Fundamental of
Analytical Chemistry
What is Analytical Chemistry?
The First lecture covers the following points:-
What is Analytical Chemistry
Classification of Analytical Methods
Plane of Analysis
-
Introduction
Analytical chemistry is a measurement
science consisting of a set of powerful ideas and
methods that are useful in all fields of science and
medicine. Qualitative analysis establishes the
chemical identity of the species in the sample.
Quantitative analysis determines the relative
amounts of these species or analytes in numerical
terms.
Analytical chemistry plays a vital role in the
development of science. The concentration of
oxygen and of carbon dioxide are determined in
millions of blood samples every day and use to
diagnose and treat illnesses. Quantities of
hydrocarbons, nitrogen oxides and carbon
monoxide present in automobile exhaust gases are
measured to determine the effectiveness of smog
control devices. Quantitative measurements of
ionized calcium in blood serum help diagnose
parathyroid disease in humans…. etc
Quantitative analytical measurements also
play a vital role in many research areas in
chemistry, biochemistry, biology, geology, physics
and the other sciences.
Classifying Quantitative Analytical Methods
We compute the results of a typical
quantitative analysis from two measurements. One
is the mass or the volume of sample to be
analyzed. The second is the measurement of some
quantity that is proportional to the amount of
analyte in the sample, such as mass, volume,
intensity of light or electrical change.
We classified quantitative methods into:
1- Volumetric Methods
2- Gravimetric Methods
3- Electro analytical Methods
4- Spectroscopic Methods
5- Miscellaneous Methods
Plane of Analysis
Atypical quantitative analysis involves the
sequence of steps:
 1- Defining the problem.
At first present a case study to illustrate steps in
solving an important and practical analysis problem
2- Sampling
The next step in a quantitative analysis is to acquire
the sample. To produce meaningful information, an
analysis must be performed on a sample whose
composition faithfully represents that of the bulk of
material from which it was taken. Where the bulk is large
and heterogeneous, great effort is required to get
representative sample.
3- Processing the sample
The third step in an analysis is to process the sample in
any of a variety of different ways. The first step in
processing the sample is often the preparation of a
laboratory sample.
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a-Drying the sample.
b-Dissolving the sample.
c-Performing the required separations.
4-Making the appropriate measurements.
All analytical results depend on a final measurement
of a physical or chemical property of the analyte. This
property must vary in a known and reproducible way
with the concentration of the analyte.
 5-Making data presentation.
Analytical results are incomplete without an estimate
of their reliability. The experimenter must provide some
measure of the uncertainties associated with computed
results if the data are to have any value.
Fundamental of
Analytical Chemistry
Titrimetric Methods of Analysis
Lecture Two covers the following points:- Titrimetric Methods
- -Some General Aspects
-Definition of Some Terms
Titrimetric methods include a large and powerful
group of quantitative procedures that are based upon
measuring the amount of reagent of known concentration
that is consumed by the analyte. Volumetric titrimetry
involves measuring the volume of a solution of known
concentration that is needed to react essentially
completely with the analyte.
Titrimetric methods are widely used for routine
analyses because they are rapid, convenient, accurate and
readily automated.
Titration:

It is a process where Titrant slowly add to
Analyte until the reaction between the two solution
is judged complete
The volume of reagent needed to complete the
titration is determined from the difference between
the initial and final volume readings
Equivalence point
This point in a titration is reached when the
amount of added titrant is chemically equivalent to the
amount of analyte in the sample.
It is a theoretical point that cannot be determined
experimentally.
End Point:
It is the point at which physical change
associated with the condition of equivalence.
Titration Error
The difference in volume
equivalence point and the end point
between
the
Indicator
 They are reagents often added to the analyte
solution in order to give an observable physical
change at the end point or near the equivalence
point.
Fundamental of
Analytical Chemistry
Some General Aspects
Lecture Three covers the following points:-Some General Aspects
-Definition of Some Terms
Standard Solution:

It is a reagent of known concentration that is
used to carry out a titrimetric analysis
A Primary Standard Substance:

It is a highly purified compound that serves as
a reference material in all volumetric titrimetric
methods.
Important properties of a primary standard substance:
 The substance should be readily purified.
 It should not be hygroscopic.
 It should not be contain hydrate water.
 It should be readily soluble.
 It should have a relatively high equivalent weight.
 It should be available and not too expensive.
 It should be stable toward air.
 It should be stable on storage.
 It should not undergo any side-reaction.
Compounds that meet or even approach these criteria
are very few, and only a limited number of primary
standard substances are available to the chemist.
The compounds that are less pure must sometimes be
employed in lieu of a primary standard. The purity of
such a secondary standard must be established by careful
analysis.
Ideal Standard Solution:
 Be sufficiently stable .
 React rapidly with the analyte.
 React more or less completely with the analyte.
 Undergo a selective reaction with the analyte.
The standard solution is prepared by dissolving an
accurate weight of pure reagent in water this is then
diluted to an exactly known volume.
Fundamental of
Analytical Chemistry
Concentration Unites
Lecture Four covers the following points:-
Concentration unites
Method of Calculation
Methods for expressing the concentration of standard
solutions:
The concentration of standard solutions are
generally expressed in units of either molarity M or
normality N .
Molarity

It is the number of gram moles of reagent that is
contained in one liter solution. (mol/L)

Molarity

M = No. of mole / Volume ( L)

No. of mole = Wt./ M. Wt.

M= Wt. / [M. Wt. X

Wt. (g) = M x M.Wt. x V (L)
M = No. of mole / Volume ( L)
V ]
Normality
•
It is the No. of gram equivalents contained in one
liter.
•
•
N = No. of gram equivalent / Volume ( L)
It is the number of milliequivalents of solute in one
milliliter of solution.
Strength
• It is the number of grams of solute in one liter
solution .
• Wt. (g) = M x M.Wt. x V (L)
• Strenth = M x M.Wt. x 1
• Strenth = M X M. Wt.
g/L
• Strenth = N X eq.Wt.
g/L
Percent concentration
123-
weight percent
volume percent
weight/volume percent
Part per million
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Cppm = mass solute (mg) / volume solution
(L)
=
mg
/ 1000 g
=
mg
/ 1000 x 1000 mg
mg / Kg
Part per billion


ppm
ppb
Cppb = ug / 1000 x 1000 x 1000 ug
1 ppb = 10-3 ppm
Fundamental of
Analytical Chemistry
Neutralization Titration
Lecture Five covers the following points:-

Types of neutralization reaction
Types of indicator used
Which one will be used in Quantitative titration

Types of neutralization reaction

Strong Acid + Strong Base = Salt + Water
HCl
+ NaOH
= NaCl + H2O
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
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
Strong Acid + Weak Base = Salt + Water
HCl
+ NH4OH
= NH4Cl + H2O
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Weak Acid + Strong Base = Salt + Water
CH3COOH + NaOH = CH3COONa + H2O
Weak Acid + Weak Base = Salt + Water
CH3COOH + NH4OH = CH3COONH4 + H2O
Neutralization Curves
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Preequivalence
Equivalence
Postequivalence
Titrating a strong acid with a strong base:
We will be calculate hypothetical titration curves of
pH versus volume of titrant


50 ml of 0.05 M HCl with 0.1 M NaOH

Initial point:
pH= - log [H3O+]
= - log 0.05 = 1.3
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Preequivalance:
After adding 10 ml of NaOH
CHCl = no. mmol HCl –no. mmol NaOH/ total V
= (M x V)HCl
_ ( M x V )NaOH
/ Vtotal
= (0.05 x 50 ) _ ( 0.1 x 10 ) / 60
= 2.5 x 10-2
pH = - log (2.5 x10-2) = 1.6

After addition of 24.9 ml NaOH
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CHCl = (M x V )HCl - ( M x V )NaOH /
Vtotal
= ( 0.05 x 50 ) – ( 0.1 x 24.9) / 74.9
= 2.5 - 2.49 / 74.9 = 0.01/ 74.9
= 1.33 x10-4
pH = - log ( 1.33 x 10-4 )
= 3.87
Equivalence point:
[ H3O+] = ( Kw )1/2 = ( 1 x 10-14)1/2 =
1 x 10-7
pH = - log ( 1 x 10-7 ) = 7
Postequivalence :
After addition of 25.1 ml NaOH
CNaOH = (M x V )NaOH - ( M x V )HCl / Vtotal
= ( 0.1 x 25.1 ) – ( 0.05 x 50) / 75.1
= 0.01/ 75.1 = 1.33 x 10-4
pOH = - log ( 1.33 x 10-4 )
= 3.87
pH = 14 – 3.87 = 10.13
After addition of 40 ml NaOH
CNaOH = (M x V )NaOH - ( M x V )HCl / Vtotal
= ( 0.1 x 40 ) – ( 0.05 x 50) / 90
= 4 – 2.5 / 90 = 1.5 / 90 = 0.0166
pOH = - log ( 0.0166 )
= 1.77
pH = 14 – 1.77 = 12.23
Acid –Base Indicator
Many substances both naturaly occurring and synthetic,
display colors that depend upon the pH of the solutions
in which they are dissolved.
An acid – base indicator is a weak organic acid or
weak organic base whose undissociated form differs in
color from its conjugate base or its conjugate acid
form. For example, the behavior of an acid type
indicator, HIn , is described by the equilibrium:
HIn + H2O = In- + H3O+
Here , internal structural changes accompany
dissociation and cause the color change. The
equilibrium for a base type indicator In, is:
In + H2O = InH+ + OHIn the following paragraphs we focus on the behavior
of acid type indicators. The discussion, however, can
be readily extended to base type indicator as well.
THE EQUILIBRIUM CONSTANT EXPRESSION
FOR DISSOCIATION OF AN ACID INDICATOR
TAKES THE FORM:
Ka = [H3O+] [ In- ] / [ HIn]
Rearranging leads to
[H3O+] = Ka
[HIn ] / [ In- ]
- log [H3O+] = - log Ka - log [HIn ] / [ In- ]
pH = pKa - log [HIn ] / [ In- ]
The human eye is not very sensitive to color
differences in a solution containing a mixture of In -
and HIn , particularly. When the ration [In-] / [HIn] is
greater than about 10 or smaller than 0.1.
Consequently, the color imparted to a solution by a
typical indicator appears to the average observer to
change rapidly only within the limited concentration
ratio of approximately 10 to 0.1.
To obtain the indicator pH range:
pH ( acid color) = pKa - 1
pH ( basic color) = pKa + 1
Fundamental of
Analytical Chemistry
Neutralization of Weak Acid with
Strong Base
Lecture Six covers the following points:-
Titration of weak acid and strong base
Titration of weak base and strong acid
-
Titration of weak acid and weak base
Neutralization of a weak acid with a strong base
•
•
•
•
CH3COOH + NaOH = CH 3COONa + H2O
Titration of 50 ml of 0.1 M acetic acid with 0.1
M NaOH
( Ka = 1.75 x 10-5 ).
Initial point:
pH= 2.88
•
•
•
•
•
The effect of reaction completeness:
Titration curves for 0.1 M solutions of acids
with different dissociation constants are shown
in Fig. Note that the pH change in the
equivalence point region becomes smaller as
the acid becomes weaker that is ,as the reaction
between the acid and the base becomes less
complete.
Indicator choice:
The choice of indicator for the titration of
weak acid and strong base is more limited than
that for a strong acid and strong base.
Neutralization of a weak base with a strong acid



NH4OH + HCl = NH 4Cl + H2O
Neutralization of a weak base with a weak acid
The titration curve for the neutralization of
weak acid and weak base shows that the change
of pH near the equivalence point and during the
whole of neutralization curve is very gradual.
There is no sudden change in pH and hence no
sharp end point can be found with any simple
indicator.
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Fundamental of
Analytical Chemistry
Neutralization of polyprotic Acid
with a strong base
Lecture Seven covers the following points:-
Reaction of polyprotic acid
Titration curves of diprotic and triprotic acid
Displacement titration
Titration of mixtures
Diprotic Acids
H2A + 2 OH- → A2- + 2 H2O
H2A + OH- → HA- + H2O
HA+ OH- → A2- + H2O
If
K1 / K2 > 10000
The solution behaves like a mixture of two acids
pH = ½ ( pK1 + pK2 )
Triprotic Acids
• H3PO4 + OH- = H2PO4+ H2O
K1= 7.5x10-3
• H2PO4- +
OH- = HPO42- + H2O
K2=6.2x10-8
• HPO42- +
OH- = PO43+ H2O
K3=5.0x10-13
•
•
•
•
•
•
K1/K2> 10000
K2/K3 > 10000
Phosphoric acid can be titrated
monoprotic and diprotic acid
pH = ½ (pK1 + pK2)
= ½ (2.23 +7.21) = 4.66
as
pH = ½ ( pK2 + pK3 )
= ½ ( 7.21 + 12) = 9.6
Analysis of Mixtures
Na2CO3 and NaHCO3
Na2CO3 and NaOH
HCl and CH3COOH
Fundamental of
Analytical Chemistry
Redox Reaction
Lecture Eight covers the following points:-
Redox reaction
Types of reactions
Types of oxidant and reductant
There are a very large number of oxidation
reduction reactions that are used as the basis
for analytical methods.
For example:
CeCl4
+ FeCl2
= CeCl3 + FeCl3
Ce4+ + 4 Cl- + Fe2+ + 2 Cl- =
Ce3+ + 3 Cl- + Fe3+ + 3
ClCe4+ + Fe2+
= Ce3+ + Fe3+
Ce4+
= Ce3+
Fe2+
=
Fe3+
Oxidation Reaction: is the loss of electrons
Reduction Reaction: is the gain of electrons
Oxidation/Redouction reaction is one in which
electrons are transferred from one reactant to
another.
Red : Reductant
It is an electron donor
OX: Oxidant
It is an electron acceptor
Standard redox potential: Eo
The higher Eo
The stronger OX
the weaker Red
Eo of Cl2/Cl- = 1.36 V
Eo of Fe3+/Fe2+ = 0.77 V
Cl2 : OX
Fe2+ : Red
2 Fe2+ + Cl2 = 2 Fe3+ + 2 ClAn oxidizing agent with a higher potential can
oxidize any reducing agent with a lower potential
A reducing agent with a lower potential can reduce
any oxidizing agent with a higher potential
E = Eo + 0.059/n log [OX]/[Red]
Fundamental of
Analytical Chemistry
Titration with Oxidizing Agents
Lecture Nine covers the following points:-
Oxidizing agent
Potassium Permanganate
Iodine
Redox indicator
Potassium Permanganate
• Very strong oxidizing agent Eo=1.51 V
• Self indicator
• Standardized
by primary standard sod.
Oxalate solution
• 5C2O42- + 2MnO4- + 16H+ =
•
10CO2 + 2Mn2+ + 8H2O
• Determination of ferrous in presence of Cl• 2MnO4- + 10 Cl- + 16 H+ =
•
2 Mn2+ + 5 Cl2 + 8 H2O
by using Zimmermann reagent which consists
of
• MnSO4 + H2 SO4 + H3PO4
•
•
•
Mn II: reduce the potential of permanganate
so it will not oxidize chloride ions
H3PO4: form a complex with ferric and
decrease the potential of iron
•
Iodine
Iodimetry
2 S2O32- + I2 = S4O62- + 2 I-
This titration are usually performed in neutral
solution.
In alkaline solution:
I2 + OH- = HIO + I- + H2O
hypoiodite
3 HIO + 3 OH- = IO3- + 2I- + 3 H2O
iodate
•
•
•
consumption of iodine
In acid solution:
1- starch tends to hydrolyze
so the end point affected.
•
2- reducing power is increased in neutral
solution
•
3-iodid produced in the reaction tends to be
oxidized
•
4 I- + O2 + 4 H+ = 2 I2 + 2 H2O
• 2 Cu2+ + 2 I= 2 Cu+ + I2
Iodometry
2 S2O32- + I2 = S4O62- + 2 Immole of Cu2+= mmole of I2 =mmole of S2O32• Most iodometric titration are performed in
strong acid solution
• To minimize air oxidation of iodide
• Thiosulphate is decomposed in acid solution
•
S2O32- + 2H+ = H2SO3 + S
•
stirring
Oxidation – Reduction indicators
 Self indicator:
permanganate
 Specific indicator : starch
 External indicator: sod. Thiosulphate
 Internal indicator

They are substances that change color
upon being oxidized or reduced
Fundamental of
Analytical Chemistry
Precipitation Titrations
Lecture Tin covers the following points:-
Precipitation titration
Indicators
The formation of a precipitate can be used as the basic of
a titration.
Volumetric method based on the formation of a slightly
soluble product are Precipitation Titrations
Solubility Rules
1- All common salts of alkali metal are soluble.
2- Acetates, chlorates, nitrates, nitrites, perchlorate are
soluble and silver nitrites is insoluble.
3- Cl, Br, I, SCN, of most cations are soluble except
Cu, Pb, Hg.
4- Sulfate, thiosulfate are soluble except Sr, Ba, Pb,
silver thiosulfate, mercuric sulfate are insoluble.
5-Arsenate, arsenite, borate, carbonate, chromate,
cyanide, oxalte, phosphate, sulfite are insoluble except
for those of alkali metal.
6- Hydroxide of alkali metal are soluble.
7- The floride of all metal cations are insoluble except
for ammonium. silver, mercury and alkali metal
cations.
AgX = Ag+ + XK = [Ag + ] [ Cl - ] / [ AgCl ]
[AgCl] = 1
because AgCl is present as a pure solid
( ionic concentration are small)
K sp = [Ag + ] [ Cl - ]
Where K sp is called the solubility product
Argentometric method.
1- The ppt must be practically insoluble.
2- Precipitation should be instantaneous.
3- The results of titration should not be effected by
adsorption.
4- The end point should be easily detected.
5- At equivalent point
K sp = [Ag + ] [ Cl - ]
Where K sp is called the solubility product
Chemical Indicators for Precipitation Titrations
Formation of a colored precipitate ( Mohr Method )
Formation of a colored complex ( Volhard Method )
Adsorption Indicators Method ( The Fajans Method
Fundamental of
Analytical Chemistry
Complex Formation Titrations
Lecture Eleven covers the following points:- Complex formation reaction
-Types of ligand
-Advantage of multidentate ligands
-Metalochromic Indicators
-Requirement of Metallochromic Indicators
A complexation reaction involves a reaction
between a metal M and another molecules (
Ligand) L containing at least one atom with an
unshared pair of electron
Types of ligand
 Unidentat Ligands
NH3 H2O Cl SO2 COOH
 Bidentat Ligands
NH2CH2CH2NH2 (en)
 Tetradentat Ligands
NH2CH2CH2
NH
CH2CH2
NH
CH2CH2NH2
(Trien)
 Polydentat Ligands
Ethylene Diamin Tetra Acetic acid EDTA
Advantage of multidentate ligands
 They generally react more completely with
cations and thus provide sharper end point
 They react with metal ions in a single step
Metalochromic Indicators
Indicators are highly colored organic dyes that
form complexes of a different color with the metal
ions.
M + In = MIn
complex
MIn + EDTA = In + M-EDTA complex
Requirement of Metallochromic Indicators
1- The metal indicator complex must be less stable
than the metal-EDTA complex.
2- The indicator must be very sensitive towards the
metal ions.
3-The metal indicator complex must be formed
under the same pH as the metal-EDTA complex.
4-Metal indicator change color over a range of 2
pM units. To obtain a sharp and accurate end point.
Fundamental of
Analytical Chemistry
Qualitative Methods of Analysis
Lecture Twelve covers the following points:-
Qualitative Analysis
Buffer Solution
Solubility
Solvent effect
Group of Analysis
Acid Radical
Basic Radical
The procedures of inorganic qualitative analysis
involve mainly the separation of various ions
through selective precipitation, that is, first
precipitating a group of several different kinds
of cations with one anion, then redissolving that
precipitate, and finally detecting each ion in the
group by tests- often the formations of other
precipitates – on the resulting solution. Acid –
base chemistry is important in all of these
precipitation reactions and throughout the
qualitative analysis scheme.
Therefore, a firm grasp of the principles of acid
– base chemistry is important in understanding
qualitative analysis and in carrying out a
successful analysis.
I- Buffer Solutions
• Whenever A WEAK ACID IS TITRATED
WITH A STRONG BASE OR A WEAK
BASE WITH A STRONG ACID A BUFFER
SOLUTION
CONSISTING
OF
A
CONJUGATE ACID/BASE PAIR IS
FORMED
Calculating the pH of buffer solutions
• Buffers formed from a weak acid and its
conjugate base
• A solution containing a weak acid HA and its
conjugate base A- , may be acidic , neutral or
basic depending on the position of two
competitive equilibria:
If the first equilibrium lies farther to the right
than the second, the solution is acidic. If the
second equilibrium is more favorable, the
solution is basic
•
HA + H2O ↔ H3O+ + AKa = [H3O+ ] [A- ] / [HA ]
•
A- + H2O ↔ OH- + HA
•
•
•
•
•
•
•
•
•
Kb = Kw/ Ka = [OH- ] [HA ] / [A- ]
Buffers formed from a weak base and its
conjugate base
A solution containing a weak base and its
conjugate acid , may be acidic , neutral or
basic depending on the position of two
competitive equilibria:
HA = H+ + AKa = [ H+ ] [ A- ] / [HA]
[H+ ] = Ka [ HA] / [A-]
- log [H+ ] =- log Ka – log [ HA] / [A-]
- log [H+ ] =- log Ka + log [ A-] / [HA ]
•
•
pH = pKa + log [Salt]/[Acid]
•
Calculate the pH of 50 ml 0.1 M acetic acid
and 49.9 ml 0.1 M NaOH ( Ka = 1.75 x 10-5
).
HA + NaOH = Na A + H2O
pH = pKa + log [Salt]/[Acid]
•
•
•
•
•
•
•
•
•
•
•
•
CHA = no. mmol of HA - no. mmol of base /
V
= (M x V)HA – (M x V)NaOH / Vtotal
=(0.1 x 50) – ( 0.1 x 49.9) / 99.9
= 0.01 / 99.9 =0.0001
CA- = no. mmol of base / Vtotal
= (M x V)NaOH/Vtotal =(0.1 x 49.9)/99.9
=0.0499
pH= pKa + log [Salt] / [Acid]
= 4.75 + log 0.0499/0.0001
=4.75 + 2.69
=7.44
II- Solubility and Complex Equilibria
Precipitation reactions are crucial in inorganic
qualitative analysis, particularly in the analysis of
cations. Cations are separated from the sample by
precipitating them in groups. These group
precipitates then are dissolved and the ions are
separated out from the resulting solution, usually
by further precipitations. Many of these
precipitation reactions depend strongly on the
acidity of the solution and thus the prior study of
acid-base reactions is quite important to the
understanding of precipitation reactions. Although
many precipitation reactions can be predicted with
simple solubility rules, often it is necessary to
calculate the final concentrations of ions present
after precipitation to determine whether subsequent
reactions are feasible.
Complex ion formation reactions are used in
qualitative analysis for two purposes. First , many
cations are identified through the distinctive colors
of the complex ions they form. These colors are
explained, and in some instances predicted by
crystal field theory. Second, when two cations are
presen in solution, one many interfere with the
detection of the second. In some of these instances,
the interference can be eliminated by forming a
complex ion of the offending ion, sequestering it in
solution, and preventing it from reacting with the
identifying reagent and masking the test results.