SUCCESS CHEMISTRY

6 In a weak acid solution, a big portion of the Strong and Weak Alkalis SPM
weak acid exists as molecules and only a
’08/P1,
small portion dissociates to ions. ’09/P1

1 A strong alkali is a chemical substance that

dissociates completely (100%) to hydroxide

ions, OH– in water.

2 Examples of strong alkalis are sodium

A concentrated acid does not mean that it is a strong hydroxide and potassium hydroxide.
acid. For example, concentrated ethanoic acid is still a
weak acid. NaOH → Na+ + OH–
Consequently, a dilute acid does not mean that it is a KOH → K+ + OH–
weak acid. For example, dilute hydrochloric acid is still
a strong acid. (The one–way arrow → indicates complete

7 dissociation)
3 A weak alkali is a chemical substance that
7 For two different acids of the same concentration,
dissociates partially (incomplete dissociation)
the acid with the lower pH value is the stronger to hydroxide ions, OH– in water.
4 Examples of weak alkalis are aqueous
acid. ammonia, calcium hydroxide and
8 Partial dissociation of a weak acid results in a magnesium hydroxide.

low H+ ion concentration. NH3 + H2O NH4+ + OH–
9 Strong and weak acids have the same Ca(OH)2 Ca2+ + 2OH–

chemical properties, but the rate of reaction
and electrical conductivity of a weak acid is
lower as shown in Table 7.8.

Table 7.8 Comparison between a strong acid (HCl) (The two–way arrow indicates partial dissociation)
and a weak acid (CH3COOH)

Solution 0.1 mol dm–3 0.1 mol dm–3 5 Partial dissociation of a weak alkali results in
Test HCl CH3COOH a low concentration of OH– ions. Hence, the
pH value of a weak alkali is lower than that of
Universal Red colour, pH~1 Orange red a strong alkali with the same concentration.
Indicator colour, pH~4
6 Table 7.9 shows the comparison of properties
of a strong alkali (sodium hydroxide solution)
with a weak alkali (aqueous ammonia
solution).

Magnesium Hydrogen gas Hydrogen gas Table 7.9 Comparison of properties of a strong alkali
ribbon evolves evolves slowly
vigorously (NaOH) with a weak alkali (NH3)

Solution 0.1 mol dm–3 0.1 mol dm–3

Test NaOH solution aqueous NH3

Calcium CO2 gas evolves CO2 gas evolves Universal Purple colour, Blue colour,
carbonate vigorously slowly Indicator pH~13 pH~10

Electrical Light bulb lights Light bulb lights Electrical Light bulb lights Light bulb lights

conductivity up brightly up dimly conductivity up brightly up dimly

Conclusion • High • Low Conclusion • High • Low
concentration concentration concentration concentration
of H+ ions of H+ ions of OH– ions of OH– ions

• HCl is a strong • CwHea3kCOacOidH is a • NaOH is a • AisqauweoeuaskNH3
acid strong alkali alkali

Acids and Bases 194

Decreasing pH Increasing pH

1 7 14

Strong acids: Weak acids: Weak alkalis: Strong alkalis:

• Complete ionisation • Partial ionisation • Partial ionisation • Complete ionisation

• High H+ ion concentration • Low H+ ion concentration • Low OH– ion concentration • High OH– ion concentration

• pH value: 1–2 • pH value: 3–6 • pH value: 8–12 • pH value: 13–14

To measure the pH values of solutions with the same SPM
concentration
’10/P1

Apparatus Conclusion Activity 7.4 7
50 cm3 beaker and pH meter.
1 Acids have pH values of less than 7.
Materials 2 Different acids with the same concentration have

0.1 mol dm–3 hydrochloric acid, 0.1 mol dm–3 different pH values. The pH of hydrochloric
ethanoic acid, 0.1 mol dm–3 aqueous ammonia and acid is lower than ethanoic acid of the same
0.1 mol dm–3 sodium hydroxide solution. concentration.
3 Alkalis have pH values of more than 7.
Procedure 4 Different alkalis with the same concentration
have different pH values. The pH of sodium
1 About 15 cm3 of 0.1 mol dm–3 hydrochloric acid hydroxide solution is higher than aqueous
is placed in a small beaker. ammonia of the same concentration.

2 The probe of a pH meter is rinsed with distilled Discussion
water.
1 Hydrochloric acid, HCl, and ethanoic acid,
3 The pH meter probe is then immersed in the acid CH3COOH, are both acidic as their pH values
in the beaker. The reading registered on the pH are less than 7.
meter after it has stabilised is recorded.
2 However, the pH value of HCl is less than that
4 Steps 1 to 3 of the experiment are repeated using of CH3COOH with the same concentration
ethanoic acid, ammonia solution and sodium indicating that the concentration of hydrogen
hydroxide solution to replace the hydrochloric ions in HCl is higher than that in CH3COOH.
acid.
3 HCl is an example of a strong acid which
Results undergoes complete ionisation. CH3COOH is an
example of a weak acid which undergoes partial
Solution of pH Inference ionisation.
0.1 mol dm–3 value
4 Ammonia, NH3, and sodium hydroxide solution
Hydrochloric 1 • High concentration are both alkaline as their pH values are more
acid, HCl of H+ ions than 7.

Ethanoic acid, • A strong acid 5 However, the pH value of NaOH is more
CH3COOH than that of NH3 with the same concentration
3 • Low concentration of indicating that the concentration of hydroxide
H+ ions ions in NaOH is higher than that in NH3.

• A weak acid 6 NaOH is an example of a strong alkali which
undergoes complete ionisation. NH3 is an
Aqueous 10 • Low concentration of example of a weak alkali which undergoes
partial ionisation.
ammonia, NH3 OH– ions
• A weak alkali 7 The strength of an acid (strong or weak) and the
strength of an alkali (strong or weak) depends on
Sodium 13 • High concentration the degree of dissociation.
hydroxide, of OH– ions
NaOH
• A strong alkali

195 Acids and Bases

7 3 ’03 7.2

Information about two solutions is given below: 1 Six solutions A, B, C, D, E and F with concentration
of 1.0 mol dm–3, have pH values as shown in the
Concentration of nitric acid = 1.0 mol dm–3 table below:
Concentration of carbonic acid = 1.0 mol dm–3
Solution A B C D E F
Which of the following statements are true about
the two solutions given above? pH values 13 7.0 10 4.0 1.0 3.5
I Nitric acid is a stronger acid than carbonic acid.
II The pH value of nitric acid is higher than carbonic (a) Which of the above solutions has
(i) the highest concentration of H+ ions?
acid. (ii) the highest concentration of OH– ions?
III The degree of dissociation of nitric acid in
(b) Which of the above solutions is
water is higher than that of carbonic acid. (i) a strong acid? (iii) a weak acid?
IV The concentration of H+ ions in nitric acid is (ii) a strong alkali? (iv) a weak alkali?

higher than that in carbonic acid. (c) Which of the above solutions may be
A I and II only (i) sodium chloride solution?
B III and IV only (ii) hydrochloric acid?
C I, III and IV only (iii) aqueous ammonia?
D I, II, III and IV (iv) sodium hydroxide solution?

Comments 2 Using suitable examples, explain the terms strong
Nitric acid is a stronger acid (I correct), has a acid and weak acid. Predict the difference in pH
higher degree of dissociation (III correct) and values of the two acids with the same concentration.
hence a higher degree of H+ ions concentration (IV
correct) but a lower pH value than carbonic acid, 3 The degree of dissociation of ethanoic acid is
which is a weak acid (II incorrect). higher than that of propanoic acid but is lower
Answer C than that of methanoic acid.
(a) Arrange the above three acids in ascending
4 ’07 order of the strength of acidity.
(b) If the pH value of 1.0 mol dm–3 ethanoic acid
Which of the following statements describe a is 3, predict the pH value of 1.0 mol dm–3
strong alkali? methanoic acid and propanoic acid.
I Has a high pH value
II Ionises completely in water 7.3 Concentration of Acids
III Has a high concentration of hydroxide ions and Alkalis
IV Exists as molecules in water
The Meaning of Concentration and
A I and II only Molarity, and Their Relationship
B III and IV only
C I, II and III only 1 A solution is formed when a solute dissolves
D I, II, III and IV in a solvent.
solute + solvent → solution
Comments For example, when sodium hydroxide (solute)
A strong alkali ionises completely in water to dissolves in water (solvent), sodium hydroxide
produce a high concentration of hydroxide ions in solution is formed.
water and hence a high pH value. (I, II and III are
correct) 2 Concentration and molarity are measurements
A strong alkali exists as ions in water. (IV is of the amount of solutes dissolved in a given
incorrect) volume of solvent when a solution is formed.
Answer C
3 The amount of a solute can be measured in
the unit of ‘gram’ or ‘mole’. The quantity of a
solution produced is usually measured in the
unit of volume, dm3.

Acids and Bases 196

4 The concentration of a solution is the mass Solution
(in grams) or the number of moles of solute
dissolved in a solvent to form 1.0 dm3 (1000 Mass of copper(II) sulphate = 5.00 g Convert
cm3) of solution. Hence the concentration of Volume of solution = 500 cm3 volume from
cm3 to dm3
a solution can be defined in two ways:
= —15—0—0—0—00 dm3 = 0.5 dm3

Hence, concentration of copper(II) sulphate solution
Concentration = —M——a—s—s—o——f —s—o—l—u—te——d—i—s—s—o—lv—e—d——(—g—) = —0—5.5—.0—d0—mg—3
Volume of solution (dm3) = 10.0 g dm–3 Concentration (g dm–3)
(g dm–3) = —M——a—s—s—o—f—s—o—l—u—te——d—i—s—s—o—lv—e—d——(g—)
Volume of solution (dm3)
Number of moles of

Concentration = ————————s—o—l—u—te——(—m——o—l—)——————
(mol dm–3) Volume of solution (dm3)
2
7

5 For example, What is the mass of sodium carbonate required to
• a 23.0 g dm–3 NaOH solution has 23.0 g of dissolve in water to prepare a 200 cm3 solution that
NaOH in 1.0 dm3 solution. contains 50 g dm–3?
• a 0.5 mol dm–3 NaOH solution has 0.5 mol
of NaOH in 1.0 dm3 solution. Solution

6 Concentration in terms of mol dm–3 is more Volume of solution = 200 cm3
commonly known as molarity. In chemistry,
the measurement of concentration in mol = —12—00—0—00 dm3 = 0.2 dm3 Convert volume
dm–3 (molarity) is more useful because all from cm3 to dm3
changes in chemical reactions occur in terms
of moles. C oncentration = —m—a—vs—os—luo—mf—N—e—ao—2fC—s—Oo—l3u—dt—ii—oss—no—(l—dv—me—d3—)(—g—)
(g dm–3)
7 Concentration (in g dm–3) can be converted
SPM to molarity by dividing concentration (in Mass of Na2CO3 required
’08/P1 g dm–3) by the molar mass. The molar mass is = 50 g dm–3  0.2 dm3 Mass = Concentration (g dm–3) 
= 10 g Volume of solution (dm3)
the mass of 1 mol of substance.

Molarity (mol dm–3) = C——o—n—c—e—n—t—r—a—ti—o—n——(—g——d—m——–—3) 3
Molar mass (g mol–1)
Calculate the number of moles of ammonia in 150
cm3 of 2 mol dm–3 aqueous ammonia.

Solution M = molarity, V = volume in cm3

The relationship between the number of mols with N umber of moles = —1—M0—0V—0
molarity, M and volume, V can be represented by the
formula below: Number of moles of ammonia = 2  —11—05—00—0 = 0.3
Number of moles = —1——M—0——0—V——0—
where M = molarity of solution (mol dm–3) 4
V = volume of solution (cm3).
A 250 cm3 solution contains 0.4 mol of nitric acid.
Calculations Involving Concentrations Calculate the molarity of the nitric acid.
and Molarity
Solution
1 N umber of moles = —1—M0—0V—0 = M  —12—05—00—0

5.00 g of copper(II) sulphate is dissolved in water to M olarity of nitric acid, M = —0—.4——2—5—10—0—0—0
form 500 cm3 solution. Calculate the concentration = 1.6 mol dm–3
of copper(II) sulphate solution in g dm–3.

197 Acids and Bases

5 M olarity = —C—Mo—n—oc—lae—rn—mt—ra—at—siso—n(—g—(g—m—do—ml—–1–—)3)

Calculate the volume in dm3 of 0.8 mol dm–3 = —11—06—6 Convert g dm–3
sulphuric acid that contains 0.2 mol of H2SO4. to mol dm–3

Solution = 0.15 mol dm–­3

Number of moles = —1M—0—0V—0 Number of moles

Volume of sulphuric acid = —0—.8—0—m.2—o—ml—od—lm——–3 8

= 0.25 dm3 Molarity, M The concentration of a potassium hydroxide
solution is 84.0 g dm–3. Calculate the number of
7 moles of potassium hydroxide present in 300 cm3
of the solution.
6 [Relative atomic mass: H, l; O, 16; K, 39]

Dilute hydrochloric acid used in school laboratories Solution
usually has a concentration of 2.0 mol dm–3. Molar mass of KOH = 39 + 16 + 1 = 56
Calculate the mass of hydrogen chloride in 250 cm3
of the hydrochloric acid? Molarity of KOH Molarity (mol dm–3)
[Relative atomic mass: H, l; Cl, 35.5] = —85—46—.0 = C——o—n—c—e—n—t—ra——ti—o—n—(—g—d——m—–—3—)
= 1.5 mol dm–3 Molar mass (g mol–1)
Solution

Number of moles of HCl Number of moles of KOH

= —2—.—01—0—0—20—5—0 Number of moles = —1M—0—0V—0 = —1—.5—1—0—0—30—0—0 N umber of moles = —1M—0—0V—0
= 0.45
= 0.5

Molar mass of HCl = 1 + 35.5 9
= 36.5 g mol–1
Mass of HCl = 0.5  36.5 g Calculate the number of moles of hydrogen ions
= 18.25 g present in 200 cm3 of 0.5 mol dm–3 sulphuric acid.
Mass = Number of moles  Molar mass

Solution

7 Number of moles of H2SO4
= —0—.5—1—0—0—20—0—0
4.0 g of sodium carbonate powder, Na2CO3, is Number of moles = —1—M0—0V–—0
dissolved in water and made up to 250 cm3. = 0.1
What is the molarity of the sodium carbonate
solution? H2SO4 → 2H+ + SO42–
[Relative atomic mass: C, 12; O, 16; Na, 23]
Sulphuric acid is a diprotic acid, which means 1 mol
Solution of sulphuric acid will produce 2 mol of H+ ions.
Hence, 0.1 mol of sulphuric acid will produce
Volume of sodium carbonate solution 0.1  2 = 0.2 mol of H+ ions.

= 250 cm3 = 0.25 dm3 Convert volume

Concentration = ———M——a—s—s—(g—)— — from cm3 to dm3
Volume (dm3)
Preparation of Standard Solutions
= —04—.2.—05 Convert mass to
concentration 1 A standard solution is a solution with a
known concentration.
=16 g dm–3
2 A volumetric flask (also known as standard
Molar mass of Na2CO3 = (2  23) + 12 + (3  16) flask) is an apparatus with a known volume.
= 106 g mol–1 Examples are: 100 cm3, 200 cm3, 250 cm3,
500 cm3 and 1000 cm3.

Acids and Bases 198

3 Volumetric flasks are used to prepare standard The steps involved in the preparation of a Activity 7.5 7
solutions. Beakers are not suitable for this standard solution
purpose because volumes measured by
beakers and measuring cylinders are not very 1 Calculate the mass (m g) of the chemical
accurate. required to prepare v cm3 of solution where
v is the volume of the volumetric flask.
4 A volumetric flask can measure the volume of
a liquid accurately, up to one decimal point. 2 Weigh out the exact mass (m g) of the
chemical accurately in a weighing bottle
Figure 7.10 A 100 cm3 volumetric flask using an electronic balance.

3 Dissolve m g of the chemical in a small
amount of distilled water.

4 Transfer the dissolved chemical into the
volumetric flask.

5 Add enough water until the graduation
mark.

To prepare 100 cm3 of 2.0 mol dm–3 aqueous sodium SPM
hydroxide solution
’06/P2
Q4

Apparatus 4 Using a filter funnel and
a glass rod, the dissolved
Electronic balance, 100 cm3 volumetric flask, filter sodium hydroxide is
funnel, dropper and washing bottle. transferred to a 100 cm3
volumetric flask.
Materials
Sodium hydroxide solid and distilled water. 5 The small beaker, the
weighing bottle and
Procedure the filter funnel are all
1 The mass of sodium hydroxide (NaOH) required rinsed with distilled
water and the contents
to prepare 100 cm3 of 2.0 mol dm–3 aqueous are transferred into the
sodium hydroxide is calculated as follows: volumetric flask.
Mass of NaOH required
= Number of moles  molar mass of NaOH 6 Distilled water is then
= (—1M—0—0V—0 )  (23 + 16 + 1) added slowly until
the water level is near
= —2—.0—1—0—0—01—0—0  40 the level mark of the
= 8.0 g volumetric flask. A
dropper is then used
2 8.0 g of sodium hydro­ to add water drop by
xide, NaOH solid is drop to finally bring the
weighed accurately in volume of solution to
a weighing bottle using the 100 cm3 graduation
an electronic balance. mark.

3 Sodium hydroxide solid 7 The volumetric flask is
is transferred to a small closed with a stopper.
beaker. Sufficient dis­ The volumetric flask
tilled water is added to is then shaken several
dissolve all the solid times to mix the solution
sodium hydroxide. completely. The solution
prepared is 100 cm3 of
2.0 mol dm–3 aqueous
sodium hydroxide.

199 Acids and Bases

The Correct Techniques Used in the Preparation 2 When a solution is diluted, the volume of
solvent increases but the number of moles of
of Standard Solution solute remains constant. Hence the concentration
of the solution decreases.
1 The chemical is weighed in a weighing bottle
and not on a piece of filter paper. Some 3 If a solution with volume of V1 cm3 and molarity
chemicals such as sodium hydroxide can of M1 mol dm–3 is diluted to become V2 cm3,
absorb moisture from the air and become wet the new concentration of the diluted solution,
and may stick to paper. M2 mol dm–3 can be determined as follows:

2 After transferring the dissolved solute to the sNoulumtebebreofofrme doilleustioofn = —M1—0—1—0V—01
volumetric flask, the weighing bottle, the
Activity 7.6 small beaker that contained the solution Number of mdiloulteisoonf = —M1—0—20—V—02
7 as well as the filter funnel are rinsed with solute after
distilled water. The content is then transferred
to the volumetric flask to ensure that all the However, the number of moles of solute
mass of the chemical that has been weighed is before dilution is the same as the number of
transferred to the volumetric flask. moles of solute after dilution,

3 The addition of distilled water to the —M1—0—10—V—01 = —M1—0—20—V0—2 or M1V1 = M2V2
volumetric flask must be carried out carefully
so that the level of the solution does not The steps involved in the preparation of a
exceed the graduation mark of the volumetric standard solution by dilution
flask. The last few cm3 of water should be
added drop by drop using a dropper. 1 The volume of the stock solution, V1
required is calculated.
4 A volumetric flask and not a beaker must be
used to prepare a standard solution because a 2 The required volume of stock solution is
volumetric flask is calibrated to a high degree pipetted into a volumetric flask.
of accuracy.
3 Enough distilled water is added to the
5 The volumetric flask is stoppered after the volumetric flask to the required volume, V2.
standard solution is prepared to prevent the
evaporation of water which can change the The formula used in dilution is M1V1 = M2V2
concentration of the solution prepared. where M1 = Initial molarity of solution
M2 = Final molarity of solution
Preparation of a Solution with a Specified V1 = Initial volume of solution
V2 = Final volume of solution
Concentration Using the Dilution Method

1 Dilution is a process of diluting a concentrated
solution by adding a solvent such as water to
obtain a more diluted solution.

To prepare 100 cm3 0.2 mol dm–3 sodium hydroxide from a
2.0 mol dm–3 sodium hydroxide solution by the dilution method

Apparatus M1V1 = M2V2
100 cm3 volumetric flask, 10 cm3 pipette, pipette 2.0 3 V1 = 0.2 3 100
filler, filter funnel, dropper and washing bottle.
V 1 = —0—.2——13—.0—1—0—0 = 10 cm3
Materials
2.0 mol dm–3 sodium hydroxide solution and distilled where M1 = Initial molarity of alkali
water. M2 = Final molarity of alkali
V1 = Initial volume of alkali
Procedure V2 = Final volume of alkali
(A) To calculate the volume of sodium hydroxide

solution that is required for dilution

Acids and Bases 200

(B) To prepare 100 cm3 0.2 mol dm–3 sodium 3 The flask is stoppered and is inverted several
hydroxide by the dilution method times to mix the solution. The solution prepared
is 0.2 mol dm–3 sodium hydroxide solution.
1 Using a pipette and a pipette filler, 10.0 cm3
of 2.0 mol dm–3 sodium hydroxide solution is Conclusion
transferred to a 100 cm3 volumetric flask. A 0.2 mol dm–3 sodium hydroxide solution can be
prepared by diluting 10 cm3 of 2.0 mol dm–3 of
2 Using a washing bottle, distilled water is added to the sodium hydroxide solution to 100 cm3.
alkali in the volumetric flask until near the graduation
mark. A dropper is then used to add water slowly in
the volumetric flask up to the graduation mark.

Relationship between pH Values and (a) The higher the degree of dissociation of Experiment 7.3 7
Molarities of Acids or Alkalis an acid, the lower the pH value of the acid.

1 The pH value of an acid or an alkali depends (b) The higher the degree of dissociation of an
on two factors, that is alkali, the higher the pH value of the alkali.
(a) degree of dissociation and
(b) molarity or concentration. 3 For an acid or alkali, its pH value depends on

2 At the same concentration, the pH value of the molarity of the solution.
an acid or an alkali depends on the degree of (a) The higher the molarity of an acid, the
dissociation.
lower the pH value.
(b) The higher the molarity of an alkali, the

higher the pH value.

7.3 SPM
’09/P3

To investigate the relationship between pH values and the molarity of an acid or an alkali

Problem statement 0.001 mol dm–3 hydrochloric acid as shown in

What is the relationship between pH values and the Figure 7.11.
molarity of an acid or an alkali?
3 The pH value
Hypothesis
shown on the pH
(A) When the molarity of an acid increases, its pH
value decreases. meter is recorded.

(B) When the molarity of an alkali increases, its pH 4 The pH values
value increases.
of hydrochloric
Variables
acid solutions with Figure 7.11 Using a pH
(a) Manipulated variable : Molarity of acid or alkali different molarities meter to measure the pH
(b) Responding variable : pH values are measured one value
(c) Constant variable : Type of acid or alkali used by one in dry

Apparatus beakers as in Steps 1 to 3.

pH meter, 100 cm3 beakers and 100 cm3 measuring 5 The experiment is repeated using sodium
cylinders.
hydroxide solutions with different molarities to

replace the hydrochloric acid.

Materials Results

Hydrochloric acids and sodium hydroxide solutions Molarities of 0.001 0.01 0.05 0.08 0.10
with molarities of 0.001 mol dm–3, 0.01 mol dm–3, HCl (mol dm–3)
0.05 mol dm–3, 0.08 mol dm–3 and 0.10 mol dm–3.
pH values 3.0 2.0 1.3 1.1 1.0
Procedure

1 30 cm3 of 0.001 mol dm–3 hydrochloric acid is Molarities of 0.001 0.01 0.05 0.08 0.10
put in a dry beaker. NaOH (mol dm–3)

2 The probe of a pH meter that has been washed pH values 11.0 12.0 12.7 12.9 13.0

with distilled water is immersed in 30 cm3 of the

201 Acids and Bases

Discussion 4 The graph of pH values versus molarity of
an alkali is an increasing curve as shown in
1 The graph of pH values versus molarity of Figure 7.13.
an acid is a decreasing curve as shown in
Figure 7.12. 5 When the molarity of an alkali increases, the
concentration of OH– ions in the alkali increases
2 When the molarity of an acid increases, the and the solution becomes more alkaline. Hence
concentration of H+ ions in the acid increases the pH value increases.
and the solution becomes more acidic. Hence the
7 pH value decreases.

3 From the graph, we can predict
(a) the pH value, if the concentration of H+ ions
of the solution is known.
(b) the concentration of H+ ions in the solution, if
the pH value is known.

Figure 7.12 Graph of pH versus molarity of HCl Figure 7.13 Graph of pH versus molarity of NaOH

Conclusion

1 The higher the molarity of hydrochloric acid,
the lower the pH value. The pH value of an acid
decreases with the increase in molarity.

2 The higher the molarity of an alkali, the higher
the pH value. The pH value of an alkali increases
with the increase in molarity.
The hypothesis is accepted.

Numerical Problems Involving Molarity M 2 = —0—.8—1—0—0—20—50 = 0.2 mol dm–3
of Acids and Alkalis
Molarity of potassium hydroxide solution produced
1 The molarity of an acid will change when is 0.2 mol dm–3.
(a) water is added to it,
(b) an acid of a different concentration is 11
added to it,
(c) an alkali is added to it. What is the volume of distilled water required to be
added to 60 cm3 of 2.0 mol dm–3 sulphuric acid to
10 produce a 0.3 mol dm–3 sulphuric acid?

What is the molarity of the potassium hydroxide Solution Calculate the total volume
solution produced when 750 cm3 of distilled water M1V1 = M2V2 of acid produced
is added to 250 cm3 of potassium hydroxide solution
of 0.8 mol dm–3? 2.0  60 = 0.3  V2

Solution V2 = —2—.—00—.—3——6—0
Final volume of alkali, V2 = 250 cm3 + 750 cm3 = 400 cm3
= 1000 cm3
Volume of distilled water needed to be added to
M1V1 = M2V2 60 cm3 H2SO4 = (400 – 60) cm3 = 340 cm3.
0.8  250 = M2  1000

Acids and Bases 202

12

500 cm3 of a solution that contains 2.0 mol sodium Two units for concentration
hydroxide is added to 1500 cm3 of a solution that
contains 4.0 mol sodium hydroxide. Calculate the
molarity of the sodium hydroxide solution produced.

Solution g dm–3  molar mass mol dm–3
 molar mass
Total number of moles of NaOH

= 2.0 + 4.0 Calculate the total number of moles determined by determined by
= 6.0 of alkali from the two solutions.

Total volume of NaOH —V——o———l—u——M—m———a—e—s——s—o——of———sf——os——ol—u——l—ut—i—ot——e—n——(——(g——d—)—m————3) N———V—o——o.——l—ou——f—m——m——e——o——ol—e—f——s—s——oo——l—fu——s—t—oi—o——l—un——t——(e——d—(—m—m———3—o—)—l)

= (500 + 1500) cm3 7

= 2000 cm3 Calculate the total volume of alkali
= 2 dm3 from the two solutions

Molarity of = —N—V—u—mo—lbu—em—r—eo—f(—dm—m—o3—l)e— s 7.3
NaOH produced
1 Calculate the mass of potassium hydroxide required
= —6—.0——m—o—l to produce
(a) 2.0 dm3 of 46.4 g dm–3 solution
2 dm3 (b) 100 cm3 of 2.0 mol dm–3 solution
[Relative atomic mass: H, 1; O, 16; K, 39]
= 3.0 mol dm–3
2 2.12 g of sodium carbonate is dissolved in 500 cm3
13 of distilled water. What is the concentration of the
solution in
200 cm3 of 2.0 mol dm–3 hydrochloric acid is added to (a) g dm–3?
300 cm3 of 0.5 mol dm–3 hydrochloric acid. Calculate (b) mol dm–3?
the molarity of the hydrochloric acid produced. [Relative atomic mass: C, 12; O, 16; Na, 23]

Solution 3 The concentration of sodium hydroxide solution is
8.0 g dm–3.
Number of moles in 200 cm3 of 2.0 mol dm–3 HCl (a) What is the molarity of the solution?
(b) What is the molarity of the solution produced
= —1M—0—0—V0 = —2—.—01—0—0—20—0—0 when 100 cm3 of distilled water is added to
100 cm3 of this solution?
= 0.4 Calculate the total (c) What is the molarity of the solution produced
number of moles of when 20 cm3 of 2.0 mol dm–3 sodium hydroxide
Number of moles in 300 cm3 acid from the two solution is added to 100 cm3 of this solution?
of 0.5 mol dm–3 HCl solutions. [Relative atomic mass: H, 1; O, 16; Na, 23]

= —1M—0—0V—0 = —0—.5—1—0—0—30—0—0 7.4 Neutralisation

= 0.15 The Meaning of Neutralisation and the
Equation for Neutralisation
Total number of moles of HCl = 0.4 + 0.15
= 0.55 1 Neutralisation is the reaction between an
acid and a base to produce salt and water
Total volume of solution = (200 + 300) cm3 only.

= 500 cm3 Calculate the total volume of 2 In a neutralisation reaction, the acidity of an
= 0.5 dm3 acid from the two solutions. acid is neutralised by an alkali. At the same
time, the alkalinity of the alkali is neutralised
Molarity of = —N——u—m—Vb—oe—rl—uo—mf—me——oo—fle—Hs—Co—fl—H——C—l by the acid. Salt and water are the only
HCl produced products of neutralisation.

= —0—0.—.55—5—dm—m—o—3l = 1.1 m ol dm–3


203 Acids and Bases

3 Some examples of neutralisation reactions are The ionic equation for neutralisation between
as follows: strong acids and strong alkalis is

acid base salt water H+(aq) + OH–(aq) → H2O(l)

HCl + NaOH → NaCl + H2O 7 In a neutralisation reaction, H+ ions from
H2SO4 + CuO → CuSO4 + H2O the acid react with OH– ions from the base to
2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O produce water. The pH value for water is 7,
and a neutral condition is achieved.
4 In the reaction between ammonia and acid
such as hydrochloric acid, ammonium salt is 5 ’03
formed.
Which of the following pairs of compounds will
7 NH3 + HCl → NH4Cl react in a neutralisation reaction?
I Hydrochloric acid and potassium hydroxide
Although there is no water formed in the II Sulphuric acid and solid copper(II) oxide
III Nitric acid and solid calcium carbonate
above equation, in actual fact there is a little IV Hydrochloric acid and zinc metal

water formed together with NH4Cl. This is A I and II only
because a portion of ammonia, NH3 exists as B III and IV only
NH4+ ions and OH– ions in aqueous solution. C I, III and IV only
OH– ions and H+ ions (from acid) react to D I, II, III and IV

5 produce water, H2O. dissociate to form free Comments
Acids, bases and salts Neutralisation is a reaction between an acid and a
base to form salts and water only. Reactions in I
ions. Only water exists as molecules. and II are neutralisation because the two reactants
react to form salts and water only. Reaction III is
For example, not neutrali­s­a­tion because carbon dioxide gas is
formed in addition to salt and water. Reaction IV is
H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → not neutrali­sa­tion because hydrogen gas is formed
from HCl from NaOH in addition to salt.

Na+(aq) + Cl–(aq) + H2O(l) Answer A
water

from NaCl molecule

6 Since Na+ ions and Cl– ions do not undergo
any changes in the reaction, these ions can be
omitted from the equation. Thus, the equation
can be simplified.

Applications of Neutralisation in Daily Life
Neutralisation is used in various fields such as agriculture, health and industries.

In agriculture SPM

’09/P1,
’10/P1

1 Controlling the acidity of water is important

in the rearing of freshwater fish and prawns.

Lime (consisting of calcium oxide, CaO)

which produces calcium hydroxide in water is

used to control the acidity in aqua farming.

2 Plants do not grow well in acidic soil or basic

soil. Lime or calcium carbonate is used to

neutralise acidic soil. Basic soil is treated

with compost which can release acidic gas to Calcium oxide and calcium hydroxide are used to

neutralise the alkali in basic soil. neutralise acidic soil

Acids and Bases 204

In health SPM

’08/P1

Toothpastes contain magnesium hydroxide to produced. This will prevent the corrosion of 7
neutralise acids in teeth teeth enamel.
2 Antacids are medicine which contain bases
1 Food trapped in gaps between teeth decompose such as magnesium hydroxide, aluminium
into organic acids by bacteria. An alkaline hydroxide, calcium carbonate and calcium
compound such as magnesium hydroxide bicarbonate. Magnesia milk contains
in toothpastes neutralises the organic acids magnesium hydroxide. Antacids and milk of
magnesia are used to neutralise the excess
hydrochloric acid in the stomachs of
gastric patients.
3 Alkaline creams or baking powder are applied
to cure bee stings and ant bites which are
acidic. Vinegar which contains ethanoic acid
is used to cure alkaline wasp stings.

In industries

1 Bacteria in latex produces organic acids which Acidic gas emitted by factories must be removed
coagulate latex. Ammonia is used to neutralise before being discharged

the organic acids produced by bacteria to
prevent coagulation, so that latex can remain
in the liquid state.
2 Calcium carbonate is used as a base to remove
acidic gas such as sulphur dioxide emitted by

power stations and industries.
3 Effluent from factories which is acidic is

treated with lime, which will neutralise the

acids in it before being discharged.
4 Neutralisation reaction is also used in the

industry to produce manufactured products

such as fertilisers, soaps and detergents.

Acid-base Titration (a) The use of acid–base indicators such as
methyl orange, phenolphthalein and
1 Titration is a quantitative analysis that involves litmus paper which changes colour at the
the gradual addition of a chemical solution end point.
from a burette to another chemical solution of
known quantity in a conical flask. (b) Measurement of the pH values during
titration using a computer interface.
2 In acid-base titration, the volume of the alkali
is measured using a pipette and transferred 5 Titration technique can be used to determine
into a conical flask. The acid solution from a the concentration of an acid (or an alkali).
burette is then added slowly to the alkali in
the conical flask until neutralisation occurs. The Use of Acid-base Indicators in
Titrations
3 The end point of a titration is when neutralisation
occurs, that is, when the acid has completely 1 Acid–base indicators are chemicals that show
neutralised the alkali. different colours when the pH value of the
solution changes.
4 Since both the reactants (acid and alkali) and
the products formed (salt and water) are all 2 Table 7.10 shows the colour of three common
colourless, the end point of neutralisation is types of indicators at different pH values.
determined as follows:

205 Acids and Bases

Table 7.10 Colours of three common types of 3 The pH meter records changes in pH values
indicators in alkaline, neutral and during titration. The information is linked to
acidic conditions a computer by the interface. The change of
pH values along the progress of titration is
Indicator Colour in Colour in Colour in displayed on the computer screen.
alkali neutral acid
solution In computer interface, the pH is displayed on a screen
4 A graph of pH value versus volume of alkali
Methyl orange Yellow Orange Red
added is shown as in Figure 7.14. It is found
Phenolphthalein Light pink Colourless Colourless that the pH value of the solution changes
sharply at the end point of neutralisation.
Litmus Blue Purple Red The neutralisation point can be determined
from the midpoint of the sharp pH change.
Activity 7.7 3 Methyl orange shows a yellow colour
7 Figure 7.14 Graph of pH value versus volume of
in an alkaline solution. At the point of alkali added

neutralisation, the colour changes from yellow
to orange.
4 Phenolphthalein shows a light pink in an alkaline
solution. The first drop of acid that decolourises
the light pink colour of phenolphthalein

indicates the end point of titration.

The Use of Computer Interface in
Titration

1 When an alkaline solution is added slowly
from a burette to an acid in a conical flask,
the pH value of the mixture solution increases
slowly.

2 In titrations using a computer interface, the
probe of a pH meter, immersed in the solution
to be titrated, is connected to the pH module
of the computer interface.

To find the end point of an acid-base titration during SPM
neutralisation using an acid-base indicator
’10/P2

Apparatus 3 A 50 cm3 burette is rinsed with distilled water and

25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, then rinsed with a little of the sulphuric acid.
retort clamp, conical flask, filter funnel and white tile. 4 The burette is then filled with sulphuric acid

Materials and is clamped to a retort stand. The initial
burette reading is recorded.
Sulphuric acid of unknown concentration, 1.0 mol dm–3 5 The conical flask containing 25 cm3 of potassium
potassium hydroxide and methyl orange.
hydroxide is placed below the burette. A piece
Procedure
of white tile is placed below the conical flask
1 A clean 25 cm3 pipette is rinsed with distilled
water and then rinsed with a little of the for clearer observation of the colour change
potassium hydroxide solution.
(Figure 7.15).
2 25 cm3 of 1.0 mol dm–3 potassium hydroxide is 6 Sulphuric acid is added slowly from the burette
transferred using the pipette to a clean conical
flask. Three drops of methyl orange indicator to the potassium hydroxide solution in the
are added to the alkali and the colour of the conical flask while swirling the flask gently.
solution is noted. 7 Titration is stopped when the methyl orange
changes colour from yellow to orange. The
final burette reading is recorded.

Acids and Bases 206

8 Steps 1 to 7 are repeated until accurate titration = —1—9—.9—5——+2—2—0—.—0—5 = 20.00 cm3
values are obtained, that is, until the difference Hence, 20.00 cm3 of H2SO4 is required to
in the volumes of sulphuric acid used in two completely neutralise 25.0 cm3 of 1.0 mol dm–3
consecutive experiments is less than 0.10 cm3. KOH.

Figure 7.15 Titration of sulphuric acid with Discussion Activity 7.8 7
potassium hydroxide
1 In this experiment, the pipette has to be rinsed with
Results potassium hydroxide solution so that water droplets
on the inner wall of the pipette do not dilute the
Volume of sulphuric acid Rough Accurate concentration of the potassium hydroxide used.

Final burette reading (cm3) 21.00 40.95 20.15 2 The burette is rinsed with sulphuric acid so that
water droplets at the inner wall of the burette do not
Initial burette reading (cm3) 0.00 21.00 0.10 dilute the concentration of the sulphuric acid used.

Volume of sulphuric acid 21.00 19.95 20.05 3 The conical flask does not need to be rinsed with
used (cm3) potassium hydroxide so that the volume of the
potassium hydroxide in the conical flasks will
Conclusion accurately be 25.0 cm3. Otherwise, droplets of
potassium hydroxide in the conical flask may
1 The volume of sulphuric acid used is calculated cause the volume of potassium hydroxide to
as follows: exceed 25.0 cm3.
Volume of sulphuric acid used
= Final burette reading – Initial burette reading 4 The end point of titration is when the colour of
the indicator changes sharply. The colour of
2 Average volume of sulphuric acid used methyl orange is yellow in potassium hydroxide
solution (because pH > 7). At the end point,
the colour of methyl orange changes to orange
(pH = 7). If methyl orange changes to a red
colour, excess sulphuric acid has been added.

5 In acid-base titrations, only 2 or 3 drops of
indicator should be used. This is because most
of the indicators are weak acid or base that will
affect the pH of the solution if used in excess.

To find the end point of acid–base titration during
neutralisation using a computer interface

Apparatus 3 A 50 cm3 burette is rinsed with distilled water and
then with a little of the sodium hydroxide solution.
25 cm3 pipette, pipette filler, 50 cm3 burette, retort
stand, retort clamp, 250 cm3 beaker, filter funnel, 4 The burette is then filled with sodium hydroxide
magnetic stirrer, magnetic stirrer bar, pH meter, solution and is clamped to a retort stand.
computer interface and computer.
5 A magnetic stirrer bar is placed in the beaker
Materials containing 25 cm3 of hydrochloric acid. The beaker
1.0 mol dm–3 hydrochloric acid and 1.0 mol dm–3 is then placed on a magnetic stirrer below the burette.
sodium hydroxide.
6 A pH meter is connected to a computer using
Procedure a computer interface. The pH meter probe is
1 A 25 cm3 pipette is rinsed with distilled water then dipped into the acid. The magnetic stirrer
is switched on and the computer is set to record
and then with a little of the hydrochloric acid. and display the pH (Figure 7.16).
2 25 cm3 of 1.0 mol dm–3 hydrochloric acid is
7 Sodium hydroxide solution is added drop by
transferred using the pipette to a clean beaker. drop from the burette at a constant rate, into the
acid in the beaker.

207 Acids and Bases

8 A graph of pH change against the volume of sodium hydroxide in cm3 is printed using the computer printer
when 50 cm3 of sodium hydroxide is added to the beaker.

7 Figure 7.16 Using a pH meter and a computer interface to measure pH changes during neutralisation

Results Conclusion
1 A graph of pH against the volume of sodium
1 The pH value of hydrochloric acid is 1.0 at the
hydroxide in cm3 as displayed by the computer beginning of titration. As sodium hydroxide is
is shown in Figure 7.17. added to the acid, the pH value of the solution
increases.
Figure 7.17 Graph of pH against the volume of
sodium hydroxide in cm3 2 The pH value increases sharply at the end point
of neutralisation. The midpoint of the sharp pH
change is 7. At pH 7, the volume of sodium
hydroxide used from the graph is 25.0 cm3.

3 When 1.0 mol dm–3 sodium hydroxide is titrated
against 25.0 cm3 of 1.0 mol dm–3 hydrochloric
acid, the end point of titration during neutralisation
occurs at pH 7 when 25 cm3 of 1.0 mol dm–3 of
sodium hydroxide solution has been added.

4 By using a computer interface to measure
pH changes, the end point of titration during
nuetralisation can be determined accurately.

6 ’95

Describe an experiment to determine the concentration The equation for the neutralisation reaction between
of a solution of sulphuric acid by titrating it with a potassium hydroxide and sulphuric acid is
1.0 mol dm–3 potassium hydroxide solution.
2KOH + H2SO4 → K2SO4 + 2H2O
Solution
A titration experiment similar to Activity 7.7 is carried According to the equation, 2 mol of KOH requires
out in which sulphuric acid of unknown concentration 1 mol of H2SO4 for complete neutralisation.
is titrated against 25.0 cm3 of 1.0 mol dm–3 potassium 0.025 mol KOH will require
hydroxide using methyl orange (or phenolphthalein) 0 .025 3 —21 = 0.0125 mol H2SO4
as an indicator. N umber of moles of H2SO4 = —1M0—0V—0
Let's say the volume of sulphuric acid required to
neutralise completely 25.0 cm3 of 1.0 mol dm–3 potassium Molarity of sulphuric acid, M = 0.0125 3 1000/V
hydroxide is V cm3. = —1—2V.—5 mol dm–3
Number of moles of KOH in 25.0 cm3 of
1.0 mol dm–3 solution = —1M0—0—V0 = —1.—0—130—0—20—5.—0 = 0.025

Acids and Bases 208

Calculation Involving Neutralisation SPM 14 SPM
Using Balanced Equations ’11/P1
’09/P2
In an experiment, 25.0 cm3 of a sodium hydroxide
1 Say, in a balanced equation, a mol of acid solution of unknown concentration required
reacts with b mol of base as represented by the 26.50 cm3 of 1.0 mol dm–3 sulphuric acid for
equation below: complete reaction in titration. Calculate the molarity
of sodium hydroxide.

aA + bB → products Solution

2 Say, the molarity of an acid iMs BMmA oml odlmd–3m. –I3f b=2
and the molarity of a base is
2NaOH + H2SO4 → Na2SO4 + 2H2O

in a titration, VA cm3 of acid neutralises VB cm3 — MM——BAVV—AB = —12 a=1 where
MB = molarity of NaOH
of base MA = molarity of H2SO4 7
VB = volume of NaOH
Number of moles of acid = —1M—0—A0V—0A — 1M—.0—B———22—65—..50—0 = —21 VA = volume of H2SO4
Number of moles of base = —M——BV—B
1000 M B = 2  —22—65—.—.5—00
= 2.12 mol dm–3

In the stoichiometry equation, a mol of acid Hence, the molarity of sodium hydroxide solution
HA reacts completely with b mol of base, is 2.12 mol dm–3.

M(OH)X. Hence the mole ratio of acid to base 15 SPM
is
’10/P1,
——1M—0———0A—V—0A = —a o r ’11/P2
M— 1—0B—0V—0B b
— M———AV—A = —a What is the volume of 1.5 mol dm–3 aqueous
MBVB b ammonia required to completely neutralise 30.0 cm3
of 0.5 mol dm–3 sulphuric acid?

Solution

3 From the above relationship, the ratio of a b=2
and b can be obtained from the balanced
equation. Any one of the four variables: MA, 2NH3 + H2SO4 → (NH4)2SO4
VA, MB, VB, can be determined if three of the
other variables are known. — MM——BAVV—AB = —12 a = 1 where
— 0—1.5—.5———3—V0—.B0 = —21 MB = molarity of NH3
MA = molarity of H2SO4
1 The volume of a solution in a burette is read from VB = volume of NH3
VA = volume of H2SO4
the top to the bottom.
2 The accuracy of a burette reading is until 2 decimal V B = —2————3—01—..05————0—.5 = 20 cm3

places. Hence, the volume of aqueous ammonia required is
3 The accuracy of a pipette reading is until 1 decimal 20 cm3.

place.

209 Acids and Bases

16

Calculate the volume (cm3) of 2.0 mol dm–3 • In an aqueous solution, the concentration of H+
hydrochloric acid that is required to react completely ions in CH3COOH is lower than HCl of the same
with 2.65 g of sodium carbonate. concentration. This is because CH3COOH dissociates
[Relative atomic mass: C, 12; O, 16; Na, 23] partially in water.

Solution • However, both CH3COOH and HCl of the same
concentration require the same amount of NaOH
Molar mass of Na2CO3 for complete neutralisation. This is because both are
= (23  2) + 12 + (16  3) = 106 g monoprotic (monobasic) acids. In the presence of
N umber of moles of Na2CO3 = —m——o—mla—ra—sm—s—a——ss NaOH, CH3COOH dissociates completely to react
with NaOH.
7 = —21—.06——65 = 0.025
• In an acid-base titration, the preferred chemical to be
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O put in the burette is the acid. This is because alkali
can react with silica in glass to form silicate, thus
dissolving the thin glass wall of the burette slowly.

From the equation, 1 mol of Na2CO3 reacts with
2 mol of HCl.

Hence, 0.025 mol of Na2CO3 will react with 7.4
0.025  2 = 0.05 mol of HCl.

N umber of moles =—1—M0—0V—0 where 1 Complete the blanks in the equations below:
M = molarity of HCl

0 .05 = —2—1.0—0—0——0V— V = volume of HCl (a) 2HCl + Mg → +

V = —0—.0—5——2—.0—1—0—0—0 = 25 cm3 (b) H2SO4 + Zn(OH)2 → +

(c) + NaOH → CH3COONa + H2O

Hence, the volume of hydrochloric acid required is (d) + → CaSO4 + 2H2O
25 cm3.
2 Nitric acid reacts with magnesium hydroxide solution
17 to produce magnesium nitrate, Mg(NO3)2 and
water.
15 cm3 of an acid with the formula HaX of 0.1 mol dm–3 (a) Write a balanced equation for the reaction
required 30 cm3 of 0.15 mol dm–3 sodium hydroxide between nitric acid and magnesium hydroxide.
solution for complete neutralisation. Calculate the value (b) 10.0 cm3 of nitric acid is required to neutralise
of a and hence determine the basicity of the acid. 0.001 mol of magnesium hydroxide. Calculate
the concentration of the nitric acid in mol
Solution dm–3.
HaX + aNaOH → aH2O + NaaX
—M——A—V—A = —1 3 Sodium carbonate reacts with hydrochloric acid as
MBVB a represented by the equation below:
—00—.1.—15——33——13—50 = —1a
Na2CO3(s) + 2HCl(aq) →
a=3 2NaCl(aq) + H2O(l) + CO2(g)

Hence HaX is a tribasic acid. Calculate the volume (cm3) of 1.25 mol dm–3
hydrochloric acid that is required to react completely
with 25.0 cm3 of 1.0 mol dm–3 sodium carbonate.

4 15.0 cm3 of sulphuric acid neutralises 25.0 cm3
of 2.0 mol dm–3 aqueous ammonia. Calculate the
molarity of the sulphuric acid used.

Acids and Bases 210

1 An acid is a chemical compound that produces (c) Reacts with a metal carbonate to produce a
hydrogen ions, H+ o­ r hydroxonium ions, H3O+
when it dissolves in water. salt, carbon dioxide gas and water.

2 Dry acids without water do not show any acidic 8 Chemical properties of an alkali:

property because they do not dissociate to H+ ions. (a) Reacts with an acid to produce a salt and
3 A base is defined as a chemical substance that can
water.
neutralise an acid to produce a salt and water.
(b) When heated with an ammonium salt,
Most bases are not soluble in water. Bases that are
soluble in water are known as alkalis. ammonia gas is produced.
4 An alkali is defined as a chemical compound that
dissolves in water to produce free moving hydroxide 9 The pH scale is a set of numbers range from (0
ions, OH–.
5 Dry alkalis do not show alkaline property. to 14) used to measure acidity or alkalinity of a
6 Physical properties of acids and alkalis:
substance. 7

10 The pH value of an acid or alkali depends on

(a) the degree of dissociation (strength of acid or

alkali),

(b) the concentration of the acid or alkali.

Acids Alkalis 11 A standard solution is a solution of known

Acids are sour in taste Alkalis are bitter in concentration.
taste and feel soapy
12 The concentration of a solution is measured in

g dm–3 or mol dm–3.

pH values of less pH values of more 13 Neutralisation is the reaction between an acid and
than 7 than 7
a base to produce a salt and water only.

Changes blue litmus Changes red litmus 14 When a mol of acid reacts completely with b mol of
paper to red paper to blue
alkali in a reaction:

7 Chemical properties of an acid: —MM—BAVV—AB– = —ab— , where MA = molarity of acid
(a) Reacts with a base to produce a salt and water. VA = volume of acid
(b) Reacts with a reactive metal to produce a salt
and hydrogen gas is evolved. MB = molarity of alkali
VB = volume of alkali

Multiple-choice Questions 7

7.1 Characteristics and D X dissolves in water to Which of the following is solution
Properties of Acids and produce H+ ions. Z?
Bases A Glacial ethanoic acid
2 Which of the following B Ethanoic acid in
1 Chemical X is an acid. Which of statements is true of all bases?
the following may not be true methylbenzene
about the property of X? ’07 A Dissolve in water C Concentrated ethanoic acid
A A pink colour is produced B Contain hydroxide ions D Hydrogen chloride dissolved
when phenolphthalein is C Have a pH value of between
added to a solution of X. 12 and 13 in propanone
B Hydrogen gas is evolved D Produce ammonia gas when
when zinc powder is added heated with ammonium salts Acids and Bases
to a solution of X.
C A solution of X reacts with an 3 The diagram shows the set-up
alkali to produce salt and water. of apparatus for the reaction

’06 between solution Z and
magnesium ribbon.

211

4 The colours of indicator X in with magnesium powder to 13 Ethanoic acid is a weak acid
solutions of different pH values liberate hydrogen gas? because
are shown below. A W and X A it is an organic acid.
B X and Y B it dissolves slightly in water.
pH value 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C X and Z C it is a weak conductor of
←⎯→ ←⎯→ ←⎯⎯→←⎯→ ←⎯⎯⎯⎯⎯→ ←⎯→ ←⎯→ D Y and Z electricity.
D it ionises partially to form
Colour red orange yellow green bluish-green blue purple 9 Sulphuric acid is known as a hydrogen ions in water.
diprotic acid because
Indicator X will show a green A there are two hydrogen atoms 14 Which of the following solutions
colour in a solution of 0.1 mol in one molecule of sulphuric has the highest pH value?
dm–3 of acid. A Ethanoic acid, 0.01 mol dm–3
A sodium hydroxide solution B one mole of sulphuric acid B Hydrochloric acid, 0.01 mol
B sodium chloride solution contains two moles of dm–3
C aqueous ammonia hydrogen atoms. C Aqueous ammonia, 0.01 mol
D ethanoic acid C one mole of sulphuric acid dm–3
dissociates into two moles of D Sodium hydroxide solution,
7 hydrogen ions in water. 0.01 mol dm–3
D sulphuric acid can be neutralised
5 Which of the following by two types of bases. 15

substances can change red 10 Which of the following particles Which of the following
in a solution of ammonia is statements is true of the two
litmus paper to blue when responsible for its alkaline aqueous solutions shown above?
properties? A Both solutions are strong
dissolved in water? BA ONHH3–
A Carbon dioxide gas C H+ acids.
B Glacial ethanoic acid D NH4+ B The pH of both solutions are
C Solid sodium oxide
D Solid sodium sulphate equal.
C Both solutions are strong
6 The acidity of hydrogen chloride 7.2 The Strength of Acids
and Alkalis electrolytes.
gas cannot be shown when D 25.0 cm3 of each solution
11 A chemical that dissociates
it dissolves in the following completely in water to produce requires 25.0 cm3 of 1.0 mol
hydroxide ions is a dm–3 sodium hydroxide to be
solvents: A strong acid neutralised.
B weak acid
I Water C strong alkali 16 Which of the following solutions
D weak alkali contains the highest number of
II Ethanol
’11 hydrogen ions?
III Methylbenzene A 50 cm3 of 2 mol dm–3
ethanoic acid
IV Propanone B 40 cm3 of 1 mol dm–3
A I and II only sulphuric acid
B III and IV only C 30 cm3 of 2 mol dm–3 nitric
C I and III only acid
D II, III and IV only D 50 cm3 of 1 mol dm–3
hydrochloric acid
7 Zinc carbonate powder is added 12 The table shows the degree of
dissociation of four solutions 17 Which of the following is true
to liquid X. A gas which turns of acids which have the same when water is added to an
limewater milky is evolved. X concentration aqueous sodium hydroxide
solution?
could be Solution Degree of
A glacial ethanoic acid. dissociation
B aqueous citric acid. W
C hydrogen chloride gas X High
Y
dissolved in methylbenzene. Z Medium
D hydrogen chloride gas
Very high
dissolved in propanone.
Low
8 The table shows the pH for four
aqueous solutions, W, X, Y and Z.

Aqueous WX Y Z
solutions 2 7 12 13

pH

Which solution has the highest

When two of the solutions with pH value?
the same volume are added
together, which mixture will react A W C Y

B X D Z

Acids and Bases 212

A The pH value decreases. 21 Calculate the mass of potassium 25 When 2.8 g of potassium
B The degree of ionisation hydroxide is dissolved in 250
hydroxide that is required to cm3 of distilled water, which of
decreases. the following are true about the
C The hydroxide ion prepare 250 cm3 of 2.0 mol solution produced? [Relative
atomic mass: H, 1; O, 16; K, 39]
concentration increases. dm–3 solution. [Relative atomic
D The alkalinity increases. I It has a molarity of 0.05 mol
mass: H, 1; O, 16; K, 39] dm–3.
18 An aqueous solution of Q has a
pH value of 1. Q may be A 22.4 g C 56 g II It contains 11.2 g in 1 dm3.
III The solution produces 0.2
I 0.1 mol dm–3 hydrochloric B 28 g D 112 g
acid mol of hydroxide ions.
22 0.2 mol dm–3 sulphuric acid is IV It contains 2 mol in 10 dm3 of
II 0.001 mol dm–3 sulphuric
acid added slowly to a conical flask solution.
A I and II only
III 0.1 mol dm–3 nitric acid containing 20.0 cm3 of 0.2 mol B I and III only
IV 0.1 mol dm–3 ethanoic acid C III and IV only
dm–3 potassium hydroxide and D II and IV only
A I and III only
B II and IV only 10 cm3 of water. What is the total 26 What is the mass of sodium
C I, II and III only hydroxide contained in 50 cm3 of
D I, III and IV only volume (in cm3) of the solution 0.4 mol dm–3 sodium hydroxide
solution? [Relative atomic mass:
19 Which of the following statements in the flask when the solution is H, 1; O, 16; Na, 23] 7
is true of both nitric acid and A 0.4 g
sulphuric acid? completely neutralised? B 0.8 g
A Both are organic acids. C 1.6 g
B Both are diprotic acids. A 30 C 50 D 3.2 g
C Both undergo complete
dissociation in water. B 40 D 60 27 Calculate the number of moles
D Both react with copper to of hydroxide ions in 2 dm3 of
produce hydrogen gas. 23 The table shows four different calcium hydroxide solution with
test tubes P, Q, R and S a concentration of 14.8 g dm–3.
7.3 Concentration of Acids containing different acids. [Relative atomic mass: H, 1; O,
and Alkalis 16; Ca, 40]
Test tube Content A 0.20
20 Steps I to V below show the B 0.26
five steps that are involved in P 15 cm3 of 0.5 mol dm–3 C 0.40
the preparation of a standard hydrochloric acid D 0.44
solution of sodium hydroxide,
NaOH which may not be Q 15 cm3 of 1.0 mol dm–3 28 Which of the following
arranged in correct order. ethanoic acid substances will react with glacial
ethanoic acid?
I Add distilled water until the R 10 cm3 of 1.0 mol dm–3 A Zinc metal
graduation mark nitric acid B Ammonia gas
C Potassium hydroxide solid
II Weigh the mass of sodium S 10 cm3 of 0.5 mol dm–3 D Aqueous sodium carbonate
hydroxide sulphuric acid solution

III Transfer the solid sodium Which test tube will produce the 7.4 Neutralisation
hydroxide into the volumetric
flask biggest volume of hydrogen gas 29 The ionic equation for the reaction
between nitric acid and sodium
IV Rinse the weighing bottle and with excess magnesium? hydroxide is represented by
pour the solution into the CABD 22HNHH+a+2+++++ONOOHO22––→3→→–→2HHHN22O2OaONO3
volumetric flask A P C R
Acids and Bases
V Shake the volumetric flask B Q D S
Which of the following is the
correct order of steps in the 24 Which of the following sodium
preparation?
A I, II, III, IV, V hydroxide solutions have a
B II, III, I, IV, V
C II, III, IV, I, V concentration of 0.5 mol dm–3?
D II, I, III, IV, V
[Relative atomic mass: H, 1; O,

16; Na, 23]

I 5 g NaOH in 250 cm3 of water

II 20 g NaOH in 1 dm3 of water

III 250 cm3 of 2 mol dm–3

NaOH to which distilled water

is added until it becomes

1 dm3

IV 1 mol dm–3 NaOH diluted to

twice its volume
A I and III only
B II and III only
C III and IV only
D I, II, III and IV

213

30 Plants do not grow well in acidic 34 Fe + 2HCl → FeCl2 + H2 IV CuO(s) + H2(g) →
soil. Which of the following are Cu(s) + H2O(l)
used to neutralise acidic soil? Based on the equation above, A I and III only
calculate the mass of iron
I Sodium hydroxide that will react with excess B II and III only
II Calcium hydroxide hydrochloric acid to produce
III Potassium hydroxide 60 cm3 of hydrogen gas at C I, II and III only
IV Calcium oxide room temperature. [Relative
atomic mass: Fe, 56; 1 mol of D II, III and IV only
A I and III only gas occupies 24 dm3 at room
B II and IV only temperature] 38 The reaction between dilute
C I, II and III only A 0.14 g hydrochloric acid and calcium
D I, II, III and IV B 0.28 g carbonate is represented by the
C 3.36 g equation as follows.
31 Antacid is used to neutralise D 22.4 g
excess acid in the stomach. CCaaCCOl2(3a(qs)) + 2HCl(aq) →
Which of the following chemicals + H2O(l) + CO2(g)
is found in antacid?
7 A Sodium hydroxide
B Potassium hydroxide
C Magnesium hydroxide 35 2NaOH(aq) + 2HH2S2OO(4(l)aq) → What is the minimum volume
D Ammonia Na2SO4(aq) +
of 2 mol dm–3 hydrochloric
32 20 cm3 of 0.5 mol dm–3 The equation above shows
sulphuric acid is added to 20 acid that is required to react
cm3 of 0.5 mol dm–3 sodium the neutralisation reaction
hydroxide. Which of the following completely with 5 g of calcium
statements are true? between sodium hydroxide and
carbonate?
I Neutralisation reaction takes sulphuric acid. Calculate the
place. [Relative atomic mass: C, 12;
number of moles of sodium
II The solution produced is O, 16; Ca, 40]
acidic. hydroxide that is required to A 5 cm3
B 25 cm3
III 0.02 mol of water is neutralise 25 cm3 of 2.0 mol C 50 cm3
produced. D 100 cm3
dm–3 sulphuric acid.
IV The solution contains sodium A 0.05 mol 39 In a titration process, 0.1 mol
sulphate and water only. B 0.10 mol
C 0.50 mol dm–3 sulphuric acid from a
A I and II only D 1.00 mol
B III and IV only ’03 burette is added slowly to
C II and III only 36 10.0 cm3 of a certain 0.5 mol
D I, III and IV only dm–3 acid requires 50.0 cm3 20 cm3 of 0.1 mol dm–3
of 0.3 mol dm–3 sodium
33 The equation shows the reaction hydroxide solution for complete aqueous ammonia solution in a
between calcium carbonate and neutralisation.
hydrochloric acid. Which of the following is the conical flask with methyl orange
possible molecular formula for
CaCO3 + 2HCl → CaCl2 + CO2 + H2O this acid? indicator until neutralisation
A HNO3
What is the mass of calcium B H2SO4 occurs. What is the total volume
carbonate required to react C H3PO4
completely with 10 cm3 of 2.0 D CH3COOH in the conical flask at the end
mol dm–3 hydrochloric acid?
[Relative atomic mass: C, 12; O, 37 Which of the following reactions point of titration?
16; Ca, 40] represent neutralisation? A 10 cm3
A 0.5 g B 20 cm3
B 1.0 g I CaCO3(s) + 2HCl(aq) → C 30 cm3
C 2.0 g CaCl2(aq) + H2O(l) + CO2(g) D 40 cm3
D 4.0 g
II H+(aq) + OH–(aq) → H2O(l) 40 The equation below represents
III Ba(OH)2(aq) + H2SO4(aq) the neutralisation reaction of

→ BaSO4(s) + 2H2O(l) ’10 aqueous W hydroxide and
hydrochloric acid.

W(OH)2 + 2HCl → WCl2 + 2H2O

What is the volume of 0.5 mol

dm–3 hydrochloric acid needed

to neutralise 25 cm3 of 0.2 mol

dm–3 aqueous W hydroxide?
A 10 cm3
B 20 cm3
C 30 cm3
D 40 cm3

Acids and Bases 214

Structured Questions [Relative molecular mass of NaOH = 40]
[2 marks]
1 Diagram 1 shows the arrangement of apparatus used
to prepare hydrogen chloride in methylbenzene and (c) (i) Explain if a measuring cylinder is suitable to
in water respectively.

be used to measure the volume of water in

the preparation of the standard solution.

[1 mark]

(ii) Name a suitable apparatus that is required

to be used in the preparation of the

standard solution. [1 mark]

(d) What are the two parameters that should be 7
measured accurately to prepare the standard
solution of sodium hydroxide?
Parameter I:
Parameter II: [2 marks]

(e) State two steps that should be taken to ensure
that the standard sodium hydroxide solution is
exactly 100 cm3 with a molarity of 0.5 mol dm–3.

[2 marks]

Diagram 1 3 An experiment is carried out in the laboratory to

determine the concentration of a strong diprotic acid,

(a) What is the purpose of using the filter funnels in Hin2dAic,atboyr titration. A few drops of phenolphthalein
is added to 25.0 cm3 of 0.5 mol dm–3
Diagram 1? [1 mark]

(b) (i) What is observed when a piece of magnesium potassium hydroxide solution and then titrated with

ribbon is placed in beakers A and B the acid, Har2eAs, hoofwunnkinnoTwanbleco1n.centration. The results
obtained
respectively? [2 marks]

(ii) State the reason for your answer in (i). Experiment

[2 marks] Volume of I II III
[2 marks] H2 A 26.55 36.15 27.20
(c) Name the particles present in
(i) beaker A Final burette reading (cm3)
(ii) beaker B

(d) The magnesium ribbon is removed. Water is Initial burette reading (cm3) 0.50 10.00 1.10

added to the solution in beaker A and the

mixture is then shaken. When sodium carbonate Volume of H2 A used (cm3) … … …

powder is added, effervescence occurs.

(i) Name the gas and suggest a suitable test to Table 1

identify the gas evolved. [1 mark] (a) What is meant by a diprotic acid? [1 mark]

(ii) State the role of water in the reaction that

caused the evolution of the gas. [1 mark] (b) Write an equation for the neutralisation reaction
between the acid, H2A and potassium hydroxide.
(iii) Write an ionic equation for the reaction [1 mark]

involving the evolution of the gas. [1 mark]

(iv) What is the conclusion that can be made (c) State the colour change of the phenolphthalein
indicator at the end point of titration. [1 mark]
from the observation? [1 mark]

2 A student is required to prepare a standard solution (d) (i) Calculate the volume of the acid, H2A used
of 100 cm3 sodium hydroxide solution with a molarity in the titration and complete Table 1.

’06 of 0.5 mol dm–3 in the laboratory. He is given all the [1 mark]
necessary apparatus required.
(ii) Calculate the average volume of the acid,
(a) State the meaning of
(i) a standard solution. [1 mark] H2A used in the experiment. [1 mark]
(ii) molarity of the solution. [1 mark]
(e) Calculate the concentration of the acid H2A used
(b) Calculate the mass of sodium hydroxide that the in the experiment. [2 marks]
student needs to prepare a 100 cm3 solution
with a molarity of 0.5 mol dm–3. (f) Draw a diagram to show the arrangement of the
apparatus used in the above experiment.

[2 marks]

215 Acids and Bases

4 Experiment I Excess magnesium ribbon is
Experiment II put in 10.0 cm3 of 1.0 mol dm–3
sulphuric acid

Excess magnesium ribbon is
put in 10.0 cm3 of 1.0 mol dm–3
ethanoic acid

Table 2

Table 2 shows two experiments carried out.

7 (a) The reactions in experiment I and II can be
represented by the same ionic equation, involving
a certain particle in both acids. Diagram 2

(i) Write the formula of this particle. [1 mark]
(ii) Write the ionic equation for the reactions

that occur in both experiments. [1 mark]

(b) The rates of reactions are different in experiments (a) Mark the pH value on the graph in Diagram 2
I and II.
when complete neutralisation occurs. What is the
(i) Which reaction is more vigorous? [1 mark]
(ii) Give a reason for your answer in (i). pH value? [2 marks]

[3 marks] (b) Mark the volume of the nitric acid required for

(c) Compare and explain the difference between the complete neutralisation on the graph in Diagram

pH values of sulphuric acid and ethanoic acid. 2. What is this volume? [2 marks]

[2 marks] (c) Write a balanced equation for the reaction
between nitric acid and sodium hydroxide.
(d) What will be the change in pH value if 10.0 cm3
[1 mark]
of water is added to the sulphuric acid before
(d) If methyl orange is added to the sodium
the magnesium ribbon is added in experiment I? hydroxide solution at the initial stage of the
experiment, what is the change in colour that will
Explain your answer. [2 marks] take place at the end point of titration? [1 mark]

5 An experiment was carried out to determine the (e) Calculate the concentration of nitric acid used in
end point of titration during neutralisation using a this experiment. [2 marks]
computer interface. Nitric acid is added 0.5 cm3 by
0.5 cm3 from a burette to 25.0 cm3 of 0.5 mol dm–3 (f) If the experiment is to be repeated by titrating
sodium hydroxide in a beaker. The graph in Diagram sodium hydroxide against nitric acid, sketch a graph
2 shows the change in pH value of the solution in of the change in pH value against the volume of
the beaker against the volume of nitric acid used in sodium hydroxide that will be obtained. [2 marks]
cm3 as measured by the computer interface.

Essay Questions

1 (a) Using suitable examples, explain what is meant by the chemical formula of magnesium hydroxide
neutralisation. [4 marks]
and explain its function in antacid. Name another

(b) Explain why sodium hydroxide solution and aqueous chemical found in antacid. [4 marks]

ammonia of the same concentration have different (b) Diagram 1 shows two beakers containing 0.1
mol dm–3 solution X and solution Y and their pH
pH values. [6 marks] readings.

(c) Explain how you would prepare 250 cm3 of 1.0
mol dm–3 potassium hydroxide, starting from solid
potassium hydroxide. Subsequently, explain how
you would prepare 250 cm3 of 0.1 mol dm–3
potassium hydroxide from the above solution.
[Relative atomic mass: H, 1; O, 16; K, 39]

[10 marks]

2 (a) Magnesium hydroxide is one of the chemical Diagram 1
compounds found in antacid medicine. Write

Acids and Bases 216

(i) Compare and contrast the two solutions X (iii) Calculate the concentration of the solution
that will be produced when 80 cm3 of
and Y in terms of their physical and chemical water is added to 20 cm3 of solution X.

properties. Give a suitable example for each [2 marks]

of the solutions X and Y. [10 marks]

(ii) Predict and explain, with suitable ionic 3 (a) Using a suitable example, explain the role of water

equations, what will happen when equal in causing the acidic properties of an aqueous

volumes of solution X and solution Y are solution of an acid. [8 marks]

mixed together. [4 marks] (b) Briefly describe three tests (other than the use

of an indicator) that can be used to confirm an

acidic solution. Explain your tests with suitable

equations. [12 marks]

Experiments 7

1 An experiment is carried out to determine the relationship between the concentrations of H+ ions and
the pH values of nitric acid solutions. The pH values of six nitric acid solutions with concentrations of
0.100 mol dm–3, 0.060 mol dm–3, 0.040 mol dm–3, 0.025 mol dm–3, 0.015 mol dm–3 and 0.010 mol
dm–3 are each measured using a pH meter. The corresponding pH values and the concentrations of the
nitric acid solutions are shown in Diagram 1.

Diagram 1

(a) State the variables involved in this experiment. [3 marks]
• Manipulated variable:
• Responding variable:
• Constant variable:

(b) State the hypothesis for this experiment. [3 marks]

(c) Construct a table to record the results of this experiment. [3 marks]

(d) Based on the results of this experiment, draw a graph of pH value versus concentration of H+ ions on a

graph paper. [3 marks]

(e) Using the graph you have drawn in (d), predict the pH value of a 0.020 mol dm–3 nitric acid solution. [3 marks]

2 The volume of a sample of sulphuric acid required to neutralise a potassium hydroxide solution can [17 marks]
be determined by titration. Design a laboratory experiment to determine the volume of the sulphuric
acid required to neutralise 25.0 cm3 of 0.5 mol dm–3 potassium hydroxide solution. In designing your
experiment, the following items must be included.
(a) Problem statement
(b) All the variables involved
(c) Statement of the hypothesis
(d) List of materials and apparatus
(e) Procedure
(f) Tabulation of data

217 Acids and Bases

8CHAPTER FORM 4

THEME: Interaction between Chemicals

Salts

SPM Topical Analysis

Year 2008 2009 2010 2011

Paper 1 2 31 2 31 2 31 2 3

Section ABC ABC ABC ABC

Number of questions 2 – – —12 – – – – —31 – 3 1 – – – 3 – – – 1

ONCEPT MAP

SALTS

preparation qualitative analysis

Soluble salts Insoluble salts Anions Cations

by by test by test by

Reaction of acids with Precipitation in double (a) Heating and (a) Sodium hydroxide
(a) alkalis decomposition reactions identifying (b) Aqueous ammonia
(b) metal oxides (i) the gases (c) Specific reagents
(c) metal carbonates evolved
(d) metals ( ii) the colour change
of products
formed

(b) Reagents such as
barium chloride, silver
nitrate and brown ring
test

8.1 Salts

The Meaning of Salts Generally, the formula of salt is

1 Salt is an ionic compound that is formed S alt = (othCerattihoanns H+) + (other Anions OH–)
when the hydrogen ion in an acid is replaced than O2– or
by a metal ion or ammonium ion (NH4+).
3 Diprotic acids and triprotic acids contain
2 The chemical formula of a salt is comprised more than one H+ ions that can be replaced.
of a cation (other than hydrogen ion) and an Hence, it is possible for these acids to form more
anion (other than oxide ion and hydroxide than one type of salt as shown in Table 8.2.

ion). Table 8.2 Examples of diprotic and triprotic salts
3 The cations and the anions of a salt are
Type of Types of Example 8
bonded by strong ionic bonds. acid Example salts that of salt
of acid can be formed
Examples of Salts

1 Examples of salts formed from their Diprotic H2SO4 2 NaHSO4,
corresponding acids are shown in Table 8.1. acid H3PO4 Na2SO4

Table 8.1 Examples of salts formed from their Triprotic 3 NaH2PO4,
corresponding acids acid Na2HPO4,
Na3PO4
Acid General Example of salt
name of salt

Hydrochloric Chloride NaCl, KCl, CuCl2,
ZnCl2, NH4Cl
acid, HCl salts

Nitric acid, Nitrate salts NaNO3, KNO3, A salt consists of cations and anions, but the anions
HNO3 Mg(NO3)2, Pb(NO3)2, must not be oxide ion or hydroxide ion. This is
NH4NO3
because if the anions are O2– or OH– ions, the

compound is a base, not a salt.

Sulphuric acid, Sulphate Na2SO4, K2SO4,
FeSO4, CaSO4,
H2SO4 salts (NH4)2SO4

Carbonic acid, Carbonate Na2CO3, CaCO3, Generally, the type of salts can be classified according
MgCO3, ZnCO3, to the cation or anion. For example, sodium chloride is
H2CO3 salts PbCO3 known as a type of sodium salt or alternatively it can
also be classified as a chloride salt.

2 Chloride salts are formed when H+ ions in hydrochloric acid, HCl is replaced by a metal
ion or ammonium ion (NH4+).

KCl (potassium chloride)

H+ replaced by K+

NaCl H+ replaced HCl H+ replaced NH4Cl
(sodium chloride) by Na+ (hydrochloric acid) by NH4+ (ammonium chloride)

H+ replaced by Mg2+ H+ replaced by Zn2+

MgCl2 ZnCl2
(magnesium chloride) (zinc chloride)

219 Salts

8 Uses of Salts in Daily Life Sodium chloride is
used to flavour food
In food preparation
Salts play an
1 Sodium chloride (NaCl), table salt, is important role
used for seasoning food. in our daily life.
Here are few
2 Monosodium glutamate (M.S.G.) examples of salts
is used to enhance the taste of food. and their uses.

3 Self-raising flour contains sodium
bicarbonate (NaHCO3) which helps
breads and cakes to rise.

In food preservation so that food can be
kept longer without spoiling.

1 Sodium chloride is used as a food
preservative in food such as salted fish and
salted eggs.

2 Sodium benzoate (C6H5COONa) is used
as a food preservative in food such as
tomato sauce, oyster sauce and jam.

3 Sodium nitrite
(NaNO2) is used to
preserve processed
meat such as burgers,
sausages and ham.

Sodium benzoate is used as
a food preservative in sauce

In agriculture: to increase the production of food. Copper(II) sulphate is a pesticide
used to kill fungi
1 Nitrate salts such as potassium nitrate
(KNO3), sodium nitrate (NaNO3) Fertilisers used in agriculture are
and ammonium salts such as ammonium salts
ammonium sulphate (NH4)2SO4,
ammonium nitrate (NH4NO3),
ammonium phosphate (NH4)3PO4
are nitrogenous fertilisers.

2 Salts such as copper(II) sulphate
(CuSO4), iron(II) sulphate
(FeSO4) and mercury(I) chloride
(HgCl) are used as pesticides.

Salts 220

In medicine patients to be seen clearly in X-ray 8
films.
1 Antacid medicine contain calcium carbonate 7 Iron pills containing iron(II)
(CaCO3), and calcium hydrogen carbonate sulphate are taken to increase the
Ca(HCO3)2 that are used to reduce acidity supply of iron for anaemic patients.
in the stomachs of gastric patients.
Plaster of Paris consists of a calcium sulphate
2 Smelling salts contain ammonium chloride
(NH4Cl).

3 Plaster of Paris, used to support fractured
bones, contains calcium sulphate.

4 Epsom salts (magnesium sulphate
heptahydrate) and Glauber salt (sodium
sulphate decahydrate) are used as laxatives
to clear the intestines.

5 Potassium permanganate (KMnO4) is used
as an antiseptic to kill germs.

6 Barium sulphate, BaSO4, enables the
intestines of suspected stomach cancer

Salts play an Other uses
important role
in our daily life. 1 Fluoride toothpaste contains tin(II) fluoride,
Here are few SnF2, that is used to prevent tooth decay.
examples of salts Silver bromide, AgBr, is used to produce black
and their uses.
2 and white photographic films.
Sodium hypochlorite, NaOCl, is used as

3 a bleaching agent in soap powders and
detergents.

Fluoride salt in toothpaste prevents tooth decay

Naturally occuring salts

• Lead(II) sulphide (galena, PbS), calcium fluoride

M(flguSoOrit4e),eCxisatF2a)s and magnesium sulphate (Epsime,
minerals in the earth’s crust.

• Corals, stalactites, stalagmites and limestone Stalactites and
stalagmites
consist of calcium carbonate, (CaCO3).
Salts
221

Soluble Salts and Insoluble Salts

1 Solubility is the ability of a compound to
dissolve in a solvent. Some salts are soluble in
water while others are not.

2 The solubility of a salt in water depends on
the types of cations and anions present as
shown in Table 8.3.

Table 8.3 Types of salt and their solubility in water Figure 8.1 Separation of a soluble salt and an
insoluble salt by filtration
Type of salt Solubility in water
5 The methods of preparing salts depend on the
Experiment 8.1 Sodium, potassium All are soluble solubility of salts.
8 and ammonium salts
6 Soluble salts can be prepared in the laboratory
Nitrate salts All are soluble by four methods as follows:
Chloride salts (a) Reaction between an acid and an alkali
All are soluble except (b) Reaction between an acid and a metal
Sulphate salts PbCl2, AgCl and HgCl (c) Reaction between an acid and a metal
carbonate
Carbonate salts All are soluble except (d) Reaction between an acid and a metal
PbSO4, BaSO4 and CaSO4 oxide or hydroxide

All are insoluble except 7 Insoluble salts can be prepared by precipitation
(NNaH2C4)O2C3,OK32CO3 and in double decomposition reactions.

3 Information on the solubility of salts is useful • If a salt is soluble in water, a solution is formed. The
solution may be coloured if the cation is cop­per(II),
in iron(II) or iron(III). Other­w­ ise, a colourless solution is
(a) the separation of salts in a salt mixture. formed.
(b) choosing the methods to prepare a salt.
• If a salt is insoluble in water, a cloudy mixture is
(c) identifying the types of ions in a salt in formed when stirred. The undissolved salt will settle
the qualitative analysis of salts. down as a precipitate.

4 Filtration can be used to separate an insoluble

salt (as the residue) from a soluble salt (as the

filtrate) as shown in Figure 8.1.

8.1

To study the solubility of nitrate, sulphate, carbonate and chloride salts

Problem statement Materials

Are nitrate, sulphate, carbonate and chloride salts Various types of salts and distilled water.
soluble in water?
Procedure
Hypothesis
Some salts are soluble in water while some are not. 1 0.2 g of copper(II) nitrate is put in a test tube
using a spatula.
Variables
(a) Manipulated variable : Types of salts 2 5 cm3 of distilled water is added to the above test
( b) Responding variable : Solubility in water tube. The mixture is stirred and the solubility of
( c) Constant variable : Quantity of salts, volume the salt is noted.

and temperature of water 3 Steps 1 and 2 are repeated using magnesium
nitrate, zinc nitrate, lead(II) nitrate, calcium
Apparatus nitrate, copper(II) sulphate, magnesium sulphate,
Test tubes, glass rods, spatulas and test tube holder. zinc sulphate, lead(II) sulphate, barium sulphate,
calcium sulphate, copper(II) chloride, magnesium

Salts 222

chloride, zinc chloride, lead(II) chloride, silver Type of salt Formula of salt Solubility
chloride, mercury(I) chloride, sodium carbonate, in water
potassium carbonate, ammonium carbonate,
copper(II) carbonate, magnesium carbonate and Carbonate Na2CO3, K2CO3, Soluble
zinc carbonate to replace the copper(II) nitrate. (NH4)2CO3 Insoluble
CuCO3, MgCO3, ZnCO3
Results

Type of salt Formula of salt Solubility Conclusion
in water
1 All nitrate salts are soluble in water.
Nitrate Cu(NO3)2, Mg(NO3)2, Soluble
Sulphate Zn(NO3)2, Pb(NO3)2, 2 All sulphate salts are soluble in water except
Chloride Ca(NO3)2 Soluble
CuSO4, MgSO4 and Insoluble 3 lABeaallSdcO(IhI4l)oarnisddueclspalhscaaiulttesm, asPrueblpSshoOlau4t,be,lbeCaairnSiuOwm4a. tesurlpexhcaetpet, 8
ZnSO4 Soluble
PbSO4, BaSO4, CaSO4 Insoluble lead(II) chloride, PbCl2, silver chloride, AgCl,
CuCl2, MgCl2, ZnCl2 and mercury(I) chloride, HgCl.
PbCl2, AgCl, HgCl
4 All carbonates are insoluble in water except

5 sodium acnadrbaomnamteo,nNiua2mCOca3,rpbootnaasstieu, m(NcHar4)b2oCnOat3e. ,
K2CO3
Some salts are soluble while some are not. The

hypothesis is accepted.

Preparation of Soluble Salts Table 8.4 Some examples of acids and alkalis used in
the preparation of soluble salts
The preparation of soluble salts can be divided
into two categories. Type of Type of Example Type of
(a) Soluble salts of sodium, potassium and soluble salt alkali used of salt acid used

ammonium. Salts of Sodium NaCl HCl
(b) Soluble salts which are not salts of sodium, sodium
hydroxide, NaOH CH3COONa CH3COOH
potassium and ammonium.
K2SO4 H2SO4
Soluble Salts of K+, Na+ and NH4+ Salts of Potassium KNO3 HNO3
potassium hydroxide, KOH

1 The cation of a salt comes from the alkali Salts of Aqueous NH4NO3 HNO3
while the anion comes from the acid. Hence ammon­ ium ammonia, NH3 (NH4)2SO4 H2SO4
the type of salt produced depends on the acid
and alkali used. 5 Impure soluble salts can be purified by
For example: recrystallisation. When an impure soluble salt is
dissolved in enough distilled water, the insoluble
Na2SO4 impurities can be removed by filtration. The
filtrate is evaporated to remove excess water to
from NaOH from H2SO4 form a saturated solution. When the saturated
solution is cooled to room temperature, the salt
2 Soluble salts of sodium, potassium and will recrystallise to form pure crystals.
ammonium can be prepared from the reaction
between an acid and an alkali, NaOH, KOH 6 Soluble salts in a mixture of a few salts can
or NH3(aq) as in Table 8.4. also be purified by recrystallisation. This is
because salts have different solubility in water.
3 Titration method is used to ensure that all the A salt with a lower solubility will recrystallise
acid is completely reacted with the alkali. earlier than a salt with a higher solubility. The
process of recrystallisation may be repeated a
4 The flowchart in Figure 8.2 shows the steps few times to obtain a pure salt.
involved in the preparation of soluble salts of

sodium, potassium and ammonium.

223 Salts

8 Acid + alkali 1
The alkali in the conical flask is titrated with acid 2
in the burette

1 titration method

Dilute salt solution
After titration, the dilute salt solution is heated to
hasten evaporation

2 evaporation

Saturated salt solution
The saturated salt solution is then cooled to
precipitate out the salt

3 cooling 3

Salt crystals in saturated salt solution
The salt crystals are then filtered out from the
solution using filter paper

4 filtration 4

Salt crystals
The salt crystals are then dried with more filter
paper

5 drying

Dry salt crystals 5

The resulting salt crystals of sodium, potassium
or ammonium are produced

Figure 8.2 Preparation of soluble salts of sodium, potassium and ammonium

Salts 224

Preparation of Soluble Salts of Sodium, Potassium and Ammonium

To prepare potassium chloride by the reaction between SPM
an acid and an alkali
’04/P2

Apparatus by dropping a drop of the solution on a piece Activity 8.1 8
25 cm3 pipette, pipette filler, 50 cm3 burette, retort of glass plate. If crystals are formed, then the
stand, retort clamp, conical flask, filter funnel, filter solution is saturated.
paper, beaker, tripod stand, wire gauze and Bunsen 9 The saturated solution is then cooled to allow
burner. crystallisation to occur.
1 0 The white crystals formed are then filtered,
Materials rinsed with a little distilled water and dried by
2 mol dm–3 hydrochloric acid and 2 mol dm–3 pressing between filter paper.
potassium hydroxide and phenolphthalein indicator.

Procedure Figure 8.3 Titration of potassium hydroxide
with hydrochloric acid
1 25 cm3 of potassium hydroxide is pipetted into a
clean conical flask. Discussion
1 In the preparation of potassium chloride, the acid
2 Three drops of phenolphthalein indicator are added
to the alkali and the colour of the solution is used is hydrochloric acid and the alkali used is
noted. potassium hydroxide.

3 A 50 cm3 burette is then filled with hydrochloric KOH + HCl → KCl + H2O
acid and is then clamped to a retort stand. The
initial burette reading is recorded. 2 Phenolphthalein is used as an indicator at the
beginning of the experiment to determine the
4 Hydrochloric acid is added gradually from the volume of hydrochloric acid that is required
burette to the potassium hydroxide solution in to react with 25 cm3 of potassium hydroxide.
the conical flask while swirling the flask gently. However, the experiment is repeated without
using phenolphthalein so that the salt prepared
5 Titration is stopped when phenolphthalein changes will not be contaminated by the indicator.
from a light pink colour to colourless. The final
burette reading is recorded. 3 The salt solution is not heated until dry because
the salt may decompose when heated strongly.
6 The volume of hydrochloric acid used is calculated
as follows:

V cm3 = Final burette – Initial burette

reading reading

7 The experiment is repeated by adding V cm3 of
hydrochloric acid to 25 cm3 of potassium hydroxide
in a beaker without using phenolphthalein as an
indicator.

8 The colourless solution in the beaker is
evaporated to form a saturated solution (to about

—31 of the original volume). This can be tested

In the preparation of soluble salts of Na+/K+/NH4+, neutral salt using an indicator. Once the exact volumes
titration is carried out to determine the exact amount of acid and alkali required are known, the indicator is
of acid required to neutralise all the alkali to form a not required to prepare the pure salt.

225 Salts

To purify potassium chloride by recrystallisation

Activity 8.2 & 8.3 Apparatus 3 The hot solution is filtered into a clean conical
8 flask to remove the impurities.
Beaker, glass rod, Bunsen burner, conical flask,
spatula, filter funnel and filter paper. 4 The filtrate is evaporated until a saturated
solution is formed.
Materials
Impure potassium chloride and distilled water. 5 The saturated solution is then allowed to cool to
room temperature for crystallisation.
Procedure
1 Impure potassium chloride is placed in a beaker. 6 The filtered crystals are then rinsed with distilled
2 A little distilled water, enough to cover the water and dried between two pieces of filter
paper.
crystals is added. The mixture is heated while
stirring, and more distilled water is added slowly Conclusion
until all the crystals are dissolved.
Impure potassium chloride can be purified by
recrystallisation.

Soluble Salts which are not salts of Na+, K+, NH4+ ensure that all the acid is reacted completely,
excess solids are used. The non-reacted excess
1 Three methods are used to prepare soluble salts solids can be removed by filtration.
which are not salts of sodium, potassium and
ammonium. This involves the reaction between Table 8.5 Examples of some salts and chemicals used
(a) an acid and a metal in their preparation
(b) an acid and a metal carbonate
(c) an acid and a metal oxide or hydroxide Example Type of Chemicals that react with
of salt acid the acid
2 Table 8.5 shows the chemicals suitable for used
the preparation of soluble salts which are not Metal Metal Metal
salts of sodium, potassium and ammonium. oxide carbonate

3 In the reaction of an acid with a metal, metals ZnCl2 HCl ZnO Zn ZnCO3
that are less electropositive than hydrogen such as Mg(NO3)2 HNO3 MgO Mg MgCO3
copper and silver do not react with dilute acids.
CuSO4 H2SO4 CuO _ CuCO3
4 Metals, metal oxides and metal carbonates are
solids that do not dissolve in water. Hence, to

Preparation of Soluble Salts which are Not Salts of Sodium, Potassium and Ammonium

To prepare copper(II) nitrate by the reaction between SPM
an acid and a metal oxide
’04/05
P2

Apparatus

Beaker, glass rod, 100 cm3 measuring cylinder, wire gauze, tripod stand, Bunsen burner, conical flask, spatula,
filter funnel and filter paper.

Materials
1 mol dm–3 nitric acid and copper(II) oxide powder.

Salts 226

Procedure 1 8
1 About 30 cm3 of 1 mol dm–3 nitric acid is put in 2
3
a beaker and is heated. 4
2 Using a spatula, copper(II) oxide powder is 5

added a little at a time, to the hot nitric acid
while stirring continuously with a glass rod. The
addition of copper(II) oxide is stopped when
some black solids remain undissolved.
3 The mixture is filtered to remove the excess
copper(II) oxide.
4 The filtrate is evaporated until a saturated
solution is formed.
5 The saturated solution is then allowed to cool to
room temperature.
6 The blue crystals formed are removed by
filtration, rinsed with a little distilled water and
dried between filter paper.

Discussion
1 Copper(II) oxide is a black powder. It dissolves

in nitric acid to form a blue solution. The
equation for the reaction is

CuO(s) + 2HNO3(aq) →
Cu(NO3)2 (aq) + H2O(l)

2 Neutralisation reaction takes place between
copper(II) oxide and nitric acid. The ionic
equation for the reaction is

CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l)

3 The salt solution is not heated until dry because
the salt may decompose when heated strongly.

4 The copper(II) nitrate crystals prepared may be
purified by recrystallisation.

5 Copper(II) nitrate can also be prepared by the
reaction between nitric acid and copper(II)
carbonate. However, copper metal does not react
with dilute nitric acid because copper is below
hydrogen in the electrochemical series.

Conclusion
Copper(II) nitrate can be prepared by the reaction
between copper(II) oxide and nitric acid.

227 Salts

To prepare iron(II) sulphate by the reaction of an acid and a
metal

Activity 8.4 & 8.5 Apparatus Discussion
8 1 Iron metal is grey in colour. It dissolves in
Beaker, glass rod, 100 cm3 measuring cylinder,
Bunsen burner, conical flask, spatula, filter funnel sulphuric acid to form a green solution with
and filter paper. effervescence. The gas evolved is hydrogen gas.
The equation for the reaction is
Materials
Fe(s)+ H2SO4(aq) → FeSO4(aq) + H2(g)
Iron powder and 2 mol dm–3 sulphuric acid.
2 The ionic equation for the reaction between iron
Procedure and H+ ion in acid is

1 30 cm3 of 2 mol dm–3 sulphuric acid is put in a Fe(s)+ 2H+(aq) → Fe2+(aq) + H2(g)
beaker.
3 Iron(II) sulphate can also be prepared by the
2 Iron powder is gradually added to the sulphuric reaction of sulphuric acid with iron(II) oxide or
acid while stirring continuously with a glass rod, iron(II) carbonate.
until a slight excess of iron is present.
Conclusion
3 The mixture is filtered to remove the excess iron Iron(II) sulphate can be prepared by the reaction
powder. between iron metal and sulphuric acid.

4 The filtrate is evaporated until a saturated
solution is formed. The saturated solution is then
allowed to cool to room temperature.

5 The green crystals formed are removed by
filtration, rinsed with a little distilled water and
dried between filter papers.

To prepare magnesium chloride by the reaction of an acid and
a metal carbonate

Apparatus Discussion
1 Magnesium carbonate is white in colour.
Beaker, glass rod, 100 cm3 measuring cylinder,
Bunsen burner, conical flask, spatula, filter funnel Effervescence occurs when it dissolves in
and filter paper. hydrochloric acid to form a colourless solution.
The gas evolved is carbon dioxide gas. The
Materials equation for the reaction is

Magnesium carbonate powder and 2 mol dm–3 MgCO3(s) + 2HCl(aq) →
hydrochloric acid. MgCl2(aq) + CO2(g) + H2O(l)

Procedure 2 The ionic equation for the reaction between
magnesium carbonate and H+ ion in acid is
1 Magnesium carbonate powder is added a little
at a time, to 30 cm3 of 2 mol dm–3 hydrochloric MgCO3(s) + 2H+(aq) →
acid in a beaker while stirring continuously. Mg2+(aq) + CO2(g) + H2O(l)
The addition is stopped when there is no more
effervescence, and a little magnesium carbonate 3 Magnesium chloride can also be prepared by the
powder remains undissolved. reaction of hydrochloric acid with magnesium
oxide or magnesium metal.
2 The mixture is filtered to remove the excess
magnesium carbonate powder. Conclusion
Magnesium chloride can be prepared from the
3 The filtrate is evaporated until a saturated solution reaction between magnesium carbonate and
is formed. hydrochloric acid.

4 The saturated solution is then allowed to cool to
room temperature.

5 The white crystals formed is removed by filtration,
rinsed with a little distilled water and dried between
filter papers.

Salts 228

1 Preparation of Insoluble Salts SPM

’10/P1

Can sodium nitrate be prepared by adding sodium 1 Insoluble salts can be prepared by precipitation
chloride solution to nitric acid? in double decomposition reactions.

Comments 2 In the precipitation method, an insoluble salt
There is no reaction between sodium chloride is precipitated when two aqueous solutions
solution and nitric acid.
containing the cations and the anions are
NaCl(aq) + HNO3(aq) NaNO3(aq) + HCl(aq)
Solution consist of Solution consist of mixed together. The precipitate is then obtained
by filtration.
Na+, Cl–, H+ and Na+, Cl–­, H+ and 3 In double decomposition, one of the aqueous
solutions contains the cations of the insoluble
NO3– ions NO3– ions 8
salt, while the other aqueous solution contains
the anions of the salt.

Sodium nitrate is usually prepared by the reaction Cation M+ Anion X– Insoluble salt,
between nitric acid (HNO3) and sodium hydroxide
(NaOH). H2O formed does not dissociate. (from a + (from a MX
soluble salt soluble salt → (formed as

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) solution) solution) precipitate)

Physical Characteristic of Salt Crystals 4 In double decomposition, the ions of the two
aqueous solutions interchange to produce a
1 Salt crystals are formed when a saturated new compound which is insoluble.
salt solution is cooled.
5 The general equation can be represented as
2 A salt is made up of positive ions and follows
negative ions. When these ions are packed
closely with a regular and repeated MY(aq) + NX(aq) → MX(s) + NY(aq)
arrangement in fixed positions, a solid solution solution precipitate solution
with definite geometry known as crystal
lattice is formed. 6 The precipitate produced is obtained by
filtration. The residue is the insoluble salt,
3 The repeating basic unit in this orderly which is then rinsed with distilled water to
structure is called a unit cell. remove any other ions as impurities.
Example
4 All crystals have these physical
characteristics: Pb(NO3)2(aq) + 2NaCl(aq) →
(a) Fixed geometrical shapes (e.g. cubic, PbCl2(s) + 2NaNO3(aq)
hexagonal or rhombic). precipitate
(b) Flat surface, straight edges and sharp
angles. 7 Precipitation of lead(II) chloride can be
(c) Fixed angle between two adjacent simplified as follows:
surfaces.
Pb2+(aq) + 2Cl–(aq) → PbCl2(s)
5 All crystals of the same salt have the same
shape although the sizes may be different. Sodium ions, Na+ and nitrate ions, NO3–
do not undergo any change in the reaction.
6 The size of a crystal formed depends on the They are known as spectator ions and can be
rate of crystallisation. Fast crystallisation ignored in the ionic equation.
(from fast cooling) will yield smaller
crystals than slow crystallisation.

7 Crystals are hard and brittle, and can be
cut into different shapes. This is because
the particles of salt crystals are arranged in
regular layers.

229 Salts

8 The following guidelines show the steps used in writing an ionic equation for the formation
of an insoluble salt.

Step 1 Step 2 Step 3

Identify the insoluble salt Separate the cations and Balance the charges of
and write it as the product anions of the salt and write the cations and anions
on the right-hand side of them as the reactants on by adding the correct
the equation. the left-hand side of the coefficient as the number
equation. of moles reacting.
→ PbCl2(s)
Pb2+(aq) + Cl–(aq) → Pb2+(aq) + 2Cl–(aq) →
PbCl2(s) PbCl2(s)

8 9 The following guidelines show the steps used in the selection of aqueous solutions in the
preparation of an insoluble salt.

Step 1 Step 2 Step 3

Identify the cation and Identify a soluble salt that Identify a soluble salt that
anion of the insoluble can supply the cation, can supply the anion,
salt. example, a nitrate salt (all example, a sodium or
Example nitrate salts are soluble). potassium salt (all
PbCl2: Cation = Pb2+ Example sodium or potassium
Pb(NO3)2 solution or salts are soluble in water).
Anion = Cl– Pb(CH3COO)2 solution Example
NaCl or KCl solution
PbCl2

from Pb(NO3)2 or Pb(CH3COO)2 from NaCl or KCl or HCl

Table 8.6 Some examples of insoluble salts

Lead(II) salt Barium salt Silver salt

Name Formula Name Formula Name Formula

Lead(II) chloride PbCl2 Barium sulphate BaSO4 Silver chloride AgCl
Lead(II) bromide PbBr2 Barium chromate(VI) BaCrO4 Silver bromide AgBr
Lead(II) iodide PbI2 Barium carbonate BaCO3 Silver iodide AgI
Lead(II) sulphate PbSO4 Silver carbonate Ag2CO3
Lead(II) chromate(VI) PbCrO4
Lead(II) carbonate PbCO3

2 ’04

Can lead(II) sulphate be prepared by adding sulphuric two aqueous solutions, one containing the lead(II) ions
acid to lead(II) oxide? (example, lead(II) nitrate) and the other containing the
sulphate ions (example, sodium sulphate).
Comments
Both lead(II) sulphate and lead(II) oxide are insoluble Pb(NO3)2(aq) + Na2SO4(aq) →
in water. Hence, when lead(II) sulphate is formed, PbSO4(s) + 2NaNO3(aq)
it cannot be separated from the mixture of lead(II)
sulphate and lead(II) oxide. Similarly, lead(II) sulphate cannot be prepared by adding
Lead(II) sulphate as an insoluble salt, is usually sodium sulphate solution to lead(II) chloride. Both lead(II)
prepared from double decomposition reaction between sulphate and lead(II) chloride are insoluble in water.

Salts 230

To prepare insoluble salts: lead(II) iodide, lead(II) SPM
chromate(VI) and barium sulphate by precipitation reaction ’08/P2

Apparatus 4 The residue is rinsed with distilled water and
Beakers, glass rods, conical flasks, filter funnels and dried using filter papers.
filter paper.
Discussion
Materials 1 The chemical equation for the reaction that occur
0.5 mol dm–3 solutions of lead(II) nitrate, potassium
iodide, potassium chromate(VI), sodium sulphate in the preparation of lead(II) iodide is
and barium chloride.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

Procedure The ionic equation for the reaction is Activity 8.6 8

(A) Preparation of lead(II) iodide Pb2+(aq) + 2I–(aq) → PbI2(s)
1 20 cm3 of 0.5 mol dm–3 potassium iodide solution
Lead(II) iodide is a yellow precipitate.
is added to 20 cm3 of 0.5 mol dm–3 lead(II) 2 The chemical equation for the reaction that occurs
nitrate solution in a beaker.
2 The mixture is stirred thoroughly with a glass in the preparation of lead(II) chromate(VI) is
rod. A yellow precip­ itate is formed immediately.
3 The mixture is filtered to obtain the yellow Pb(NO3)2(aq) + K2CrO4(aq) →
solids of lead(II) iodide as the residue. PbCrO4(s) + 2KNO3(aq)
4 The residue is rinsed with distilled water to
remove any trace of other ions in it. The ionic equation for the reaction is
5 The yellow solid is dried by pressing between
two pieces of filter papers. Pb2+(aq) + CrO42–(aq) → PbCrO4(s)

(B) Preparation of lead(II) chromate(VI) Lead(II) chromate(VI) is a yellow precipitate.
1 20 cm3 of 0.5 mol dm–3 lead(II) nitrate solution 3 The chemical equation for the reaction that

is added to 20 cm3 of 0.5 mol dm–3 potassium occurs in the preparation of barium sulphate is
chromate(VI) solution in a beaker.
2 The mixture is stirred thoroughly with a glass BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
rod. A yellow precipitate is formed immediately.
3 The mixture is filtered to obtain the yellow solids The ionic equation for the reaction is
of lead(II) chromate(VI) as the residue.
4 The residue is rinsed with distilled water and Ba2+(aq) + SO42–(aq) → BaSO4(s)
dried using filter papers.
Barium sulphate is a white precipitate.
(C) Preparation of barium sulphate
1 20 cm3 of 0.5 mol dm–3 barium chloride solution Conclusion
Insoluble salts of lead(II) iodide, lead(II) chromate(VI)
is added to 20 cm3 of 0.5 mol dm–3 sodium and barium sulphate can be prepared by precipitation
sulphate solution in a beaker. in double decomposition reactions.
2 The mixture is stirred thoroughly with a glass
rod. A white precipitate is formed immediately.
3 The mixture is filtered to obtain the white barium
sulphate as the residue.

Preparation of a Specified Salt 2 The following flowchart shows the procedure
for the selection of the methods of preparing
1 The method of preparing a salt depends on a specified salt (Figure 8.4).
(a) whether or not the salt is soluble in water,
(b) whether the salt is a salt of sodium,
potassium or ammonium for soluble salts.

231 Salts

Method of preparing a salt
Is the salt soluble in water?

Yes No
Is it the salt of sodium, potassium or ammonium?

8 Yes No Precipitation by double
decomposition reaction
Neutralisation Reaction of an acid with
using titration • a metal Select two aqueous solutions
• a metal oxide/metal hydroxide that can supply the cations and
• a metal carbonate
the anions of the salt

Filtration to remove excess solids (keep the filtrate)

Salt solution Filtration (keep the residue)

1 evaporation
2 cooling (crystallisation)
3 filtration
4 recrystallisation (if necessary)

Salt crystals

1 rinse with distilled water 2 dry with filter paper

Pure salt crystals

Figure 8.4

3 ’03

You are supplied with sodium carbonate solution, Step 1 Magnesium nitrate is converted to magnesium
sulphuric acid and magnesium nitrate solution. Plan carbonate by double decomposition between
a scheme to prepare a sample of magnesium sulphate sodium carbonate solution and magnesium
using the above chemicals. Write equations for the nitrate.
reactions involved.
Mg(NO3)2(aq) + Na2CO3(aq) →
Comments MgCO3(s) + 2NaNO3(aq)
Magnesium sulphate is a soluble salt which is not a
salt of sodium, potassium and ammonium. Step 2 Magnesium carbonate that is produced is
then added to sulphuric acid until in excess.
Magnesium sulphate can be prepared by the reaction
of sulphuric acid with magnesium metal/magnesium MgCO3(s) + H2SO4(aq) →
oxide/magnesium carbonate. Hence, a scheme of MgSO4(aq) + CO2(g) + H2O(l)
preparing magnesium sulphate is as follows:

Salts 232

Ionic Equations of Insoluble Salts Table 8.7 Formation of ionic equations from the mole
ratio of ions
1 An ionic equation for the formation of a salt
can be written if No. of moles No. of moles Ionic equation
(a) the formula of the salt is known (from of cation of anion
the charges of cation and anion),
(b) the number of moles of ions required to 1 mol Pb2+ 2 mol Cl– Pb2+(aq) + 2Cl–(aq) →
form the salt is known. 1 mol Pb2+ PbCl2(s)
2 mol Ag+
2 The following guidelines show the construction 1 mol CrO42– Pb2+(aq) + CrO42–(aq)
of the ionic equation for the formation of an → PbCrO4(s)
insoluble salt from the charges of cation and
anion. 1 mol CrO42– 2Ag+(aq) + CrO42–(aq)
→ Ag2CrO4(s)

4 The number of moles of the cation and anion 8
can be calculated if the volume and molarities
Step 1
are known using the formula —1M—0—0V—0 .
If the charge of cation M is b and the charge
of anion X is a, the formula of the salt is MaXb, 1

MaXb

charge of X charge of M 6.0 cm3 of 0.2 mol dm–3 Xn+ solution reacts completely
with 4.0 cm3 of 0.1 mol dm–3 Ym– solution to form a salt
For example, the formula of iron(III) carbonate XmYn. Write the ionic equation and hence determine
is Fe2(CO3)3. the empirical formula of the salt in this reaction.

Step 2 Solution —1M—0—0V—0
N umber of moles of Xn+ ions = –0–1.2–0–0–0–6– —1M—0—0V—0
This show that a mol of Mb+ ions has combined = 0.0012
with b mol of Xa– ions.
Fe2(CO3)3 shows that 2 mol of Fe3+ ions N umber of moles of Ym– ions = –0–1.1–0–0–0–4–
combines with 3 mol of CO32– ions. = 0.0004

Step 3 Mole ratio of Xn+ ions : Ym– ions

Thus the ionic equation for the formation of = 0.0012 : 0.0004
MaXb is
aMb+(aq) + bXa–(aq) → MaXb(s) = –00––..00–00–10–24–– : –00–..0–0–00–00–44–
The ionic equation for the formation of
Fe2(CO3)3 is
2Fe3+(aq) + 3CO32–(aq) → Fe2(CO3)3(s) = 3 : 1

3 The examples in Table 8.7 show the method of Hence 3 mol of Xn+ react with 1 mol of Ym–.
writing ionic equations based on the simplest
mole ratio of cations to anions combined to Ionic equation is : 3Xn+ + 1Ym– → X3Y
form the salts. Empirical formula of the salt is X3Y.

Constructing Ionic Equations Using the
Continuous Variation Method

1 The mole ratio of ions that react to form a
salt can be determined from an experiment
through the continuous variation method.

2 In this method, fixed volumes of a reactant
X are added to varying volumes of a second
reactant Y in different test tubes. If the salt
formed is an insoluble salt, the amount of

233 Salts

precipitate produced will increase until all of the ions in solution X have reacted completely.
The height of the precipitate will remain constant despite the increasing volumes of solution
Y.
3 The flowchart of Figure 8.5 shows the steps involved in the continuous variation method.

To determine the ionic equation of the reaction between X ions and Y ions

Carry out an experiment to investigate the reaction between
• fixed volumes of solution X and
• different and varying volumes of solution Y

Experiment 8.2 Determine the volume of Y ions that reacts with all of the X ions
8
Calculate the number of moles of X ions and the number of moles of Y
ions that have reacted using the formula:
Number of moles = —1M—0—0V—0

Determine the simplest mole ratio of X ions to the Y ions in the reaction
to construct the ionic equation

Figure 8.5 Flowchart for the steps in the continuous variation method

8.2 SPM

’11/P3

To construct a balanced ionic equation for the precipitation of lead(II) chromate(VI)
using the continuous variation method

Problem statement Procedure

How to determine the ionic equation for the 1 A burette is filled with 0.5 mol dm–3 lead(II)
precipitation of lead(II) chromate(VI)? nitrate solution and another burette is filled with
0.5 mol dm–3 potassium chromate(VI) solution.
Hypothesis
2 Eight test tubes are labelled 1 to 8 and placed in
The height of precipitate will increase with the a test tube rack.
increase in volume of lead(II) nitrate solution until
all the potassium chromate(VI) has reacted. 3 5.00 cm3 of potassium chromate(VI) solution
from the burette is placed in every test tube.
Variables Potassium chromate(VI) solution is yellow in
colour.
(a) Manipulated variable : Volumes of lead(II)
nitrate solution 4 Using another burette, 1 cm3 of 0.5 mol dm–3 of
lead(II) nitrate solution is added to the first test
(b) Responding variable : Height of yellow preci- tube. Progressively increase the volume of the
pitate lead(II) nitrate solution by 1 cm3 to the rest of the
test tubes until 8 cm3 of lead(II) nitrate solution is
(c) Constant variable : Volume of potassium added to the eighth test tube (Figure 8.6(a)).
chromate(VI) solution
and the size of test tubes 5 Every test tube is well shaken in order to mix
the solutions completely. The test tubes are then
Apparatus allowed to stand for 20 minutes for the yellow
precipitate, lead(II) chromate(VI) to settle
Test tubes of the same size, test tube rack, 50 cm3 (Figure 8.6(b)).
burette, retort stand with clamp and ruler.

Materials

0.5 mol dm–3 lead(II) nitrate solution and 0.5 mol
dm–3 potassium chromate(VI) solution.

Salts 234

6 The height of the precipitate formed in every test tube is measured accurately using a ruler. The colour of the

solution above the precipitate is noted.
7 The result obtained is recorded in Table 8.8.

1.1

Figure 8.6 Continuous variation method 8

Results Table 8.8 67 8
12345 5.0 5.0 5.0
Test tube number 5.0 5.0 5.0 5.0 5.0
Volume of potassium 6.0 7.0 8.0
chromate(VI) solution (cm3) 1.0 2.0 3.0 4.0 5.0
Volume of lead(II) nitrate 2.8 2.8 2.8
solution (cm3) 0.6 0.9 1.8 2.2 2.8 colourless
Height of precipitate (cm) yellow yellow yellow yellow
Colour of solution

Calculation = ­—­1­­—­M0—0V—0 = —0—.—51—0—0—05—.—0 = 2.5 3 10–3

1 A graph showing the height of precipitate versus Number of moles of CrO42– ions in 5.0 cm3 of
the volume of lead(II) nitrate solution is drawn 0.5 mol dm–3 potassium chromate(VI) solution
(Figure 8.7).
= ­—­1­­—­M0—0V—0 = —0—.—51—0—0—05—.—0 = 2.5 3 10–3

4 Hence, 2.5  10–3 mol of Pb2+ ions react completely

with 2.5  10–3 mol of CrO42– ions.
∴ 1.00 mol of Pb2+ ions will react completely

with 1.00 mol of CrO42– ions.
The ionic equation for the reaction is

Figure 8.7 Graph of height of precipitate versus the Pb2+(aq) + CrO42–(aq) → PbCrO4(s)
volume of lead(II) nitrate solution 1 mol 1 mol 1 mol

2 From the graph, it is found that the height of 5 Consequently, the balanced chemical equation for
precipitate increases as the volume of lead(II) the reaction is
nitrate increases. However, a constant height is
reached when 5.0 cm3 of lead(II) nitrate solution Pb(NO3)2(aq) + K2CrO4(aq) →
is added. Thereafter, the height remains constant PbCrO4(s) + 2KNO3(aq)
despite further increases in the volume of lead(II)
nitrate solution. Conclusion

3 This means that when 5.0 cm3 of 0.5 mol dm–3 1 Since the diameter of the test tubes are the
lead(II) nitrate solution is used, all the chromate(VI) same, the height of the precipitate is directly
ions in 5 cm3 of 0.5 mol dm–3 potassium proportional to the mass of precipitate formed.
chromate(VI) solution has been precipitated.
2 The ionic equation for the precipitate of lead(II)
Number of moles of Pb2+ ions in 5.0 cm3 of chromate(VI) is Pb2+ + CrO42– → PbCrO4. The
0.5 mol dm–3 lead(II) nitrate solution hypothesis is accepted.

235 Salts

Discussion

From test tubes 1 to 4, the 1 3 From test tubes 6 to 8, the heights of precipitate
formed remains constant because all the
increase of Pb2+ ions from the chromate(VI) ions in the test tubes have been
precipitated. There is an excess of Pb2+ ions
increase in volumes of lead(II) in the test tubes. The clear solution above the
precipitate which is colourless contains Pb2+ ions,
nitrate solution added, increases K+ ions and NO3– ions.

the mass of precipitate formed.

There are excess (unr­eacted)

CrO42– ions in the test tubes which
produces the yellow colour of the

solutions above the precipitate.

The yellow solution contains

8 CrO42– ions, K+ ions and NO3–
ions. The yellow colour became

paler as more CrO42– ions have
reacted.

2 In test tube 5, the reaction is completed when the precipitate formed reaches a
maximum height. All the chromate(VI) ions have reacted with all the lead(II)
ions. The clear solution contains K+ ions and NO3– ions.

4 ’97

You are supplied with a lead(II) ions solution and a 0.1 From the graph, V cm3 of lead(II) ions solution is
mol dm–3 potassium chromate(VI) solution. Explain how required to react completely with 5 cm3 of potassium
you can determine the concentration of the lead(II) ions chromate(VI) solution, when the height of the precipitate
solution using the precipitation method. b ecomes constant.
Calculation:
Comments • Number of moles of CrO42– ions = —0—.110—30—0—5
An experiment using the continuous variation method = 0.0005
of precipitating lead(II) chromate(VI), using a constant • N umber of moles of Pb2+ ions = —M—1—030—0—V,
volume of potassium chromate(VI) solution and
different volumes of lead(II) ions as in Experiment where M is the concentration.
8.2 is carried out. A graph of the height of precipitate • Ionic equation of the reaction
against the volume of lead(II) ions solution will be Pb2+ + CrO42– → PbCrO4
obtained as follows: • From the equation, 1 mol of Pb2+ ions react with 1 mol
of CrO42– ions.
Mole ratio of Pb2+ : CrO42– = 1:1,
t hat is —0—0.0.—000—10M—5—V = —11
• Concentration of Pb2+ ions, M = 0.5/V mol dm–3

Salts 236

Numerical Problems Involving SPM H ence, 0.2 mol of H2SO4 produce 0.2  —31
Calculation of Quantities of Reactants ’09/P1 = 0.067 mol of Al2(SO4)3.
or Products in Stoichiometric Reactions
Step 3: Relate the number of moles of
1 A balanced equation gives information chemicals in the equation to that in the
regarding the number of moles of reactants question.
in a reaction and the number of moles of
products formed. Type 2: Calculation involving quantities in mass
If the quantities of reactants/products are given in
2 For example, the equation for the reaction terms of mass in gram, the quantity of solid can be
between magnesium and hydrochloric acid is converted to moles by the following relationship

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) SPM
1 mol 2 mol 1 mol 1 mol
’08/P1 Number of moles = ———————M—a—s—s——(—g—)————— 8
Molar mass (g mol–1)
The coefficients before the reactants and products
indicate the number of moles of chemicals 3
involved in the reaction. The equation above
shows that 1 mol of magnesium reacts with 2 2.0 g of sodium hydroxide reacts with excess
mol of hydrochloric acid to produce 1 mol of sulphuric acid. What is the mass of sodium sulphate
magnesium chloride and 1 mol of hydrogen gas. produced?
3 A balanced chemical equation (stoichiometric [Relative atomic mass: H, l; O, 16; Na, 23; S, 32]
reaction) can be used to calculate the
stoichiometric quantities of the reactants or Solution
products in terms of:
• Mass 2NaOH + H2SO4 → Na2SO4 + 2H2O
• Volume and concentration (of aqueous
Molar mass of NaOH Step 1: Write a
solution) = 23 + 16 + 1 balanced equation.
• Volume (of gas) = 40 g mol–1
4 Since the quantities of chemicals involved in a
reaction are in terms of moles, the quantities 2 g of NaOH = —4—0—g—2—mg—o—l—–‑1 Step 2: Convert mass to mol.
of reactants or products (the quantity of = 0.05 mol
chemicals in terms of volume of gas, volume
of solution, mass, numbers of molecules or From the equation, 2 mol of NaOH produces 1 mol
atoms), must first be converted to moles in
the initial step of the calculation regarding of Na2SO4.
quantities of reactants and products.

Examples of Calculation Step 3: Get the mole ratio of NaOH
Type 1: Calculation involving quantities in moles and Na2SO4

2 Hence, 0.05 mol of NaOH produce —21  0.05

Calculate the number of moles of aluminium = 0.025 mol of Na2SO4 Step 4: Relate the number
sulphate produced by the reaction of 0.2 mol of Molar mass of Na2SO4 of moles of chemicals in
sulphuric acid with excess aluminium oxide. = 2(23) + 32 + 4(16) the equation to that in the
question.

Solution Step 1: Write a balanced equation.

3H2SO4 + Al2O3 → Al2(SO4)3 + 3H2O = 142 g mol–1

3 mol 1 mol Step 2: Get the mole ratio 0.025 mol of Na2SO4 Step 5: Convert mol
= 0.025 mol  142 g mol–1 to mass.
of H2SO4 and Al2(SO4)3. = 3.55 g

From the equation, 3 mol of H2SO4
produce 1 mol of Al2(SO4)3

237 Salts

Type 3: Calculation involving volumes of gas Or: Number of moles = molarity of solution
If the quantities of reactants or products are given in (mol dm–3)  volume of solution (dm3)
terms of volumes of gas, the volume of gas can be
converted to moles by the following relationship: 5

At s.t.p. (0°C and 1 atm): What is the mass of magnesium required to react
with 20 cm3 of 2 mol dm–3 hydrochloric acid to
Number of moles = —V—o—l—u—m——e—o—f—g—a—s—(—d—m——3) produce 120 cm3 of hydrogen at room temperature?
22.4 dm3 mol–1 [Relative atomic mass: Mg, 24; 1 mol of gas
occupies 24 dm3 at room temperature]
At room conditions (25°C and 1 atm) :

Number of moles = —V—o—l2—u4—m.—0—e—do—mf—g3—a—ms—o(—dl—–m—1—3) Solution Step 1: Write a
Mg + 2HCl → MgCl2 + H2 balanced equation.

8

1 mol of gas occupies 24 dm3. Step 2: Convert
volume to mol.

4 1 20 cm3 gas = —2—4——1—2—10—0—0—0 mol = 0.005 mol

What is the volume of carbon dioxide gas evolved From the equation, 1 mol of H2 is produced by
at s.t.p. when 2.1 g of magnesium carbonate reacts 1 mol of Mg.
with excess nitric acid?
[Relative atomic mass: C, 12; O, 16; Mg, 24; 1 mol Step 3: Get the mole ratio of Mg and H2.
of gas occupies 22.4 dm3 at s.t.p.]
Hence, 0.005 mol of H2 is produced by 0.005 mol
Solution Step 1: Write a balanced equation. of Mg

MgCO3 + 2HNO3 → Mg(NO3)2 + CO2 + H2O Step 4: Relate the number of moles of chemicals
in the equation to that in the question.

Molar mass of MgCO3 Step 2: Convert 0.005 mol of Mg Step 5: Convert mol to mass.
= 24 + 12 + 3(16) = 84 g mol–1 mass to mol. = 0.005  24 g
2.1 g MgCO3 = —28—.41 = 0.025 mol = 0.12 g

From the equation, 1 mol of MgCO3 produces 1 mol 6

of CO2. Step 3: Get the mole ratio of MgCO3 and CO2. What is the volume of 2 mol dm–3 hydrochloric
acid required to dissolve 10 g of marble (calcium
Hence, 0.025 mol of MgCO3 produce 0.025 mol of carbonate)?
CO2. [Relative atomic mass: H, 1; O, 16; C, 12; Ca, 40]
Step 4: Relate the number of moles of chemicals
in the equation to that in the question.
Solution
1 mol of gas occupies 22.4 dm3. Step 5: Convert Step 1: CaCO3 + 2HCl → CaCl2 + CO2 + H2O
mol to volume.
Step 2: 10 g of CaCO3 = —4—0—+——1—21—0+— —3—(—1—6) = —1—10—00
Hence, 0.025 mol of gas CO2 gas occupies = 0.1 mol
0.025  22.4 = 0.56 dm3 or 560 cm3.

Type 4: Calculation involving volumes and Step 3: From the equation, 1 mol of CaCO3 requires
molarities of solutions 2 mol of HCl for a complete reaction.
If the quantities of reactants or products involves
solutions, the quantity of chemicals in a solution Step 4: Hence, 0.1 mol of CaCO3 requires 0.1  2
can be converted to moles using the following = 0.2 mol of HCl for a complete reaction
relationship
N umber of moles = —1—M0—0V—0 S tep 5: Number of moles = molarity  volume (dm3)

Where M = molarity of solution (mol dm–3) Volume of HCl
V = volume of solution (cm3)
= —N——u—m—Mb—e—or—lao—rf—itm—y—oo—lfe—Hs——oC—fl— H—C—l = —2——0m—.2o—lm—d—om—l—–3


= 0.1 dm3 = 0.1  1000 cm3 = 100 cm3

Salts 238

7 SPM 5 ’02
’08/P1
5.0 cm3 of a potassium iodide solution requires
4.0 g of magnesium oxide is added to 30.0 cm3 of 20.0 cm3 of 0.25 mol dm–3 lead(II) nitrate solution
2.0 mol dm–3 hydrochloric acid. to react completely according to the equation below.
What is the mass of magnesium oxide that does not
dissolve in this reaction? Pb2+(aq) + 2I–(aq) → PbI2(s)
[Relative atomic mass: O, 16; Mg, 24]

Solution Number of moles = —1M—0—0V—0 What is the molarity of the potassium iodide

MgO + 2HCl → MgCl2 + H2O solution? Number of moles = —1M—0—0V—0
Comments

N umber of moles of HCl = ——2—.10—0—0—0—3—0 = 0.06 N u mber of moles of Pb2+ ions = 0.25  —12—00—0.—00 8
From the equation, 2 mol of HCl will dissolve
= 0.005

1 mol of MgO. From the equation, 1 mol of Pb2+ ions reacts with
2 mol of I– ions.
Hence, 0.06 mol of HCl will dissolve

0.06  —21 = 0.03 mol of MgO. Hence, 0.005 mol of Pb2+ ions react with (0.005  2)
= 0.01 mol of I– ions. M = Number of moles 3 —1—0V—0—0
0.03 mol of MgO = 0.03  (24 + 16) g = 1.2 g
M olarity of KI solution = —0—.0—1——5——1—0—0—0 = 2 mol dm–3
∴ Mass of MgO that does not dissolve

= Initial mass of MgO – mass of MgO dissolved
= 4.0 g – 1.2 g = 2.8 g

8 Mass, m
m  molar mass
What is the mass of copper(II) carbonate that is n  molar mass
produced when 60 cm3 of 1 mol dm–3 sodium carbonate
is added to 50 cm3 of 2 mol dm–3 copper(II) sulphate? Mole, n
[Relative atomic mass H, 1; O, 16; Cu, 64]

Solution V  molar V1  molarity n  molarity
CuSO4 + Na2CO3 → CuCO3 + Na2SO4 volume
n  molar
Number of moles of CuSO4 Calculate the number of volume
= —21—0—0—50—0 = 0.1 moles of both reactants
to check which of the Volume of gas, V Volume of solution, V1
Number of moles of Na2CO3 reactants is used up in the
= —1—1—0—06—00 = 0.06 reaction. The number of 8.1
From the equation, moles of product formed
depends on the number 1 Suggest suitable methods and reactants for the
of moles of reactant that preparation of the following salts.
is used up (the limiting (a) Na2SO4
reactant). (b) (NH4)2SO4
(c) Al2(SO4)3
1 mol of CuSO4 reacts with 1 mol of Na2CO3 to (d) Pb(NO3)2
produce 1 mol of CuCO3. (e) ZnCl2
(f) PbSO4
Hence, 0.06 mol of Na2CO3 will react completely, (g) AgCl
while 0.1 mol of CuSO4 is in excess.
2 Suggest chemicals that can react with nitric acid
Thus, 0.06 mol of Na2CO3 will Na2CO3 is the to produce magnesium nitrate. Write equations for
produce 0.06 mol of CuCO3. limiting factor. the reactions that take place.

0.06 mol of CuCO3 = 0.06  (64 + 12 + 3(16)) g
= 0.06  124 g = 7.44 g

239 Salts

3 State the mole ratio of the ions in the following Inference from the Colours of Salts or
Salt Solutions
salts:

(a) CaSO4 (b) Al(OH)3

4 Write the ionic equations for the following 1 Initial observation of the physical properties
of a salt such as colour and solubility in water
reactions: enables us to make inferences regarding the
possible cations or anions present. However,
(a) 2 mol of silver ions react with 1 mol of the presence of the cations or anions needs to
be confirmed by other tests.
chromate(VI) ions
2 Most salts are white in colour and when dissolved
(b) 0.3 mol of lead(II) ions react with 0.6 mol of in water, will form colourless aqueous solutions.

bromide ions 3 Cations of transition elements have specific
colours.
8 5 5 cm3 of 0.2 mol dm–3 barium chloride solution
reacts completely with 10 cm3 of 0.1 mol dm–3 4 Table 8.9 below gives the colours of different
sodium chromate(VI) solution. Calculate the mole cations in the solid form or in aqueous
ratio of the ions involved in the formation of solutions.
barium chromate precipitate.

Subsequently, write the balanced chemical
equation for the reaction that occurs.

6 In an experiment, 10 cm3 of 0.5 mol dm–3 silver Table 8.9 Colours of cations
nitrate reacts completely with 5 cm3 of 0.5 mol
dm–3 potassium carbonate. Determine the ionic Colour Solid Solution
equation for the precipitation above.

7 Magnesium oxide reacts with excess phosphoric White or Salts of Na+, K+, Na+, K+, NH4+,
acid to produce 1.2 mol of magnesium phosphate. colourless NH4+, Mg2+, Ca2+, Mg2+, Ca2+, Ba2+,
(a) Write a balanced equation for the reaction that Ba2+, Al3+, Pb2+, Al3+, Pb2+, Zn2+
occurs. Zn2+ (if the anions
(b) Calculate the number of moles of magnesium are colourless)
oxide that is used in the reaction above.

8.2 Qualitative Analysis of Yellow PbO, PbI2, Fe3+, CrO42–
Salts Blue PbCrO4, BaCrO4 Cu2+
Hydrated Cu2+ salt
The Meaning of Qualitative Analysis
Green Hydrated Fe2+ Fe2+
1 Qualitative analysis is a chemical technique Black salt, CuCO3 and –
used to determine the identities of chemical CuCl2
substances present in a mixture but not their
quantity. Cu2+, Fe2+ oxide
or sulphide
2 Qualitative analysis of salt is a scheme of tests
carried out to identify the cation and anion Brown/ Hydrated Fe3+ salt Fe3+, CrO72–
present in the salt. orange

3 The technique of qualitative analysis includes: 5 Table 8.10 shows the solubility of different
(a) Observing the colour of the salt or colour types of salts in water.
of the aqueous salt solution.
(b) Observing the solubility of the salt in water. Table 8.10 Solubility of salts in water
(c) Observing the effect of heat on the salt.
(d) Identifying the gas evolved when a test is Type of Salt Solubility in water
performed on the salt.
(e) Identifying the precipitate formed when a Salts of Na+, All are soluble
specific chemical reagent is added to the K+, NH4+ All are soluble
aqueous salt solution.
(f) Carrying out confirmatory tests, which are Nitrate
specific chemical tests to confirm the identity
of a cation or an anion present in a salt. Sulphate All common sulphates are soluble
Chloride except BaSO4, PbSO4 and CaSO4

All common chlorides are soluble
except AgCl, HgCl and PbCl2(soluble
in hot water)

Salts 240

Type of Salt Solubility in water Tests of Gases

Carbonate All common carbonates are 1 Certain gases may be evolved when a chemical
insoluble except Na2CO3, K2CO3 and substance is
(NH4)2CO3 (a) heated,
(b) reacted with a dilute or concentrated acid,
Oxide All oxides are insoluble except Na2O, (c) heated with an alkali.
K2O and CaO (slightly soluble)
2 Based on the gas evolved, information about
Hydroxide All hydroxides are insoluble except the types of ions present can be deduced. For
KOH, NaOH, Ca(OH)2 and Ba(OH)2
instance, if carbon dioxide gas is evolved in
Lead halides PcoblCdlw2,aPtbeBr br2uat nsodlPubbIl2eairnehinostowluabtelerin 8
a reaction, carbonate ions are present in the

salt.
3 The physical properties and chemical tests for

a few gases are summarised in Table 8.11.

Table 8.11 Physical properties and tests on gases

Name of gas Colour of gas Smell of Effect on damp Confirmatory test on gas
gas litmus

Oxygen, O2 Colourless No smell No effect When a glowing wooden splint is lowered into the
Hydrogen, H2 Colourless test tube of oxygen, the glowing splint is lighted
No smell No effect
Carbon Colourless When a lighted wooden splint is pla­ced near the
dioxide, CO2 Colourless No smell Moist blue litmus mouth of the test tube of hydrogen, a ‘pop’ sound
turns to red is produced
Ammonia,
NH3 Pungent Moist red litmus When carbon dioxide gas is bubbled into limewater
turns to blue using a delivery tube, the limewater becomes milky
Chlorine, Cl2 Greenish-
yellow Choking Decolourises moist When a glass rod dipped into concentrated
Hydrogen red or blue litmus hydrochloric acid is placed near the mouth of the
chloride, HCl Colourless test tube with ammonia, white fumes are formed
Pungent Moist blue litmus
Sulphur Colourless turns to red –

dioxide, SO2 Pungent Moist blue litmus When a glass rod dipped into concentrated
turns to red ammonia is placed near the mouth of the test tube
Nitrogen Brown with hydrogen chloride, white fumes are formed
Pungent Moist blue litmus
turns to red When sulphur dioxide gas is bubbled into
acidified potassium manganate(VII) solution,
the purple colour is decolourised (or when it is
bubbled into acidified potassium dichromate(VI)
solution, the colour changes from orange to green)

–

dioxide, NO2

Heating Test on Salts (NH4)2CO3(s) → 2NH3(g) + H2O(l) + CO2(g)

1 All ammonium, carbonate, nitrate and some (NH4)2SO4(s) → 2NH3(g) + H2SO4(l)
sulphate salts will decompose when heated.
3 All carbonates except potassium carbonate and
2 All ammonium salts liberate ammonia gas sodium carbonate produce carbon dioxide
when heated. gas when heated. Table 8.12 shows the effect
Examples of heating on metal carbonates.

NH4NO3(s) → NH3(g) + HNO3(g)

241 Salts

8 Carbonate salt Table 8.12 Effect of heat on carbonate salts
Potassium carbonate
Sodium carbonate Effect of heat

Calcium carbonate Will not decompose on heating
Magnesium carbonate
Aluminium carbonate Decompose to metal oxide and carbon dioxide gas
Zinc carbonate CaCO3(s) → CaO(s) + CO2(g)
Iron(III) carbonate MgCO3(s) → MgO(s) + CO2(g)
Lead(II) carbonate Al2(CO3)3(s) → Al2O3(s) + 3CO2(g)
Copper(II) carbonate ZnCO3(s) → ZnO(s) + CO2(g)
Fe2(CO3)3(s) → Fe2O3(s) + 3CO2(g)
Mercury(II) carbonate PbCO3(s) → PbO(s) + CO2(g)
Silver carbonate CuCO3(s) → CuO(s) + CO2(g)
Gold(I) carbonate Decompose to metal, carbon dioxide gas and oxygen gas
2HgCO3(s) → 2Hg(l) + 2CO2(g) + O2(g)
Ammonium carbonate 2Ag2CO3(s) → 4Ag(s) + 2CO2(g) + O2(g)
2Au2CO3(s) → 4Au(s) + 2CO2(g) + O2(g)
Decompose to carbon dioxide gas, ammonia and water vapour without any residue
(NH4)2CO3(s) → 2NH3(g) + H2O(g) + CO2(g)

4 All nitrates decompose when heated. Table 8.13 shows the effect of heating on metal nitrates.
(a) Sodium nitrate and potassium nitrate produce oxygen gas and nitrites when heated.
(b) Other metal nitrates produce oxygen gas, nitrogen dioxide gas and metal oxides when heated.

Nitrate Salt Table 8.13 Effect of heat on nitrate salts

Potassium nitrate Effect of heat
Sodium nitrate
Decompose to metal nitrite and oxygen gas
Calcium nitrate
Magnesium nitrate 2KNO3(s) → 2KNO2(s) + O2(g)
Aluminium nitrate 2NaNO3(s) → 2NaNO2(s) + O2(g)
Zinc nitrate Decompose to metal oxide, oxygen gas and nitrogen dioxide gas
Iron(III) nitrate
Lead(II) nitrate 2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g)
Copper(II) nitrate 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)
4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g)
Mercury(II) nitrate 2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g)
Silver nitrate 4Fe(NO3)3(s) → 2Fe2O3(s) + 12NO2(g) + 3O2(g)
Gold(I) nitrate 2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)
2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g)
Ammonium nitrate Decompose to metal, nitrogen dioxide gas and oxygen gas

Hg(NO3)2(s) → Hg(l) + 2NO2(g) + O2(g)
2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g)
2AuNO3(s) → 2Au(s) + 2NO2(g) + O2(g)
Decompose to nitrous oxide gas, water vapour without any residue

NH4NO3(s) → N2O(g) + 2H2O(g)

Salts 242

5 Most sulphate salts do not decompose when Figure 8.8
heated. Only a few sulphates such as iron(II)
sulphate, zinc sulphate and copper(II) sulphate Table 8.14 Deduction of types of ion present from gas
decompose to sulphur dioxide or sulphur produced
trioxide gas when heated.
Type of gas Type of ion
Examples produced
2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g)
CO2 Carbonate ion, CO32– (except Na2CO3 Activity 8.7 8
ZnSO4(s) → ZnO(s) + SO3(g) and K2CO3)
O2 Nitrate ion, NO3–
CuSO4(s) → CuO(s) + SO3(g) NO2 and O2 Nitrate ion, NO3– (except NaNO3 and
KNO3)
6 All chloride salts are stable on heating except SO2 Sulphate ion, SO42–
ammonium chloride. Ammonium chloride NH3 Ammonium ion, NH4+
sublimes and decomposes to produce ammonia
gas and hydrogen chloride gas. 9 Most salts that decompose produced metal
oxides as residue. The change of colour during
NH4Cl(s) → NH3(g) + HCl(g) heating gives a good indication towards the
type of metal oxide formed as shown in Table
7 The deduction of the types of ions present based 8.15.
on the gas produced is shown in Table 8.14.

8 When a salt is heated,
(a) the type of gas evolved has to be identified.
This will give information to the type of
anion (or cation, NH4+) present.
(b) the colour change of the solid in the test tube
must be recorded. This will give information
regarding the type of cation present.

Original colour of salt Table 8.15 Colour change of salts on heating Cation present in salt
White Colour of residue after heating Metal oxide produced Zn2+

Yellow when hot, white when cold ZnO

White Brown when hot, yellow when cold PbO Pb2+

Blue/green Black CuO Cu2+

Green/yellow Brown Fe2O3 Fe2+/Fe3+

To study the effect of heat on carbonate salts SPM

’10/P2

Apparatus Procedure
Boiling tubes, test tubes, test tube holder, delivery
tube with rubber stopper, spatula and Bunsen burner. 1 One spatula of potassium carbonate powder is
placed in a dry boiling tube and the colour of the
Materials solid is recorded.
Potassium carbonate, sodium carbonate, calcium
carbonate, magnesium carbonate, zinc carbonate, 2 The boiling tube is fitted with a stopper with a
lead(II) carbonate, copper(II) carbonate and limewater. delivery tube.

3 The carbonate salt is heated slowly and then
strongly.

243 Salts