3 Chemical Equations and Stoichiometry 3.1 Formulae of

1
Chemical Equations and Stoichiometry
3.1 Formulae of Compounds 3.2 Derivation of
Empirical Formulae 3.3 Derivation of Molecular
Formulae 3.4 Chemical Equations 3.5 Calculations
Based on Chemical Equations 3.6 Simple Titrations
2
Stoichiometry (?????)p.19
Deals with quantitative relationships (a) among
atoms, molecules and ions RAM / RMM / Formula
Mass
3
Stoichiometry (?????)p.19
Deals with quantitative relationships (b) among
the constituent elements of a compound Empirical
/ Molecular Formulae
4
Stoichiometry (?????)p.19
Deals with quantitative relationships (c) among
the substances participating in chemical
reaction Calculations involving chemical
equation
5
Formulae of Compounds
6
Empirical formula
3.1 Formulae of compounds (SB p.43)
Shows the simplest whole number ratio of the
atoms or ions present
E.g. Methane, CH4 Sodium chloride, NaCl
7
Molecular formula
3.1 Formulae of compounds (SB p.43)
Shows the actual number of each kind of atoms
present in one molecule
E.g. CH4 methane Ionic compounds do not
have molecular formulae
8
Structural formula
3.1 Formulae of compounds (SB p.43)
Shows the bonding order of atoms in one molecule
9
3.1 Formulae of compounds (SB p.44)
Different types of formulae of some compounds
10
Derivation of Empirical Formulae
11
3.2 Derivation of empirical formulae (SB p.45)
From composition by mass
Mass of N used (0.623-0.450)g 0.173 g
12
3.2 Derivation of empirical formulae (SB p.45)
From composition by mass
Mg3N2
13
3.2 Derivation of empirical formulae (SB p.45)
Water of Crystallization Derived from Composition
by Mass
Example 2 Q.14
14
Q.14
heat
CuSO4.xH2O CuSO4 xH2O
10.0 g
6.4 g
(10.06.4) g 3.6 g
x 5
15
From combustion data
Vitamin C
CxHyOz O2 ? CO2(g) H2O(g)
excess
0.2000 g
0.2998 g
0.0819 g
16
CxHyOz O2 ? CO2(g) H2O(g)
Mass of C in sample mass of C in CO2 formed
Mass of H in sample mass of H in H2O formed
Mass of O in sample total mass of sample
mass of C mass of H
17
Mass of O in sample (0.2000-0.0818-0.0091) g
0.109 g
18
O
H
C
0.109
0.0091
0.0818
Mass (g)
Number of moles (mol)
3
4
3
Simplest whole no. ratio
C3H4O3
19
Derivation of Molecular Formulae
20
What is molecular formulae?
3.3 Derivation of molecular formulae (SB p.49)
Molecular formula
(Empirical formula)n
21
From empirical formula and known relative
molecular mass
3.3 Derivation of Molecular Formulae (SB p.49)
Empirical formula
Molecular mass
22
Q.15
142 g mol?1
Relative molecular mass 142
23
Q.15
473 K, 1.00 atm
CnH2n2(l) CnH2n2(g)
3.28 dm3
12.0 g
RMM 12n 2n2 142
The molecular formula is C10H22
24
Determination of Chemical Formulae
Ethanoic acid
25
Structural formula bond-line structure
26
Calculate the by mass of the constituent
elements of soda alum.
Na2SO4Al2(SO4)324H2O
27
Calculate the by mass of the constituent
elements of soda alum.
Na2SO4Al2(SO4)324H2O
28
A certain compound was known to have a formula
which could be represented as PdCxHyNz(ClO4)2. A
nalysis showed that the compound contained 30.15
carbon and 5.06 hydrogen. When converted to the
corresponding thiocyanate, PdCxHyNz(SCN)2, the
analysis was 40.46 carbon and 5.94
hydrogen. Calculate the values of x, y and
z. (Relative atomic masses C 12.0, H 1.0, N
14.0, O 16.0, Cl 35.5, S 32.0, Pd
106.0)
29
Let M be the formula mass of PdCxHyNz(ClO4)2
Then, the formula mass of PdCxHyNz(SCN)2
M 2(32.012.014.0) 2(35.54?16.0)
M 83.0
by mass of C in PdCxHyNz(ClO4)2
by mass of C in PdCxHyNz(SCN)2
30
Solving by simultaneous equations, x 14, M 557
by mass of H in PdCxHyNz(ClO4)2
y 28
557 106.0 12.0?14 1.0?28 14.0z
2(35.54?16.0) z 4
31
Chemical Equations
32
3.4 Chemical equations (SB p.53)
Chemical equations
a A b B ? c C d D
a, b, c, d are stoichiometric coefficients
33
3.5 Calculations Based on Equations (SB p.65)
Calculations based on equations
Calculations involving reacting masses
34
3.4 Chemical equations (SB p.53)
Example (a) Excess oxygen
2Mg(s) O2(g) ? 2MgO(s)
35
3.4 Chemical equations (SB p.53)
Example (a) Excess oxygen
2Mg(s) O2(g) ? 2MgO(s)
2.43 g
excess
?
4.03 g
36
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s) O2(g) ? 2MgO(s)
2.43 g
1.28g
?
37
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s) O2(g) ? 2MgO(s)
0.100 mol
0.040 mol
?
Mg is in excess, O2 is the limiting reagent
38
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s) O2(g) ? 2MgO(s)
0.040 mol
0.080 mol
3.22 g
Q.16, 17
39
Q.16
4.00 g
6.00 g
40
0.0323 mol
0.188 mol
O2 is in excess
41
Q.16
0.0323 mol
2?0.0323 mol
9.17 g
42
Q.17
2 3
2Al(s) 3Cu2(aq) ? 2Al3(aq) 3Cu(s)
43
3.5 Calculations Based on Equations (SB p.66)
Calculations involving volumes of gases

• Gases not at the same conditions
• Gases at the same conditions
• - Gay Lussacs Law

44
CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
2.8 dm3 25?C 1.65 atm
35.0 dm3 31?C 1.25 atm
? dm3 125?C 2.50 atm
O2 is in excess and CH4 is the limiting
reactant
45
CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
0.189 mol
0.189 mol
? dm3 125?C 2.50 atm
2.47 dm3
46
Gay Lussacs law When gases reacts, they do so
in volumes which bears a simple whole number
ratio to one another, and to the volumes of
gaseous products, all volumes being measured
under the same conditions of temperature and
pressure.
Watch video
47
20 cm3
10 cm3
29.5 cm3
2NH3(g) ? 1N2(g) 3H2(g)
48
Gay Lussacs law is an application of the
Avogadros law.
a A(g) b B(g) ? c C(g) d D(g)
At fixed T P, n ? V
49
Gay Lussacs law
a A(g) b B(g) ? c C(g) d D(g)
At fixed T P
50
Determination of Molecular Formula from Reacting
Volumes of Gases
For CxHy
51
Determination of Molecular Formula from Reacting
Volumes of Gases
For CxHyOz
52
Q.18
15 cm3
75 cm3
45 cm3
Volume of CO2 formed volume of gas absorbed by
NaOH (70 25) cm3 45 cm3
53
Q.18
15 cm3
75 cm3
45 cm3
C3H8
54
Q.19
50 cm3
100 cm3
100 cm3
100 cm3
C2H4O2
55
Simple Titrations
56
Simple titrations
3.6 Simple titrations (SB p.58)
Simple titrations
Acid-base titrations
Redox titrations
57
Simple titrations
3.6 Simple titrations (SB p.58)
Acid-base titrations
58
End point - The point at which the indicator
undergoes a sharp colour change in a titration
Equivalence point - The point at which the
reaction is just complete without excess of any
reactant.
59
3.6 Simple titrations (SB p.62)
Titration without an indicator

• By following the change
• in pH value (pH titration)
• in temperature (thermometric titration)
• in electrical conductivity
• (conductometric titration)
• in the course of the reaction

60
pH Titration Curves
61
pH Titration Curves
62
pH Titration Curves
63
pH Titration Curves
Slow change in pH around equivalence point
Most indicators are not suitable
64
Thermometric Titration
65
Temperature / K
Volume of acid added / cm3
66
Conductometric Titration NaOH vs HCl
Electrical Conductivity
Volume of acid added /cm3
67
Conductometric Titration NaOH vs HCl
Electrical Conductivity
Volume of acid added /cm3
68
Conductometric Titration NaOH vs HCl
Beyond the equivalence point, conductivity ?
sharply due to excess H (most mobile) Cl?
Electrical Conductivity
Steeper slope
Volume of acid added /cm3
69
At the equivalence point
Electrical Conductivity
Volume of acid added /cm3
70
Conductometric Titration NH3 vs HCl
Electrical Conductivity
Volume of acid added /cm3
71
Electrical Conductivity
Volume of acid added /cm3
72
Electrical Conductivity
Volume of acid added /cm3
73
Beyond the equivalence point, conductivity ?
sharply due to excess H (most mobile) Cl?
Electrical Conductivity
steeper slope
Volume of acid added /cm3
74
Conductometric Titration NaOH vs CH3COOH
Electrical Conductivity
Volume of acid added /cm3
75
Electrical Conductivity
Volume of acid added /cm3
76
Beyond the equivalence point, conductivity ?
slowly because the excess acid is weak
Electrical Conductivity
smaller slope
Volume of acid added /cm3
77
Conductometric Titration NH3 vs CH3COOH
Electrical Conductivity
Volume of acid added /cm3
78
Electrical Conductivity
Volume of acid added /cm3
79
CH3COOH (aq) NH3(aq)
Electrical Conductivity
CH3COO?(aq) NH4(aq)
Volume of acid added /cm3
80
Beyond the equivalence point, conductivity ?
slowly because the excess acid is weak
Electrical Conductivity
smaller slope
Volume of acid added /cm3
81
Conductometric Titration Ba(OH)2 vs H2SO4
Electrical Conductivity
At equivalence point, conductivity ? 0 because
BaSO4 is insoluble in water
Volume of acid added / cm3
82
Simple titrations
3.6 Simple titrations (SB p.58)
6B
Simple titrations
Acid-base titrations
Redox titrations
83
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
During the course of the titration
brown ? yellow
84
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
During the course of the titration
yellow ? dark blue
85
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
At the end point
dark blue ? colourless
86
3.6 Simple Titrations (SB p.66)
2. Titrations involving potassium manganate(VII)
MnO4-(aq) 8H(aq) 5Fe2(aq) ?Mn2(aq)
H2O(aq) 5Fe3(aq)
purple
pale green
yellow
colourless
During the course of the titration
87
3.6 Simple Titrations (SB p.66)
2. Titrations involving potassium manganate(VII)
MnO4-(aq) 8H(aq) 5Fe2(aq) ?Mn2(aq)
H2O(aq) 5Fe3(aq)
purple
pale green
yellow
colourless
At the end point
88
(No Transcript)
89
NaHCO3 HCl ? NaCl H2O CO2
90
The END
91
3.1 Formulae of compounds (SB p.45)
Back
Check Point 3-1
Give the empirical, molecular and structural
formulae for the following compounds (a)
Propene (b) Nitric acid (c) Ethanol (d) Glucose
Answer
92
3.2 Derivation of empirical formulae (SB p.46)
Example 3-2A
A hydrocarbon was burnt completely in excess
oxygen. It was found that 1.00 g of the
hydrocarbon gives 2.93 g of carbon dioxide and
1.80 g of water. Find the empirical formula of
the hydrocarbon.
Answer
93
3.2 Derivation of empirical formulae (SB p.46)
Example 3-2A
94
3.2 Derivation of empirical formulae (SB p.46)
Back
Example 3-2A
Therefore, the empirical formula of the
hydrocarbon is CH3.
95
3.2 Derivation of empirical formulae (SB p.46)
Example 3-2B
Compound X is known to contain carbon, hdyrogen
and oxygen only. When it is burnt completely in
excess oxygen, carbon dioxide and water are given
out as the only products. It is found that 0.46 g
of compound X gives 0.88 g of carbon dioxide and
0.54 g of water. Find the empirical formula of
compound X.
Answer
96
3.2 Derivation of empirical formulae (SB p.47)
Example 3-2B
97
3.2 Derivation of empirical formulae (SB p.47)
Back
Example 3-2B
Therefore, the empirical formula of
compound X is C2H6O.
98
3.2 Derivation of empirical formulae (SB p.47)
Check Point 3-2A

• 5 g of sulphur forms 10 g of an oxide on complete
  combustion. What is the empirical formula of the
  oxide?

Answer
99
3.2 Derivation of empirical formulae (SB p.47)
Check Point 3-2A

• (b) 19.85 g of element M combines with 25.61 g of
  oxygen to form an oxide. If the relative atomic
  mass of M is 31.0, find the empirical formula of
  the oxide.

Answer
100
3.2 Derivation of empirical formulae (SB p.47)
Check Point 3-2A
(c) Determine the empirical formula of copper(II)
oxide using the following results. Experimental
results Mass of test tube 21.430 g Mass of
test tube Mass of copper(II) oxide 23.321
g Mass of test tube Mass of copper 22.940
g
Answer
101
3.2 Derivation of empirical formulae (SB p.47)
Back
Check Point 3-2A
102
3.2 Derivation of empirical formulae (SB p.48)
Example 3-2C
Compound A contains carbon and hydrogen atoms
only. It is found that the compound contains 75
carbon by mass. Determine its empirical formula.
Answer
103
3.2 Derivation of empirical formulae (SB p.48)
Back
Example 3-2C
104
3.2 Derivation of empirical formulae (SB p.48)
Example 3-2D
The percentages by mass of phosphorus and
chlorine in a sample of phosphorus chloride are
22.55 and 77.45 respectively. Find the
empirical formula of the phosphorus chloride.
Answer
105
3.2 Derivation of empirical formulae (SB p.48)
Back
Example 3-2D
106
3.2 Derivation of empirical formulae (SB p.49)
Check Point 3-2B

• Find the empirical formula of vitamin C if it
  consists of 40.9 carbon, 54.5 oxygen and 4.6
  hydrogen by mass.

Answer
107
3.2 Derivation of empirical formulae (SB p.49)
Back
Check Point 3-2B
(b) Each 325 mg tablet of aspirin consists of
195.0 mg carbon, 14.6 mg hydrogen and 115.4 mg
oxygen. Determine the empirical formula of
aspirin.
Answer
108
3.3 Derivation of molecular formulae (SB p.50)
Example 3-3A
A hydrocarbon was burnt completely in excess
oxygen. It was found that 5.0 g of the
hydrocarbon gave 14.6 g of carbon dioxide and 9.0
g of water. Given that the relative molecular
mass of the hydrocarbon is 30.0, determine its
molecular formula.
Answer
109
3.3 Derivation of molecular formulae (SB p.50)
Example 3-3A
110
3.3 Derivation of molecular formulae (SB p.50)
Back
Example 3-3A
Therefore, the empirical formula of the
hydrocarbon is CH3. Let the molecular formula of
the hydrocarbon be (CH3)n. Relative molecular
mass of (CH3)n 30.0 n ? (12.0 1.0 ? 3)
30.0 n 2 Therefore, the molecular formula of
the hydrocarbon is C2H6.
111
3.3 Derivation of molecular formulae (SB p.50)
Example 3-3B
Compound X is known to contain 44.44 carbon,
6.18 hydrogen and 49.38 oxygen by mass. A
typical analysis shows that it has a relative
molecular mass of 162.0. Find its molecular
formula.
Answer
112
3.3 Derivation of molecular formulae (SB p.50)
Example 3-3B
113
3.3 Derivation of molecular formulae (SB p.50)
Back
Example 3-3B
Let the molecular formula of compound X be
(C6H10O5)n. Relative molecular mass of (C6H10O5)n
162.0 n ? (12.0 ? 6 1.0 ? 10 16.0 ? 5)
162.0 n 1 Therefore, the molecular formula of
compound X is C6H10O5.
114
3.3 Derivation of molecular formulae (SB p.51)
Example 3-3C
The chemical formula of hydrated copper(II)
sulphate is known to be CuSO4 xH2O. It is found
that the percentage of water of crystallization
by mass in the compound is 36 . Find the value
of x.
Answer
115
3.3 Derivation of molecular formulae (SB p.51)
Example 3-3C
Back
116
3.3 Derivation of molecular formulae (SB p.52)
Check Point 3-3A

• Compound Z is the major ingredient of a healthy
  drink. It contains 40.00 carbon, 6.67
  hydrogen and 53.33 oxygen.
• (i) Find the empirical formula of compound Z.
• (ii) If the relative molecular mass of compound
  Z is 180, find its molecular formula.

Answer
117
3.3 Derivation of molecular formulae (SB p.52)
Check Point 3-3A

• (i) Let the mass of compound Z be 100 g.
• Therefore, the empirical formula of compound Z is
  CH2O.

118
3.3 Derivation of molecular formulae (SB p.52)
Check Point 3-3A
(ii) Let the empirical formula of compound Z be
(CH2O)n. n ? (12.0 1.0 ? 2 16.0)
180 30n 180 n 6 Therefore, the
molecular formula of compound Z is C6H12O6.
119
3.3 Derivation of molecular formulae (SB p.52)
Check Point 3-3A
(b) (NH4)2Sx contains 72.72 sulphur by mass.
Find the value of x.
Answer
120
3.3 Derivation of molecular formulae (SB p.52)
Back
Check Point 3-3A
(c) In the compound MgSO4 nH2O, 51.22 by mass
is water. Find the value of n.
Answer
121
3.3 Derivation of molecular formulae (SB p.52)
Example 3-3D
The chemical formula of ethanoic acid is CH3COOH.
Calculate the percentage of mass of carbon,
hydrogen and oxygen respectively.
Answer
122
3.3 Derivation of molecular formulae (SB p.52)
Example 3-3D
Back
123
3.3 Derivation of molecular formulae (SB p.53)
Back
Example 3-3E
Calculate the mass of iron in a sample of 20 g of
hydrated iron(II) sulphate, FeSO4 7H2O.
Answer
124
3.3 Derivation of molecular formulae (SB p.53)
Check Point 3-3B

• Calculate the percentages by mass of potassium ,
  chromium and oxygen in potassium dichromate(VI),
  K2Cr2O7.

Answer
125
3.3 Derivation of molecular formulae (SB p.53)
Check Point 3-3B

• Find the mass of metal and water of
  crystallization in
• (i) 100 g of Na2SO4 10H2O
• (ii) 70 g of Fe2O3 8H2O

Answer
126
3.3 Derivation of molecular formulae (SB p.53)
Check Point 3-3B
Back
127
3.4 Chemical equations (SB p.54)
Check Point 3-4
Back

• Give the chemical equations for the following
  reactions
• Zinc steam ?? zinc oxide hydrogen
• (b) Magnesium silver nitrate ?? silver
  magnesium nitrate
• (c) Butane oxygen ?? carbon dioxide water

Answer
128
3.5 Calculations based on chemical equations
(SB p.55)
Example 3-5A
Calculate the mass of copper formed when 12.45 g
of copper(II) oxide is completely reduced by
hydrogen.
Answer
129
3.5 Calculations based on chemical equations
(SB p.55)
Back
Example 3-5A
130
3.5 Calculations based on chemical equations
(SB p.55)
Example 3-5B
Sodium hydrogencarbonate decomposes according to
the following chemial equation 2NaHCO3(s) ??
Na2CO3(s) CO2(g) H2O(l) In order to obtain
240 cm3 of CO2 at room temperature and pressure,
what is the minimum amount of sodium
hydrogencarbonate required? (Molar volume of gas
at R.T.P. 24.0 dm3 mol-1)
Answer
131
3.5 Calculations based on chemical equations
(SB p.55)
Back
Example 3-5B
132
3.5 Calculations based on chemical equations
(SB p.56)
Example 3-5C
Calculate the volume of carbon dioxide formed
when 20 cm3 of ethane and 70 cm3 of oxygen are
exploded, assuming all volumes of gases are
measured at room temperature and pressure.
Answer
133
3.5 Calculations based on chemical equations
(SB p.56)
Back
Example 3-5C
2C2H6(g) 7O2(g) ?? 4CO2(g) 6H2O(l) 2 mol 7
mol 4 mol 6 mol (from equation) 2
volumes 7 volumes 4 volumes - (by
Avogadros law) It can be judged from the
chemical equation that the mole ratio of CO2
C2H6 is 4 2, and the volume ratio of CO2 C2H6
should also be 4 2 according to the Avogadros
law. Let x be the volume of CO2(g) formed. Number
of moles of CO2(g) formed number of moles of
C2H6(g) used 4 2 Volume of CO2(g) volume
of C2H6(g) 4 2 x 20 cm3 4 2 x 40
cm3 Therefore, the volume of CO2(g) formed is 40
cm3.
134
3.5 Calculations based on chemical equations
(SB p.57)
Example 3-5D
10 cm3 of a gaseous hydrocarbon was mixed with 80
cm3 of oxygen which was in excess. The mixture
was exploded and then cooled. The volume left was
70 cm3. Upon passing the resulting gaseous
mixture through concentrated sodium hydroxide
solution (to absorb carbon dioxide), the volume
of the residual gas became 50 cm3. Find the
molecular formula of the hydrocarbon.
Answer
135
3.5 Calculations based on chemical equations
(SB p.57)
Example 3-5D
Let the molecular formula of the hydrocarbon be
CxHy. Volume of hydrocarbon reacted 10
cm3 Volume of O2(g) unreacted 50 cm3 (the
residual gas after reaction) Volume of O2(g)
reacted (80 50) cm3 30 cm3 Volume of CO2(g)
formed (70 50) cm3 20 cm3 CxHy O2 ?? xCO2
H2O 1 mol mol x mol 1
volume volumes x volumes
136
3.5 Calculations based on chemical equations
(SB p.57)
Back
Example 3-5D
137
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5

• Find the volume of hydrogen produced at R.T.P.
  when 2.43 g of magnesium reacts with excess
  hydrochloric acid.
• (Molar volume of gas at R.T.P. 24.0 dm3 mol-1)

Answer
138
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5
(b) Find the minimum mass of chlorine required to
produce 100 g of phosphorus trichloride
(PCl3).
Answer
139
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5
(c) 20 cm3 of a gaseous hydrocarbon and 150 cm3
of oxygen (which was in excess) were exploded in
a closed vessel. After cooling, 110 cm3 of gases
remained. After passing the resulting gaseous
mixture through concentrated sodium hydroxide
solution, the volume of the residual gas became
50 cm3. Determine the molecular formula of the
hydrocarbon.
Answer
140
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5
141
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5
142
3.5 Calculations based on chemical equations
(SB p.58)
Check Point 3-5
Back

• Calculate the volume of carbon dioxide formed
  when 5 cm3 of methane was burnt completely in
  excess oxygen, assuming all volumes of gases are
  measured at room temperature and pressure.

Answer
143
3.6 Simple titrations (SB p.61)
Example 3-6A
25.0 cm3 of sodium hydroxide solution was
titrated against 0.067 M sulphuric(VI) acid using
methyl orange as an indicator. The indicator
changed colour from yellow to red when 22.5 cm3
of sulphuric(VI) acid had been added. Calculate
the molarity of the sodium hydroxide solution.
Answer
144
3.6 Simple titrations (SB p.61)
Back
Example 3-6A
2NaOH(aq) H2SO4(aq) ?? Na2SO4(aq) 2H2O(l)

? Number of moles of NaOH(aq) Number
of moles of H2SO4(aq) Number of moles of
H2SO4(aq) 0.067 mol dm-3 ? 22.5 ? 10-3 dm3

1.508 ? 10-3 mol Number of moles of
NaOH(aq) 2 ? 1.508 ? 10-3 mol
3.016 ?
10-3 mol Molarity of NaOH(aq)

0.121 mol dm-3 Therefore, the molarity
of the sodium hydroxide solution was 0.121 M.
145
3.6 Simple titrations (SB p.61)
Example 3-6B

• 2.52 g of a pure dibasic acid with formula mass
  of 126.0 was dissolved in water and made up to
  250.0 cm3 in a volumetric flask. 25.0 cm3 of this
  solution was found to neutralize 28.5 cm3 of
  sodium hydroxide solution.
• Calculate the molarity of the acid solution.

Answer
146
3.6 Simple titrations (SB p.61)
Example 3-6B

• (b) If the dibasic acid is represented by H2X,
  write an equation for the reaction between the
  acid and sodium hydroxide.

Answer
(b) H2X(aq) 2NaOH(aq) ?? Na2X(aq) 2H2O(l)
147
3.6 Simple titrations (SB p.61)
Back
Example 3-6B

• (c) Calculate the molarity of the sodium
  hydroxide solution.

Answer
148
3.6 Simple titrations (SB p.62)
Example 3-6C
0.186 g of a sample of hydrated sodium carbonate,
Na2CO3 nH2O, was dissolved in 100 cm3 of
distilled water in a conical flask. 0.10 M
hydrochloric acid was added from a burette, 2
cm3 at a time. The pH value of the reaction
mixture was measured with a pH meter. The results
were recorded and shown in the following figure.
Calculate the value of n in Na2CO3 nH2O.
Answer
149
3.6 Simple titrations (SB p.63)
Example 3-6C
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150
3.6 Simple titrations (SB p.63)
Example 3-6D
5 cm3 of 0.5 M sulphuric(VI) acid was added to
25.0 cm3 of potassium hydroxide solution. The
mixture was then stirred and the highest
temperature was recorded. The experiment was
repeated with different volumes of the
sulphuric(VI) acid. The laboratory set-up and the
results were as follows
151
3.6 Simple titrations (SB p.63)
Example 3-6D

• Plot a graph of temperature against volume of
  sulphuric(VI) acid added.

Answer
152
3.6 Simple titrations (SB p.63)
Example 3-6D
(b) Calculate the molarity of the potassium
hydroxide solution.
Answer
153
3.6 Simple titrations (SB p.63)
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Example 3-6D
(c) Explain why the temperature rose to a maximum
and then fell.
Answer
(c) Neutralization is an exothermic reaction.
When more and more sulphuric(VI) acid was added
and reacted with potassium hydroxide, the
temperature rose. The temperature rose to a
maximum value at which the equivalence point of
the reaction was reached. After that, any excess
sulphuric(VI) acid added would cool down the
reacting solution, causing the temperature to
drop.
154
3.6 Simple titrations (SB p.66)
Example 3-6E
When excess potassium iodide solution (KI) is
added to 25.0 cm3 of acidified potassium iodate
solution (KIO3) of unknown concentration, the
solution turns brown. This brown solution
requires 22.0 cm3 of 0.05 M sodium thiosulphate
solution to react completely with the iodine
formed, using starch solution as an indicator.
Find the molarity of the acidified potassium
iodate solution.
Answer
155
3.6 Simple titrations (SB p.66)
Example 3-6E
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IO3-(aq) 5I-(aq) 6H(aq) ?? 3I2(aq) 3H2O(l)
(1) I2(aq) 2S2O32-(aq) ?? 2I-(aq)
S4O62-(aq) (2)
Molarity of IO3-(aq) 7.33 ? 10-3 M Therefore,
the molarity of the acidified potassium iodate
solution is 7.33 ? 10-3 M.
156
3.6 Simple titrations (SB p.67)
Example 3-6F
A piece of impure iron wire weighs 0.22 g. When
it is dissolved in hydrochloric acid, it is
oxidized to iron(II) ions. The solution requires
36.5 cm3 of 0.02 M acidified potassium
manganate(VII) solution for complete reaction to
form iron(III) ions. What is the percentage
purity of the iron wire?
Answer
157
3.6 Simple titrations (SB p.67)
Example 3-6F
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MnO4-(aq) 8H(aq) 5Fe2(aq) ?? Mn2(aq)
4H2O(l) 5Fe3(aq) Number of moles of MnO4-(aq)
Number of moles of Fe2(aq) 1 5 Number of
moles of Fe2(aq) 5 ? Number of moles of
MnO4-(aq)
5 ? 0.02 mol dm-3 ? 36.5 ? 10-3 dm3
3.65 ? 10-3
mol Number of moles of Fe dissolved Number of
moles of Fe2 formed
3.65 ? 10-3 mol Mass
of Fe 3.65 ? 10-3 mol ? 55.8 g mol-1 0.204 g
158
3.6 Simple titrations (SB p.67)
Check Point 3-6

• 5 g of anhydrous sodium carbonate is added to 100
  cm3 of 2 M hydrochloric acid. What is the volume
  of gas evolved at room temperature and pressure?
• (Molar volume of gas at R.T.P. 24.0 dm3 mol-1)

Answer
159
3.6 Simple titrations (SB p.67)
Check Point 3-6
Na2CO3(s) 2HCl(aq) ?? 2NaCl(aq) H2O(l)
CO2(g)
Since HCl is in excess, Na2CO3 is the limiting
agent. No. of moles of CO2 produced No. of
moles of Na2CO3 used
0.0472 mol Volume of CO2
produced 0.0472 mol ? 24.0 dm3 mol-1
1.133 dm3
160
3.6 Simple titrations (SB p.67)
Check Point 3-6
(b) 8.54 g of impure hydrated iron(II) sulphate
(formula mass of 392.14) was dissolved in water
and made up to 250.0 cm3. 25.0 cm3 of this
solution required 20.76 cm3 of 0.020 3 M
acidified potassium manganate(VII) solution for
complete reaction. Determine the percentage
purity of the hydrated iron(II) sulphate.
Answer
161
3.6 Simple titrations (SB p.67)
Check Point 3-6
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(b) MnO4-(aq) 8H(aq) 5Fe2(aq) ?? Mn2(aq)
4H2O(l) 5Fe3(aq)
No. of moles of Fe2 ions 5 ? No. of moles of
MnO4- ions
2.107 ? 10-3 mol No. of moles of Fe2 ions in
25.0 cm3 solution 2.107 ? 10-3 mol No. of moles
of Fe2 ions in 250.0 cm3 solution 0.02107
mol Molar mass of hydrated FeSO4 392.14 g
mol-1 Mass of hydrated FeSO4 0.02107 mol ?
392.14 g mol-1 8.26 g