Liquid-liquid Solution Systems

1
SOLUBILITY OF LIQUIDS IN LIQUIDS

2
SOLUBILITY OF LIQUIDS IN LIQUIDS
There are liquids:
 Showing complete miscibility e.g. water & ethyl alcohol
 Showing complete immiscibility e.g. water & mercury
 Showing partial miscibility e.g. water & phenol

3
 The components of an ideal solution are miscible in all
proportions. Such complete miscibility is also observed in
some real binary systems, e.g. ethanol and water, under
normal conditions.
 attractions between the molecules of one component are
greater than those between its molecules and those of the
other component,
i.e. if a positive deviation from Raoult's law occurs, the
miscibility of the components may be reduced.
 The greater the strength of the self-association
the greater the immiscibility
the greater the degree of +ve deviat. from Raoult's law.
e.g. of large +ve deviat.: water and Hg binary system.

4
In cases of partial miscibility
Degree of miscibility may be dependent on the temperature
1. Solubility  with  in temperature (water-phenol)
2. Solubility  with  in temperature (water-
triethylamine)
3. Solubility  with  &  in temperature (water-
nicotine)
4. Solubility not affected by temperature
In case of three component system the third liquid may
influence the degree of solubility of the 2 liquid
systems.

5
Effect of temperature variation on the degree of
miscibility in these systems is described by means
of phase diagrams.
Phase diagrams
= graphs of temperature versus composition at
constant P

6
1. Systems showing an increase in miscibility
with rise in temperature
A +ve deviation from Raoult's law due to difference in the
cohesive forces that exist between the molecules of each
component in a liquid mixture.
T +ve deviation miscibility
Each phase consists of a saturated solution of one
component in the other liquid.
Such saturated solutions are known as conjugate solutions

7
Phenol and water system phase diagram.
Temperature fixed at 50 °C
Point a, system containing
100% pure water.
Addition of phenol to water
will result in the formation of
a single liquid phase until the
point b is reached.
At point b, appears a second
phase.
Phase A: water rich phase
containing 11% phenol
Phase B: phenol rich phase
containing 63% phenol

8
 increasing quantities
of phenol,
i.e., as we proceed across
the diagram from point b to
point c, we form systems
in which the amount of the
phenol-rich phase (B)
continually increases
 At the same time the
amount of the water-rich
phase (A) decreases.
Once the total conc. of
phenol exceeds 63 % at 50
0C a single phenol-rich
liquid phase is formed.

9
At 50°C
Aqueous phase saturated with phenol:
contains 11% phenol (point b)
Phenolic phase saturated with water:
contains 63% phenol (point c)

10
 The line bc drawn across
the region containing two
phases is termed a tie line;
it is always parallel to the
base line in two component
systems.
 all systems prepared on a
tie line at 50° C will
separate into phases of
constant composition
whose composition is b
and c. These phases are
termed conjugate phases.

11
 All combinations of phenol and
water above this temperature
are completely miscible and
yield one-phase liquid systems.
The critical solution temperature
(upper consolute temperature):
Is the maximum temperature at which the two phase
region exists. In the case of the phenol-water system
this is 66.8° (point h in Figure).

12
Example:
A mixture of phenol and water at 20° C has a total composition of
50 % phenol. The tie line at this temperature cuts the curve at
points equivalent to 8.4 and 72.2 per cent w/w phenol.
 What is the weight of the aqueous layer and of the phenol
layer in 500 g of the mixture?
 How many g of phenol are present in each of the two layers?
Solution:
Let Z be the weight in grams of the aqueous layer.
(500- Z) is the weight in grams of the phenol layer.
The sum of the percentages of phenol in the two layers =
500 X 50/100 = 250 g
Z (8.4/100) + (500 - Z)(72.2/100) = 250
1} weight of aqueous layer Z = 174 g
weight of phenol layer (500 - Z) = 326 g
2} The weight of phenol in the aqueous layer
174 X 8.4/100 = 15 grams
The weight of phenol in the phenolic layer
326 X 72.2/100 = 235 grams
Z
500-Z

13
 All combinations of phenol
and water above this
temperature are completely
miscible and yield one-
phase liquid systems.
The critical solution temperature
(upper consolute temperature):
Is the maximum temperature at which the two
phase region exists. In the case of the phenol-
water system this is 66.8° (point h in Figure).

14
The solubility of liquid
pairs may increase as
the temperature is
lowered
The system will exhibit a
lower consolute temp.
Below which the two
members are soluble in
all proportions
Above which two
separate layers are
formed.
2. Systems showing a decrease in miscibility
with rise in temperature
TRIETHYLAMINE & WATER

15
Mixtures such as nicotine &
water show both an upper and
a lower consolute temperature
with an intermediate
temperature region in which
the two liquids are only
partially miscible.
4. Systems with no critical solution temperature
The pair, ethyl ether and water, has neither an upper nor a
lower consolute temperature and shows partial miscibility
over the entire temperature range at which the mixture exists.
0 20 40 60 80 100
040100160220280
Temperature(°C)
2 phases
1 phase
1 phase
upper CST
lower CST61
208
Nicotine in water
% by weight
3. Systems showing upper and lower
consolute temperature
NICOTINE & WATER

16
The effects of added substances
on critical solution temperatures
Critical solution temperatures are very sensitive to
impurities or added substances.
The addition of a substance to a binary liquid system
produces a ternary system.

17
A) If the added material is soluble in only 1 of the 2
components or if the solubility in the two liquids are
markedly different, the solubility of the liquid pair is
decreased due to salting-out
If the original binary mixture has an upper CST, the T
if it has a lower CST, the T by the addition of the third
component.
Example:
If 0.1 M naphthalene is added to a mix. of phenol and water
it dissolves only in the phenol and raises the CST about 20°;
If 0.1 M KCl is added to a phenol-water mix, it dissolves
only in water and raises the CST approximately 8°.

18
B)
If the added material is soluble in both of the liquids to about
the same extent, the solubility of the liquid pair is increased;
The increase in solubility of two partially miscible solvents by
an additive is referred to as blending.
an upper CST is lowered and a lower CST is raised.
Example:
The addition of succinic acid or Na oleate to a phenol-water
system.

19
For solid- liquid mixtures in which the two components
are completely miscible in the liquid state and
completely immiscible as solid.
e.g. salol and thymol, and salol and camphor.
In phase diagram for salol thymol system are four
regions:
(i) a single liquid phase,
(ii) region containing solid salol & a conjugate liquid
phase,
(iii) region in which solid thymol & a conjugate liquid
phase
(iv) region in which both components are present as pure
solid phases.
Two-Component Systems Containing Solid
and Liquid Phases

20
0 10 20 30 40 50 60 70 80 90 100
01020304050
Eutectic
point
Solid thymol + solid salol (iv)
liquid + solid
thymol (iii)
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Thymol in salol % weight
Temperature°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

21
Regions containing 2 phases (ii, iii, and iv) are comparable to
the 2-phase region of the phenol-water system. Thus it is possible to
calculate the composition and relative amount of each phase from the
tie lines and the phase boundaries.
0 10 20 30 40 50 60 70 80 90 100
010203040
50
Eutectic
point
liquid + solid thymol (iii)
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Temperature°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

22
A system of 60% thymol in salol:
temperature of the mixture to 50°C (point x).
On cooling: single liquid until 29°C at which a minute amount of
solid thymol separates out to form a 2-phase solid-liquid system.
At 25°C (x1), liquid phase, a1 (53% thymol in salol) and pure solid
thymol, b1.
At 20°C (x2), liquid phase, a2(45% thymol in salol) and pure thymol
b2.
0 10 20 30 40 50 60 70 80 90 100
0102030
4050
Eutectic
point
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Temperature
°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

23
At 15°C (x3), liquid phase, a3 (37% thymol in salol and pure thymol
b3.
Below 13°C liquid phase disappears and the system contains 2
solid phases of pure salol and pure thymol.
At 10°C (x4), system contains an equilibrium mixture of pure
solid salol (a4) and pure solid thymol (b4)
0 10 20 30 40 50 60 70 80 90 100
010203040
50
Eutectic
point
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Temperature°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

24
13°C is the lowest temperature at which one liquid
phase coexists in the thymol salol system.
It occurs in a mixture containing 34% thymol in salol.
This point on the phase diagram is known as:
Eutectic point
0 10 20 30 40 50 60 70 80 90 100
0102030
4050
Eutectic
point
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Temperature
°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

25
At the eutectic point, three phases (liquid, solid salol, and
solid thymol) coexist
The eutectic point therefore denotes an invariant
system in a condensed system (i.e. constant pressure,
F= C-P+1)
F = 2 - 3 + 1 = O
Several other substances form eutectic mixtures (e.g.,
camphor, chloral hydrate, menthol, and betanaphthol).
0 10 20 30 40 50 60 70 80 90 100
01020
304050
Eutectic
point
liquid +
solid salol (ii)
ONE liquid
phase (i)
Mp pure salol
Mp pure thymol
x
Temperatu
re°C
x1
x2
x3
b1
b2
b3
a1
a2
a3
a4 x4 b4

26
Three component systems containing one pair of
partially miscible liquids
Phase Equilibria in three component
systems
Water & benzene are partially
miscible
•Alcohol is completely miscible with
both
•Addition of sufficient alcohol gives
one phase system in which all the 3
components are miscible (alcohol
acts as heat in phenol water system)
•It breaks cohesive force between
mol.
Benzene saturated
with water
water saturated
with Benzene

27
3 components & one phase
In non condensed system
F = C – P + 2
F = 3 - 1+ 2 = 4
In condensed system and constant temp
F = C – P + 0
F = 3 – 1 + 0 = 2 (conc of 2 component)
A C
B
1
phase
2
phases
a
f
d e
i
c
F may be:
1- T 2- P 3,4 – conc of 2 C
conc of the third component =
total conc - sum of 2 comp. conc.

29
The area within the triangle
represents all the possible
combinations of A, B, and C to
give three component systems.
The location of a particular three
component system within the
triangle, e.g. point x may be
undertaken as follows:
 The line AC, opposite apex B , represents systems,
containing A and C. B is absent, i.e., B.= 0 .
 The horizontal lines running across the triangle parallel to
AC indicate increasing percentages of B from B = 0 (on line
AC) to B = 100 (at point B).
 The line parallel to AC which cuts point x is equivalent to
15 % B; consequently, the system contains 15 percent of B
and 85 percent of A and C together.
A C
B
x

30
Applying similar arguments to
the other two components in the
system, we can say that along the
line AB, C=0.
 As we proceed from the line AB
towards C across the diagram, the
concentration of C increases until
at the apex, C = 100 percent.
 The point x lies on the line parallel to AB, that is equivalent
to 30 percent of C.
Therefore the concentration of A is
100 - (B + C) = 100 - (15 + 30) = 55 per cent.
A C
B

31
A, B and C represent water,
alcohol, and benzene, respectively.
 The line AC represents binary
mixtures of A and C.
 The curve afdeic termed a binodal
curve marks the extent of the two
phase region. The remainder of the
triangle contains 1 liquid phase.
Ternary systems water, benzene and alcohol contains one
pair of partially miscible liquids.
Water and benzene are miscible only to a slight extent and
so a mixture of the 2 produces a 2 phase system, water
saturated with benzene and benzene saturated with water.
 On the other hand, alcohol is completely miscible with
both benzene and water. Thus, the addition of alcohol to a
two phase system of benzene and water would produce a
single liquid phase in which all 3 components are miscible.
A C
B
1 phase
2 phases
a
f
d
e
i
c

32
Tie line fi
Systems g and h prepared along the tie line fi both rise to two
phases having the compositions denoted by the points f and i.
For example, system g, after reaching equilibrium, will separate
into two phases, f and i. The ratio of phases f to phase i, on a
weight basis, is given by the ratio gi : fg.
A C
B
1 phase
2 phases
a
f
d
e
i
c
g h

33
Effect of Temperature
The area of binodal decrease as the temp. is
raised & miscibility increase till it is completely
miscible & the binodal disappear.

Practical Demonstration
 Solubilization is the process of bringing into
solution substances which are insoluble or
sparingly soluble in water.
 Methods of Solubilization:
1) The use of mixed solvents (blending and Co-
solvency).
2) The use of surface active agents (Micelle formation).
3) The use of hydrotropic salts e.g. iodine and potassium
iodide.
4) Complex formation e.g. caffeine and barbiturate.

Determination of Solubility of Benzoic acid
• Procedure:
1. In a glass bottle, add 0.5 gm benzoic acid & 20 ml
distilled water.
2. Shake for 10 min and set aside for another 10 min.
3. Filter the contents of the bottle, take 10 ml of the
filterate and titrate with 0.1 N NaOH using Ph.Ph. as
indicator (3-5 dps).
4. The End Point is determined by the appearance of
the first permanent faint pink colour.
5. Repeat the above steps using surfactant solutions
(1%, 3%, 5%, 7%) instead of distilled water and
determine the end points.

E.P = Volume of titrant used
0.5 gm benzoic acid
20 ml dist. Water or sample soln
Shake for 10 minSet aside for 10 min
10 ml of
filterate
3 dps
Ph.Ph
0.1 N
NaOH
Procedure
Filter

Calculation of Solubility
)/(
10
1000.
lub lg
FPE
ilitySo


Factor (F): is the amount of benzoic acid (in grams) neutralized by
1 ml of titrant (0.1 N NaOH).
Calculation of Factor (F):
+ NaOH + H2O
1 mole benzoic acid ≡ 1 mole NaOH
122 gm ≡ 1000 ml 1 M NaOH
122 gm ≡ 1000 ml 1 N NaOH
122 /10 ≡ 1000 ml 0.1 N NaOH
F ≡ 1 ml 0.1 N NaOH
0122.0
100010
122


F

Calculation of Solubility
)/(
10
10000122.0.
lub lg
PE
ilitySo


Results:
Solubility (g/l)E.PSurfactant Concn
1 %
3 %
5 %
7 %
Then draw the relationship between surfactant concn (on x-axis)
and solubility (on y-axis).

Study Questions
 Define the following terms:
[Miscibility, Osmotic pressure, vapour pressure, ideal solution, real solution, conjugate solution, saturated
solution, consulate temperature point, molar mass, electrolyte, weight mass, gravity, molality, molarity,
solubilization, etc]
 Respond to the following questions:
 Considering a practical process, illustrate the procedural steps that can be applied to determine the
solubility of such an item as Benzoic acid
 Group work discussional questions:
 Give a detailed account of the variables that can be considered in the quantification of liquid-
liquid solution system
 A mixture of phenol and water at 20° C has a total composition of 50 % phenol. The tie line at
this temperature cuts the curve at points equivalent to 8.4 and 72.2 per cent w/w phenol.
• What is the weight of the aqueous layer and of the phenol layer in 500 g of the mixture?
• How many g of phenol are present in each of the two layers?

Liquid-liquid Solution Systems

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