Analytical pharmaceutical Chemistry1-1.ppt

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1. 1
Analytical Chemistry (1)

2. 2
Chemistry Science:
is a branch of sciences which that import for studying the
chemical substances to know their physical and
chemical properties.
 Chemistry Departments:
there are two main departments:
1. Organic Chemistry
2. Inorganic Chemistry which was divided to:
2.1 Physical Chemistry and
2.2 Analytical Chemistry

3. What Is Analytical Chemistry?
 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.
 Analytical chemistry is often described as the area of
chemistry responsible for characterizing the composition of
matter, both qualitatively (what is present) and
quantitatively (how much is present).
3

4. 4
.
Classification of Analysis:
 Qualitative Analysis
determination of chemical identity of the species in
the
sample.
 Quantitative Analysis
determination of the amount of species or analysts,
in
numerical terms. Hence, Math is heavily involved.
 In order to perform quantitative analysis,
typically
one needs to complete qualitative analysis.
 One needs to know what it is and then select the
means to determine the amount.

5. Classification of Quantitative Methods of
Analysis
 Gravimetric Method: mass is measured.
 Volumetric Method: volume is measured or used
to determine amount of sample via concentration.
 Instrumental Method: use an instrumental
technique to assay the amount of sample:
Such as: Electro-analytical based upon electron-
transfer and Spectroscopy including mass
spectrometry .
5

6. Roles of Analytical Chemistry
 Play a vital roles in all sciences, just to name of a few:
chemistry, biology, biotechnology, forensic science,
food science, material science, environmental science,
medicinal chemistry, clinical chemistry, toxicology,
geochemistry etc.
6

7. Numbers in Analytical Chemistry:
Analytical chemists make measurements and perform
calculations. Measurements usually consist of a unit
and a number expressing the quantity of that unit.
SI units:
Stands for Système International Unités.
These are the internationally agreed on units for
measurements.
7

8. • Scientific notation
A shorthand method for expressing very large or very small
numbers by indicating powers of ten; for example, 1000 is
1 x 10
3
. Chemists frequently work with measurements
that are very large or very small. For example, a mole
contains 602,213,670,000,000,000,000,000 particles, and some
analytical techniques can detect as little as
0.000000000000001 g of a compound. For simplicity, we
express these measurements using scientific notation; thus,
amole contains 6.0221367 x 10
23
particles, and the stated
mass is 1 x10
-15
g.
8

9. 9

10. 1 mL = 1x10
-3
L
1 L = 1x10
3
mL
1 μL = 1x10
-6
L
1 L = 1x10
6
μL
1 attoliter = 1x10
-18
L
1 L = 1x10
18
attoliter
10

11. •Molecular Weight (MW):
MW is the summation of atomic weights for all atom in chemical
formula.
Equivalent Weight(EW):
The mass of a compound containing one equivalent.
An equivalent weight is defined as the ratio of a chemical species’ molecular
weight (MW) to the number of its equivalents (n).
EW = MW/n


11

12. 12

13. •Terms for Expressing Concentration:
Concentration: is an expression stating the relative amount
of solute per unit volume or unit mass of solution.
Solution consists of:
Solute: a minor species in a solution
Solvent: a major species in a solution
(e.g., water is a solvent for aqueous solution)
13

14. The Mole (mole):
A unit which defines the number of units of a chemical
species (molecules, atoms, ions, etc.) and from which we
can calculate the weight of the species if we have a
knowledge of the chemical formula of that species.
Number of moles = Weight/Mol.Wt.
 Avogadro Number = 6.022 x 10
23
particles is one mole.
14

15. Examples 2:
How many moles and millimoles are contained in 2.00 g
of pure benzoic acid (C
6
H
5
COOH) (M.W.= 122.1g/mol.).
Solution:
Moles of Benzoic acid = Weight/M.W.= 2.0/122.1= 0.016 mol.
mmoles of benzoic acid = moles x1000
= 0.016x 1000 =16.4 mmol.

15

16. Terms for Expressing Concentration:
1.Molarity:
The number of moles of solute dissolved per liter of solution
(M).
M = Moles of Solute/ Liters of Solution
Number of Moles of Solute (mole)=
Weight of Solute (g)/ M.W. of solute (g/mol.)
M= Weight/M.W X 1000/Vml
16

17. Example 5:
What is the molarity of a solution that has 4.5 mol. of
solute dissolved in 300.0 mL of solution?
Molarity = moles of solute/ liters of solution
Molarity = 4.5 mol. / 0.3000 L = 15 M
Example 6:
 117.0 g of NaCl dissolved in 1.00 L of water has an
analytical concentration of __ mol /L
17

18. M.W. of NaCl = (23 + 35.5) = 58.5 g/mol
Mole of NaCl = Wt. of NaCl / M.W. of NaCl
= 117.0 / 58.5= 2.00 moles
M =Mole of NaCl / V (liter)= 2.00 moles /1 liter
M = 2.00 moles /L = 2.00 M

18

19. •Dilution Role:
When solutions are prepared such that their
concentration is known directly through their
preparation by weighing a pure solid into the
container, any solution(low conc.) is prepared by
diluting another solution whose conc. is precisely
known (high conc.).
CB × VB = CA ×VA
C : concentration, V: volume, B: before dilution, A : after dilution
19

20. Example 7:
What is the molarity of a solution prepared by diluting
10.00 mL of a 4.281 M solution to 50.00 mL?
Using dilution equation: CB × VB = CA ×VA
4.281 M×10.00 mL = CA×50.00 mL
CA = 4.281M / 50.00mL = 0.8562 M
20

21. Terms for Expressing Concentration:
2. Normality:
The number of equivalents of solute per liter of solution (N).
Normality makes use of the chemical equivalent, which is
the amount of one chemical species reacting with another
chemical species.
Equivalent
The moles of a species that can donate one reaction unit.
 Relationship exists between normality and molarity is:
N = n x M
21

22. Example 8:
Calculate the equivalent weight and normality for a
solution of 6.0 M H3PO4 in the following
reactions:
(a) H
3PO
4 + 3OH
-
PO
4
3-
+ 3H
2O
(b) H
3PO
4 + 2NH
3
HPO
4
2-
+ 2NH
4
+
(c) H
3PO
4 + F
-
H
2PO
4
-
+ HF
22

23. 23

24. Terms for Expressing Concentration:
Molality (m):
The number of moles of solute per kilogram of solvent.
 Percent Concentration:
Weight Percent (%w/w): Grams of solute per 100g of
solution.
Weigh Percent (w/w)= (mass of solute/mass of solution)x
100 %
24

25. Terms for Expressing Concentration:
Volume percent (% v/v): Milliliters of solute per 100 mL of solution.
Volume percent (V/V) = (Volume of solute/ Volume of solution) x
100 %

Weight-to-Volume Percent(%w/v): Grams of solute per 100 mL of
solution.
weight/volume percent (w/v)= mass of solute (g)/ volume of solution
(mL) x100%
25

26. Example 8:
What is the v/v% of ethanol in a solution
prepared by mixing 5.00 mL of ethanol with
enough water to give 1.00 L of solution?
1) Volume of solute (ethanol) = 5.00 mL = 5.00 x10
3-
L
2) Volume of solution = 1.00 L
3) Volume percent (v/v) = volume of solute/ volume of
solution x 100% = (5.00 x10
3-
L/1L)x 100%= 0.5%
26

27. Terms for Expressing Concentration:
Strength of Concentration (S):
Number of grams of solute per one liter of solution.
Sppt = M × MW (g /L)
Sppt = N × EW (g / L)
Sppm = Sppt × 1000 (mg / L)
Sppb = Sppm × 1000 (µg /L)
Or Sppb = Sppt × 10
6
(µg / L)
27

28. Terms for Expressing Concentration:

parts per million (ppm): Micrograms of solute per gram of solution; for
aqueous solutions the units are often expressed as milligrams of

solute per liter
of solution.

ppm = mg / L = μg / mL

parts per billion(ppb): Nanograms of solute per gram of solution; for aqueous
solutions the units are often expressed as micrograms of solute per liter of
solution.

ppb = μg / L = ng / mL

28

29. Calculation of M & N for Concentrated Acids or
Bases:
M = % × d × 1000 / M.W.
 N = % × d × 1000 / E.W.
d : density of Acid
Mole Fraction:
is a ratio between gram molecules of solute or solvent to total
summation of number of gram molecules of solute and solvent.
29

30. Mole Fraction:
X1 = n1 /n1+n2 & X2 = n2 / n1+n2
X1 : Solute , X2 : Solvent
n1 : gram molecules of solute , n2 : gram molecules of solvent
Percentage of Mole Fraction:
% X1 = X1 × 100 & % X2 = X2 × 100
30

31. p-function:
A function of the form pX, where pX = -log(X)
Examples:
The pH of a solution that is 0.10 M H
+
is
pH = –log[H
+
] = –log(0.10) = 1.00
The pH of 5.0 x 10
-13
M H
+
is
pH = –log[H
+
] = –log(5.0 x 10
-13
) = 12.30
[H+]= shift log [-pH]
 pOH= -log [OH-]
 [OH-] shift log [-pOH]
31

32. Density:
Expresses the mass of a substance per unit volume.
In SI Units: density unit is g/mL or kg/L
ex. Density of water is approximately 1.00 g/mL at 4
o
C.
Weight = Volume x density
32

33. Exercises:
1.What is the w/w % of an aqueous ammonia solution at
14.3 M, with density = 0.900 g/mL (900g/L)?
 M.W. of NH
3 = 17.0 (g/mol.)
Mole of NH
3 at 14.3 M in 1.00 L=14.3 (mol/L) x 1.00 L=14.3 mol.
Weight of NH
3 at 14.3 M in 1.00 L= Mole of NH3 x M.W. of NH
3
= 14.3 mol.x 17.0 (g/mol.)= 243 g
Weight of 1.00 L solution = volume x density = 1.00 (L) x
900(g/L)
= 900 g
Weigh Percent (w/w)=(mass of solute/mass of solution )100%
=(243/900) 100%= 27 %
33

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2. What is pNa for a solution of 1.76 x 10
-3
M Na
3PO
4?
Since each mole of Na
3PO
4 contains three moles of Na
+
, the
concentration of Na
+
is
[Na
+
]= 3x 1.76 x 10
-3
= 5.28x10
-3
M


pNa = –log[Na
+
] = –log(5.28 x 10
-3
) = 2.277

35. 35
3. The maximum allowed concentration of chloride in a
municipal drinking water supply is 2.50x10
2
ppm Cl
-
When
the supply of water exceeds this limit, it often has a
distinctive salty taste. What is this concentration in moles
Cl
-
/liter?
Solution:
Number of moles Cl-/liter = moles / volume (L)
 Sppt = M × M.W. , Sppt = Sppm / 1000 , M = (Sppm / 1000) / M.W.
M = (2.50x10
2
/ 1000) / 35.453= 7.05x 10
-3
mol./L

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Mixing Rule:
is used to prepare a solution with a definite % conc.
from other solutions of known % concentrations.
a (c – b) a= % concentration of solution 1
c b= % concentration of solution 2
b (a – c) c= % concentration of the final solution

37. Example 9:
How to prepare a solution of 80% nitric acid from two
solutions of nitric acid of concentrations 96% and 75% ?
96 (80 – 75 = 5)
80
75 (96 – 80= 16)

5 g of 96% nitric acid solution is mixed with 16 g of 75%
solution to get solution of concentration 80%
37

38. 38
How to prepare 40 % (w/w) solution of nitric acid from a 96%(w/w) nitric
acid solution that has density = 1.08 g/cm
3
?
96 % 40
40%
0 % 56
40 g of the 96% solution is added to 56 g water to obtain 40%
nitric acid solution.
Since w = d x v , therefore the volume taken is:
v = w / d = 40 / 1.08 = 37.04 ml
So 37.04 ml of 96% solution is added to 56 ml of water to obtain
40%

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What is the molarity of a solution contain 8g of
NaOH in 500 ml?
M = no. of moles / Liter solution
no. of moles = Wt. / M.W.
M = (Wt. / M.W.) x 1/ volume in liter
M = (8 / 40) x (1/ 0.5) = o.4 M

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Classification of Volumetric Reactions:

42. 42
 Reaction between acids and bases or basic salts. Although
between bases and acids or acidic salts to produce a water, as the
following equation:
H
+
+ OH
-
↔ H
2
O
 Water’s dissociation constant (K
W
):
H
2
O ↔ H
+
+ OH
-

Or 2H
2
O ↔ H
3
O
+
+ OH
-

Kw = [H
3O
+
][OH
-
]

43. 43
•Value & Calculation of pH :
1. In Natural Solutions:
[H+]= [OH-] , pH = 7, pkw = pH + pOH = 14
2. In Strong Acidic Solutions:
[H+]< [OH-] , pH < 7, pH = -log [H+]
Or pH = 14-pOH
3. In Strong Basic Solutions:
[H+]> [OH-] , pH > 7, pH = -log [H+]
Or pH = 14 - pOH

44. 44
C: Concentration
In Weak Bases:
pH = pKw - ½ pKb + ½ log C
Base dissociation constant (Kb):
The equilibrium constant for a reaction in which a base accepts a
proton from the solvent.

45. 45

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•pH for Salt solution:
1- Weak Acid- Strong Base
2. Strong Acid-Weak Base

50. 50

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1.

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2.
3.

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4.

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56. 56
Theories Acid Base
1 Arrhenius Any substance ionizes in
water to give hydrogen
ions
Any substance ionizes
in water to give
hydroxyl ions
Solvent is
water
2 Franklin A solute yields the
cation of the solvent
A solute yields the
anion of the solvent
3 Bronsted-
Lowry
Any substance can
donate a proton
Any substance can
accept a proton
Transfer
protons from
an acid to a
base
4 Lowis Any substance can
accept an electron pair
Any substance can
donate an electron pair

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involves measuring the volume of a solution of known
concentration that is needed to react essentially completely with the
analyte.
• SOME TERMS USED IN VOLUMETRIC TITRIMETRY:
1.Standard solution (or a standard titrant):
is a reagent of exactly known concentration that is used in a
titrimetric analysis.
2. Equivalence point:
is the point in a titration when the amount of added standard reagent
is exactly equivalent to the amount of analyte.

58. 58
3. End point:
is the point in a titration when a physical change occurs that is
associated with the condition of chemical equivalence.
4. Primary standard:
is a highly purified compound that serves as a reference material in
volumetric and mass titrimetric methods.

59. 59
•Important requirements for a primary standard are:
1. High purity.
2. Atmospheric stability.
3. Absence of hydrate water.
4. Modest cost.
5. Reasonable solubility in the titration medium.
6. Reasonably large molar mass.
5. Secondary standard:
is a compound whose purity has been established by chemical
analysis and then, that serves as the reference material for a
titrimetric method of analysis.

60. 60
6. Indicator:
is a reagent used to indicate when the end point has
been reached.
is a weak organic compounds (acids or bases) which its
colour is change at specific degree of pH degrees.
 Weak organic acid used as indicator: phenolphthalein
 Weak organic base used as indicator: methyl orange

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Burrette
 Pipette
 Conical Flask
 Beaker
 Bottle of Distilled Water
Balance
 Measuring Flask
 Holder

62. 62
Complexometric Reactions:
are widely used in analytical chemistry. One of the first uses of these
reactions was for titrating cations.
•Complex: is a compound which is formed in a complexation
reaction between two or more species which are capable of
independent existence.
Or : is a compound which was produced with reaction between metal
ion (lewis acid) and the ligand (lewis base) by coordenates bonds.
•A ligand: is an ion or a molecule that forms a covalent bond with
a cation or a neutral metal atom by donating a pair or
more of electrons.

63. 63
 Types of Ligands:
1. Many ligand molecules contain more than one donating centre
and are called multidentate ligands. Such as, EDTA.
2.One such as ethylenediamine, which has two groups available
for covalent bonding, is called bidentate. a ring compound is
produced. Compounds of this type are known as chelates and the
ligands are chelating agents.
3. A ligand that has a single donor group, such as ammonia, is
called unidentate (single-toothed).

64. 64
example: Copper(II) in aqueous solution, is readily complexed
by water molecules to form species such as Cu(H2O) .
•Coordination number: is the number of covalent bonds that a
cation tends to form with electron donors.
•The Formation of Complexes in Solution:
Most metal ions react with electron-pair donors to form
coordination
compounds or complexes. The donor species, or ligand, must have
at least one pair of unshared electrons available for bond
formation.

65. 65
For example:
copper(II), which has a coordination number of 4:
1.forms a cationic ammine complex, Cu(NH2)4+2;
2.forms a neutral complex with glycine, Cu(NH2CH2COO)2; and
3.forms an anionic complex with chloride ion, CuCl4-2.
•Typical values for coordination numbers are 2, 4, and 6.
The species formed as a result of coordination can be electrically
positive, neutral, or negative.

66. 66
Oxidation and Reduction Reactions (Redox):
In an oxidation–reduction reactions, also known as a redox reactions,
any chemical reaction in which the oxidation number of a chemical
species change.
•Major Classifications:
Most redox processes involve the transfer of oxygen atoms, hydrogen
atoms, or electrons from one unit of matter to another, which all three
processes are illustrated below in examples of three most common types
of redox reactions:

67. 67
1.Oxygen Atom Transfer:
Carbon reacts with mercury (II) oxide (a compound in which mercury has oxidation
state +2) to produce carbon dioxide and mercury metal. This reaction can be written
in equation form:
C +2HgO → CO 2 + 2Hg
Carbon receiving oxygen is oxidized, mercury (II) oxide losing oxygen undergoes
the complementary reduction, and the net change is the transfer of two oxygen atoms
from mercury(II) oxide to carbon atom.

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2. Hydrogen Atom Transfer:
Hydrogen atoms are transferred from hydrazine (a compound of nitrogen and hydrogen)
to oxygen in the following reaction:
N2H4 + O2 → N2 + 2H2O
Hydrazine losing hydrogen is oxidized to molecular nitrogen, while oxygen gaining
hydrogen is reduced to water.
3. Electron Transfer:
Zinc metal and copper (II) ion react in water solution, producing copper metal and
aqueous zinc ion , according to the following equation:
Zn + Cu+2aq → Zn+2aq + Cu

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With the transfer of two of its electrons, the Zn metal is oxidized, becoming an
aqueous Zn ion, while the Cu++ ion gaining electrons is reduced to Cu metal.
Net change is the transfer of two electrons, lost by Zn and acquired by Cu.
•Oxidation Number: is the number of charges which found on an ion, if the
charge is negative, an ion is anion, and if the charge is positive, an ion is cation.
Or defined as the number of electrons in an atom that can be involved in forming
bonds with other atoms.
•Oxidation process: A gain of an oxygen or losing of electrons with
increasing of the oxidation number.
•Reduction process: A gain of hydrogen or electrons with decreasing of the
oxidation number.

70. 70
•Reducing agent: is a reactant that brings about the reduction and itself oxidized
by the oxidizing agent.
•Oxidizing agent: is a reactant that brings about the oxidation and itself reduced
by the reducing agent.
•States of Oxidation Number:
1. Oxid. N. = zero
Ex. I2, O2, N2, Cl2, KMnO4, H2C2O4, NaOH …etc.
2. Oxid. N. equals specific value (positive or negative charges number) for ion:
Ex. NO3-, I-, SO4-2, Fe+3, ….etc

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•Oxid. N. for ion in a compound:
Ex. Oxid. N. of Mn ion in KMnO4 = +7
Oxid. N. of Fe ion in FeSO4 = +2
Oxid. N. of S ion in Na2S2O3 = +2

72. 72
3. Precipitation Reactions:
A precipitation reaction occurs when two or more soluble species combine to form an
insoluble product that we call a precipitate.
Precipitate is an insoluble solid that forms when two or more soluble reagents
are combined.
The most common precipitation reaction is a metathesis reaction, in which two soluble
ionic compounds exchange parts. When a solution of lead nitrate is
added to a solution of potassium chloride, for example, a precipitate of lead chloride
forms.

73. 73
Solubility Product Constant(Ksp):
is the equilibrium constant for a reaction in which a solid dissociates
into its ions.
States of Solubility Product Constant (Ksp):
1. Ksp < [A]a[B]b , so, precipitate is formed.
2. Ksp > [A]a[B]b , so, precipitate is not formed.
3. Ksp = [A]a[B]b , so, precipitate is not formed.