1.1 The Atomic Structure
A. Dalton’s atomic theory Atoms are the smallest particles of an element that can take part in a chemical change. All elements are made up of atoms
Atoms of a given element are identical in chemical behaviour
Chemical compounds are formed when different atoms combine chemically;
The atoms themselves are not changed in a chemical reaction, they just reorganized.
Constituents of atoms
Definition of key terms: Atomic number
= Number of protons in an atom; or = Number of electrons in the neutral atom;
= Ordinal number of that element in the Periodic Table; It determines the chemical properties of an element. Mass Number
Number of protons + number of neutrons ~ Relative atomic mass of an atom
Isotopes
Atoms with the same number of protons but different number of neutrons. Atoms with the same atomic number but different mass number.
Mass number (A)
11 23
Na Element symbol
Atomic number (Z) Notation of an element1.2 Relative Isotopic, Atomic and Molecular Masses
A. Mass spectrometer It measures the mass to charge ratio of a particle.
Atomic masses can be determined by mass spectrometry.
Components of a mass spectrometer and their functions:
1. Vaporization chamber: To vaporize the sample
2. Ionization chamber: The vaporized sample is then bombarded by fast moving electrons to
give positive ions.
M(g) + e– → M+(g) + 2e–
3. Accelerating electric field: The ions are accelerated by an electric field.
4. Deflecting magnetic field: It deflects the moving ions along a circular path.
As a result, the ions are separated as different ions have different deflection in the
magnetic field. The lighter (lower mass/charge ratio) the positive ion, the greater the deflection will be.
By varying the strength of the deflecting magnetic field, ions of a particular mass/charge ratio are brought to the ion detector
5. Ion detector: It detects the signals generated by the ions and passes them to the recorder. 6. Recorder: A mass spectrum is recorded using the recorder.
7. Vacuum pump: The mass spectrometer is operated at low pressure to prevent ions from
colliding with other particles.
B. Uses of mass spectrometer
Example 1
The mass spectrum of naturally occurring chlorine atom is as follows:
75.77% Relative abundance 24.23%
35 37
(a) Why there are two peaks of chlorine?
There are two isotopes of chlorine atoms in nature. The two isotopes are 35Cl and 37Cl.
(b) What is the relative atomic mass of the naturally occurring chlorine atom? Relative atomic mass of chlorine = 35 x 0.7577 + 37 x 0.2423 =
In general, only ions with +1 charge are detected. It is because +2 ions are very difficult to
generate, as large amount of energy is required to remove the second electron.
The highest m/e ratio of a sample most likely represents the molecular ion.
When the sample is NOT monoatomic, many peaks may appear. We should consider different isotopes in a molecule:
e.g. Peaks of HCl in a mass spectrum
m/e ratio peaks: fragments: 36 1H – 35Cl+
37 2H – 35Cl+
38 39
Peaks of 37 and 39 may be very small due to low % abundance of 2H.
Example 2
The following mass spectrum was obtained for gaseous chlorine molecules. (Chlorine molecules are diatomic) Relative intensities 27 18 3 m/e ratio 70 72 74
(a) How many types of chlorine molecules are there? What are they?
(b) Does chlorine has an isotope with atomic mass 36, i.e. 36Cl?
(c) Calculate the relative atomic mass of chlorine in this sample.
(d) What is the molar mass of chlorine gas (i.e. the relative molecular mass for chlorine molecules)?
The covalent bond within chlorine molecules may be broken by the bombardment of fast moving electrons. Thus the ions 35Cl+ and 37Cl+ may be formed.
Example 3
The mass spectrum of an element, which is diatomic in the vapor state, gives peaks which corresponded to masses of 158, 160 and 162.
(a) What deduction may be made about this element if there were no other peaks in the vicinity of the above numbers?
(b) The heights of the peaks in (a) were in the ratio of 1:2:1.
(i) What is the relative abundance of each of the isotopes you have stated in (a)? (ii) What is the relative atomic mass of the element?
(c) What is the molar mass (i.e. molecular mass) of the diatomic molecule?
Example 4
The mass spectrum of neon consists 3 lines corresponding to relative m/e ratio of 20,21and 22 with relative intensities of 0.91:0.0026:0.088 respectively.
Relative intensities
m/e ratio 20 21 22 How many isotopes are there for neon? (b) Calculate the relative atomic mass of neon.
1.3 The mole concept
A. Mole concept revisited (P.18 – 23)
number of carbon atoms in 12 g of carbon-12. The particles may be atoms, molecules, ions or electrons.
1 mole = 6.02
×
1023 particles. L = 6.02
×
1023 is called the Avogadro constantB. Molar volume of gases (P.24 – 26)
From the Avogadro’s Law, equal volumes of gases, measured at the same temperature and pressure,
contains the same number of molecules (i.e. same no. of mole).
The gas molar volume measured at R.T.P. (room temperature and pressure) = 24.056 dm3 and
the gas molar volume measured at S.T.P. (standard temperature and pressure) = 22.414 dm3.
Gas Molar mass Density at 25°C and 1 atm Molar volume at 25°C and 1 atm.
= molar mass/density Hydrogen 2.0 0.083 24.1 Oxygen 32.0 1.333 24.0 Chlorine 71.0 2.994 23.7 Ammonia 17.0 0.706 24.1 Carbon dioxide 44.0 1.811 24.3 Exercise Calculate:
1. Volume of 2.50 moles of hydrogen at R.T.P.
2. Volume of 1.3 moles of carbon dioxide at S.T.P.
3. The number of atoms in 120 cm3 of chlorine gas at R.T.P.
Important equations!
1. Number of particles in a substance = Number of moles of the substance × L 2. mass Molar mass mole of Number = 3. ) (dm solution of Volume (mol) solute of moles of Number ) (moldm solution a of Molarity -3 = 3 4. Dilution: M1V1 = M2V2
C. Ideal gas equation (P. 27 – 31)
Foundation: quantitative properties of gas (volume, pressure, temperature and amount of gas) and their inter-relationship.
(i) Boyle’s Law: At constant amount and temperature,
P
V ∝ 1 or PV =constant
(ii) Charles’ Law: At constant amount and pressure, V ∝T or V =constant×T
(iii) Avogadro’s Law: At constant temperature and pressure, V ∝n or V =constant×n
Gases approach ideal behaviors at low pressures and high temperatures. Molecules occupy no volume
Molecules are in constant random motion No intermolecular forces between the particles
The collisions involving the gas molecules are totally elastic
Avogadro’s, Boyle’s and Charles’s laws (see textbook P.36 – 37 for details) are now known to be three special cases of a more general equation, the ideal gas equation:
Where P – Pressure (Nm–2, atm. or mmHg)
V – Volume (dm3, cm3)
n – No. of moles (mol)
R – Universal gas constant, 8.3140JK–1mol–1/ 0.0821 atm. dm3 K–1 mol–1
T – Absolute temperature (K)
Example 1
Calculate the pressure inside a television picture tube which has a volume of 5.0 dm3 at a temperature of
25°C, and it contains 0.010 mg of nitrogen.
Gas constant = 0.08206 atm dm3 K–1 mol–1
Relative atomic mass of nitrogen = 14.01
Example 2
What is the volume of 20 g of CO2 at 300°C and 2 atm?
(Given: Relative atomic mass: C = 12; O = 16; R = 8.314 JK–1mol–1; 1 atm. = 1.0130
×
105Nm–2)PV = nRT
Units P V T R
SI units Nm-2 (Pa) m3 K 8.314 JK–1mol–1
Determination of the relative molecular mass of volatile liquid by using the ideal gas law
1. A gas syringe containing a small but known volume of air is fitted with a self sealing rubber cap. A small volume of air is needed to provide some space for the volatile liquid to expand smoothly).
2. Steam is passed through the steam jacket, until the thermometer reading and the volume of air in the syringe become steady. Steady temperature (T1) and initial volume of air (V1) are recorded.
3. The hypodermic syringe is filled with about 1 cm3 of the volatile liquid (e.g. propanone).
4. Hypodermic syringe with the volatile liquid is weighed (m1) (Air bubbles must be expelled from
the hypodermic syringe, to ensure that it is completely filled.)
5. Small amount (e.g. 0.2 cm3) is then injected into the air space of the gas syringe through the
rubber cap.
6. After injection, small hypodermic syringe with small amount of volatile liquid is weighed (m2).
7. After the volatile is completely vaporized and final volume of the gas syringe (V2) is recorded.
8. The atmospheric pressure (P1) is finally recorded.
9. Calculations:
Mass of liquid injected: m=m1−m2
Volume of vapour formed by the injected liquid:V=V2 −V1
By using the equationPV =nRT , we have RT M
m
PV = , where M is the molar mass of the gas/volatile liquid P ) V V ( RT ) m m ( PV mRT M 1 2 2 1 − − = = ∴ (*)
Hint: If the density of that volatile liquid (ρ) is known, we can also determine the value of M by modifying (*) as follows: V m = ρ From (*), = P RT V m M P RT M =ρ ∴
Example 3
In order to determine the relative molecular mass of an unknown compound, an experiment was taken at 363 K by using the above method, the following data was obtained:
Volume and mass of the gas syringe:
Mass of the gas syringe/g Volume of the gas syringe/cm3
Before injection 20.120 V1
After injection 20.255 69.8
What is the relative molecular mass of the compound?
Possible sources of errors:
1. The assumption that the vapour obeys ideal gas behaviour may not be valid.
2. Errors in obtaining the mass of volatile liquid since volatile liquid may evaporate during weighing. 3. Errors in recording the temperature. (T of the steam jacket
≠
T of vapour inside the gas syringe). 4. The volatile liquid does not vapourize completely5. Error in measuring pressure.
Exercise
Using ideal gas equation to calculate:
1. The volume of 1.50 g of hydrogen. H2 at 15°C and a pressure of 750 mmHg (1 atm. = 760 mmHg).
D. Partial pressure of a gas and its relationship with mole fraction (P. 35 – 39)
According to Dalton’s law of partial pressure:
In a mixture of gases, the total pressure exerted is the sum of the pressure that each gas would exert if it were present alone under the same condition. That is,
P
total= P
A+ P
B+ …P
NWhere Ptotal is the total pressure and PA + PB + …PN are the partial pressure of gas A, B and
N respectively
The partial pressure of gas of each species can be calculated by using the concept of mole fraction. From the ideal gas law,
V RT n P A A = , V RT n P B B= V RT n P C C = ……,
Where
n
A,n
B,n
C……are the amount of the corresponding components in the mixtureThen, Ptotal =PA +PB +PC +...
=
+ + +... V RT n V RT n V RT nA B C=
( ...)( ) V RT n n nA + B + C +=
( ) V RT ntotalConsider the partial pressure of A and the total pressure. By dividing PA by Ptotal, we have
total A total A total A n n V RT n V RT n P P = = Rearrangement gives total A A total total A A P P P n n P χ = =
The quantity
χ
A is called the mole fraction of A; it is the ratio of the number of moles of A to thetotal number of moles of gases present
Note:
P
A+ P
B+ P
C+ … = P
total∴
(χ
A+ χ
B+ χ
C+…)P
total= P
total∴
χ
A+ χ
B+ χ
C+… = 1
Exercise
1. 0.25 mole of nitrogen and 0.30 mole of oxygen are introduced into a vessel of 12 dm3 at 50°C.
gases.
2. 46 dm3 of O
2 at 25°C and 1.0 atmosphere was pumped along with 12 dm3 (or litres) of He at 25°C and
1.0 atmosphere into a tank with a total volume of 5.0 dm3. Calculate the partial pressure of each gas
and the total pressure in the tank at 25°C. (Given R = 0.0821 atm dm3 K-1 mol-1)
3. A containing vessel holds a gaseous mixture of nitrogen and butane. The pressure in the vessel at 126.9°C is 3.0 atm. At 0°C the butane condenses completely and the pressure drops to 1.0 atm.
Calculate the mole fraction of nitrogen in the original gaseous mixture.
1.4 Formulae of compounds (P.43 – 53)
making up the compound or ion. It does not represent the actual number of atoms present in a compound or ion.
The molecular formula is the formula showing the actual number of atoms of each element making
up the ion or compound. It is often a simple multiple of the empirical formula of a compound, that is: Molecular formula = n x Empirical formula, where n is a positive integer.
The structural formula is used to show how the constituent atoms are joined up within the
compound or ion.
Example: Butene
Empirical formula Molecular formula Structural formula
CH2 C4H8
A. Relationship between Empirical formula and molecular formula
1. Compound X has an empirical formula CH2 and relative molecular mass 84.0, what is the molecular
formula of X?
Let the molecular formula of compound X be (CH2)n, where n is a positive integer.
Relative molecular mass of (CH2)n = 84.0
∴
(12.0 + 1.0 x 2) n = 84.0∴
n = 6
∴
The molecular formula of compound X is C6H12.B. Derivation of Empirical formula using composition by mass
2. A sample of Mg of mass 0.450 g reacts with excess nitrogen to form 0.623 g of magnesium nitride. Determine the empirical formula of magnesium nitride.
Given: relative molecular masses: Mg = 24.31; N = 14.01
Original mass of Mg = Mass of Mg in magnesium nitride = 0.450 g No. of mole of Mg present = 0.0185
31 . 24 450 . 0 =
Mass of N in magnesium nitride = 0.623 – mass of Mg = 0.623 – 0.450
= 0.173 g
∴
No. of mole of N present = 0.0123 01 . 14 173 . 0 =∴
Mg : N = 0.0185 : 0.0123 = 1.5 : 1= 3 : 2 (simplest integral ratio)
∴
The empirical formula of magnesium nitride should be Mg3N2.C. Derivation of Empirical formula using combustion data
During complete combustion (i.e. O2 is in excess), elements in a compound are oxidized (e.g. H to
H2O, C to CO2, S to SO2, etc).
From the masses of the products formed, the number of moles of these atoms originally present can
be found.
Exercise
3. (a) Vitamin C is an organic compound known to contain the elements carbon, hydrogen and oxygen only. Complete combustion of a 0.2000g sample of this compound yields 0.2998 g CO2 and 0.0819
g H2O. What is the molecular formula vitamin C?
(b) Vitamin C is found by mass spectrometry to have a relative molecular mass 176, what is its molecular formula?
(a) Mass of C in vitamin C = Mass of C in CO2 collected
= 0.2998 ×4412..010 = 0.0818 g
Mass of H in vitamin C = Mass of H in H2O collected
=0.0819 ×182..020 = 0.0092 g
Mass of O in vitamin C = Mass of vitamin C – Mass of C – Mass of H = 0.2000 – 0.0818 – 0.0092
= 0.1090 g
∴
Mole ratio of atomsC : H : O = :016.1090.0 00 . 1 0092 . 0 : 01 . 12 0818 . 0 = 1 : 1.33 :1 = 3 : 4 : 3
∴
The empirical formula of vitamin C should be C3H4O3.(b) Let the molecular formula of vitamin C be (C3H4O3)n, where n is a positive integer.
Relative molecular mass of (C3H4O3)n = 176
∴
(3 x 12.0 + 4 x 1.0 + 3 x 16) n = 176 n = 2∴
The molecular formula of vitamin C should be C6H8O6.*Hint: To deal with this type of calculation more effectively, you may consider tabulating all the things above as follows:
Carbon Hydrogen Oxygen
Mass (g) 0.0818 0.0092 0.1090 No. of moles 6.82 10 3 0 . 12 0818 . 0 − × = 0.0092 0 . 1 0092 . 0 = 6.81 10 3 0 . 16 1090 . 0 − × =
Relative no. of moles
1 10 81 . 6 10 82 . 6 3 3 ≈ × × − − 35 . 1 10 81 . 6 0092 . 0 3 ≈ × − 1 10 81 . 6 10 81 . 6 3 3 = × × − −
Simplest mole ratio 3 4 3
∴
The empirical formula of vitamin C should be C3H4O3.D. Determination of chemical formulae – A summary
To determine the chemical formula of a compound, there are many aspects of analysis to be
considered.
Qualitative analysis: e.g. observing chemical reactions with other substances, to identify the species
present in the compound.
Quantitative analysis: e.g. using combustion data, to find the empirical formula. Instrumental analysis
Determination of molecular formula: e.g. using mass spectrometer to find the relative molecular
mass and hence the molecular formula.
Determination of structural formula: using advanced instruments like infra-red (IR) spectroscopy, NMR (nuclear magnetic resonance), etc.
1.5 Chemical Equations and Stoichiometry
Stoichiometry is the calculation of quantitative relationships of the reactants and products in chemical
reactions
For a chemical reaction as shown below:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
The relative amounts (mole ratio) of reactants and products, are indicated by the coefficients in the balanced equation.
These coefficients are called the stoichiometric coefficients. They reflect the mole ratio of all the
species involved in the chemical equation.
To deal with stoichiometric calculations, writing a balanced equation for the reaction and working out
the appropriate mole ratios are often necessary.
A. Stoichiometric calculations involving reacting masses
Example
Calculate the mass of magnesium oxide formed when 2.43 g of magnesium are burnt with (a) excess oxygen
Given: relative molecular masses: Mg = 24.30; O = 15.99
The balanced equation of the reaction: 2Mg(s) + O2(g) → 2MgO(s)
No. of mole of Mg = 0.1 3 . 24 43 . 2 =
(a) From the equation, 2 moles of Mg give 2 moles of MgO after reaction. Since O2 is in excess, no. of mole of MgO = no. of mole of Mg = 0.1
∴
Mass of MgO formed = 0.1 x 40.3 = 4.03 g (b) No. of mole of 1.28 g O2 = 2 16.0 0.04 28 . 1 = ×From the equation, 1 moles of O2 can react with 2 moles of Mg.
∴
0.04 moles of O2 can react with 0.08 moles of Mg (i.e. O2 is the limiting reactant while Mg is inexcess).
∴
No. of mole of MgO formed = 0.08∴
Mass of MgO formed = 0.08 x 40.3 = 3.22 gB. Stoichiometric calculations involving volume of gases Consider a reaction: a A(g) + b B(g) → c C(g) + d D(g)
Mole ratio:
By the Avogadro’s Law, at fixed T and P, V ∝n
In the reactions involving gaseous reactants and products, mole ratio = volume ratio.
Example 1
Calcium oxide (quicklime) is produced by thermal decomposition of calcium carbonate. Calculate the volume of CO2 at R.T.P produced from the decomposition of 152 g of CaCO3, according to the reaction
CaCO3(s) → CaO(s) + CO2(g)
(Given the relative atomic mass: C = 12.0, O = 16.0 and Ca = 40.0)
No. of mole of 152 g of CaCO3 = 1.52
100 152 =
According to the given equation, 1.52 mol of CaCO3 gives 1.52 mol of CO2 after decomposition.
∴
Volume of CO2 produced at R.T.P = 1.52×
24.056 = 37.53 dm3Example 2
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 residual gas became 50 cm3. Find the molecular formula of that hydrocarbon.
d : c : b : a n : n : n : na b c d = d : c : b : a V : V : V : Va b c d = ∴
Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrocarbon reacted = 10 cm3.
Residual gas ⇒ Unreacted O2(g)
Volume of residual gas = 50 cm3
Volume of O2(g) reacted = 80 – 50 = 30 cm3 Volume of CO2(g) formed = 70 – 50 = 20 cm3 CxHy(g) + ) 4 (x+ y O2(g) → xCO2(g) + 2 y H2O(l)
For mole ratio, CxHy : O2 : CO2 = x
y x ): 4 ( : 1 +
∴
For volume ratio, CxHy : O2 : CO2 = xy x ): 4 ( : 1 + 1 x ) (g H C of Volume (g) CO of Volume y x 2 =
∴
x = 2 1 4 y x ) (g H C of Volume (g) O of Volume y x 2 = +∴
3 4 = + y x As x = 2, we have 3 4 2+y = y = 4∴
The molecular formula of the hydrocarbon is C2H4.Exercise
1. What is the volume of oxygen needed for the complete combustion of 2 dm3 of propane?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Mole ratio, 1 : 5 : 3 : 4
∴
Volume ratio, 1 : 5 : 3 : 4∴
Volume of oxygen needed = 2 x 5 = 10 dm32. 10 cm3 of a gaseous hydrocarbon was mixed with 33 cm3 of oxygen which was in excess. The mixture
was exploded and then cooled to room temperature. The remaining volume of gas occupied 28 cm3.
On adding concentrated potassium hydroxide the volume decreased to 8 cm3. Find the molecular of
that hydrocarbon.
Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrocarbon reacted = 10 cm3
Volume of O2(g) reacted = 33 – 8 = 25 cm3 CxHy(g) + ) 4 (x+ y O2(g) → xCO2(g) + 2 y H2O(l)
For mole ratio, 1 : ) 4
(x+ y : x For volume ratio, 10 : 25 : 20
20 10 1 = x
∴
x = 2 10 25 4 = + y x∴
1025 4 2+ y = y = 2∴
The molecular formula of the hydrocarbon is C2H2.3. 10 cm3 of a hydrocarbon X was exploded with excess oxygen. The mixture was exploded and then
cooled to room temperature. There was a contraction in volume of 35 cm3 (all volumes were measured
at R.T.P. After treatment with concentrated sodium hydroxide, There was another contraction of 40 cm3. Deduce the molecular formula of X.
Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrocarbon reacted = 10 cm3.
Volume of CO2(g) formed = 40 cm3
For mole ratio, 1 : ) 4
(x+ y : x For volume ratio, 1 : )
4 (x+ y : x 10 : 10 ) 4 (x+ y : 10x 10 x = 40 x = 4
By contraction (i.e. decrease) in volume of 35 cm3,
)] 40 35 4 y x ( 10 10 [ + + − = y = 10
∴
The molecular formula of the hydrocarbon is C4H10. Always remember: Molarity = ) dm (in solution of Volume solute of mole of No. 3 Example
(a) 11.5 g of solid NaOH was dissolved in distilled water to make 1.50 dm3 solution. What is the
concentration of NaOH in mol-1dm3?
(b) What mass of hydrogen chloride, HCl, would be in 100cm3 of 2M HCl(aq)?
(c) To neutralize 50 cm3 of solution NaOH in (a), what volume of 2M HCl is used?
(a) No. of mole of NaOH = 0.288 0 . 40 5 . 11 =
Concentration of NaOH solution = 0.192 5 . 1 288 . 0 = M (b) 100 cm3 = 0.1 dm3
∴
No. of mole of hydrogen chloride in this acid = 0.2∴
Mass of HCl = 0.2 x (1.0 + 35.5) = 7.3 g(c) Let V be the volume of 2M HCl required.
The balanced equation of the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
∴
No. of mole of HCl required = No. of mole of NaOH∴
0.192 x 0.05 = 2VV = 0.0048 dm3 = 4.8cm3
1.6 Titrations
Titration is the procedure for finding the molarity of an unknown solution by controlled addition of
a known volume of standard solution to another solution with unknown concentration, until complete reaction.
A. Standard solution ( 標準溶液 )(P. 68-69) Standard solution: A solution of known concentration.
Primary standard: A pure compound from which a standard solution of accurately known
concentration can be prepared directly.
Criteria to be met for a compound to be used as a primary standard:
must be available in pure form;
should be stable in air over long periods of time; should not absorb CO2 and H2O from the atmosphere; should dissolve in water easily;
should not decompose in solution;
should have a larger molar mass (usually M > 100 gmol-1), so that the weighing error in using a
Some commonly used primary standards: sodium carbonate (Na2CO3, Mr = 106), sodium
ethanedioate (i.e. sodium oxalate, Na2C2O4, Mr = 134).
B. Back Titration
In some cases, direct titration may not be easily done for the reaction between reagent X and
reagent Y, because:
1. One of the reagents is a solid and thus the reaction is very slow. (i.e. we do not know whether
the reaction has completed.) For example, we cannot titrate solid Mg(s) or CaCO3(s) with
standard HCl(aq) although they react with each other.
2. No suitable indicator for the detection of end-point. For example, titration between NH4Cl(aq)
with KOH(aq) would have NO suitable acid-base indicator.
In this case, we would use back titration.
An excess but known amount of reagent A is added to the reagent B. When the reaction is
completed, the excess reagent A can be titrated with another reagent C. (A suitable indicator can be found for the titration between reagent A and C).
For case 1:
If we want to find out the amount of CaCO3 in a sample, first, add excess with known concentration
of HCl into the CaCO3(s) sample. After complete reaction, the excess amount of acid can be ‘back
titrated’ with standard NaOH solution.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
No. of moles of HCl = 2(no. of moles of CaCO3(s)) + no. of moles of NaOH For case 2:
The amount of ammonium salt (e.g. NH4NO3) in a sample can be found by adding a known volume
of standard alkali, e.g. 50.0 cm3 of 2 M NaOH(aq). The reaction mixture is then warmed to drive off
the product NH3(g). When the reaction is completed, excess NaOH can be determined by titration
with standard acid, e.g. 2 M HCl.
NH4NO3(aq) + NaOH(aq) → NaNO3(aq) + NH3(g) + H2O(l)
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
No. of moles of NaOH = no. of moles of NH4NO3 + no. of moles of HCl
C. Types of titrations
1. Acid-Base titrations ( 酸鹼滴定 )
Example: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Equivalence point (當量點): The point at which the amount of standard solution added is equivalent,
or equal, to the amount of substance to be analyzed present in the sample.
point is reached. An acid-base indicator is often used, which would give a sharp colour change at the end point (終點) that is supposed to be very close to the equivalence point.
End point: The point which the indicator changes colour.
Different acid-base indicator may be used according to different strength of acid and base involved.
For choices of indicator, refer to textbook P. 59 – 60 and they will be further discussed in Section 6.
The end point can also be detected by:
A change in temperature (thermometric titration). Since neutralization is exothermic, the
temperature of the reaction mixture is expected to rise to a maximum at the equivalence point.
A change in electrical conductivity. The equivalence point of a neutralization reaction should be
with electrical conductivity.
Change in pH value (using pH meter)
2. Acid-carbonate titrations
Similar to acid-base titration. For example: Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) For soluble carbonate (e.g. Na2CO3), the end-point can be detected by acid-base indicator.
For insoluble carbonate (e.g. CaCO3), back titration is used.
been completed reacted.
The amount of excess acid is determined by titration with standard alkali.
3. Redox titrations ( 氧化還原滴定 )
Redox titration means titration of two solutions which undergo redox reaction.
Some applications: determination of heavy metal content in drinking water, vitamin C content in fruits and vegetables, etc.
Case 1: Determination of thiosulphate ( 硫代硫酸根離子 ) or iodine
A standard iodine solution is used to titrate a thiosulphate solution to determine its concentration (i.e.
to standardize it).
The standardized thiosulphate solution can be used to titrate another iodine solution of unknown
concentration.
Standardization of thiosulphate solution:
1. A standard solution of iodine is prepared by dissolving a known amount of pure potassium
iodate(V) (KIO3) in an acidic medium, containing excess potassium iodide.
IO
3–(aq) + 5I
–(aq) + 6H
+(aq) → 3I
2(aq) + 3H
2O(l)
Note: No. of mole of I2(aq) formed = No. of mole of IO3-(aq) added
×
3Hint 1: Standard iodine solution cannot be prepared by dissolving iodine in water because:
iodine is only slightly soluble in water iodine may easily lost through sublimation
Hint 2: The solution of iodine is brown in colour and should be used immediately (
iodine may easily lost through sublimation).2. The standard iodine solution is then used to titrate with a thiosulphate solution of unknown
concentration. Here iodine oxidizes thiosulphate ions to tetrathionate, and forms iodide ions.
I
2(aq) + 2S
2O
32–(aq) → S
4O
62–(aq) + 2I
–(aq)
thiosulphate tetrathionate
(brown) (colourless) (colourless) (colourless)
Hint: Standardization of thiosulphate with standard I2(aq) is necessary because it is unstable:
in acidic medium,
in the presence of microorganism,
in the presence of Cu(II) ions (Cu2+),
under sunlight
3. Since the colour of iodine in iodide solution cannot be used to accurately detect the end point (
colour change: brown → pale yellow → colorless, very difficult to observe). Thus, starch is used as the indicator:Starch + iodine → blue complex (reversible)
4. Since starch would react with iodine irreversibly at high concentration of I2, so starch solution
cannot be added at the beginning of the titration as it would hinder the reaction between iodine and thiosulphate.
5. Starch solution should be added in a later stage of the titration (when the brown colour of
iodine changes from brown to pale yellow). After the adding of starch solution, the solution turns deep blue.
6. The end point is shown by the complete decolorization of the blue colour. (Iodine is
completely reacted).
7. After standardization, that thiosulphate solution can be used to titrate another solution of iodine
of unknown molarity.
Example 1 – Thiosulphate determination
2.015 g of pure potassium iodate(V)(碘酸鉀) was dissolved and made up to 250cm3. Excess potassium
iodide and dilute sulphuric acid were also added to a 25.0 cm3 portion of this standard solution. The
solution was then titrated with a solution of sodium thiosulphate, starch solution was added near the end-point. In the end, 29.8 cm3 of thiosulphate solution were required. Calculate the concentration of
thiosulphate solution. (Given Mr of KIO3 = 214)
No. of mole of IO3-(aq) added = 10
1 214 015 . 2 × = 9.42
×
10-4∴
No. of mole of I2 (aq) formed = 3×
9.42×
10-3= 2.83
×
10-3From the equation I2(aq) + 2S2O32-(aq) → S4O62-(aq) + 2I-(aq)
No. of mole of S2O32-(aq) used = 2.83
×
10-3×
2= 5.66
×
10-3 29.8 cm3 = 0.0298 dm3∴
S2O32-(aq)] = 0298 . 0 10 66 . 5 × −3 = 0.190 MExample 2 – Estimation of copper(II) ions
It is known that copper(II) ions oxidize iodide ions to iodine: 2Cu2+(aq) + 4I-(aq) → I
2(aq) + 2CuI(s)
The iodine produced can be titrated with standard thiosulphate solution, the amount of copper(II) ions can be deduced from the amount of iodine produced. A sample of 4.256 g of copper(II) sulphate-5-water was dissolved and made up to 250 cm3. A 25.0 cm3 portion is added to an excess of potassium
iodide. The iodine formed required 18.0 cm3 of a 0.0950M sodium thiosulphate solution for reaction.
Calculate the percentage of copper in the crystal. (Relative atomic mass of Cu = 63.5) I2(aq) + 2S2O32–-(aq) → S4O62–-(aq) + 2I–(aq) --- (1)
2Cu2+(aq) + 4I–(aq) → I
2(aq) + 2CuI(s) ---(2)
From (1), No. of mole of S2O32-(aq) = No. of mole of I2 (aq) formed
×
2∴
No. of mole of I2 (aq) formed = 1000 )0 . 18 ( 095 . 0 2 1 × × =8.55
×
10-4∴
No. of mole of Cu2+(aq) in 25.0 cm3 solution = 2×
8.55×
10-4∴
No. of mole of Cu2+(aq) in 250 cm3 solution = 1.71×
10-2∴
Mass of copper in the sample = 1.71×
10-2×
63.5 = 1.086 g% by mass of copper in the crystals = 100% 25.5% 256 . 4 086 . 1 = ×
Case 2 Potassium manganate(VII )titrations
Acidified potassium manganate(VII) solution is a strong oxidizing agent
Equation:
Once it has been standardized, it can be used to determine the amount of various reducing agents such as iron(II) salts.
No indicator is needed as the oxidant changes from purple to colourless during titration. At the end
point, a light but permanent purple colour (persists for at least 30 seconds) appears due to the addition of a drop of excess KMnO4(aq).
It should be noted that the redox reaction involved occurs slowly at room temperature. Thus the
reaction mixture should be heated to about 60°C so that the reaction takes places at a suitable rate.
Example 1 – Standardization of potassium manganate(VII) solution by using sodium ethanedioate*
For example, a 25.0 cm3 of sodium ethanedioate solution of concentration 0.200 moldm-3 was titrated
against a solution of acidified potassium manganate(VII). If 17.2 cm3 of potassium manganate(VII)
were used, what is the concentration of the solution?
From the equation: 2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Mole ratio, MnO4-(aq) : C2O42-(aq) = 2 : 5
No. of mole of MnO4-(aq) =
×
5
2
No. of mole of C2O42-(aq)
∴
No. of mole of MnO4-(aq) =×
5
2
5.00×
10-3 = 2.00×
10-3∴
[MnO4-(aq)] = = × × − − 3 3 10 2 . 17 10 00 . 2 0.116 moldm-3*Note 1: ethanedioate ions are also known as oxalate ions (草酸根離子).
Note 2: Standardized potassium permanganate solution can be used in a wide variety of redox reaction for the determination of many different species, for example:
oxalate content in some vegetables (e.g. spinach)
hydrogen peroxide content in commercially available hydrogen peroxide solution
iron content of iron tablets that can be obtained from any pharmacy
metal content (e.g. copper, iron and tin) in some alloys or impure metal samples
Example 2 – Iron determination
An impure sample of iron of mass 7.50 g was dissolved in dilute sulphuric(VI) acid and the solution was made up to 250 cm3. The solution contained iron(II) ions together with the impurities. 25.0 cm3 of
this solution was titrated with potassium manganate(VII) solution mentioned in the Example 1. The average volume of potassium manganate(VII) solution used was 20.00 cm3. Determine the percentage
of purity of iron in the sample (Given relative atomic mass of iron = 56.0).
From the equation: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Mole ratio, MnO4-(aq) : Fe2+-(aq) = 1 : 5
No. of mole of Fe2+-(aq) in 25.0 cm3 solution = No. of mole of MnO
4-(aq)
×
5= (0.116
×
20×
10-3)×
5= 0.0116
