Practical 1: Phase diagram

TITLE:

Determination of Phase Diagram for Ethanol/ Toluene/ Water System Theory

AIM:

To determine Phase Diagram for Ethanol/ Toluene/ Water System Theory

INTRODUCTION:

                 Figure 1: The triangular diagram for three-component systems

For three-component systems at constant temperature and pressure, the compositions may be stated in the form of coordinates for a triangular diagram.

           In the diagram above, each of the three corners or apexes of the triangle represent 100% by weight of one component (A, B, C).The three lines joining the corner points represent two-component mixtures of the three possible combinations of A, B, and C. Thus, the lines AB, BC, and CA are used for two-component mixtures of A and B, B and C, and C and A, respectively. The area within the triangle represents all the possible combinations of A, B, and C to give three-component systems. If a line is drawn through any apex to  point on the opposite side ( line DC in Figure 1), then all systems represented by points on such a line have constant ratio of two components, in this case A and B. Any line drawn parallel to one side of the triangle, for example, line HI in Figure 1, represents ternary system in which the proportion of one component (in this case, C) is constant. 

          The addition of a third component to a pair of miscible liquids can change their mutual solubility. If this third component is more soluble in one of the two different components the mutual solubility of the liquid pair is decreased. However, if it is soluble in both of the liquids, the mutual solubility is increased. Thus, when ethanol is added to a mixture of benzene and water, the mutual solubility of the liquid pair increased until it reached a point whereby the mixture becomes homogenous.

                      Figure 2 : Ternary Phase Diagram

          The benefits of preparing an oily substance as homogenous water in liquid are already clear. However, what will happen to a system like this when it is diluted should also be known and this can be explained through the understanding of the triangular phase diagram. Figure 2 is for the system containing components peppermint oil­polysorbate 20-water. A concentration of 7.5% oil, 42.5% polysorbate 20 and 50% water (point A in diagram) can be diluted for 10 times with water giving a solution that is still clear (now containing 0.75% of oil, 4.25% polysorbate 20 and 95% water). However, when 1 ml of water is added to 10ml of clear solution B (49% oil, 5% polysorbate 20, 1% water) the solution becomes cloudy, point B' (44.55% oil, 45.45% polysorbate 20 and 10% water). If I ml of water is further added, the solution becomes clear, point B" (40.5% oil, 41.3% polysorbate 20, 18.2% water) but if the original solution is diluted three times (16 1/3% water, 16 2/3% polysorbate 20, 67% water) the solution becomes cloudy.

APPARATUS:

Conical flask, burette, glass rod, white tile, thermometer

MATERIALS:

Peppermint oil, polysorbate-20, water

PROCEDURES:

1.   Ethanol/ toluene mixtures of different compositions were prepared and placed in sealed conical flasks.

2.   Each mixture contained different % volume of ethanol in 50 ml: 10, 25, 35, 50, 65, 75, 90, 95% v/v.

3.   A burette was filled with distilled water.

4.   The mixtures were titrated with water, accompanied by vigorous shaking of the conical flask.

5.   Titration was stopped when a cloudy mixture was formed.

6.   The volume of the water used was recorded.

7.  Steps 1-6 were repeated to do a second titration. The volume of water required for complete titration of each mixture was recorded.

8.  Average volume of water used was calculated.

9.  % volume of each component of the ternary system for when a second phase became separated was calculated.

10. These values were plotted on a graph paper with triangular axes to produce a triple phase diagram.

RESULTS:

DISCUSSION:

            Phase diagram for ternary systems is usually represented using a triangle. The graph accounts for the fact that only two variables are required. Along the phase boundary only one variable is required. The real curve was determined in this experiment. Water and toluene form a two-phase system because they are only slightly miscible. Ethanol is completely miscible with both toluene and water. Thus, the addition of sufficient amount of ethanol to the toluene-water system would produce a single liquid phase in which all the three components are miscible and the mixture is homogenous. This is shown in the triple phase diagram that has been plotted on the triangular diagram.

The curve of the plotted graph is termed as binomial curve. The region under the curve shows that the presence of 2 phases that is water and toluene whereas the region above the curve boundary shows one phase of homogenous solution. The bounded region is actually between the binomial curve and line of water and toluene mixture. Addition of ethanol which acts as surfactant will allow the 2 phase of solution to be in one phase.

            

            The points that are at both ends of the curve are the limits of solubility of toluene in water and water in toluene. Along the toluene-water line, which represents a binary mixture of toluene and water, the liquids are able to form a homogenous mixture as long as the first point is not exceeded. However, the second point must be exceeded for a homogenous mixture to be formed. The length of line between the two points represents the mixture of toluene and water with such composition that they cannot form a homogenous mixture. This may be due to insolubility of toluene in water or water in toluene.

            

            However, the binomial curve is incomplete. This may be due to some errors. Firstly, ethanol and toluene are volatile liquids and they will vaporize if left longer. This caused the measured volume to be less than the actual one as some of them already evaporated and thus affected the volume of water needed for titration. Secondly, parallax errors may occur due to the eyes of the observer is not perpendicular to the meniscus of the liquids. This caused inaccurate measurement of liquids and thus affecting the curve. Next, the cloudiness was hard to be judged because there was no specific range of degree of cloudiness in each of the experiment. This might affect the volume of water added to the solution and has greatly affected the percentage by volume and the curve too. Also, the temperature in the laboratory that was not constant. Temperature will change the curve pattern and this might be the cause of incomplete binomial curve. The contaminated apparatus may affect the results obtained.

           

           Therefore, precaution steps need to be taken. The volatile liquids must be used immediately when poured from the container as to avoid loss of volume of liquids. The eye of the observer must in perpendicular to the meniscus of the liquids to avoid parallax error to obtain accurate volume of liquids. We can choose the same student to observe the cloudiness throughout the experiment so that the results will be more accurate. The room temperature must be consistent. The apparatus must be cleaned before using them.

                

QUESTIONS:

1. Will a mixture containing 70% ethanol, 20% water and 10% toluene remain clear or form two phases?

The mixture will remain clear and  form one liquid phase.

2. What will happen if you dilute 1 part of the mixture with 4 parts of (a) water; (b) toluene; (c) ethanol?

1 part mixture x 70% ethanol = 1 x 70/100 = 0.7 part of ethanol

1 part mixture x 20% water = 1 x 20/100 = 0.2 part of water

1 part mixture x 10% toluene = 1 x 10/100 = 0.1 part of toluene

Therefore, there are 0.7 part of ethanol; 0.2 part of water; 0.1 part of toluene in the mixture.

(a) Water: 1 part of mixture + 4 parts of water:

Ethanol = 0.7/5 x 100% =14%

Water = (0.2+4)/5 x 100% = 84%

Toluene = 0.1/5 x 100% =2%

Therefore, from the phase diagram, this mixture is outside the area of the binodal curve. Therefore, a clear single liquid phase of solution is formed.

(b) Toluene: 1 part of mixture + 4 parts of toluene

Ethanol = 0.7/5 x 100% =14%

Water = 0.2/5 x 100% = 4%

Toluene = 0.5/5 x 100% =82%

Therefore, from the phase diagram, this mixture is outside the area of the binodal curve. Therefore, a clear single liquid phase of solution is formed.

Ethanol = 4.7/5 x 100% =94%
Water = 0.2/5 x 100% = 4%
Toluene = 0.1/5 x 100% =2%
Therefore, from the phase diagram, this mixture is outside the area of the binodal curve. Therefore, a clear single liquid phase of solution is formed.


CONCLUSION:
The phase diagram for Ethanol, Toluene and Water System (Ternary System) is determined. The binomial curve is determined, but incomplete, due to some errors in experiment.

REFERENCES:
1.      Physicochemical Principles of Pharmacy , 3^rd edition (1998) . A.T. Florence and D.Attwood. Macmillan Press Ltd.

2.      Physical Pharmacy: Physical Chemistry Principles in Pharmaceutical Sciences, by Martin, A.N.