MKS@Chemistry Module Form 4 Complete Set

mohd faisol mansor/chemistry form 4/chapter 5

(b) Table 1.2 shows some physical properties of two compounds, U
and V.

Compound Melting pt Boiling pt Solubility in Solubility in
(oC) (oC) water organic
solvent
1 420
U 800 Soluble Insoluble

V - 95 86 Insoluble Soluble
Table 1.2

i) State the physical state of the following compound at room
condition.

U : _______________________________________________________

V : _______________________________________________________

ii) State the type of compound for U.

__________________________________________________________

iii) Explain why melting point and boiling point of compound U is
higher than V?

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

99

mohd faisol mansor/chemistry form 4/chapter 6

CHAPTER 6

ELECTROCHEMISTRY

Electrolytes are Electrolysis is a process
substances that can whereby compounds in
conduct electricity when molten or aqueous state
they are in molten state
and aqueous solution. are broken down into
This due to the present of their constituent
free moving ions in the
elements by passing
electrolytes. electricity through them.

Non-electrolytes are 2NaCl (l) Example
substances that can not 2Na (s) + Cl2 (g)
conduct electricity when
they are in all state. This
because non-electrolyte
exist as molecule which
means contain no ions.

Chlorine Gas
Sodium Metal

100

mohd faisol mansor/chemistry form 4/chapter 6

Electrolytic Cell (molten state)

[ Draw the apparatus of electrolysis molten sodium chloride ]

a) Electrode attach to positive terminal (battery) =
b) Electrode attach to negative terminal (battery) =
c) Anion (negative ion) discharged at electrode =

Anion will _______________ electrons.
d) Cation (positive ion) discharged at electrode =

Cation will ______________ electrons.
e) Electron flow from ___________________ to ___________________
f) Electrolytic Cell will change the _______________ energy to

________________ energy.

101

mohd faisol mansor/chemistry form 4/chapter 6

Product of Electrolysis

Ion Observation Half-equation Product
discharged Product
at Cathode Observation Half-equation
All ion metal 102
Gas Test
except
Copper ion

Copper ion

Hydrogen
ion
Ion

discharged
at Anode

Oxide ion

Chloride ion

Bromide ion

Iodide ion

Hydroxide
ion

Hydrogen
gas

Oxygen gas

Chlorine gas

mohd faisol mansor/chemistry form 4/chapter 6

Electrolysis Molten Lead(II) Bromide

PQ

1. State the ion consists in the electrolyte.

2. Which electrode is

a) Cathode =

b) Anode =

3. Which ion will be discharged at

a) Cathode =

b) Anode =

4. State the observation at

a) Cathode =

b) Anode =

5. State the product formed at

a) Cathode =

b) Anode =

6. Write the half equation at

a) Cathode =

b) Anode =

7. Draw the electron flow on the diagram above.

103

mohd faisol mansor/chemistry form 4/chapter 6

Exercise

1. State the ion present in the following electrolyte. Predict the
products from the electrolysis of some molten compound and
write the ionic equation involved.

a) Magnesium oxide

b) Copper(II) chloride

c) Lead(II) iodide

104

mohd faisol mansor/chemistry form 4/chapter 6

2. State the meaning of the following terms.

a) Anode b) Cathode c) Electrolysis

3. A molten oxide, R2O3 is electrolysed using carbon electrodes.

a) Draw a labeled diagram to show the set-up of apparatus for
the electrolysis.

b) What ions are present in the electrolyte? Write the formulae for
the ions present in the electrolyte.

c) State the ions move to each of the electrodes during
electrolysis.

105

mohd faisol mansor/chemistry form 4/chapter 6

d) Write half equation of the reaction at each of the electrodes.
e) Name the substances formed at each of the electrodes.
f) Label the flow of electron in the diagram (a).

106

mohd faisol mansor/chemistry form 4/chapter 6

107

mohd faisol mansor/chemistry form 4/chapter 6

Electrolysis of Aqueous Solution
Aqueous solution consists of four types of ions. Two ions from the

compound and two ions from the water.
Example:

Molten sodium chloride Sodium chloride solution

Generally, there are 3 factors that may influence the selective of ions
during electrolysis of an aqueous solution

1. Position of ions in the electrochemical series
2. Concentration of ions in the electrolytes
3. Types of electrodes used in the electrolysis

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mohd faisol mansor/chemistry form 4/chapter 6

Position of ions in the electrochemical series (ECS)
The ions that are lower in the ECS will selected to be discharged.

[ Draw the apparatus of electrolysis sodium chloride solution ]

1. State the ion consists in the electrolyte.

2. Which electrode is

a) Cathode =

b) Anode =

3. Which ion will be discharged at

a) Cathode =

b) Anode =

4. State the observation at

a) Cathode =

b) Anode =

5. State the product formed at

a) Cathode =

b) Anode =

6. Write the half equation at

a) Cathode =

b) Anode =

7. Draw the electron flow on the diagram above.

109

Exercise mohd faisol mansor/chemistry form 4/chapter 6
1.
Carbon electrode
Copper(II)
sulphate solution

For the electrolysis of copper(II) sulphate solution,
(a) State all the ions that are present in the electrolyte.

(b) State the ions in (a) which discharged to the
i) anode :
ii) cathode :

(c) Write a half equation for the reaction at the
i) anode :
ii) cathode :

(d) The blue colour of copper(II) sulphate solution fades if the
electrolysis is carried for a long period of time. Explain why.

110

mohd faisol mansor/chemistry form 4/chapter 6

2.
Carbon electrode

Dilute sulphuric
acid

For the electrolysis of dilute sulphuric acid,
a) State all the ions that are present in the electrolyte

b) State the ion in (a) which discharged to
i) anode
ii) cathode

c) Write half equation for the reaction at the
i) anode
ii) cathode

d) Explain why the concentration of dilute sulphuric acid increases
gradually during the electrolysis

3. Base on the answer 1(c) and 2(c), name the process that occur
at the
a) anode
b) cathode

111

mohd faisol mansor/chemistry form 4/chapter 6

Concentration of ions in the electrolytes
If the concentrations of particular ions are high, the ion is selectively

discharged

[ Draw the apparatus of electrolysis concentrated sodium chloride solution ]

1. State the ion consists in the electrolyte.

2. Which electrode is

a) Cathode =

b) Anode =

3. Which ion will be discharged at

a) Cathode =

b) Anode =

4. State the observation at

a) Cathode =

b) Anode =

5. State the product formed at

a) Cathode =

b) Anode =

6. Write the half equation at

a) Cathode =

b) Anode =

7. Draw the electron flow on the diagram above.

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mohd faisol mansor/chemistry form 4/chapter 6

Exercise

1. Carbon
electrode

Dilute Hydrochloric
acid solution

Experiment A

Carbon
electrode

Concentrated
Hydrochloric acid

solution

Experiment B

Diagram above show the apparatus set up for the experiments of
electrolysis using two different concentration of hydrochloric acid.
a) State all the ions that are present in the electrolyte

i) Experiment A :
ii) Experiment B :

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mohd faisol mansor/chemistry form 4/chapter 6

b) State the ion in (a) which discharged to anode and cathode in
i) Experiment A :
ii) Experiment B :

c) Write half equation for the reaction at the anode and cathode in
i) Experiment A :
ii) Experiment B :

d) State the observation occur at cathode and anode in
i) Experiment A :
ii) Experiment B :

e) State the product formed at cathode and anode in
i) Experiment A :
ii) Experiment B :

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mohd faisol mansor/chemistry form 4/chapter 6

Types of electrodes used in the electrolysis

If using the active electrode at anode, ions that are present in the
electrolytes are not discharge. Instead the active electrode will
corrodes and dissolves in the electrolytes.

[ Draw the apparatus of electrolysis silver chloride solution using silver electrodes ]

1. State the ion consists in the electrolyte.

2. Which electrode is

a) Cathode =

b) Anode =

3. Which ion will be discharged at

a) Cathode =

b) Anode =

4. State the observation at

a) Cathode =

b) Anode =

5. State the product formed at

a) Cathode =

b) Anode =

6. Write the half equation at

a) Cathode =

b) Anode =

7. Draw the electron flow on the diagram above.

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mohd faisol mansor/chemistry form 4/chapter 6

Exercise

1. Carbon
electrode

Copper(II) sulphate
solution

Experiment A

Copper
plate

Copper(II) sulphate
solution

Experiment B

Diagram above show the apparatus set up for the experiments of
electrolysis using two different electrodes immersed in copper(II)
sulphate solution.
a) State all the ions that are present in the electrolyte

i) Experiment A :
ii) Experiment B :

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mohd faisol mansor/chemistry form 4/chapter 6

b) State the observation occur at anode and cathode in
i) Experiment A :
ii) Experiment B :

c) Write half equation for the reaction at the anode and cathode in
i) Experiment A :

ii) Experiment B :

d) Explain the observation on the colour of copper(II) sulphate
solution in
i) Experiment A :

ii) Experiment B :

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mohd faisol mansor/chemistry form 4/chapter 6

Electrolysis in Industry

Most common 1) Extraction of Metal
application:
Extraction of aluminium
i) Extraction of metal from aluminium oxide.
ii) Purification of metal
iii) Electroplating

2) Purification of Metal Copper nugget
In purification:
The impure metal is made to be the 118
anode
The cathode is a thin layer of pure metal

3) Electroplating
Electroplating is a process to coat
one metal onto another metal.
The purposes of electroplating
onto metal are:-
i) Make it look more attractive
ii) more resistant to corrosion
In electroplating :
 object to be electroplated
as the cathode
 anode is the metal used for plating
 Electrolyte is a solution of the
compound of the electroplating
metal

mohd faisol mansor/chemistry form 4/chapter 6

Extraction of Metal

[ Draw the apparatus of electrolysis for extraction of aluminium from aluminium oxide ]

1) Ion present in electrolyte =

2) Ion discharged

a) Cathode =

b) Anode =

3) Observation

a) Cathode =

b) Anode =

4) Half equation

a) Cathode =

b) Anode =

5)Function of cryolite, Na3AlF6

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mohd faisol mansor/chemistry form 4/chapter 6

Purification of Metal

[ Draw the apparatus of electrolysis for purification of impure copper ]

1) Ion present in electrolyte =

2) Ion discharged

a) Cathode =

b) Anode =

3) Observation

a) Cathode =

b) Anode =

4) Half equation

a) Cathode =

b) Anode =

5) Colour changes of electrolyte

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mohd faisol mansor/chemistry form 4/chapter 6

Electroplating

[ Draw the apparatus of electrolysis to electroplate key by using copper as electrode ]

1) Ion present in electrolyte =

2) Ion discharged

a) Cathode =

b) Anode =

3) Observation

a) Cathode =

b) Anode =

4) Half equation

a) Cathode =

b) Anode =

5) Colour changes of electrolyte

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mohd faisol mansor/chemistry form 4/chapter 6

Voltaic Cells

 A simple voltaic cell can be made by dipping two different
types of metals in an electrolyte

 Electron flow from one metal to another metal through the
connecting wire in the external circuit.

 More electropositive metal will release electron, thus act as
the negative terminal. Less electropositive metal will accept
electron and act as the positive terminal.

 Continuous flow of electron produces an electric current.

Simple zinc-copper  Zinc more reactive than
Voltaic copper

 Zinc will act as terminal
________________, and
copper will act as
terminal ______________.

 Zinc will release electron
to form Zn 2+.
Half equation:

 Cu 2+ ions from copper(II) sulphate solution receive
electron to form copper metal.
Half equation :

Overall equation:

 The further the distance between the position of two

metals is in ECS the bigger the cell voltage. 122


mohd faisol mansor/chemistry form 4/chapter 6

Different Types of Voltaic Cells

Two types of voltaic cell:
1) Primary cells: non rechargeable cell

Example: Daniell cell, dry cells, alkaline cell
2) Secondary cells: rechargeable cells

Example: Lead-acid accumulator, Nickel-cadmium

Daniell Cell 1

1. Used salt bridge

Salt bridge contain inert
ions or salt that does not

react with electrolyte.

Example:

2. Used porous pot Daniell Cell 2

Porous pot has fine pores 123
that allow ions flow through.

What is the function of salt
bridge and porous pot in

Daniell Cell?

mohd faisol mansor/chemistry form 4/chapter 6

Exercise

1. For the simple voltaic cell that you see at the diagram

Mg Cu
a) State how electricity was produced.
Magnesium sulphate
solution

b) What are the chemical changes that occur at the
magnesium ribbon and the copper plate.

c) Write the half equation for the changes that occur at each
the electrode.

d) What is the direction of electron flow from terminal to
another through the external circuit.

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mohd faisol mansor/chemistry form 4/chapter 6

Electrochemical series (ECS)

Tendency of Tendency of cation
metal to release to receive

electrons to electrons to form
form ions metals
increases
increases

The electrochemical series (ECS) can be constructed by two
method:

a) The potential difference (voltage difference) between pairs of
metal.

b) The ability of metal to displace another metal from its salt solution.

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mohd faisol mansor/chemistry form 4/chapter 6

The potential difference (voltage difference) between pairs of metal.
 The bigger the voltage value the further apart their position.

 The metal act as negative terminal is placed at higher position
in electrochemical series (ECS).

How to determine the positive/negative terminal?

Example: The voltaic cells are constructed as shown in the figure. The
voltmeter reading of the cell I is 1.1 V while that of cell II is
2.5 V.

QP RP

Cell 1 Cell 2

Arrange the metals in descending order in the
electrochemical series.

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mohd faisol mansor/chemistry form 4/chapter 6

The ability of metal to displace another metal from its salt solution
If the M can displace metal N from an aqueous N salt solution, then:

i) Metal M is more electropositive than metal N
ii) Metal M is placed at a higher position than metal N in the ECS
Example: Zinc and copper(II) sulphate sulphate solution

Observation :

Half-equation :

The Important of ECS
ECS can be used to determine:

 The terminal of voltaic cell
 The standard cell voltage
 The ability of a metal to displace another metal from its

salt solution.

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mohd faisol mansor/chemistry form 4/chapter 6

Exercise

1. The diagram shows an electrolytic cell. The left section of the cell
(S) is a source of electricity to drive the right section (T) of the
cell.

Copper

Aluminium Zinc

Aluminium
Sulphate

ST

Zinc Sulphate Copper(II) Sulphate

a) State the change of energy in cell S

b) i) For cell S, state the positive terminal of the cell

ii) Explain your choice for b(i)

c) i) State what has happened at the negative terminal

ii) What process has happened in this electrode

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mohd faisol mansor/chemistry form 4/chapter 6

d) Explain why the color of copper(II) sulphate remain unchanged
e) Determine the anode of cell T
f) Write down the half equation for the cathode in cell T
g) What will happened if the aluminium in cell S is replaced by copper

129

mohd faisol mansor/chemistry form 4/chapter 7

CHAPTER 7

ACIDS AND BASES

Arrhenius Theory A base defined as a
chemical substance
An acid is a chemical that can neutralise an
compound that acid to produce a

produces hydrogen ions, salt and water.
H+ or hydroxonium ions
H3O+ when dissolve in

water.

An alkali is defined as Example
a chemical HCl (g) H2O H+ (aq) + Cl- (aq)
NaOH(s) H2O Na+(aq) + OH-(aq)
compound that
dissolve in water to
produce hydroxide

ions, OH- .

The role of water

In the presence of water an acid will ionise to form hydrogen ion and
alkaline will dissociate into hydroxide ions, OH-.

Therefore, water is essential for the formation of hydrogen ions, H+
that cause acidity and hydroxide ions, OH- that cause alkalinity.

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mohd faisol mansor/chemistry form 4/chapter 7

BASICITY OF AN ACID

Is the number of ionisable hydrogen atoms per molecule of an
acid.

Diprotic Acid Triprotic Acid

Monoprotic Acid

Acid which produces
1 hydrogen ion when
one molecule of an

acid ionises in H2O.

Example: a) CH3COOH? (ethanoic acid) =

b) H2SO4? (sulphuric acid ) =

c) HNO3? (Nitric acid) =

d) H3PO4? (Phosphoric acid) =

STRENGTH OF ACID & ALKALI

131

mohd faisol mansor/chemistry form 4/chapter 7

STRONG ACID WEAK ACID

Strong acid will dissociate or ionize WEAK ALKALI
completely in water to produce
hydrogen, H+ ions.

Degree of dissociation is higher.

Thus, higher concentration of
hydrogen ions in aqueous acid

solution.

Therefore, low pH value of the acid
solution.

STRONG ALKALI

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mohd faisol mansor/chemistry form 4/chapter 7

PHYSICAL PROPERTIES OF ACID & ALKALI

pH Value Conduct Litmus paper
electricity

ACID

Taste Corrosive

Litmus paper

pH Value

ALKALI

Taste

Corrosive Conduct
electricity

133

mohd faisol mansor/chemistry form 4/chapter 7

CHEMICAL PROPERTIES OF ACID & ALKALI

Acids can react with,
i) bases to produce salts and water
eg :
ii) metal to produce salts and hydrogen gas
eg :
iii) metal carbonates to produce salts, carbon dioxide
and water
eg :
iv) alkali to produce salts and water (neutralization)
eg :

Alkali can react with,
i) acid to produce salts and water (neutralization)
eg :
ii) ammonium salt to produce salts, water and ammonia
gas
eg :

134

mohd faisol mansor/chemistry form 4/chapter 7

The concentration of acid and alkali

Concentration are measurement of the quantity of solutes dissolved
in a quantity of solvent.

Grams per dm3 Concentration Moles per dm3
( g dm-3) Unit Conversion ( mol dm-3)
÷ molar mass
Grams per *known as Molarity
dm3 ( g dm-3) × molar mass
Moles per dm3
( mol dm-3)

Example

1. The molarity of a bottle of nitric acid, HNO3 solution is 2.0 mol dm-3.
What is the concentration of the solution in g dm-3?
[RAM: H, 1 ; N, 14 ; O , 16]

135

mohd faisol mansor/chemistry form 4/chapter 7

2. Calculate the molarity of a sodium sulphate, Na2SO4 solution with
a concentration of 28.4 g dm-3. [RAM: O, 16 ; Na, 23 ; S, 32]

Calculating Involving Concentration and Molarity
No of mole = Molarity x Volume ( cm3)
1000
n = MV
1000

1. 5.00 g of copper (II) sulphate is dissolved in water to form 500
cm3 solution. Calculate the concentration of copper (II) sulphate
in g dm-3.

2. A 250 cm3 nitric acid solution contains 0.4 moles. Calculate the
molarity of the nitric acid.

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mohd faisol mansor/chemistry form 4/chapter 7

3. What is the mass of sodium carbonate required to dissolve in
water to prepare 200 cm3 solution contains 50 g dm-3.

4. Calculate the number of moles of ammonia in 150 cm3 of 2 mol
dm-3 aqueous ammonia.

5. Calculate the volume in dm3 of a 0.8 mol dm-3 sulphuric acid
that contains 0.2 mol.

6. 4.0 g sodium carbonate powder, Na2CO3 is dissolved in water
and made up to 250 cm3. What is the molarity of the sodium
carbonate solution. [RAM: C,12;O,16;Na,23]

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mohd faisol mansor/chemistry form 4/chapter 7

7. Dilute hydrochloric acid used in the school laboratories usually
has a concentration of 2.0mol dm-3. Calculate the mass of
hydrogen chloride that found in 250 cm3 of the hydrochloric
acid? [RAM : H,1; Cl,35.5]

8. The concentration of a potassium hydroxide solution is
84.0 g dm-3. Calculate the number of moles of potassium
hydroxide present in 300 cm3 of the solution. [RAM: K,39,H,1 O,16]

138

mohd faisol mansor/chemistry form 4/chapter 7

9. Calculate the number of moles of hydrogen ions present in
200 cm3 of 0.5 mol dm-3 sulphuric acid.

Preparation of Standard Solution

A solution in which its concentration is accurately known is a
standard solution.

Preparation of a solution by dilution method
Adding water to a concentrated solution changes the concentration
of the solution but does not change the amount solutes ( number of

moles)of solution present in solution.

Moles of stock solution = moles of dilute solution
n1 = n2

M1V1 = M2V2

139

mohd faisol mansor/chemistry form 4/chapter 7

Exercise

1. Find the volume of 2.0 mol dm-3 sulphuric acid, H2SO4 needed to
prepare 100 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4.

2. Calculate the volume of a concentrated solution needed to
prepare each of the following dilute solution:

a) 50 cm3 of 0.1 mol dm-3 sodium hydroxide, NaOH solution
from 2.0 mol dm-3 sodium hydroxide, NaOH solution.

b) 100 cm3 of 0.5 mol dm-3 potassium manganate(VII),
KMnO4 solution from 1.0 mol dm-3 potassium
manganate(VII), KMnO4 solution.

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mohd faisol mansor/chemistry form 4/chapter 7

3. Calculate the volume of 2.0 mol dm-3 sulphuric acid, H2SO4
needed to prepare 2.5 dm3 of 0.5 mol dm-3 of the same acid
solution.

4. Calculate the molarity of potassium hydroxide, KOH, solution if
200 cm3 of 2.0 mol dm-3 potassium hydroxide, KOH, solution is
added to 200 cm3 of water.

5. 60 cm3 of 0.5 mol dm-3 sodium hydroxide, NaOH, solution is
diluted with 30 cm3 of water. Calculate the molarity of the
solution produced.

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mohd faisol mansor/chemistry form 4/chapter 7

The pH Values and Molarity

As the molarity of an acid increases, the pH value of the acid
decreases, however the pH value of an alkali increases when the

molarity of the alkali increases.

Neutralisation

Reaction of an acid Acid-base titration An acid of known
and a base that concentration is
produce salt and Titration is a very useful carefully delivered from
water. laboratory technique in burette to completely
neutralise a known
which one solution is volume of an alkali in a
used to analyse another
conical flask.
solution.

Acid-base indicator is The point at which the
used to detect the end colour of the solution
change is called the
of titration.
end point.
Eg: methyl orange,
phenolphthalein and

litmus.

Titration using
phenolphthalein

142

mohd faisol mansor/chemistry form 4/chapter 7

NEUTRALISATION

ACID-BASE
TITRATION

143

mohd faisol mansor/chemistry form 4/chapter 7

Example

1. Write a balanced equation for the neutralization of each of the
following:
a) Sulphuric acid, H2SO4 and barium hydroxide, Ba(OH)2
solution.

b) Nitric acid, HNO3 and calcium hydroxide, Ca(OH)2 solution.

c) Ethanoic acid, CH3COOH and potassium hydroxide, KOH
solution.

Acid-base Indicator

Indicator Colour in Colour in neutral Colour in
alkalis solution acids
Methyl orange
phenolphthalein

litmus

The end-point of neutralisation also can determined by another two
method:-

i) Measurement of pH values by computer
ii) Measurement of electrical conductivity during titration

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mohd faisol mansor/chemistry form 4/chapter 7

Numerical Problem involving Neutralisation

Say the balance equation is product

aA + bB

which,

A = acid
a = no of mole of acid
B = base
b = no of mole of base

MAVA = a and MBVB = b

Therefore MAVA = a

MBVB b

Exercise

1. In an experiment, 25.0 cm3 of a sodium hydroxide solution of
unknown concentration required 26.50 cm3 of 1.0 mol dm-3
sulphuric acid to complete a reaction in titration. Calculate the
molarity of sodium hydroxide.

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mohd faisol mansor/chemistry form 4/chapter 7

2. What is the volume of 0.5 mol dm-3 sulphuric acid, H2SO4 needed
to neutralize 25.0 cm3 of 0.8 mol dm-3 ammonia, NH3 solution?

3. A sample of copper(II) oxide, CuO was found to completely
neutralize 100 cm3 of 0.5 mol dm-3 hydrochloric acid, HCl.
Calculate the mass of the sample. [RAM: O, 16 ; Cu, 64]

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mohd faisol mansor/chemistry form 4/chapter 7

4. The volume of 0.15 mol dm-3 sulphuric acid, H2SO4 required to
completely neutralize 25.0 cm3 of potassium hydroxide, KOH
solution is 30.5 cm3. Calculate the molarity of the potassium
hydroxide, KOH solution.

5. A student dissolved 3.65 g of hydrogen chloride gas, HCl in water
to make 1.0 dm3 of solution. Calculate the volume of a 0.1 mol
dm-3 barium hydroxide, Ba(OH)2 solution required to completely
neutralize 25.0 cm3 of the acid solution. [RAM : H, 1 ; Cl, 35.5]

147

mohd faisol mansor/chemistry form 4/chapter 8

CHAPTER 8

SALTS

A salt is an ionic The salt consists of two
substance produced parts, cation from base
when the hydrogen and anion from acid.

ion of the acid is NaCl
replaced by metal
ion or an ammonium NaOH HCl
(Base) (Acid)
ion.

148