Physical pharmacy 2014

Physical Pharmacy UKM 2014 ( Group B5)

Sunday, May 25, 2014

Experiment 3 - Adsorption from Solution



Date: 
25 March 2014

Objectives: 
To determine the surface area of activated charcoal via adsorption from solution.

Introduction
Adsorption occurs when particles which are ion, atom or molecules on the surface of solids are capable of attracting other molecules due to the instability of energies around the particles. The energy here refers to electrostatic, valency or Van de Waals. Meanwhile, absorption is the process in which a fluid is dissolved by a liquid or a solid. In general, adsorption is a surface-based process where a film of adsorbate is created on the surface while absorption involves the entire volume of the absorbing substance.
                The nature of forces between adsorbent (solid surface) and adsorbate (gas or dissolved solutes) influence the strength of bonds. Adsorption consists of two types which are chemisorption and physical adsorption. Chemisorption involves only chemical bonds between adsorbent and adsorbate which acquire activation energy that can be strong and not readily reversible. Physical adsorption occurs when adsorbent and adsorbatereact due to Van de Waals force which is non-specific and occur at any condition.
In this experiment, we use activated charcoal sample as adsorbent and iodine as adsorbate. Adsorption of iodine will determine the surface area of activated charcoal. This method can be applied in determination of the surface area of powder drug in pharmaceutical industries. The surface area is very important in field of pharmacy where it is one of factor that affects the rate of dissolution and bioavailability of drugs that are absorbed in gastrointestinal tract.

Materials and apparatus :
12 conical flasks, 6 centrifuge tubes, measuring cylinders, analytical balance, Beckman J6M/E centrifuge, burettes, retort stand and clamps, paster pipettes, iodine solutions, 1%w/v starch solution, 0.1M sodium thiosulphate solution, distilled water and activated charcoal.

Procedure:
1.       12 conical flasks are filled with 50ml mixtures of iodine solutions (A and B) as stated in Table 1 by using burettes or measuring cylinders.

Solution A: Iodine (0.05M)
Solution B: Potassium iodide (0.1M)

Flask
Volume of solution A (ml)
Volume of solution B (ml)
1 and 7
10
40
2 and 8
15
35
3 and 9
20
30
4 and 10
25
25
5 and 11
30
20
6 and 12
50
0
Table 1

Set 1: Actual concentration of iodine in solution A (X)

For flask 1-6
1.       1-2 drops of starch solution are added as an indicator.
2.       The solution is then titrated by using 0.1M sodium thiosulphate solution until the colour of the solution change from dark to colorless.
3.       The volume of the sodium thiosulphate used is recorded

Set 2: Concentration of iodine in solution A at equilibrium 
 For flask 7-12
1.       0.1 g of activated charcoal is added into the flasks.
2.       The flasks are capped tightly. Then, every 10 minutes the flask is swirled or shacked for 2 hours.
3.       After 2 hours, the solutions are transferred into centrifuge tubes and labeled.
4.       The solution is centrifuge at 3000 rpm for 5 minutes and the resulting supernatant is transferred into new conical flask. The conical flask is labeled accordingly.
5.       Steps 1.2 and 3 are repeated as carried out for flasks 1-6 in Set1.
  

 Result


QUESTION:


1. Calculate N for iodine in each flask.


2. Plot amount of iodine adsorbed (N) versus balance concentration of solution (C) at equilibrium to obtain adsorption isotherm.


Graph of Amount of Iodine Adsorbed (N) versus Balance Concentration of Solution at Equilibrium (C)




3. According to Langmuir theory, if there is no more than a monolayer of iodine adsorbed on the charchoal,

C/N = C/Nm + I/KNm
Where:

C = concentration of solution at equilibrium
Nm = number of mole per gram charcoal required
K = constant to complete a monolayer

Plot C/N versus C, if Langmuir equation is followed, a straight line with slope of 1/Nm and intercept of 1/KNm is obtained.

Obtain the value of Nm, and then calculate the number of iodine molecule adsorbed on the monomolecular layer. Assume that the area covered by one adsorbed molecule is 3.2 x 10-19m2.
Avogadro number = 6.023 x 1023 molecule, calculate the surface area of charcoal in m2g-1.

Answer:

C/N (M/molg-1)
C (M)
0.1957
0.0009
0.2034
0.0024
0.2086
0.0034
0.1905
0.0040
0.1862
0.0046
0.2190
0.0090






Slope = 1/Nm =
                          = 2.419

Y - intercept = 0.191

Therefore, to obtain the value of Nm:
1/Nm = 2.419
    Nm = 1 ÷ 2.419
            = 0.4134 molg-1

To calculate the iodine molecule adsorbed on the monomolecular layer:

We know that Nm = 0.4134 molg-1

That’s mean in 0.1434 mol of iodine, 1g of charcoal is adsorbed. Therefore, the number of mol of iodine molecules that adsorbed on the monomolecular layer is:

Avogadro number = 6.023 x 1023

Numbers of molecules = number of moles x Avogadro number
                                         = 0.1434 mol x 6.023 x 1023
                                                             = 8.64 x 1022 molecules/g

Area covered by one adsorbed molecules is 3.2 x 10-19 m2

Therefore, the surface area of charcoal:

Surface area of charcoal = (8.64 x 1022 molecules/g) x (3.2 x 10-19 m2)
                                            =27648 m2g-1


4. Discuss the result of the experiment. How do you determine experimentally that equilibrium has been reached after shaking for 2 hours?

By observing the colour changes in the flask. The iodine will undergo colour change from dark brown to light brown at certain period of time after the flask is shook for every 10 minutes for 2 hours. The light brown colour of the iodine will not change until the end of the reaction and this marks the equilibrium point of the reaction.

Repeat the experiment and titrate with sodium thiosulphate. If the volume stays constant then equilibrium is reached.

DISCCUSION:

Determination of the surface area of powder drug, which is related to its particles size, is important in pharmacy area. As stated earlier in the introduction, surface area is one of the factors that govern the rate of dissolution and bioavailability of the drugs that are absorbed through the gastrointestinal tract.
This experiment, adsorption of iodine from solution used to determine the measurement of surface area of the solid. Langmuir equation is used to estimate the surface area of activated charcoal sample.  

Cf/N = (1/Nm) Cf + (1/kNm)

Where (1/Nm) is the slope, and (1/kNm) is the intercept, when Cf/N is plotted versus the concentration Cf. The inverse of the slope is Nm,and this represents the moles adsorbed at monolayer coverage. Nm can be used to determine the specific surface area of a solid. The basic limitation of Langmuir adsorption equation is that it is valid at low pressure only.

In the set 1 experiment,to calculate the actual concentration of iodine in solution A (X), titration method was used to calculate the concentration of iodine. This is because the iodide ion and iodine molecule are in equilibrium in the conical flask. Starch is used as an indicator in the titration. The solution turn dark blue colour when starch is added as iodine molecules are present. Then, when sodium thiosulfate is added, the iodine molecule react with sodium thiosulfate to form sodium iodide. When there is totally no iodine molecule in the solution, the dark colour change to colourless. The equation below used to calculate the moles of iodine.

I2 + 2Na2S2O3 → Na2S4O6 + 2 NaI
     
In the set 2 experiment, 0.1 g of activated charcoal was added into flask 7-12 and capped tightly. The activated charcoal added is act as adsorbent to adsorp the iodine molecule. Adsorption of iodine molecule on the activated charcoal is a result from Van der Waal’s forces which exists between molecules.

According to Langmuir theory, there is no more than a monolayer of iodine adsorbed on the charcoal. The surface of charcoal can be calculated by plotting a graph of C/N versus C with a straight line slope of 1/Nm.


 CONCLUSION

The Nm value is 0.4134 g mol-1. The number of iodine adsorbed onto the monomolecular layer 8.64 x 1022 molecules. The surface area of charcoal is 27648 m2/g.

REFERENCES

Patrick J. Sinko, Lippincott Williams and Wilkins. Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5th Edition.

Alexander T Florence and David Attwood. (2006). Physiocochemical Principles of Pharmacy. 4th Ed. Palgrave. USA.




Tuesday, May 13, 2014

Experiment 1 - Phase Diagram (Part A)

Determination of Phase Diagram for Ethanol/ Toluene/ Water System Theory Three-Component Systems



Date

15th April 2014

Title:  

Determination of Phase Diagram for Ethanol / Toluene / Water System

Objectives:  

To determine the phase diagram of three-component systems which is ethanol, toluene and water.

Introduction: 

The above triangular diagram is use to plot the composition for the three-component systems at constant temperature and pressure. There are about five rules relating to the use of a three-phase diagram which are:-

1.      Each of three corners or apexes of the triangle represent 100% by weight of one component (A, B, or C) and 0% of the other two components.
2.      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.
3.      The area within the triangle represents all the possible combinations of, A, B, and C to give three-component systems.
4.      If a line drawn through any apex to appoint on the opposite side, then all systems represented by points on such a line have a constant ratio of two components, in this case A and B for line DC.
5.      Any line drawn parallel to one side of the triangle represents ternary systems in which the proportion of one component is constant. In this instance, all systems prepared along HI will contain 20% of C and different concentrations of A and B.

     The mutual solubility of the miscible liquids can change by the addition of a third component. If this third component is more soluble in either one from the two components, the solubility of both components will reduce. But if the third component is soluble in both components at the same time, the solubility increases. Thus, when ethanol is added into a mixture of benzene and water, the solubility of these two components will increase until a point is reached, where the mixture become homogenous. This application can be used in formulations of solutions. Examples of three-component liquid system that have been tested are castor oil / alcohol / water, peppermint oil / propylene glycol / water, peppermint oil / polyethylene glycol / water.

     By having the understanding of the triangular phase diagram, we can predict what will happen to the system when it is diluted. This experiment is to determine the phase diagram for ethanol / toluene / water system.


Apparatus : Conical flaks, Measuring cylinder, Burette, Retord stand

Materials : Ethanol, Toluene, Distilled water

Procedure:
1.      Different compositions of ethanol and toluene mixtures were prepared and placed in sealed conical flasks.
2.      Each mixture contained different % volume of ethanol in 50ml at 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.      A cloudy mixture was formed and titration was stopped.
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.      The average volume of water used was calculated.
9.      Percentage volume of each component of the ternary system for when a second phase became separated was calculated. 
10.    All data were plotted on a graph paper with triangular axes to produce a triple phase diagram.



Results:



Ethanol
(%v/v)
Toluene
(%v/v)
Volume of water used (mL)
Average volume of water used (mL)
Titration I
Titration II
Initial
Final
Initial
Final
10
90
1.00
1.30
1.30
1.60
0.30
25
75
1.60
2.40
2.40
3.20
0.80
35
65
14.20
15.10
15.10
16.30
1.05
50
50
16.30
18.00
18.00
19.90
1.80
65
35
19.90
22.60
22.60
25.20
2.65
75
25
25.20
29.20
29.20
33.40
4.10
90
10
33.40
43.70
13.70
23.50
10.05
95
5
23.50
40.70
20.10
37.30
17.20


Calculation:
Total volume (mL) (Average water + 20mL)
Ethanol
Toluene
Water
Volume (mL)
%(v/v)
Volume (mL)
%(v/v)
Volume (mL)
%(v/v)
20.30
2.00
9.85
18.00
88.67
0.30
1.48
20.80
5.00
24.04
15.00
72.12
0.80
3.84
21.05
7.00
33.25
13.00
61.76
1.05
4.99
21.80
10.00
45.87
10.00
45.87
1.80
8.26
22.65
13.00
57.40
7.00
30.90
2.65
11.70
24.10
15.00
62.24
5.00
20.75
4.10
17.01
30.05
18.00
59.90
2.00
6.66
10.05
33.44
37.20
19.00
51.08
1.00
2.69
17.20
46.23


Graph:


Question
1. Does the mixture containing 70% ethanol, 20% water and 10% toluene (volume) appear clear or does it form two layers?
A clear solution will appear and form one liquid phase.

2. What will happen if you dilute 1 part of the mixture with 4 parts of:-
a)      Water
Two phases will be observed
b)      Toluene
Two phases will be observed
c)      Ethanol
One phase will be observed

Discussion

The presence of third liquid component can sometimes be very useful in changing the mutual solubility of two other components. In this case, we can add a third component that will make the other two components less soluble in each other.

Solubilities for three component liquid mixtures are covalently displayed on an equilateral triangle. Each vertex of the triangle represents a pure component. Thus, the top vertex represent pure component of ethanol, the right vertex represents pure toluene and the left vertex represent the pure component of water.

Each of the three slides of triangle represents a binary (two-component) mixture. For example, the right side of the triangle represents mixtures of ethanol and toluene, the left side represents mixtures of ethanol and water, and the base represents mixtures of water and toluene. Inside the triangle, we have mixtures containing all three components.

This experiment was carried out at constant temperature and pressure. In this experiment, increasing weight of water has been added to a constant weight mixture of toluene and ethanol. 

Along the phase boundary only one variable is required. 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.

In a solution containing 10% ethanol and 90% toluene, an average of 0.3mL of water was added for a second phase to occur. The percentage of each component now is 9.85% ethanol, 88.67% toluene and 1.48% of water. These are represented by a point on triangular diagram.

In the solution containing 25% ethanol and 75% toluene, an average 0.80mL of water was added for a second phase to occur. The percentage of each component now is 24.04% ethanol, 88.67% toluene and 3.84% of water.


In the solution containing 35% ethanol and 65% toluene, an average 1.05mL of water was added for a second phase to occur. The percentage of each component now is 33.25% ethanol, 61.76% toluene and 4.99% of water.

In the solution containing 50% ethanol and 50% toluene, an average 1.80mL of water was added for a second phase to occur. The percentage of each component now is 45.87% ethanol, 45.87% toluene and 8.26% of water. As we can see here, the percentage of ethanol and toluene are same.

 In the solution containing 65% ethanol and 35% toluene, an average 2.65mL of water was added for a second phase to occur. The percentage of each component now is 57.40% ethanol, 30.90% toluene and 11.70% of water.

In the solution containing 75% ethanol and 25% toluene, an average 4.10mL of water was added for a second phase to occur. The percentage of each component now is 62.24% ethanol, 20.75% toluene and 17.01% of water.

 In the solution containing 90% ethanol and 10% toluene, an average 10.05mL of water was added for a second phase to occur. The percentage of each component now is 59.90% ethanol, 6.66% toluene and 33.44% of water.

In the solution containing 95% ethanol and 5% toluene, an average 17.20mL of water was added for a second phase to occur. The percentage of each component now is 51.08% ethanol, 2.69% toluene and 46.23% of water.

For overall, we can see here that the percentage of ethanol and water are keep increasing meanwhile the concentration of toluene is decreasing. That’s mean the presence of water have changed the mutual solubility between ethanol and toluene.

However, in this experiment the binomial curve is incomplete. This may be due to some errors. Firstly, the measured volume of ethanol and toluene are less than the actual value.  Ethanol and toluene are volatile liquids and they will vaporize if left longer. Some of them already evaporated and thus affected the volume of water needed for titration. Secondly, parallax errors may occur during the experiment. 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.

Conclusion
The phase diagram for ethanol / toluene / water system can be determined by using triangular diagram. Ethanol, toluene and water system is ternary system with one pair of partially miscible liquid (toluene and water)

Reference

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