Form 4 chapter 8 salts notes

1. SALTS
What is salt?
In the preparation of salts, we must identify the type of salt. This can be done by analysing
the cations and the anions that are present in salts.
Two types of salts
• Soluble salt – salts that can be dissolve in water at room temperature
• Insoluble salt – salts cannot be dissolve in water at room temperature
Type of salt Solubility in water
Sodium salts
Potassium salts
Ammonium salts
All dissolves in water
Nitrate salts All dissolves in water
Chloride salts
All dissolves in water, Except: Lead(II) chloride, PbCl2
Silver chloride, AgCl
Mercury chloride, HgCl
Sulphate salts
All dissolves in water Except: Lead(II) sulphate, PbSO4
Barium sulfat, BaSO4
Calcium sulfat, CaSO4
Carbonate salts
All did not dissolves in water, Except: Sodium carbonate, Na2CO3
Potassium carbonate, K2CO3
Ammonium carbonate, (NH4)2CO3
Special Properties of lead(II) chloride and lead(II) iodide
NOTES: Lead halide such as lead(II) chloride (PbCl2), lead(II) bromide (PbBr2), and
lead(II) iodide (PbI2) did not dissolve in cold water but dissolve in hot water.
1
White precipitate of
PbCl2
White precipitate
dissolves in hot
water
White precipitate formed
when the water is cooled
down.
Salt is an ionic compound formed when the hydrogen ion, H+
from acid
is replaced by a metal ion or ammonium ion, NH4
+
PbCl2
are soluble in
hot water.

2. Use of salts;
Item Use Example
Food preparation
Flavor
Monosodium glutamate (MSG)
Sodium chloride
Preservatives
Sodium chloride - salted fish
Sodium benzoate - sauce
Sodium nitrite - processed meat, burger
Baking powder Sodium hydrogen carbonate
Agriculture
Nitrogen fertilizers
Potassium nitrate
Sodium nitrate
Pesticide
Copper(II) sulphate
Iron(II) sulphate
Medicine
Reduce stomach acidic
(gastric)
Calcium carbonate
Calcium hydrogen carbonate
Sniff salt (fainted) Ammonium carbonate
Plaster of Paris (cement
to support broken bone)
Calcium sulphate
A. Preparation of Salt
The procedure of preparation salt depends to the type of salt.
a. Insoluble salt is prepared through precipitation reaction.
b. Soluble salt is prepared by one of these reactions;
i. Acid and alkali
ii. Acid and metal oxide
iii. Acid and metal carbonate
iv. Acid and reactive metal
2
Yellow precipitate of
PbI2
Yellow crystals formed when
the water is cooled down.
Yellow precipitate
dissolves in hot water
PbI2
are also soluble
in hot water.

3. a. Insoluble Salts
i. Preparing Insoluble Salts
1. Insoluble salts can be prepared through precipitation reactions or double decomposition reactions.
2. Precipitation or double decomposition reaction involves;
- two aquoues solutions/soluble salts were mix together
- one of the solutions contains the cations of the insoluble salt.
- one of the solutions contains the anions of the insoluble salt.
- the ions of the two aqueous solutions above interchange to produce two new compound
which is insoluble salt or precipitate, and aqueous solution.
- the precipitate produced is obtained by filtration. The residue left in the filter paper is the
insoluble salt. The filtrate is soluble salt.
- the residue/precipitate (insoluble salt) then rinsed with distilled water to remove any other
ions as impurities.
3
Anion
(Non-metal ion)
Cation
(Metal ion) nn mm

4. Chemical and ionic equations
Chemical equation : MX(aq) + NY(aq) → MY(s) + NX(aq)
solution solution precipitate solution
4
Na+
Na+
NO3
-
NO3
-
PbCl2
Pb2+
ions combined
with Cl-
ions to form
white precipitate
Na+
ions and NO3
-
ions do not
take part in the reaction and
are free to move in the solution
Ionic equation: Pb2+
+ 2Cl-
 PbCl2
Glass rod
Residue/precipitate
(Insoluble salt)
Filtrate
(Soluble salt)
Mixture of solutions
Filter funnel
Filter paper
Retort stand

5. Ionic equation : M+
(aq) + Y-
(aq) → MY(s)
Study this reaction carefully
In the formation of the precipitate of barium sulphate, BaSO4, the chemical equation can be written:
BaCl2(aq) + Na2SO4 (aq)  BaSO4(s) + 2NaCl (aq)
Ions Ba2+
+ Cl-
+ Na+
+ SO4
2-
 BaSO4 + Na+
+ Cl-
Ionic equation : Ba2+
+ SO4
2-
 BaSO4
(shows the ions that take part in the reaction to form precipitate/insoluble salts)
More examples;
Insoluble Salt Ions Ionic equation
ZnCO3 Zn2+
, CO3
2-
Zn2+
+ CO3
2-
 ZnCO3
AgCl Ag+
, Cl-
Ag+
+ Cl-
 AgCl
BaSO4 Ba2+
, SO4
2-
Ba2+
+ SO4
2
 BaSO4
PbCl2 Pb2+
, Cl-
Pb2+
+ Cl-
 PbCl2
PbSO4 Pb2+
, SO4
2-
Pb2+
+ SO4
2-
 PbSO4
CaCO3 Ca2+
, CO3
2-
Ca2+
+ CO3
2-
 CaCO3
ii. Preparation and purification of insoluble salts
Preparation of Plumbum(II) iodide
Chemical equation : Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
Ionic equation : Pb2+
(aq) + 2I-
(aq) → PbI2 (s)
Step 1: Preparation
5

6. 1. 20 cm3
lead(II) nitrate 0.1 mol dm-3
solution is measured with measuring cylinder 50 ml,
and poured into a beaker.
2. 20 cm3
potassium iodide 0.1 mol dm-3
solution is measured with measuring cylinder 50 ml
and poured into a beaker contains lead(II) nitrate solution.
3. The mixture is stirred with a glass rod. A yellow precipitate is formed.
4. The mixture is filtered to obtain the yellow solids of lead(II) iodide as the residue.
Step 2: Purification
6
+ →
20 cm3
Lead(II) nitrat
0.1 mol dm-3
20 cm3
potassium iodide
0.1 mol dm-3
Glass rod
Precipitate of lead(II) iodide
(yellow)
Sodium nitrate solution
Mixture of solutions
Filter funnel
Filter paper
Retort stand
Beaker

7. 5. The residue is rinsed with distilled water to remove other ions in it.
6. The yellow solid is dried by pressing between two pieces of filter paper.
EASY LAH !
b. Soluble Salt
7
Glass rod
Distilled water
Precipitate of lead(II)
iodide
Precipitate of lead(II)
iodide
Filter paper

8. i. Preparaing Soluble Salt
- Sodium salts
- Potassium salts Acid + alkali  salts + water
- Ammonium salts
Soluble Salts
Acid + metal oxide  salts + water
- Others salts Acid + reactive metal  salts + hydrogen gas
Acid + metal carbonate  salt + water + carbon dioxide
Notes: Reactive metal is magnesium, aluminium, and zinc
Unreactive metal is iron, lead, silver
a. Sodium, potassium or ammonium salts prepared from acid and alkali reaction.
Salt Alkali Acid Chemical equation
NaCl NaOH HCl NaOH + HCl → NaCl + H2O
K2SO4 KOH H2SO4 2KOH + H2SO4 → K2SO4 + 2H2O
NH4NO3 NH3/NH4OH HNO3 NH3 + HNO3 → NH4NO3 + H2O
CH3COONa NaOH CH3COOH NaOH + CH3COOH → CH3COONa + H2O
Note: To prepare the above salts, titration technique is use.
b. Soluble salt (except sodium, potassium and ammonium salt) is prepared using these methods
- Acid and metal
- Acid and metal oxide
- Acid and metal carbonate
Name of Salt
Acid that must be
used
Substance that can be use to react with acid
Metal Metal oxide Metal carbonate
ZnCl2 HCl Zn ZnO ZnCO3
Mg(NO)3 HNO3 Mg MgO MgCO3
CuSO4 H2SO4 × CuO CuCO3
Pb(NO3)2 HNO3 × PbO PbCO3
Write a chemical equation for each experiment above.
8

9. Remember this notes ok
1. Metal that is less reactive from hydrogen such as copper, lead and
silver/argentum did not react with dilute acid.
2. Metal, metal oxide and metal carbonate above is a solid that cannot dissolves in
water, hence during reaction that solid must be added excessively to make sure
all hydrogen ions in acid is completely reacted. Excess solid can be expelling
through filtration.
3. Impure soluble salt can be purified through crystallization process.
ii. Preparation and purification of soluble salts
A. Preparing soluble salt through reaction between acid and alkali.
Preparation of Soluble Sodium, Potassium and ammonium Salts
Soluble salts of sodium, potassium and ammonium can be prepared by the reaction between an
acid and alkali.
Acid (aq) + alkali (aq)  Salt (aq) + Water (l)
Procedure :
• Using pipette, 25 cm3
of alkali solution is measured and
transferred into a conical flask.
• Two drops of phenolphthalein are added to the alkali solution.
• Dilute acid is place in a burette. The initial reading is recorded.
• Acid is added slowly into the alkali solution while shaking
the conical flaks, until the pink solution turn colourless.
• The final reading of the burette is recorded.
• The volume of acid added, V cm3
is calculated.
• The experiment is repeated by adding V cm3
of acid to
25 cm3
of alkali solution in a beaker without using
phenolphthalein as an indicator.
• The mixture is transferred into a evaporating dish.
• The mixture is heated until saturated and the saturated solution
is allowed to cool at room temperature.
• Salt crystals formed are filtered and rinsed with a little of cold distilled water.
• Salt crystals are dried by pressing it between filter papers (or in oven)
The reaction between acid and alkali is known as what process?
Refer to acids and bases notes ok. Don’t worry I’ll help you.!
Kita bukan along kita cuma nak tolong.
9

10. Example: Preparing sodium chloride
Step 1: Preparation (Titration)
1. 25.0 cm3
sodium hydroxide solutions is pipette into conical flask.
2. Two drops of phenolphthalein indicator are added into conical flask. The colour of solution is
recorded.
3. A 50 cm3
burette is filled with hydrochloric acid. The initial burette reading is recorded.
4. Hydrochloric acid is added gradually from a burette into conical flask and swirling the conical
flask.
5. Titration is stopped when phenolphthalein changes from pink to colourless. The final burette
reading is recorded.
6. The volume of hydrochloric acid used is calculated.
7. The experiment is repeated by adding hydrochloric acid (known volume) to 25.0 cm3
sodium
hydroxide in a beaker without using phenolphthalein.
Step 2: Preparation (Crystallization)
8. The mixture is transferred into a evaporating dish.
10
Retort
stand
Burette
Hydrochloric acid
Conical flask
25 cm3
NaOH + phenolphthalein
indicator
Bunsen burner
Evaporating dishSalt solution

11. 9. The colourless solution is slowly heated/evaporated until its saturated or to about one-third
(1/3) of the original volume.
10. The saturated solution is then cooled to allow crystallization to occur.
Step 3: Purification
10. The white crystals formed are then filtered, rinsed with a little distilled
water and dried by pressing between filter paper.
Note: Phenolphthalein indicator is used at the beginning of the experiment to determine the volume
of hydrochloric acid that is required to react completely with 25 cm3
of sodium hydroxide.
However experiment is repeated without using phenolphthalein so that the salt prepared will
not contaminated by the indicator.
11
Glass rod
Distilled water
Copper(II) sulphate
Filter funnel

12. B. Preparing soluble salt through reaction between acid
i. Metal oxide. ii. Metal iii. Metal carbonate
Procedure To Prepare a Soluble Salt (not Na, K or NH4
+
)
• 50 cm3
of acid is measured using a measuring cylinder and poured into a beaker. The acid is
heated slowly.
• Using a spatula, metal / metal oxide / metal carbonate powder is added a little at a time while
stirring the mixture with a glass rod.
• The addition of the solid powder is stopped when some solids no longer dissolve anymore. (the
solid is excess and all the acid is completely neutralised by the solid)
• The mixture is filtered to remove the excess solid powder.
• The filtrate is transferred to an evaporating dish.
• The filtrate is heated until saturated. (The filtrate is evaporated to about one-third (1/3) of the
original volume)
• The saturated solution is then allowed to cool to room temperature and the salt crystals are
formed.
• The crystals are filtered and rinsed with a little cold distilled water.
• Salt crystals are then dried by pressing it between filter papers.
12
Heating
Acid
Powder
of :
Metal oxide
Metal carbonate
Metal
Excess of solid powder
Filtrate (Salt solution)
Heating
Saturated solution
Crystals
Filtrate

13. Example: Preparing copper(II) sulphate
(Sulphuric acid and copper(II) oxide powder)
Step 1: Preparation
1. 50 cm3
sulphuric acid 0.1 mol dm-3
is put in a beaker and is heated.
2. Using spatula copper(II) oxide powder is added a little at a time to the hot sulphuric acid while
stirring continuously with glass rod.
3. The addition of copper(II) oxide is stopped when solids powder remain undissolved.
4. The mixture is filtered to remove the excess copper(II) oxide.
5. The filtrate is transferred to an evaporating dish.
13
xxxxxxxxxxxxxxxx
Copper(II) oxide
Glass rod
Beaker
Wire gauze
50 cm3
sulphuric acid 0.1 mol dm-3
Tripod
Bunsen
burner
Spatula
Stir
Glass rod
Reactant mixture
Excess copper(II) oxide
Copper(II) sulphate solution

14. 6. The filtrate is slowly heated/evaporated until its saturated, or to about one-third (1/3) of the
original volume.
7. The saturated solution is then allowed to cool to room temperature.
Step 3: Purification
8. The crystals are filtered and rinsed with a little cold distilled water.
9. Salt crystals are then dried by pressing it between filter papers.
REMEMBER…. THIS NOTES OK
Unreactive metal such as lead (Pb), copper (Cu), and silver (Ag) cannot react
with dilute asid. So to prepare salt contains lead ions (Pb2+
), copper ions (Cu2+
)
or silver ions (Ag+
), we must use either oxide powder or carbonate powder only.
Example: CuO + H2SO4  CuSO4 + H2O (ok)
CuCO3 + H2SO4  CuSO4 + H2O + CO2 (ok)
Cu + H2SO4  no reaction (not ok)
14
Bunsen burner
Evaporating dish
××××××××××××
Copper(II) sulphate
solution
Glass rod
Distilled water
Copper(II) sulphate

15. B. Physical Characteristics of Crystals.
A salt is made up of positive and negative ions. When these ions are packed closely
with a regular and repeated arrangement in an orderly manner, a solid with
definite geometry known as crystal lattice is formed.
All crystals have these physical characteristics:
a) Reqular geometry shapes, such as cubic or hexagonal.
b) Flat faces, straight edges and sharp angles.
c) Same angle between adjacent faces.
d) All crystals of the same salt have the same shape although the sizes may be different.
Start to memorize the solubility of a salt in water OK.
It will help you a lot to better understand this chapter.
KNOWLEDGE IS POWER
The solubility of a salt in water depends on the types of cations and anions present.
C. Qualitative
Analysis of
Salts
15
Salt Solubility in water
Sodium, potassium and
ammonium salts
(Na+
, K+
, NH4
+
)
• All are soluble
Nitrate salt (NO3
-
) • All are soluble
Chloride salt (Cl -
)
• All chloride salts are soluble in water except
PbCl2, AgCl and HgCl2
Sulphate salt (SO4
2-
)
• All sulphate salts are soluble in water except
PbSO4, BaSO4 and CaSO4
Carbonate salt (CO3
2-
)
• All carbonate salts are insoluble except Na2CO3,
K2CO3 and (NH4)2CO3

16. What is Qualitative analysis?
In the qualitative analysis of salts, we need to identify the ions that are present in salts.
This can be done by analysing their physical and chemical properties.
Observations on the physical properties of salts
1. Colour and solubility in water
Certain physical properties of salts such colour and solubitity in water are observed to help us infer certain
cations and anions that are present in salts.
The table shows the colour of salts in solid , in aqueous solution and the solubility of salts in water
Salt Colour in solid
Solubility in
water
Colour in Aqueous solution
1. Ammonium chloride NH4Cl white soluble colourless
2. Ammonium nitrateNH4(NO3)3 white soluble colourless
3. Calcium carbonate CaCO3 white insoluble -
4. Calcium nitrate Ca(NO3)2 white soluble colourless
5. Magnesium sulphate MgSO4 white soluble colourless
6. Magnesium carbonate MgCO3 white insoluble -
7. Zinc sulphate Zn SO4 white soluble colourless
8. Zinc nitrate Zn(NO3)2 white soluble colourless
9. Lead(II) chloride , PbCl2 white insoluble -
10. Lead(II) sulphate , PbSO4 white insoluble -
11. Lead(II) carbonate , PbCO3 white insoluble -
12. Copper(II) chloride , CuCl2 Blue soluble Blue
13 Copper(II) sulphate , PbSO4 Blue soluble Blue
14. Copper(II) carbonate , PbCO3 Green insoluble -
15. Iron(II) sulphate , FeSO4 Green soluble Pale green
16. Iron(III) chloride , FeCl3 Brown / Yellow soluble Brown/Yellow/ Yellowish brown
17. Sodium nitrate , NaNO3 white soluble colourless
18, Sodium carbonate , Na2CO3 white soluble colourless
19. Potassium nitrate , KNO3 white soluble colourless
20. Potassium carbonate , K2CO3 white soluble colourless
The table shows the colour of different cations in the solid form or in aqueous solution
16
Qualitative analysis is a chemical technique used to determine what
substances are present in a mixture but not their quantities.

17. Observation Inference
Blue solution Ion copper (Cu2+
) present
Pale green solution Ion Iron(II) Fe2+
present
Yellow/Yellowish-
brown/brown solution
Ion Iron (III) Fe3+
present
Green solid Hydrated Fe 2+
, CuCO3
Brown solid Hydrated Fe 3+
salt
White solid
Salts of Na+
, K+
,NH4
+
, Mg 2+
, Ca 2+
Al 3+
, Zn 2+
, Pb 2+
(If the anions are
colourless
Colourless solution Na+
, K+
,NH4
+
, Mg 2+
, Ca 2+
, Al 3+
, Zn 2+
, Pb 2+
The table shows the solubility of different types of salts in water
Compounds Solubility in water
Sodium salts
Potassium salts
Ammonium salts
All are soluble
Nitrate salts
All are soluble
Chloride salts
All are soluble except AgCl, HgCl and PbCl2 (soluble in hot water)
Sulphate salts
All are soluble except BaSO4, PbSO4 and CaSO4
Carbonate salts All are insoluble except sodium carbonate, potassium carbonate and
ammonium carbonate
2. Tests for gases
17

18. Gases are often produced from reactions carried out during laboratory tests on salts. By identifying the gases
evolved,it is possible to infer the types of cations and anions that are present in a salt.
The table shows the test and the result of different gases
Gas Test Result
Oxygen gas, O2
Test with a glowing wooden
splinter
Wooden splinter is rekindled /lighted
Hydrogen gas , H2
Test with a lighted wooden
splinter
Gas explodes with a pop sound
Carbon dioxide gas , CO2
Bubble the gas produced into
lime water
Lime water turns milky
Ammonia gas, NH3
Test with moist red litmus
paper
Moist red litmus paper turns blue
Chlorine gas, Cl2
Test with moist blue litmus
paper
Moist blue litmus paper turns red and then
turns white
Hydrogen chlorine gas , HCl
Test with a drop of
concentrated ammonia NH3
solution
Dense white fumes
Sulphur dioxide gas , SO2
Bubble the gas produced into
purple acidified potassium
manganate (VII), KMnO4
solution
Purple acidified potassium manganate
(VII),KMnO4 solution decolourises
Nitrogen dioxide gas , NO2
Test with moist blue litmus
paper
moist blue litmus paper turns red
3. Action of heat on salts
18

19. Effect of heat on carbonate salts
Carbonaate salt
Colour of salt
before heating
Colour of residue
Effect on lime water
Hot cold
Copper (II)
carbonate, CuCO3
Green powder Black powder Black powder
The gas liberated turns
lime water milky/chalky
Zinc carbonate ,
ZnCO3
White solid Yelow solid White solid
The gas liberated turns
lime water milky/chalky
Lead(II) carbonate,
PbCO3
White solid Brown sold Yelow solid
The gas liberated turns
lime water milky/chalky
Sodium carbonate,
Na2CO3
White solid White solid White solid No change
Calcium carbonate,
CaCO3
White solid White solid White solid
The gas liberated turns
lime water milky/chalky
Potassium
carbonate, K2CO3
White solid White solid White solid No change
Magnesium
carbonate, MgCO3
White solid White solid White solid
The gas liberated turns
lime water milky/chalky
Effect of heat on nitrate salts
Nitrate Salt
Colour of salt
before heating
Colour of residue
Test on gases liberated
Hot cold
Copper (II) nitrate,
Cu(NO3)2
Blue solid
Black
powder
Black
powder
A brown gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
Zinc nitrate,
Zn(NO3)2
White solid
Yellow
solid
White solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
Lead(II) nitrate,
Pb(NO3)2
White solid Brown solid
Yellow
solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
Sodium nitrate,
NaNO3
White solid White solid White solid
A colourless gas that rekindles a
glowing splinter is liberated
Calcium nitrate,
Ca(NO3)2
White solid White solid White solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
19

20. Potassium nitrate,
KNO3
White solid White solid White solid
A colourless gas that rekindles a
glowing splinter is liberated
Magnesium nitrate,
Mg(NO3)2
White solid White solid White solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
Iron(II) nitrate,
Fe(NO3)2
Pale Green
solid
Pale Green
solid
Pale
Green
solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
Iron(III) nitrate,
Fe(NO3)3
Brown solid
Reddish-
Brown solid
Reddish-
Brown
solid
A browan gas that turns blue litmus
paper red is liberated.
The gas liberated also ignites a
glowing splinter
The table shows the comparison of the effect of heat on carbonate and nitrate salts
Metal Effect of heat on carbonate salt Effect of heat on nitrate salt
Potassium
Sodium
Are not decomposed by heat Decompose to nitrite salt and oxygen gas.
Calcium
Magnesium
Aluminium
Zinc
Iron
Tin
Lead
Copper
Decompose to metal oxide and
carbon dioxide gas.
Decompose to metal oxide, nitrogen dioxide
gas and oxygen gas.
Mercury
Silver
Gold
Decompose to metal, carbon
dioxide gas and oxygen gas.
Decompose to metal , nitrogen dioxide gas
and oxygen gas.
Most sulphate salts are not decomposed by heat. Only a few sulphate such as iron(II) sulphate,zinc sulphate
and copper sulphate decompose to sulphur dioxide or sulphur trioxide gas when heated.
All chloride salts are stable when heated except ammonium chloride. Ammonium chloride sublimes and
decomposes to produce ammonia gas and hydrogen chloride gas.
The table shows the deduction of the types of ion present based on the gas produced
Type of gas produced Type of ion present(anion)
CO2 Carbonate ion (CO3
2-
) present except Na2CO3 and K2CO3
O2 Nitrate ion (NO3
-
) present
NO2 Nitrate ion (NO3
-
) present except NaNO3 and KNO3
SO2 Sulphate ion (SO4
2-
) present
NH3 Ammonim ion (NH4
+
) present
20

21. Tests for anions
Reagent / Condition Observation Anion Ionic Equation (if any)
2 cm3
the unknown solution +
dilute hydrochloric acid / nitric
acid / sulphuric acid  pour
into a test tube  gas liberated
is immediately bubbled through
lime water.
Effervescence.
Colourless gas
turns lime water
milky.
CO3
2-
ion CO3
2-
+ 2H+
 CO2 +
H2O
2 cm3
of nitric acid + 2 cm3
of
the unknown solution  pour
into a test tube  + 2 cm3
silver nitrate solution
White precipitate
is formed.
Cl-
ion Ag+
+ Cl-
 AgCl
2 cm3
of dilute hydrochloric
acid / nitric acid + 2 cm3
of the
unknown solution  pour into
a test tube  + 2 cm3
of
barium chloride / barium nitrate
solution  shake well
White precipitate
is formed.
SO4
2-
ion Ba2+
+ SO4
2 -
 BaSO4
2 cm3
of the unknown solution
 pour into a test tube 
2 cm3
of dilute sulphuric acid +
2 cm3
of iron(II) sulphate
solution  shake well.
Then drop carefully and slowly
a few drops of concentrated
sulphuric acid along the side of
a slanting test tube into the
mixture without shaking it.
Brown ring is
formed at the
boundary
between the
concentrated
H2SO4 (top layer)
and aqueous
solution of the
mixture (bottom
layer)
NO3
-
ion -
21

22. Tests for cations
Confirmatory Test for Fe2+
, Fe3+
, Pb2+
, NH4
+
Ions
Confirmatory Test for Fe2+
Reagent Observation Conclusion
Potassium
hexacyanoferrate(II) solution
Pale blue precipitate Fe2+
ion is present
Dark blue precipitate Fe3+
ion is present
Potassium
hexacyanoferrate(III) solution
Dark blue precipitate Fe2+
ion is present
Greenish-brown solution Fe3+
ion is present
Potassium thiocyanate
solution
Pale red colouration Fe2+
ion is present
Blood red colouration Fe3+
ion is present
Confirmatory Test for Pb2+
Method Observation Ionic Equation
• Using aqueous solution of chloride
- 2 cm3
of any solution of Cl-
+
2 cm3
of any solution of Pb2+

dilute with 5 cm3
of distilled water 
heat until no further change occurs 
allow the content to cool to room
temperature using running water from
the tap
- A white precipitate is
formed
When heated – dissolve in
water to form colourless
solution
When cooled – white
precipitate reappear
Pb2+
+ 2Cl-
 PbCl2
• Using aqueous solution of iodide
- 2 cm3
of any solution of I-
+
2 cm3
of any solution of Pb2+

dilute with 5 cm3
of distilled water 
heat until no further change occurs 
allow the content to cool to room
temperature using running water from
the tap
- A yellow precipitate is
formed
When heated – dissolve in
water to form colourless
solution
When cooled – yellow
precipitate reappear
Pb2+
+ 2I-
 PbI2
Confirmatory Test for NH4
+
Method Observation
• 2 cm3
of any solution of NH4
+
+ 2 cm3
of
NaOH / KOH / Ca(OH)2  heat  put a
piece of moist red litmus paper at the
mouth of the test tube
- Moist red litmus paper turns blue
Reaction with Nessler’s Reagent
• 2 cm3
of any solution of NH4
+
+ 2 cm3
of
Nessler’s Reagent  shake well
- A brown precipitate is formed
22

23. Reaction of Cations With NaOH
Pb2+
Zn2+
Al3+
: White precipitate dissolves/larut in excess NaOH
Ca2+
Mg2+
: White precipitate insoluble/tidak larut in excess NaOH
23
Cations
+ NaOH (aq)
Precipitate producedNo precipitate
White precipitate Coloured precipitate
Green Blue Brown
Fe2+
Cu2+
Fe2+
NH4
+
K+
Na+
Dissolve in excess
NaOH (aq) to form
colourless solution
Sodium hydroxide solution is poured
slowly into 2 cm3
of the solution to be
tested in a test tube, until in excess.
Insoluble in
excess
NaOH (aq)
NH3
gas produced
warm
Zn2+
Al3+
Pb2+
Ca2+
Mg2+
Easylah

24. Reaction of Cations With NH3
Zn2+
: White precipitate dissolves/larut in excess NH3
Pb2+
Al3+
Mg2+
: White precipitate insoluble/tidak larut in excess NH3
24
Cations
Precipitate producedNo precipitate
White precipitate Coloured precipitate
Green Blue Brown
Fe2+
Cu2+
Fe2+
NH4
+
K+
Na+
Aqueous ammonia solution is poured
slowly into 2 cm3
of the solution to be
tested in a test tube until in excess.
NH3
gas produced
warm
Zn2+
Al3+
Pb2+
Ca2+
Mg2+
Easylah
+ NH3
(aq)
+ excess
NH3
(aq)
Dark blue
solution
Dissolve in excess
NH3
(aq) to form
colourless solution
Insoluble in
excess NH3
(aq)

25. Zn2+
ion is the only cation that form white precipitate and dissolves in
both excess NaOH and NH3 solutions.
Mg2+
ion is the only cation that form white precipitate and insoluble in
both excess NaOH and NH3 solutions.
Ca2+
ion in the only cation that form white precipitate in NaOH
solutions, but no precipitate in NH3 solution.
Fe2+
, Fe3+
and Cu2+
ions is easy to spot because the ions shows
coloured precipitate.
Pb2+
ion and Al3+
ion form white precipitate and dissolves in excess
NaOH
solution, but insoluble in excess NH3 solutions.
Example: lead(II) nitrate solution and aluminium nitrate solution
Sodium sulphate solution is added slowly into 2 cm3
of the solution to be tested in a
test tube.
If a white precipitate is formed, then then the solution tested is lead(II) nitrate.
If no change occurs, then the solution tested is aluminium nitrate.
25
Based from the observation, I can conclude that
How to differentiate between Pb2+
and Al3+
?
A chemical tests can be carried out in
the laboratory to differentiate between
Pb2+
and Al3+
.
(Please refer to Confirmatory Test for
Pb2+
, in ealier notes).
Now let see some questions about salt.
Try to solve it by yourself first and then
compare with the answers provided.

26. Example 1:
Describe chemical tests that can be carried out in the laboratory to differentiate between
(a) lead(II) nitrate solution and aluminium nitrate solution
Sodium sulphate solution is added slowly into 2 cm3
of the solution to be tested in a test tube.
If a white precipitate is formed, then then the solution tested is lead(II) nitrate.
If no change occurs, then the solution tested is aluminium nitrate.
(b) aluminium nitrate solution and zinc nitrate solution
Aqueous ammonia solution is poured slowly into 2 cm3
of the solution to be tested in a test tube
until in excess.
If a white precipitate that dissolves in excess aqueous ammonia solution is formed, than the
solution tested is zinc nitrate.
If a white precipitate that is insoluble in excess aqueous ammonia solution is formed, than the
solution tested is aluminium nitrate.
(c) ammonium chloride solution and potassium chloride solution
Nessler’s Reagent is added to 2 cm3
of the solution to be tested in a test tube.
If a brown precipitate is formed, then the solution tested is ammonium chloride.
If no change occurs, then the solution tested is potassium chloride
(d) iron(II) sulphate solution and iron(III) sulphate solution
Potassium hexacyanoferrate(II) solution is poured into 2 cm3
of the solution to be tested in a
test tube.
If a dark blue precipitate is formed, then the solution tested is iron(III) chloride.
If no change occurs, then the solution tested is iron(II) chloride.
Or
Potassium hexacyanoferrate(III) solution is poured into 2 cm3
of the solution to be tested in a
test tube.
If a greenish-brown solution is formed, then the solution tested is iron(III) chloride.
If no change occurs, then the solution tested is iron(II) chloride.
Or
Potassium thiocyanate solution is poured into 2 cm3
of the solution to be tested in a test tube.
26

27. If a blood red colouration is formed, then the solution tested is iron(III) chloride.
If no change occurs, then the solution tested is iron(II) chloride.
(e) sodium chloride and sodium sulphate
Silver nitrate solution is poured into 2 cm3
of the solution to be tested in a test tube.
If a white precipitate is formed, then the solution tested is sodium chloride.
If no change occurs, then the solution tested is sodium sulphate.
Or
Barium chloride solution is poured into 2 cm3
of the solution to be tested in a test tube.
If a white precipitate is formed, then the solution tested is sodium sulphate.
If no change occurs, then the solution tested is sodium chloride.
Example 2:
1. State three examples of
a) soluble salts b) insoluble salts
Potassium carbonate Magnesium carbonate
Lead(II) nitrate Lead(II) sulphate
Ammonium chloride Argentum chloride
2. Which of the following salts is soluble
3. Identify the gas that turns moist red litmus paper blue
Ammonia gas
4. Gas X has the following properties
Gas X is carbon dioxide gas
27
Lead(II) chloride Sodium carbonate
Calcium sulphate Barium sulphate
• Colourless
• Acidic gas
• Turns lime water milky
Salt P Metal oxide X Gas Y

28. 5. Heat +
Colour of metal oxide X is yellow when hot and white when cold. Gas Y turns lime water milky.
a) Name gas Y : carbon dioxide gas
b) Name metal oxide X : zinc oxide
c) Name salt P : zinc carbonate
d) Write an equation to represent the action of heat on salt P
ZnCO3 (s) ZnO (s) + CO2 (g)
6. A sample of copper(II) nitrate, Cu(NO3)2 was heated strongly. Write down the expected observation.
Copper(II) nitrate decompose to produce black colour of residue when hot and cold. A brown
gas that changed moist blue litmus paper to red and colourless gas that lighted up a glowing
wooden splinter are produced.
28

29. D. Numerical problem involving stoichiometric reaction in the
preparation of salt
Example 1;
Ammonium phosphate, (NH4)3PO4 is use as a fertilizer. 29.8g of this salt is prepared by neutralizing
phosphoric acid, H3PO4 with ammonium gas, NH3. Calculate the volume of ammonium gas, NH3
reacted at room conditions.
( Relative atomic mass; H, 1: N, 14: P, 31; O, 16; Molar volume; 24 dm3 mol-1
at room conditions)
Solutions;
a. Calculate the number of moles
2.88 g
[3(14) + 12(1) + 31 + 4(16)
= 0.2 mol
b. Write a balanced chemical equation
Compare the mole ratio of NH3 and
(NH4)3PO4
H3PO4(aq) + 3NH3(aq)  (NH4)3PO4(aq)
c. Calculate the number of moles of NH3 base on
the mole ratio
= 3 X 0.2 mol
= 0.6 mol
d. Calculate the volume of NH3
Volume = number of mole X volume
= 0.6 mol X 24 dm3
mol -1
= 14.4 dm3
Example 2:
29
A balanced chemical equation for a reaction in preparation of a salt
can be used to calculate the stoichiometric quantities of the following
Masses of reactants
Volumes and concentrations of reactants
Masses of products
Volumes of products
=
3 mol 1 mol

30. 3.9 g of potassium is burnt completely in the air as shown in the following equation;
4K(s) + O2(g) → 2K2O(s)
What is the mass of potassium oxide produced?
[Relative atomic mass: K, 39; O, 16]
Solutions
Tip: Solve the question step by step
Step 1: Write Chemical Equation
4K(s) + O2(g) → 2K2O(s)
4 mol of K react with 1 mol of O2 produce 2 mol K2O
Step 2: Calculate the number of mole
[Get the information from the question]
Step 3: Find the coefficient From Balance Chemical Equation
FBCE;
4 mol of K produce 2 mol K2O
Thus;
0.1 mol of K produce 2/4 mol K2O = 0.2 mol K2O
FBCE;
[Sebelah kiri] [Sebelah kanan]
Bil. mol yang telah dikira Bil. Mol yang hendak ditentukan
4 mol K = 2 mol K2O
0.1mol K = 2/4 x 0.1mol K2O = 0.05 mol K2O
No. of mol of K2O = 0.05 mol
Step 4: Solve the questions
Thus;
Mass of K2O = 0.05 mol × Molar mass
= 0.05 mol× 55 g mol-1
= 2.75 g
Example 3:
30
No. of mol K =
mass
Molar mass
=
3.9 g
39 gmol-1
0.1 mol=

31. Acids reacts with calcium carbonate, CaCO3 in limestone to form a salt and carbon dioxide, CO2.
A piece of limestone reacted completely with 100 cm3
of 31.5 g dm-3
nitric acid, HNO3.
[Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Ca, 40. Molar volume: 24 dm3
mol-1
at room
conditions]
a. Calculate the mass of salt produced.
b. What is the volume of carbon dioxide, CO2 liberated at room conditions?
Step 1: Write Chemical Equation
Chemical Equation: 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
Step 2: Calculate the number of mole
Get the information from the question;
FBCE; 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
2 mol HNO3 = 1 mol Ca(NO3)2
0.05 mol HNO3 = ½ x 0.05 mol Ca(NO3)2 = 0.025 mol Ca(NO3)2
No. of mol of Ca(NO3)2 = 0.025 mol
Mass of Ca(NO3)2 = 0.025 mol × 40 + 2[14 + 3(16)] g mol-1
= 4.1 g
31
No. of mole of HNO3 =
Molarity × Volume
1000
=
0.5 mol dm-3
× 100 cm3
1000
= 0.05 mol
Concentration of HNO3 = 31.5 g dm3
=
Molar mass of HNO3
31.5 g dm3
= 0.5 mol dm-3
=
31.5 g dm3
1 + 14 + 48 g mol-1
Change the
concentration
given in g dm-3
to mol dm-3
first

32. FBCE; 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
2 mol HNO3 = 1 mol CO2
0.05 mol HNO3 = ½ x 0.05 mol CO2 = 0.025 mol CO2
No. of mol of CO2 = 0.025 mol
Volume of CO2 = 0.025 mol × 12 + 2(16) dm3
mol-1
= 1.1 dm3
Example 4:
Pb(NO3)2 compound decomposes when heated as shown in the following equation.
If 6.62 g of Pb(NO3)2 compound is heated, calculate;
[Relative atomic mass: N, 14; O, 16; Pb, 207; 1 mol of gas occupies 22.4 dm3
at s.t.p.]
(i) mass of PbO that is produced
(ii) volume of nitrogen dioxide produced at s.t.p
(ii) volume of oxygen produced at s.t.p
Solution:
FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 2 mol PbO
0.02 mol Pb(NO3)2 = 0.02 mol PbO
No of mol PbO = 0.02 mol
Mass of PbO = 0.02 x 223 = 4.46 g
32
No of mol Pb(NO3
)2
=
mass
Molar mass
=
6.62 g
331 gmol-1
0.02 mol=
2Pb(NO3)2 → 2PbO + 4NO2 + O2

33. FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 4 mol NO2
0.02 mol Pb(NO3)2 = 4/2 x 0.02 mol O2 = 0.04 mol O2
No of mol O2 = 0.04 mol
Volume of O2 = 0.04 x 22.4 dm3
= 0.896 dm3
// 896 cm3
FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 1 mol O2
0.02 mol Pb(NO3)2 = ½ x 0.02 mol O2 = 0.01 mol O2
No of mol O2 = 0.01 mol
Volume of O2 = 0.01 x 22.4 dm3
= 0.224 dm3
// 224 cm3
Numerical Problems involving stoichiometric reactions in the precipitation of salts
Question 1:
A student prepare copper(II) nitrate, Cu(NO3)2 by reacting copper(II) oxide, CuO with 200 cm3
of 2.0
moldm-3
nitric acid, HNO3. Calculate the mass of copper(II) oxide, CuO needed to react completely
with the acid.
[Relative atomic mass: Cu, 64 ; O, 16]
Question 2:
X cm3
of 0.5 moldm-3
sulphuric acid, H2SO4 is added to 100 cm3
of 1.0 moldm-3
lead(II) nitrate
solution to produce lead(II) sulphate, PbSO4.
[Relative atomic mass: Pb, 20; O, 16; S, 32]
a. Calculate the value of X.
b. Calculate the mass of lead(II) sulphate obtained.
Start to do exercises from any book.
I will help and guide you to master this topic.
Prepared by;
Kamal Ariffin Bin Saaim
SMKDBL
33