted eggs: NaCl
Tomato sauce, oyster sauce and jam: C6H5COONa
Processed meat: NaNO2

Food seasoning: NaCl
Enhance food taste: Monosodium glutamate
Self-raising flour: NaHCO3

Fluoride toothpaste: SnF2
Black and white photography films: AgBr
Bleaching agent: NaOCl

6.9 Preparation of Salts

Soluble Salts and Insoluble Salts

1. Table below shows the type of cations and anions present in salts and their
solubilities.

Types of salt Solubility in water

Sodium, potassium, All are soluble
ammonium and nitrate salts

Chloride salts All are soluble except PbCl2, AgCl, HgCl

Sulphate salts All are soluble except BaSO4, CaSO4, PbSO4

Carbonate salts All are insoluble except Na2CO3, K2CO3, (NH4)2CO3

Preparation of Soluble Salts

1. Preparation of soluble sodium, potassium and ammonium salt can be done

from the reaction between acids and alkalis(using titration method). Acid

forms the anion of the salt while alkali forms the cation of the salt.

Acid Alkali Salt
H2SO4 NaOH Na2SO4
NH3(aq) NH4Cl
HCl KNO3
HNO3 KOH

2. Preparation of soluble sodium, potassium and ammonium salts.

3. Preparation of soluble salts other than sodium, potassium and ammonium
salts
Reaction involves
Acid + metal (Metals that are less electropositive than hydrogen do not

react with dilute acids)

Acid + metal oxide (or metal hydroxide)
Acid + metal carbonate

4. Purification of soluble salts involve the process of recrystallisation.

5. Mixture of different soluble salts can be purified by recrystallisation too as
different salts have different solubilities in water. A lower solubility salt will
recrystallise faster than a higher solubility salt.

Preparation of Insoluble Salts
1. Preparation of insoluble salts can be done by precipitation from the

reaction of double decomposition. Two different aqueous solutions form
the anion and cation of the insoluble salt.

MY(aq) + NX(aq) → MX(s) + NY(aq)
(solution) (solution) (salt) (solution)
2. For example,
Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)
Simplified equation: Pb2+(aq) + 2Cl-(aq) → PbCl2(s)

3. Preparation of insoluble salts (lead(II) chloride).

4. Guide to selecting aqueous solutions in preparing insoluble salts.
For the cations in insoluble salts, nitrate solutions with the cations
needed are used as all nitrate salts are soluble in water
For the anions in insoluble salts, potassium or sodium solutions with
the anions needed are used as all potassium and sodium salts are
soluble in water

To Construct Ionic Equations for the Formation of Salts
1. To construct the ionic equation for the formation of a salt, we need to know

either the formula of the salt or the mole ratio of the ions reacted.

2. For example, if the formula of a salt is MaXb, then a is the charge of anion
X and b is the charge of cation M. The ionic equation formed is
aMb+(aq) + bXa-(aq) → MaXb(s)

3. If the molarity and volume of the ions are known, then the mole ratio of the
ions forming a salt can be calculated using the formula below.

Example

2.0cm3 of 0.6mol dm-3 Mn+ solution reacts completely with 2.0cm3 of 0.2mol
dm-3 Nm- solution to form MmNn. Determine the ionic equation and empirical
formula of the salt.

Solution: Mn+ Nm-
Number of mole

Mole ratio

Ionic equation: 3Mn+ + Nm- → M3N
Empirical formula: M3N

Construction of Ionic Equations Using the Continuous Variation Method
1. Continuous variation method can be used to determine the mole ratio of

ions in forming a salt.
2. In the formation of insoluble salt, a fixed volume of solution X is added with

a varying volume of solution Y until the ions in solution X is completely
react. The amount of precipitate formed after the ions in solution X reacted
completely remains constant despite the increasing volume of solution Y.
3. The mole ratio of the ions forming the salt can be calculated using the the
volume and molarity of the salt solution.

6.10 Effect of Heat on Salts

Tests for Gases
1. Gases may evolved when a chemical substance is either heated, reacted

with dilute or concentrated acid or heated with alkalis. From the gas
released, we can identify the type of ions present in the substance.
2. Table below shows the physical properties and the test for gases.

Type of gas Colour of Smell of Effect on Confirmation test on gas
gas gas litmus paper
Oxygen, O2 Lighted up glowing wooden splinter
Hydrogen, Colourless No smell No effect Produced a ‘pop’ sound when tested
Colourless No smell
H2 No smell No effect with a lighted wooden splinter
Carbon Colourless Pungent
dioxide, CO2 Colourless Pungent Blue litmus Limewater becomes milky
Ammonia, Greenish- Pungent turns red
Red litmus Forms white fumes of hydrogen chloride
NH3 yellow Pungent turns blue gas.
Colourless Decolourises
Chlorine, Cl2 Pungent -
Colourless litmus
Hydrogen Blue litmus Forms white fumes of ammonia gas
chlorine, HCl Brown turns red
Decolourises purple colour of acidified
Sulphur Blue litmus potassium permanganate solution or
dioxide, SO2 turns red changes the orange colour of acidified
potassium dichromate(IV) solution to
Nitrogen Blue litmus
dioxide, NO2 turns red green
-

Effect of Heat on Salts
1. Heating a solid salt may cause it to decompose and produce metal oxides.
2. Decomposition of a solid may cause the following changes:

 Change in colour: determine the type of cation present

Original Colour change Metal oxide Cation
colour of salt produced present in salt
Yellow when hot, white when
White cool ZnO Zn2+

White Brown when hot, yellow when PbO Pb2+
Blue/green cool CuO Cu2+
Green/yellow Black Fe2O3 Fe2+/Fe3+
Brown

 Gas evolved: determine the type of anion present
3. Table below shows the product of heating and the examples of salts.

Ammonium salt Action of heating Example
All ammonium salt NH4Cl(s) → NH3(g) + HCl(g)
Produce ammonia gas
Carbonate salt Action of heating Example
Potassium carbonate
Sodium carbonate Do not decompose -

Other carbonate salt Produce carbon dioxide CuCO3(s) → CuO(s) + CO2(g)
gas MgCO3(s) → MgO(s) + CO2(g)
Nitrate salt
Sodium nitrate Action of heating Example
Potassium nitrate Produce oxygen gas and 2NaNO3(s) → 2NaNO2(s) +
nitrites O2(g)
Other nitrate salt Produce oxygen gas,
nitrogen dioxide gas and 2Zn(NO3)2(s) → 2ZnO(s) +
Sulphate salt metal oxides 4NO2(g) + O2(g)

Iron(II) sulphate Action of heating Example
Zinc sulphate 2FeSO4(s) → Fe2O3(s) +
Copper(II) sulphate Produce metal oxide and SO2(g) + SO3(g)
sulphur dioxide/sulphur ZnSO4(s) → ZnO(s) + SO3(g)
Other sulphate salt trioxide CuSO4(s) → CuO(s) + SO3(g)
Chloride salt
Do not decompose -
Ammonium chloride Action of heating Example

Other chloride salt Produce ammonia gas NH4Cl(s) → 2NH3(g) + HCl(g)
and hydrogen chloride gas
Do not decompose -

6.11 Qualitative Analysis

1. To identify the cation and anion ions present in a salt, a chemical technique
called qualitative analysis is used by analysing the physical and chemical
properties of salt.

2. The technique of qualitative analysis are
 Examining the colour of the salt and solubility of the salt in
water Observing the effect of heat on the salt
 Identifying the gas evolved when a test is carry out on the salt
 Identifying the colour and solubility of the precipitate formed
when a chemical reagent is added

3. Based on the gas produced, we can deduce the type of ions present in a
substance.

Gas produced Ions present
CO2 Carbonate ion (CO32-)
O2
Except Na2CO3 and K2CO3
O2 and NO2
SO2 or SO3 Nitrate ion (NO3-)

NH3 Nitrate ion (NO3-)

Except NaNO3 and KNO3

Sulphate ion (SO42-)
Ammonium ion (NH4+)

4. Table below shows the tests for anions in aqueous solutio. n

Anion Method Observation Inference
Carbonate
ion (CO32-) Dilute acid (HCl, H2SO4 or HNO3) Effervescence All carbonate salts react
is poured into aqueous carbonate with dilute acid to produce
Nitrate ion solution occurs and carbon dioxide gas
(NO3-) CO32- + 2H+ → CO 2 +H2O
limewater turns
Sulphate
ion (SO42-) milky

Chloride a. Dilute H2SO4 followed by Nitrate ion NO3- is present
ion (Cl-) Fe2SO4 solution are added in the solution
into aqueous nitrate solution A brown ring is

b. A few drops of concentrated formed
H2SO4 are added along the
side of test tube

Dilute HCl acid followed by BaCl2 White precipitate The white precipitate is

(or Ba(NO3)2) is poured into the (BaSO4) is barium sulphate
Ba2+ + SO42- → BaSO 4
aqueous sulphate solution formed

Dilute nitric acid followed by White precipitate The white precipitate is
AgNo3 solution is poured into (AgCl) is formed
aqueous chloride solution silver chloride
Ag+ + Cl- → AgCl

Test for Cations
1. The cations are tested to identify Al3+, Pb2+, Zn2+, Mg2+, Ca2+, Fe2+, Fe3+,

Cu2+, NH4+.

2. To prepare a cation aqueous solution,
 Salt is dissolved in water (soluble salt)
 Salt is dissolved in dilute acid (insoluble salt). Solution formed is
filtered and the filtrate contains cations.

3. Three ways to identify the cations,
 Test with sodium hydroxide solution, NaOH
 Test with aqueous ammonia solution, NH3(aq)
 Test with a specific reagent as a confirmatory test

4. NaOH and NH3(aq) supply OH- ions to produce metal hydroxide precipitate
with cation solutions except Na+, K+, and NH4+ ions.
Mn+(aq) + nOH-(aq) → M(OH)n(s)

5. Table below shows the deduction of cations when NaOH is added.

Observation Types of cations
Al3+ or Pb2+ or Zn2+
White precipitate formed soluble in excess NaOH
solution to form complexes Mg2+ or Ca2+
Fe2+ (green), Fe3+
White precipitate formed insoluble in excess NaOH (brown), Cu2+ (blue)
solution
NH4+
Coloured precipitate (blue, green or brown) formed
insoluble in excess NaOH solution

No precipitate formed, only NH3 gas is released when
heated

6. Table below shows the deduction of cations when NH3(aq) is added.

Observation Types of cations
Al3+ or Pb2+ or Mg2+
White precipitate formed insoluble in excess NH3(aq)
Zn2+
White precipitate formed soluble in excess NH3(aq) to
form complexes Cu2+

Light blue precipitate is soluble in excess NH3(aq)
forming a dark blue solution

Coloured precipitate (green or brown) formed insoluble Fe2+ (green), Fe3+
in excess NH3(aq) (brown)

No precipitate formed Ca2+ or NH4+

Confirmatory Tests for Fe2+, Fe3+, Pb2+, and NH4+ ions
Table below shows the confirmatory tests for Fe2+, Fe3+, Pb2+, and NH4+ ions.

Reagent Observation Cation
KI present
Yellow precipitate formed soluble in hot water
and recrystallise when cooled

KCl, NaCl, HCl White precipitate formed soluble in hot water Pb2+
and recrystallise when cooled

K2SO4, Na2SO4, White precipitate formed insoluble in hot water NH4+
H2SO4, Brown precipitate

Nessler’s reagent

Potassium Light blue precipitate
hexacyanoferrate(II), Dark blue precipitate

K4Fe(CN)6 Fe2+

Potassium
hexacyanoferrate(III),

K3Fe(CN)6

Acidified potassium Purple colour decolourises
permanganate,
KMnO4

Potassium Blood red colour
thiocyanate, KSCN

Potassium Dark blue precipitate Fe3+
hexacyanoferrate(II),

K4Fe(CN)6

Potassium Greenish brown solution
hexacyanoferrate(III),

K3Fe(CN)6