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237
Solution
1. Association of solute in one of the solvents: Consider the distribution of
benzoic acid between benzene and acidified water phase. It forms
dimer in benzene but remains as monomer in acidified water.
Therefore, the distribution law mentioned as in Equation (1) can-
not be used for the determination of partition coefficient. Let c
1
be its concentration in water and c
2
be its total concentration in
benzene. Applying law of chemical equilibrium between the sim-
ple molecule HA and associated molecule (HA)
n
HA HA
n
() ()
Associated
molecule
Simple
molecule
n
K
d
o
n
n
HA
HA
[]
()
⎡
⎣
⎢
⎤
⎦
⎥
(12)
and the equilibrium constant
HA HA
o
d
n
n
n
[] []
K
Since, benzoic acid exists largely in the form of dimers in benzene C
o
may be replaced by [(HA)
2
], where C
o
is total molar concentration
of solute in benzene (organic layer). Equation (12) may be written as
[]HA
oo
constant ≈×C
(13)
As we know the true distribution coefficient is always expressed in
terms of simple species common to both phases, i.e. in terms of [HA]
w
and [HA]
o
. In this case of benzene-water system [HA]
o
is given
by Equation (14) and modified distribution constant can be writ-
ten as
′′
[]
[] []
K
C
HA
HA HA
0
ww
0
(14)
Fig. 8.4
Representation of
distribution of benzoic
acid between organic and
aqueous phases
Organic phase Aquous phase
C
o
C
w
C
65
C
2
A
n
C
65
C
A
n
d
C
65
CC
65
C
a
AA
238
Practical Pharmaceutical Analytical Chemistry
2. Dissociation of solute in one of the solvents: Consider distribution of benzoic
acid in water and peanut oil. In this case benzoic acid will dissociate in
water depending upon pH and dissociation according to the equation
(15)
Let C
o
be the total concentration of benzoic acid in organic phase and can
be taken equal to [HA]
o
the monomer concentration in the organic phase.
The distribution is expressed as
K
C
HA
HA HA
w
o
w
>@
>@ >@
(16)
where K is the true distribution coefficient, [HA]
o
C
o
is molar con-
centration of the simple benzoic acid molecules in the organic phase and
[HA]
w
is molar concentration of the undissociate acid molecules in the
water phase. But the total concentration of acid in water phase is different
because benzoic acid also exists in dissociated form in water. Therefore
C
w
w
w
HA A
>@
⎡
⎣
⎤
⎦
(17)
and the apparent distribution coefficient will be
′
[]
[]
⎡
⎣
⎢
⎤
⎦
⎥
−
K
C
C
HA
HA A
o
w
o
w
(18)
The distribution coefficient of benzoic acid in water and peanut oil
depends on two equilibria—the distribution of the undissociated acid
between the immiscible phases as expressed by Equation (18) and the spe-
cies distribution of the acid in the aqueous phases which depends upon
hydrogen ion concentration and the dissociation constant K
a
of acid. Sup-
pose a is the degree of dissociation of the solute (X) in aqueous phase.
Then, the concentration of the various species would be as
XAB
CC
+
−
()
w
1 αα
w
C
w
α
(19)
Evidently concentration of X in water phase will be C
w
( 1 D), therefore
according to distribution law
CC K
ow
/1−
()
=α
d
(20)
Note: Benzoic acid forms dimer in all organic solvents except peanut oil.
3. Solvation of solute (solutes enter into chemical combination with any
of the solvent): When solvation of solute (S) in any phase occurs, no
239
Solution
change in general equation for distribution coefficient is observed. if
C
o
is the concentration in organic phase and C
w
is the concentration
in water phase and the solute enters in chemical combination with it
forming complex molecule
S ns S.ns (21)
Suppose a is the portion that enters in chemical combination with solvent, the
concentration of uncombined solute molecule will be S C
w
(1 a) and the
concentration of combined solute molecule ‘S.ns’ formed will be C
w
a
Applying law of mass action,
K
C
C
w
w
n
1 solvent
α
α−
()[ ]
(22)
Since, solvent is in excess, its concentration may be taken as constant
Constant
1
w
w
C
C
α
α−
()
(23)
According to the distribution law, equilibrium must exist between concentra-
tions of similar molecular species, i.e. single molecule of ‘S’ in both phases
C
C
o
w
1
Constant
−
()
α
(24)
Subtracting Equation (24) from Equation (23)
C
C
o
w
Constant
B
(25)
‘a’ is the fraction of solute that combines with the solvent and is constant at
constant temperature Equation (25) therefore reduces to
C
C
o
w
Constant
(26)
Application The concept of partition coefficient is useful in cosmetic for-
mulations, since it is a measure of the ability of a material to penetrate the hair
and skin. Partition coefficient is also the scientific principle used in liquid
liquid chromatographic separations. Octanolwater partition coefficient logP
is used in QSAR studies and rational drug design as a measure of molecular
hydrophobicity. Hydrophobicity affects drug absorption, bioavailability, hydro-
phobic drugreceptor interactions, metabolism of molecules, as well as their
toxicity. LogP has become also a key parameter in studies of the environmental
fate of chemicals.
240
Practical Pharmaceutical Analytical Chemistry
Experiment 12 Study distribution of benzoic acid between acidified water
and benzene.
Requirements
Benzoic acid, benzene, acidified distilled water, 0.05 N sodium hydroxide solu-
tion, phenolphthalein, three 250 mL stoppered flasks, 100 conical flasks, burette,
pipette, and thermometer
Procedure
1. Take 190 mL of distilled water and add 10 mL concentrated hydrochloric
acid to make acidified distilled water.
2. Weigh 10 g of benzoic acid and dissolve in 120 mL benzene.
3. Take three stoppered conical flasks and label them as 1, 2 and 3.
4. Place 20, 30 and 50 mL of benzoic acid solution in flasks 1, 2 and 3,
respectively.
5. Add 30 and 20 mL of benzene to flasks 1 and 2, respectively, to make total
volume of 50 mL in each conical flask.
6. Now add 50 mL of acidified distilled water to each conical flask. Cork them
and shake them vigorously.
7. Place all the three bottles in a thermostat or trough of water at room tem-
perature.
8. Record the temperature of water bath.
9. Pipette out 10 mL of acidified distilled water in 100 mL conical flask and
titrate with 0.05 N NaOH. Note down the concordant reading.
10. Let the conical flasks stand in the thermostat for about half an hour while
shaking them after every 4–5 min in turns.
11. After final shaking, swirl the bottles so that any drops of liquid adhering to
sides of the conical flasks settle down.
12. Take the flask and separate the two layers in two conical flasks by using
separating funnel.
13. Pipette out 10 mL of aqueous layer in a 100 mL conical flask and titrate with
0.05 N NaOH using phenolphthalein as indicator.
14. Record concordant reading for flask 1 and repeat the procedure for flasks 2
and 3.
15. Now pipette out 10 mL of organic layer in a 100 mL conical flask contain-
ing 20 mL of water and titrate with 0.05 N NaOH using phenolphthalein
as indicator.
16. Record concordant reading for all the three flasks 1, 2 and 3.
241
Solution
Observations and Calculations
Temperature ______˚C
Volume consumed by 10 mL acidified water x mL
Titration of Upper layer (aqueous layer)
Flask
Volume taken
(mL)
Burette reading
Concordant
reading
123
Flask 1
Flask 2
Flask 3
Titration of Lower layer (organic layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Gram equivalent weight of is 122 g.
Subtract the volume of sodium hydroxide consumed by acidified water from
the readings of aqueous layer of flasks 1, 2 and 3 and then perform calculations.
Concentration (mol/L) of benzoic acid in two layers is calculated as in
Experiment 10.
As we know the true distribution coefficient is always expressed in terms of
simple species common to both phases, i.e. in terms of [
aq
]
w
and [
org
]
o
. In
this case of benzene-water system [
org
]
o
is given by Equation (25) benzoic
acid forms dimer in benzene, therefore modified distribution constant Equa-
tion (26) is used to calculate partition coefficient.
Partition coefficient
D
org
aq
K
C
C
(27)
Result The partition coefficient of benzoic acid between benzene and
water is ________.
242
Practical Pharmaceutical Analytical Chemistry
Judge yourself
1. Define solubility, solution, solute and solvent.
2. What are the factors that affect solubility of a solute in a solvent?
3. Give methods of expressions of solubility.
4. How is solubility determined in a laboratory?
5. What are different classes of solutions? Give examples.
6. Differentiate between saturated and supersaturated solutions.
7. How is saturated solution prepared?
8. Define molarity, molality and mole fraction.
9. Discuss the effect of temperature on solubility.
10. How is density of a liquid determined?
11. How will you prepare 0.2 N NaOH and 0.05 N HCl solutions?
12. What is meant by heat of the solution?
13. Give the equation for the determination of heat of the solution.
14. Define solubility rules and explain giving examples.
15. Define solubility product principle.
16. What is the effect of common ion on solubility product?
17. Write the equilibrium constant expression for solubility product.
18. What is effect of temperature on solubility product?
19. Solubility product principle does not hold true for dilute solutions
of strong electrolytes. Why?
20. Give the equation for the determination of solubility product.
21. Give the application of solubility product determination.
22. What do you mean by temperature coefficient and how is it deter-
mined?
23. Define binary system.
24. What do you understand by phase diagram?
25. What is CST, upper CST and lower CST?
26. Name a liquid pair which shows upper CST, lower CST and both.
27. What are factors that affect CST of a liquid pair?
28. What is effect of pressure on CST of a pair of liquid having both
upper and lower CST?
29. What is the CST of phenol-water system and percentage of phe-
nol at CST?
243
Solution
30. What is the CST of nicotine-water system and percentage of
nicotine at CST?
31. How is CST determined in a laboratory.
32. What is the effect of the presence of impurity in water on CST of
the phenol-water system?
33. Explain the distribution coefficient.
34. Give the units of partition coefficient.
35. How partition coefficient and absorption of a substance related?
36. What are the factors which affect distribution of solute in two
adjacent phases?
37. What is the importance of partition coefficient determination?
38. Distribution of benzoic acid in acidified water and benzene does
not follow the partition coefficient principle. Explain why?
39. Give the equation for the determination of partition coefficient.
40. Explain the role of partition coefficient in the dosage formulation.
41. What do you understand by Log P?
42. Define the distribution law.
43. How is partition coefficient of a substance measured in a labora-
tory?
44. What equation is used for calculation of partition coefficient when
dissociation of solute in one of the solvents takes place?
45. How does solvation of a solute in any phase affects the partition
coefficient?
7KLVSDJHLQWHQWLRQDOO\OHIWEODQN
Chapter
9
Chapter
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical reactions and the
factors that influence the rates. It is a process related to various fundamentals
that are involved in pharmaceutical development of drugs to dosage forms. It is
essential to study the stability of the drug formulations, their shelf life and other
conditions related to the development of various dosage forms like dissolution
rate of tablets, incompatibility of substances with each other, etc.
9.1 Rate of Reaction
Rate of reaction is the velocity with which a reactant or reactants undergo
chemical change. The rate, velocity or speed of a reaction is given by the expres-
sion dc/dt, where dc is the increase or the decrease of concentration over an
infinitesimal time interval dt. Chemical reaction takes place through collision
between molecules of reactants. A chemical reaction involves breaking of bonds
in reacting species and formation of new bonds in product. Since the number
and the nature of bonds vary in different substances, the rates of reactions also
vary. Ionic reactions are almost instantaneous, e.g. precipitation of AgCl takes
4
-10
s
because no bonds are to be broken or formed.
According to the law of mass action, the rate of a chemical reaction is propor-
tional to the product of the molar concentration of the reactant each raised to a
power usually equal to the number of molecules undergoing reaction.
aA bB ........... product
(1)
Rate of reaction
d
d
d
d
r
a
A
t
b
B
t
kA B
ab
()
=
[]
=
[]
=
[][]
1
1
(2)
where k is the proportionality constant known as rate constant or velocity constant.
The sum of the exponents of the concentration terms involved in the rate
equation is known as order of reaction.
246
Practical Pharmaceutical Analytical Chemistry
rkAB
ab
=
[][]
(3)
Order of reaction a b.
Rate of equation can be written in a number of ways depending whether we
are interested in the rate of disappearance of the reactant species or in the rate of
appearance of the product of species. Thus, for a simple equation
AB
k
⎯→⎯⎯⎯
r
A
t
B
t
kA=−
[]
=+
[]
=
[]
d
d
d
d
(4)
Units of basic rate constant
Zero-order rate constant
k
A
t
=−
[]
=
d
d
mol/L s
First-order rate constant
k
A
tA
=−
[]
[]
=
d
d
1
L/s
Second-order rate constant
k
A
t
A
=−
[]
[]
=
d
d
1
2
L/mol s
[A] is molar concentration of reactant.
9.2 Molecularity
Molecularity is the number of molecules involved in the steps leading to the
chemical reaction. It is defined as the number of molecules of reactants taking
part in a chemical reaction. It is a theoretical value derived from the mechanism
of reaction. Molecularity can be neither zero nor fractional. For most of the
reactions, molecularity is less than 3.
Example of unimolecular reaction
Br Br
2
2⎯→⎯⎯⎯
Bimolecular reaction
HI HI
22
2+⎯ →⎯⎯⎯
The process is bimolecular as two molecules, one of H
2
and one of I
2
, must come
together to form two molecules of HI.