Wednesday, 7 May 2014

PRACTICAL 2

PRACTICAL 2 : PHASE DIAGRAM

PART B  
MUTUAL SOLUBILITY CURVE FOR Phenol and water

Introduction:

A few types of liquid are miscible at all different composition, one of the example is: ethanol and water. On the other hand other liquids they are partially miscible with one another in limited proportions, for example: ether-water system and phenol-water system. (Even though phenol is not really a liquid, but we consider it as a liquid because as we add in the first part of water, the melting point is reduced until it becomes below room temperature to form a liquid-liquid system.)

Generally, two liquids will be more miscible when the temperature is increased until it reaches the critical temperature or consolute point of the solution and above this point, the two liquid is miscible at any proportion. Any pair of liquid is able to form a closed system, however the critical temperature of the two solutions is not easy to be determined (i.e. the temperature before the substance solidify or evaporate) with the exception of nicotine and water system.

At any temperature below the critical point, composition of the two layers of liquid in equilibrium is always a constant and it does not depend on the different amount of the two phases. Miscibility of a pair of liquid that is partially miscible with each other is affected by the presence of a third component in the mixture.


OBJECTIVE:

  1. To determine the mutual solubility of phenol and water.
  2. To determine the critical temperature of phenol-water system.

MATERIAL: 

Phenol, Distilled Water, Parafilm, Aluminium foil.
                  
APPARATUS: 

Measuring cylinder, Beaker, Test tube, Thermometer, Water bath, hot plate stirrer and pipette.

procedure

  1. 20% weight by weight of distilled water is added into a clean test tube. Then, 80% weight by weight of phenol is added into that test tube. The total volume of liquid in the test tube in 10mL.  A thermometer is placed into the test tube. The mouth of the test tube is sealed immediately using parafilm and aluminium foil.

  1. The mixture is heated in a water bath. The test tube is shaken while heating in the water bath to allow the solutions to mix well. The temperature when the turbid liquid becomes clear is recorded.

  1. The test tube from the water bath is removed and allowed to cool. The temperature when the liquid becomes turbid and two layers are separated is recorded.
  2. Procedures 1 to 3 for different composition of 8%, 11%, 37% and 63% weight by weight of phenol with water is repeated. The total liquid in the test tube was 10mL.

Phenol concentration %
8
11
37
63
80
Amount of phenol
0.8
1.1
3.7
6.3
8.0
Amount of water
9.2
8.9
6.3
3.7
2.0


  1. Using the data obtained, a graph of temperature (°C) against phenol in water (% by weight) is plotted.



RESULT


% by weight of Phenol
Temperature (°C)
Length of phenol (cm)
Length of water
(cm)
When the liquid turns clear
When mixture turns cloudy
Average
temperature
8.00
40.00
38.00
39.00
0.00
4.50
11.00
60.00
58.00
59.00
0.40
4.60
37.00
70.00
67.00
68.50
2.50
2.50
63.00
62.00
50.00
56.00
3.90
0.50
80.00
34.00
35.00
34.50
5.40
0.00






DISCUSSION

In a phenol-water system, the tie line is 50°C. Tie line is parallel to the base line. It shows the conditions where the two phases can exist at equilibrium with each other.



Between 13% weight by weight (% w/w) and 63 % w/w in water, the c. Region outside this curve contains systems having one liquid phase, which is less than 12% w/w in water or above 63 % w/w in water. During this phase, phenol and water are completely miscible.

When the system contains 100% water at 50°C, the system exists as one liquid phase. When 13 % w/w of phenol is added to the water at 50°C, a minute amount of a second phase would appear. During this phase, phenol and water are partially miscible. 2 liquid phases would still exist until the amount of phenol in water is gradually increased to 63 % w/w in water at 50°C. When the amount of phenol is more than 63 % w/w in water at 50°C, the system consists of only one liquid phase.

Between 13% w/w and 63 % w/w of phenol in water at 50°C, the test tube would contain a turbid mixture. 2 layers can be clearly seen. The bottom layer is the phenol-rich region. The top layer is the water-rich region because water is less dense than phenol. Density of water is 1.00 g cm-3 while the density of phenol is 1.07g cm-3. When the phenol increases from 13 % w/w to 63 % w/w in water at 50°C, the phenol-rich region in the test tube increases. When the system is at equilibrium, these 2 phases are known as conjugate phases.

When the temperature is higher than 68°C, the system would consist of one phase. This is the maximum temperature at which the 2 phase region exists. This temperature is called critical solution temperature or upper consolute temperature. At this temperature, the test tube would form one liquid phase when placed in the water bath. But once the test tube is taken out from the water bath, two liquid phases appear immediately.

According to theory, the system exists as one liquid phase when the system contains 13% w/w to 63 % w/w of phenol in water at 50°C. The experimental results are higher because measuring errors may occur. Parallax error may occur during the measurement of phenol and water.

Phenol is a volatile agent. Phenol is also carcinogenic. As a precaution, face mask, goggles and gloves are worn to protect ourselves. The heating of phenol and water is carried out in a fume cupboard to remove the poisonous air. Parafilm and aluminium foil are used to seal the test tube containing phenol to prevent the evaporation of phenol.


condition of phenol after taken out from water bath





                  measured length of phenol after heating

QUESTIONS AND ANSWERS

1) Discuss the diagrams with reference to the phase rule.

The diagrams obtained is a phase diagram for a two component condensed system having one liquid phase. Phenol and water are miscible with each other at a particular condition. The phase rule is expressed as F = C – P + 2,  where F is the number of degrees of freedom in the system, C is the number of components and P is the number of phases present. Therefore, by using phase rule, the degree of freedom, F = 2 − 1 + 2 = 3. Since the pressure is fixed for this system, which is 1 atm, therefore F is reduced to 2. Thus, only two independent variables are required for us to define the phenol-water system completely which is temperature and concentration. From the graph we obtained, if the temperature is given, the composition of the mixture can be determined easily through the graph.

2) Explain the effect of adding foreign substances and show importance of this effect in pharmacy.

The effect of adding foreign substance is known as contamination. If the foreign substances reduces the miscibility of the two liquid, the dispensed medicine may changes its nature and no longer suitable for consumption. It may also cause allergies, infections or toxicity in the patient. The product may be deemed therapeutically ineffective. Contamination may arise especially in extemporaneous preparation when the place of medicine preparation is not hygienic and properly maintained.

CONCLUSION:


In this experiment, it is concluded that the solubility of two partially miscible liquids could be determined by constructing a mutual solubility curve for the pair of liquids. The heating and cooling temperatures gradually increased but as it reached a certain constant high temperature, it gradually decreased. The critical temperature that these two liquids can exist in a two phase system is 68.5˚C. Phenol is partial miscible with water and produce one liquid phase system at certain temperature and concentration when pressure is fixed.

REFERENCES
3)      Sinko, Patrick J, Martin’s Physical Pharmacy and Pharmaceutical Sciences 5 th editon, Lippincott Williams & Wilkins, 2005, page 51.






PRACTICAL 1

Practical 1 : Phase Diagram
PART A
Determination of Phase Diagram For Ethanol / Toluene / Water System Theory
Tree-Component System
OBJECTIVE:

To study the miscibility of ethanol / toluene / water system and to investigate the behavior of system of the three liquid which is ethanol, toluene and water.

APPARATUS:

Burette, conical flask, retort stand, measuring cylinder, test tubes

MATERIALS:

Ethanol, Toluene, Water

INTRODUCTION:
Phase diagrams for ternary systems are usually represented using a triangle shown in Fig. 2.





                                                                      Figure 1

A three-component phase diagram has four degrees of freedom: F = 3 − 1 + 2 = 4. In this case, temperature and pressure are two of the conditions and the concentrations of two of the three components make up the rest. In this diagram, each of three corners or apexes of the triangle represent 100% by weight of one component (in this experiment ethanol, toluene, water). As the result the other two apex will represent 0% of the other two components.Only two concentrations are required because the third will be the difference between 100% and the sum of the other two components.
These systems are used for determining miscibility/solubility, coacervation regions, gel-forming regions for multicomponent mixtures, etc. The lines joining the corner points forming the triangle each represent two component mixtures of the three possible combinations (AB, BC, and CA). If two of the components are known, the third is known by difference. Any combination of the three components is described by a single point on the diagram. 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 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 become homogenous.

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:

Ethanol
Toluene
Water
Volume of Ethanol /mL
% of Ethanol
Volume of Toluene/mL
% of Toluene
Volume of Water/mL
% of Water
2.0
9.9
18.0
88.7
0.3
1.5
5.0
24.0
15.0
72.1
0.8
3.9
7.0
33.3
13.0
61.8
1.1
5.0
10.0
45.6
10.0
45.6
2.0
8.9
13.0
55.6
7.0
30.0
3.4
14.5
15.0
60.0
5.0
20.0
5.0
20.0
18.0
59.0
2.0
6.6
10.6
34.5
19.0
49.0
1.0
2.6
18.8
48.5




QUESTIONS:
1.       Does the mixture containing 70% ethanol, 20% water, and 10% toluene (volume) appear clear or does it form two layers?

No. The solution appear clear and form one liquid phase.

2.       What will happen if you dilute one part of the mixture with four part of (a) water (b) toluene (c) ethanol?

The mixture contain 70% ethanol, 20% water and 10% toluene

 1 part x 70/100 = 0.7 part of ethanol
 1 part x 20/100 = 0.2 part of water
 1 part x 10/100 = 0.1 part of toluene
~There are 0.7 part of ethanol, 0.2 part of water, 0.1 part of toluene in the                                    mixture.

(a)    One part mixture + four part water
Ethanol =         0.7
                    --------- × 100%
                      1+4
               = 14%

Water =       0.2 + 4
                  ---------- × 100%
                     1+4
            = 84%

Toluene =       0.1
                 ----------× 100%
                      1+4
                = 2%
̃ based on phase diagram, the mixture is outside the curve. Thus single liquid      phase will formed.

(b)   One part of mixture + four part toluene
        Ethanol =       0.7
                         ----------- × 100%
                             1+4     
               = 14%

                           
                                 Water =    0.2
                     ---------- × 100%
                              1+4
                = 4%

            Toluene =    0.1 + 4
                              ----------× 100%
                                 1+4
                    =82%
̃ based on phase diagram, the mixture is outside the curve. Thus, single liquid       phase will formed.

(c)    One part mixture + four part ethanol

        Ethanol =     0.7 + 4
                         ----------- × 100%
                             1+4     
               = 94%

                           
                                 Water =     0.2
                     ---------- × 100%
                              1+4
                = 4%

            Toluene =        0.1
                              ----------× 100%
                                 1+4
                    =2%
̃ based on phase diagram, the mixture is outside the curve. Thus, single liquid      phase will formed.

    DISCUSSION:

For the diagram, each corner represent 100% of each component, which is 100% toluene, 100% ethanol, and 100% water while the other two component 0%. In going along the line bounding the triangle represent the concentration in two-component system, it does not matter whether we proceed in a clockwise or counterclockwise direction. But the more usual convention is clockwise. In this case, as we move along water to ethanol in the direction of ethanol, we assume the system of containing increasing concentration of ethanol and smaller amounts of water. Moving along ethanol to toluene will results in increasing amount of toluene and the closer we approach water, the more its concentration will be. For example, at the first point the amount of ethanol increases as we read it in clockwise. Ethanol is 9.9%, water 1.5% and toluene 88.7%. Result obtained in this experiment was not accurate due to some error occurred during the experiment. Some of the error occurred are parallax error during recorded and measured the scale. Furthermore, the temperature also is not constant causing the measurement become slightly deviated and some of the glassware have been contaminated causing the chemical mixed with other compound. All this error causing result to be inaccurate and slightly affect the graph obtained.

Some precaution need to be taken such as make sure our eyes are parallel with the meniscus so that measurement is more accurate. We also need to make sure the temperature is constant so that the reaction of the chemical are almost the same for all set and clean the apparatus before use to reduce contamination.





CONCLUSION:

Water and toluene form a two-phase system because they are only slightly miscible. Ethanol is completely miscible with both toluene and water. The rule that relate to the use of triangular coordinates fully explained the three-component system.

REFERENCES:



3.       Martin’s Physical Pharmacy And Pharmaceutical Sciences, 6th edition (2011). Patrick J. Sinko. Lippincott Williams and Wilkins