AIM:
ü To learn the methods of determining the
solubility limits in a ternary system (water, ethanol and toluene) with
components of different miscibility.
ü To practice the correct method to plot and
obtain the solubility curve of the ternary system on a triangular diagram.
DATE OF
EXPERIMENT:
7
NOVEMBER 2016
INTRODUCTION:
When we are preparing pharmaceutical
formulation, the knowledge on mutual
solubility and phase
diagram can be applied as it often involve the mixing of more than one components.
At the same time, a homogeneous form of resulting formulation is formed. This requirements
can be made possible by knowing the exact ratio of each component needed to be
mixed where the temperature and pressure of the surrounding has to be taken
into consideration.
Ethanol,
Toluene and Water were the three components involved in this experiment. If water
and toluene mixed together with ethanol in suitable proportion, a homogeneous solution
is formed at equilibrium. The theory behind this phenomena is that solutions
are homogeneous because the ratio of solute to solvent remains unchanged throughout
the solution even if homogenized with multiple sources, and stable because the
solute will not settle out after any period of time, and it cannot be removed
by a filter or centrifuge. There are 3 components but only 1 phase exists.
At
constant pressure and temperature, the compositions of the three components can
be stated in the form of coordinates in triangular diagram.
Figure
1 shows the example of a triangular diagram. Triangular diagram is very
convenient in determining the compositions of three components as each apex
will represent one type of pure component, with an amount of 100%. Meanwhile,
each side of the triangle represents the mixture of two components. For
instance, the side connecting A and B shows the mixture of A and B. Within the
triangle, all three components are shown.
For any line that is parallel to any side of
the triangle, constant percentage of a component is shown. For example, DE ,which is parallel to BC, shows 20% of A with
varying amount of B and C. Also, FG, which is
parallel to AC, shows 30% of B with varying amount of A and C. The point of
intersection of the three parallel lines will show the composition of each
components in a mixture at the instance. For example, K which is the
intersection point of DE, FG and HI shows
that there are 20% A, 50% B and 30% C. It is appropriate to measure in this way
because the sum of all distances from K, which are KI, KC, KE, KH, KG and KD
equals to the length of any one side of the triangle.
When
a third component is added into a pair of miscible liquids, for instance when
water is added into ethanol and toluene, their mutual solubility will change.
If the third component is more soluble in one of the two components, the mutual
solubility will decrease. On the other hand, if the third component is soluble
in both two components, the mutual solubility will increase. The mutual
solubility will only increase until the mixture becomes homogeneous, for
example become clear.
EXPERIMENTAL
METHOD – APPARATUS:
Burette
Funnel
Pipette 10ml
Pipette pump
Conical Flask
Beaker 100ml
.
EXPERIMENTAL
METHOD – CHEMICALS:
Toluene
Ethanol
Distilled water
EXPERIMENTAL
METHOD – EXPERIMENTAL PROCEDURE:
- 20 mL of mixtures of ethanol and toluene are prepared in conical flasks as below according to fixed ratio:
Beaker
|
Percentage
of ethanol (%)
|
Percentage
of toluene (%)
|
Volume
of ethanol (mL)
|
Volume
of toluene (mL)
|
A
|
10
|
90
|
2
|
18
|
B
|
25
|
75
|
5
|
15
|
C
|
35
|
65
|
7
|
13
|
D
|
50
|
50
|
10
|
10
|
E
|
65
|
35
|
13
|
7
|
F
|
75
|
25
|
15
|
5
|
G
|
90
|
10
|
18
|
2
|
H
|
95
|
5
|
19
|
1
|
- Distilled water is titrated into mixture into conical flask A and shaken until cloudiness (due to the existence of second phase) is observed.
- The amount of distilled water titrated is recorded in the Table 1 below. The room temperature is recorded.
- Step 2 to step 4 are repeated for conical flask B, C, D, E, F, G and H.
- Step 1 to step 5 are repeated for second measurement. The result of the second titration is recorded in Table 2 below.
- Average volume of water added into each conical flask is calculated from the first and second titration.
- New percentage of each components in the mixture after titration is recalculated.
- All results calculated in step 6 and 7 are tabulated in Table 3.
- A solubility curve is drawn on a triangular diagram.
- RESULT
Initial
Percentage of ethanol (%)
|
Volume
of ethanol (mL)
|
Volume
of Toluene (mL)
|
Volume
of water added (mL)
|
||
Titration
I
|
Titration
II
|
Average
|
|||
10
|
2
|
18
|
4.50
|
3.70
|
4.10
|
25
|
5
|
15
|
0.50
|
0.70
|
0.60
|
35
|
7
|
13
|
0.90
|
1.10
|
1.00
|
50
|
10
|
10
|
1.70
|
1.60
|
1.65
|
65
|
13
|
7
|
2.00
|
1.90
|
1.95
|
75
|
15
|
5
|
3.80
|
3.50
|
3.65
|
90
|
18
|
2
|
11.10
|
10.90
|
11.0
|
95
|
19
|
1
|
15.10
|
14.80
|
14.95
|
Volume (mL)
|
Percentage (%)
|
||||||
Ethanol
|
Toluene
|
Water
|
Total
|
Ethanol
|
Toluene
|
Water
|
Total
|
2.00
|
18.00
|
4.10
|
24.10
|
8.30
|
74.70
|
17.00
|
100
|
5.00
|
15.00
|
0.60
|
20.60
|
24.30
|
72.80
|
2.90
|
100
|
7.00
|
13.00
|
1.00
|
21.00
|
33.30
|
61.90
|
4.80
|
100
|
10.00
|
10.00
|
1.65
|
21.65
|
46.20
|
46.20
|
7.60
|
100
|
13.00
|
7.00
|
1.95
|
21.95
|
59.20
|
31.90
|
8.90
|
100
|
15.00
|
5.00
|
3.65
|
23.65
|
63.42
|
21.14
|
15.43
|
100
|
18.00
|
2.00
|
11.0
|
31.00
|
58.10
|
6.45
|
35.50
|
100
|
19.00
|
1.00
|
14.95
|
34.95
|
54.36
|
2.86
|
42.78
|
100
|
Triangular Diagram Plotted:
DISCUSSION:
For a three-component
systems, the composition of all three phases can be expressed in the ternary
phase diagram. Each side of the triangular diagram corresponds to one of the
three components in the system. By following the phase rule, that is applied for
prediction number of stable phases may exist in equilibrium for a particular
system, it is determine that a single phase in three components system may
possess four degree of freedom.
F = C-P + 2
= 3-1+2
= 4
F = degree of freedom
C = number of components
P = number of phases
Since there were four
degree of freedom, four of the intensive variables which are pressure,
temperature and two out of three concentration of the components may vary
independently to achieve the equilibrium. Only concentration of two components
are required to define the system at equilibrium as the concentration of the
third component can be known by subtracting the concentration of the first two
components given from the total concentration.
For this experiment, the
water and toluene usually form two-phase system as they are just partially
miscible. However, ethanol is completely miscible with both of the water and
toluene, it is therefore expected to act as a surfactant and the increased
concentration of ethanol in the partially miscible, two-phase system of toluene
and water would eventually produce a single-phase where all 3 liquid are
miscible. Hence, with the sufficient amount of ethanol in the toluene-water
system, a single-phase system is produced where all the components are miscible
and the mixture is homogeneous. This is shown in the ternary phase diagram that
has been plotted in the triangular diagram.
In the experiment, water
is added into the ethanol-toluene mixuture, and the solubility of ethanol in
water and toluene in water are markedly different. Toluene is insoluble in
water while ethanol is soluble in water due to the presence of hydroxyl group.
The mutual solubility of the original homogenous pair (ethanol and toluene) is
decreased when water is added into the system. Therefore, cloudiness is
observed due to the presence of two phase liquid.
From the triangular
paper, the curve of the plotted graph is known as binodal curve. The region
bounded by the binodal curve indicates the presence of two phases while the
region above the curve shows one phase of homogeneous solution. The mixture
within the bounded region appears cloudy as there are phase separation due to the insufficient amount of ethanol to produce
homogeneous mixture. Meanwhile at the upper region, the addition of ethanol
allows the two phase solution to be in one phase.
There were errors arised
in this experiment. First error is degree of cloudiness. Requirements for the
degree of cloudiness in the experiment is not specific leading the inaccurate
reading amount of water added and the result obtained is affected. Next error
is both of the ethanol and toluene are volatile liquids which some of them might
vapourize and the measured volume may be less than the actual volume that has
been measured earlier. The total volume of water needed during the titration is
influenced. Moreover, the apparatus also may affect the result obtained. As
some of the conical flask used is not completely dried and can cause a slight
dilution. In obtaining a good result, precautions need to be taken to minimize
the errors.
CONCLUSION
:
Toluene, ethanol and
water system is a ternary system with one pair of partially miscible liquid (
toluene and water). The addition of sufficient amount of ethanol to the
toluene-water system able to produce a single liquid phase in which all the
three components are miscible and the mixture is homogeneous.
REFERENCE :
1. Martin's
Physical Pharmacy and Pharmacautical Science, Sixth Edition, Patrick J. Sinko,
Wolters Kluwer, Lippincott Williams & Wilkins.
2. Physicochemical
Principles of Pharmacy , 4th edition (1998) . A.T. Florence and
D.Attwood. Macmillan Press Ltd.
QUESTIONS :
- Does
the mixture containing 70% ethanol, 20% water and 10% toluene (volume)
appear a clear or does it form two layer?
Following the phase
diagram, at these concentration, the solution appear clear.
2.
What
will happen if you dilute 1 part of the mixture with 4 parts of (a) water (b)
toluene (c) ethanol ?
1 part × 70%
ethanol = 1 part × 70/100 = 0.7 part of ethanol
1 part × 20% water
= 1 part × 20/100 = 0.2 part of water
1 part ×10% toluene
= 1 part × 10/100 = 0.1 part of toluene
(a) 1 part of mixture＋4 parts of water
Ethanol =(0.7/1+4) x 100%
=14%
Water =(0.2+4/1+4) x 100%
=84%
Toluene=(0.1/1+4) x 100%
=2%
According to the phase diagram drawn, this
mixture lies outside the area of
bounded by the binodal curve. Therefore, a
clear homogenous liquid phase of
solution is formed.
(b) 1 part of mixture + 4 parts of toluene
Ethanol =(0.7/1+4) x 100%
=14%
Water =(0.2/1+4) x 100%
=4%
Toluene=(0.1+4/1+4) x 100%
=82%
According to the phase diagram drawn, this
mixture lies outside the area of
bounded by the binodal curve. Therefore, a
clear homogenous liquid phase of
solution is formed.
(c) 1 part of mixture + 4 parts of ethanol
Ethanol =(0.7+4/1+4) x 100%
=94%
Water =(0.2/1+4) x 100%
=4%
Toluene=(0.1/1+4) x 100%
=2%
According to the phase diagram drawn, this
mixture lies outside the area of
bounded by the binodal curve. Therefore, a
clear homogenous liquid phase of
solution is formed.
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