SALUTATION :D
INTRODUCE  OUR PHARMACY TEAM 

NUR NADIANA BT ADNAN A139791

MUHAMMAD SHAZERIN BIN KAMARUDIN A139798

MALA A/P BALAKRISHNAN A140050

CHAN MING SEE A140285

SONIA YAP YUNG HWEI A140284

KO WEI CHENG A140250

PRACTICAL 1 : PHASE DIAGRAM

Title:

Phase diagram for the ethanol/toluene/water system theory

Objective:

To determine the phase diagram for the ethanol/toluene/water system

Introduction:

At constant temperature and pressure, the composition of a ternary system can be presented in the form of triangular coordinates.

In the diagram above, each corner of the triangle represents a pure component, which are 100% A, 100% B and 100% C. Each side of the triangle represents a binary mixture and any line drawn parallel to one of the sides shows the percentage of a particular component. For example, DE show a mixture at point K is composed of 20% A, 50% B and 30% C.

In fact , addition of the third component to a pair of miscible liquids can alter their miscibility. If the third component is more soluble in one of the liquids than in the other, then the miscibility between that pair of liquids decreases. However, if it is equally soluble in both liquids, the miscibility of the two liquids increases. Thus, when ethanol is added to a mixture of the benzene and water, the miscibility of the two liquids rises to a point where the mixture becomes homogenous .This approach can be used in formulating solutions.

Material:

1. Ethanol
2. Toluene
3. Distilled water

Apparatus:

1. Measuring cylinder
2. Beaker
3. Conical flask
4. Pipette
5. Burette
6. Retort stand and clamp

Procedures:

1. Ethanol-toluene mixture of different composition were prepared and placed in lid-closed conical flasks.
2. Each mixture contained different % volume of ethanol in 20mL: 10, 25, 35, 50, 65, 75, 90 and 95% v/v.
3. A burette was filled with distilled water.
4. The mixture was titrated with water, accompanied by vigorous shaking of conical flask using a hand.
5. Titration was stopped right away 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 become separated was calculated.
10. These values were plotted on a graph paper with triangular axes to produce a triple phase diagram.

Results:

Calculations:

Questions:

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

From the graph, the mixture will appear clear at that composition.

2. What will happen if you dilute one part of the mixture with 4 parts of

a)      Water

                 One part of the mixture contains 0.7 part of ethanol, 0.2 part of water and 0.1

                  part of toluene. Thus,

              

               From the graph, the mixture will remain clear.

b)     Toluene

                

                 From the graph, the mixture will become cloudy.

c)      Ethanol

                

                  From the graph, the mixture will remain clear.

Discussion:

            Ethanol/toluene/water system is a three-component system. In this system which containing three components but one phase, F= 3-1+2= 4. the four degree of freedom are temperature, pressure, and the concentrations of two of the three components. Only two concentrations term are required

            Factually speaking, water and toluene are miscible only to a slight extent, and so a mixture of the two usually produces a two-phase system. The heavier phase consists of water saturated with toluene, while the lighter phase is toluene saturated with water. On the other hand, ethanol, being the third component, is completely miscible with both water and toluene. Therefore, the addition of sufficient ethanol to a two-phase system of water and toluene would produce a single liquid phase in which all three components are miscible. This situation is illustrated by a ternary system. This is the solvent effect. A ternary system is a system containing three components which is usually held under condensed condition and constant temperature. Therefore, F=3-3+2=2 if three phases are present like ethanol / toluene / water system. The degree of freedom is 2, meaning that the concentration of two of the three components, must be required. This is because the third component’s concentration can be obtained by substraction of the addition of the concentration of the two components from the known total concentration. The various phase equilibria that exist in three-component systems are frequently complex, so the vapour phase is ignored.

            Furthermore, in order to display this system, triangular coordinate graph paper called triangular diagram is used. The rules relating to the triangular phase diagram has been discussed in Introduction.

            In the practical, the ternary system involves a pair of partially miscible liquids-water and toluene. Water and toluene are only miscible to a slight extent, thus a mixture of the two liquids usually produces a two-phase system. Ethanol is used in such system because it is miscible in toluene and water. Therefore, the addition of a sufficient amount of ethanol to the toluene/water system would produce a single liquid phase in which all the three components area miscible, and the mixture is termed homogenous. However, the ethanol / toluene / water system in this experiment involves adding water (as a third component) into a miscible mixture of ethanol and toluene. Thus, the explanation can be adjusted by the appearance of cloudiness in the mixtures of three components.

            From the graph plotted, a curve-shaped graph bounds a region of B, the two-liquid phases within the curve and the line toluene-water. The region of A, which is outside the tull-shaped region B, is the single-liquid phase. This is the region which denotes a homogenous miscible liquid of the three components. The curve is termed binoda or binodal curve, of which the points of intersection between its two ends and the line toluene-water, are the limits of the two-liquid phase mentioned before. Prediction for the ternary mixture of other concentrations can be made more easily if with the aid of the graph.

            Besides, 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. To form a homogenous mixture, the second point has to be exceeded. Due to insolubility of toluene in water or water in toluene, 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.

            Moreover, changes in temperature will cause the area of immiscibility, which is A, to change. Temperature increases promotes miscibility and in turn cause the area of the binodal decrease. A point will be reached with continuous increase in temperature where complete miscibility is achieved. The binodal system will be destroyed at this point.

Conclusion:

            Ethanol, toluene and water system is a ternary system with one pair of partially miscible liquid (toluene and water).

                             

References:

1. Physical Pharmacy: Physical Chemistry Principles in Pharmaceutical Sciences, by Martin, A.N.
2. Martin,A.N.2006. Physical Pharmacy: Physical Chemistry Principles in Pharmaceutical Sciences. 5^th Ed. Philadelphia: Lea & Febiger.

3.      Alfonso R. Gennaro al.1995. Remington: The Science & Practice of Pharmacy.19^th Edition. Easton, Pennsylavania: Mack Publishing Company.

PRACTICAL 4 : DETERMINATION DIFFUSION COEFFICENT

PRACTICAL 4 : DETERMINATION OF A DIFFUSION COEFFICIENT 

INTRODUCTION:

Introduction:

            Diffusion is a process where the spontaneous movement of certain molecules from a high concentration gradient to a low concentration gradient. This phenomenon can be explained by Fick’s law of diffusion. Fick’s law state that the flowing of a substance (amount, dm in time, dt) through certain dimensions (area A) is directly proportional to concentration gradient dc/dx.

dm = -DA(dc/dx)dt

 where D is the diffusion coefficient or diffusion force for solute that has unit as m²sֿ¹.

            If a solution which have neutral molecules with concentration, Mo, put in a slim tube next to a water tube, diffusion can be stated as

M = M0 eksp (-x²/ 4 Dt)

where M is the concentration at x distance from the level between water and solution that measured at time t.

            By changing equation (II) to logarithm form, we can obtain

ln M = (ln M0 )(-x²/4Dt)

or 2.303 x 4D (log ₁₀ M₀ –log ₁₀ M) t = x² …….(III)

            Thus, one x² versus t graph can produce a straight line which cross the origin with its gradient 2.303 x 4D (log ₁₀ M₀–log ₁₀ M). From here, D can be counted.

            If the molecules in the solution are assumed to be a sphere shape, then the size and mass of the molecules can be counted from Stokes-Einstein equation.

D = kT/6пŋa

(D =kT/9 and 9 = 6пŋa)

where k is the Boltzmann constant 1.38 x 10²³ Jk^-1 , T is the temperature in Kelvin, ŋ is the viscosity of the solute, in Nm-2s and a half diameter of molecule in M. The volume for that certain sphere molecule is 4/3пa³, thus the mass of M is equal to 4/3пa³р where p is the density of the molecule.

As we know that the molecular mass of M = mN, where N is the Avogadro’s number 6.02 x 10²³ mol^-1  .

M = 4/3пa³Nр

Diffusion for molecules with charges, equation (III) has to be changed to insert the gradient force effect that exists between the solution and the solvent. However, this can be overcome by adding a little of sodium chloride into the solvent to prevent the forming of this gradient force.

Agar gel contains a semi-solid molecular net that can be interfering by water molecules. The water molecules will form a continuous phase in the agar gel. By this, the solute molecules can be diffused freely in the water, if not there will be no chemical interaction and diffusion occur. Thus, these agar gels provide a suitable supportive system that can be used in the experiment for diffusion of certain molecules in a aqueous medium

APPARATUS :

1.  Test tubes
2.  pestle and mortar
3.  stirrer
4.  beaker
5.  electronic heater
6.  test tube holder
7.  test tube 
8. stoppers
9.  measuring cylinders.

PROCEDURES:

1.      7g of agar powder was weighed and mixed with 450ml of Ringer solution.

2.      The mixture in step 1 was stirred and boiled on a hot plate until transparent yellowish solution.

3.      20ml of the agar solution was pour into each 6 test tubes. The test tubes ware put in the fridge to let them cool.

4.      An agar test tube which contained 5ml of 1:500,000 crystal violet was being prepared for standardize the colour distance that cause by the diffusion of crystal violet.

5.      After the agar solutions in the test tubes were become solidify, 5ml of 1:200, 1:400, 1:600 crystal violet solution were pour into each test tubes and 3 test tubes were to be put in room temperature while another 3 were been put in 37ºc water bath.

6.      The test tubes were been closed immediately to prevent the vaporization of the solutions.

7.      Step 2 to 6 was repeated for Bromothymol Blue solutions.

DISCUSSION:

            Diffusion is a passive process by which a concentration difference is reduced by a spontaneous flow of matter. The solute will spontaneously diffuse from a region of high chemical potential to a region of low chemical potential. This means that it is from a region of high concentration to a region of low concentration; in which the solvent molecules move in the reverse direction. The expression which relates the flow of material to the concentration gradient is referred to as Fick’s first law equation.

                      dm = -DA(dc/dx)dt ………….(I)This equation states that the amount of dm substance diffuse in the direction (x) at time (dt) go through an area (A) is proportional with the concentration gradient (dc/dx) in the area. While D is the diffusion coefficient(or diffusivity). The negative sign indicates that the flux is in the direction of decreasing concentration. Fick’s first law equation describes the diffusion process under conditions of steady state; that is the concentration gradient does not change with time.In this experiment, the diffusion particles are neutral with Mo concentration, the agar medium is considered homogenous and with constant concentration, hence the diffusion can be expressed as 

                      M = Mo eksp ^(x²/4Dt)  ………….(II)

Change the equation II to logarithm form, we have

                      ln M = ln Mo – (x²/4Dt)

                      2.303 x 4D (log ₁₀ Mo- log ₁₀ M) t = x² ………….(III)

Therefore, when a graph x² against time t is plotted, a straight line is obtained with the gradient of 2.303 

x 4D (log ₁₀ Mo- log ₁₀ M) . From here D can be calculated. We can know the both 28ºC and 37 ºC 

system, the rate of diffusion from the result that is 1:200 > 1:400 > 1:600.M is the system with the dilution 1:500,000. It acts as a constant system during the experiment. When Mo is increased, (log ₁₀ Mo- log ₁₀  M) will increased. This causes the concentration gradient become larger, therefore the driving force for the occurrence of diffusion would be larger and the diffusion process will become faster.

         Other than that, the size of the solute particles also influences the diffusion rate. In this experiment,

 the diffusion rate in Bromothymol Blue system is faster than Crystal Violet system. Thus we can

conclude that the size particle of Bromothymol Blue molecules is smaller than the size particle of Crystal 

Violet molecules.The diffusion rate at 37ºC is higher than at 28ºC. This can be explained by the Stokes-Einstein equation—the relationship between the the radius of the drug molecules and its diffusion coefficient 

(assuming spherical particles or molecules)  in which the particles are spherical.

                         D = kT/6пŋa

                       (D = kT/9 and 9 =6пŋa)

     From this equation, we know that D, the diffusion coefficient is directly proportional to the T, 

temperature. This is because when the temperature is increased, the particles gain more energy and 

become more active, hence they can diffuse more rapidly into the agar medium.

   Stokes-Einsteins equation also shows the influence of the viscosity of the medium, ŋ , on the diffusion 

coefficient. Besides that, the agar gel also can influence the rate of diffusion. When the concentration of 

gel substance is increase, the size of the hole will decrease and the diffusion rate will decrease too as the 

hole size same with the size of the diffuse molecule. Moreover, the viscosity of the solution in the hole 

also can influence the diffusion rate. When the crystallinity of the gel medium is increased, the 

diffusion rate will decrease. The larger the volume fraction of crystalline material, the slower the 

movement of diffusion molecules. This can be happened because crystalline regions of the gel medium 

represent an impenetrable barrier to the movement of solute particles where it have to circumnavigate 

through it.

   There were errors in answer for bromothymol blue system in 37 ºC where the diffusion rate observed 

was lower than in 28 ºC . That was because of the way everyone observed the length of the diffused 

solution into agar was not same, causing inaccurate of results being recorded.

QUESTION:

1) From the trial value for Crystal Violet system D   , estimate the value of D   by using the following equation:

                        D_{28ºC      =}      T_28ºC

                       D_37ºC     =     T_37ºC

Where n₁ and n₂ are viscosity of water at 28ºC and 37ºC.Is the calculated value of D_37ºC value the same as the value from the experiment? Give some explanation if it is different. Is there any difference between the calculated molecular weight with the real molecular weight?

          

      The D_37ºC value is 5.497 x10^-11 cm²/s while the trial value is  5.631 x10^-11 cm²/s .There is a difference between these two values, where it is less of 1.34 x10^-12   cm²/s of D_37ºC value from the trial value. This is because there are some errors occurred during the trial. For example, parallax error occurred when reading the measurements, the temperature at the room is not constant at 28 ºC and the viscosity of agar in the test tube is not uniform.

2) Between Crystal Violet and Bromothymol Blue, which will diffuse much faster? Explain it if there are some differences in the diffusion coefficients.

Bromothymol Blue will diffuse faster than Crystal Violet as the diffusion coefficient of Bromothymol Blue is larger than diffusion coefficient of Crystal Violet. The smaller size of molecule, the easier it needs to penetrate through agar.

CONCLUSION:

Diffusion coefficient, D for Crystal Violet system at 28ºC is 5.338 x10^-11 cm²/s while at 37ºC is
5.497 x10 ^-11 cm²/s. The diffusion coefficient, D for Bromothymol Blue system at 28ºC is 4.06x10^-11 cm²s^-1..while at 37ºC is 4.181 x10^-11 cm²/s. The temperature and concentration of diffusing molecules are the main factors for this experiment. In both 28ºC and 37ºC system, diffusion rate is faster in the concentration of diffusing molecules 1:200> 1:400> 1:600.

REFERENCES:

A.T.Florence and D.Attwood. (1998). Physicochemical Principals of Pharmacy, 3^rd Edition. Macmillan Press Ltd.

PRACTICAL 3 : ADSORPTION

PRACTICAL 3 : ADSORPTION

OBJECTIVE :

To study the adsorption of iodine from solution and Langmuir equation to estimate the surface area of activated charcoal sample.

THEORY :

            Adsorption is a surface phenomenon. Adsorption is a process that occurs when a gas or liquid solute accumulates on the surface of a solid or a liquid (adsorbent), forming a molecular or atomic film. It is different from absorption, in which a substance diffuses into a liquid or solid to form a solution. The term sorption encompasses both processes, while desorption is the reverse process.

Adsorption is operative in most natural physical, biological, and chemical systems, and is widely used in industrial applications such as activated charcoal, synthetic resins and water purification. It occurs when particles such as ion, atom or molecules on the surface of solids are capable of attracting other molecules due to the instability of energies (electrostatic, valency or van der waals) around the particles resulting to the adsorption phenomenon.

Similar to surface tension, adsorption is a consequence of surface energy. In a bulk material, all the bonding requirements (be they ionic, covalent or metallic) of the constituent atoms of the material are filled. But atoms on the (clean) surface experience a bond deficiency, because they are not wholly surrounded by other atoms. Thus it is energetically favourable for them to bond with whatever happens to be available. The exact nature of the bonding depends on the details of the species involved, but the adsorbed material is generally classified as exhibiting physisorption or chemisorption.

The physical adsorption is categorized by low heat of adsorption, which is between 20 to 40kJ per mole gas. It occurs at low temperature when shaking of thermal molecule is not enough to cause complete evaporation at adsorbed layer on the surface of solid. The forces of attraction between the molecules of the adsorbate and the adsorbent are of the weak van der Waals' type. Since the forces of attraction are weak, the process of adsorption can be easily reversed by heating or decreasing the pressure of the adsorbate (as in the case of gases).

            However, chemical adsorption involves combination of chemical substance adsorbed to the surface of adsorbent. It normally occurs at high adsorbent heat, which is between 40 to 400 kJ/mol and it is irreversible chemical adsorption. The forces of attraction between the adsorbate and the adsorbent are very strong; the molecules of adsorbate form chemical bonds with the molecules of the adsorbent present in the surface. Chemical adsorption generally produces adsorption of a layer of absorbate (monolayer adsorption). On the other hand, physical adsorption can produce adsorption of more than one layer of absorbate (multilayer adsorption). Nevertheless, it is possible that the chemical adsorption can be followed by physical adsorption on subsequent layer.

           Factors that affecting adsorption are solute concentration, temperature, pH and surface area of absorbent. With the increased solute concentration will increase the amount of adsorption occurring at equilibrium until a limiting value is reached.. Besides, increase in temperature will decrease adsorption. And pH influences the rate of ionization of the solute. Lastly, an increase in surface area will increase the extent of absorption.

Adsorption measurement can be used to determine the surface area of a solid. With rough surfaces and pores, the actual surface area can be large when compared to geometric apparent surface area. Irving Langmuir published an isotherm for gases adsorbed on solids, which retained his name. It is an empirical isotherm derived from a proposed kinetic mechanism.It is based on four hypotheses, that is the surface of the adsorbent is uniform, that is, all the adsorption sites are equal. Adsorbed molecules do not interact. All adsorption occurs through the same mechanism. Lastly, at the maximum adsorption, only a monolayer is formed. Molecules of adsorbate do not deposit on other, already adsorbed, molecules of adsorbate, only on the free surface of the adsorbent.

MATERIAL:

Iodine solutions (specified in Table 1), 1% w/v starch solution, 0.1M sodium thiosulphate solution, distilled water and activated charcoal

APPARATUS:

1. 12 conical flask
2.  6 centrifuge tubes
3.  measuring cylinders
4.  analytical balance.
5.  Beckman J6M/E centrifuge
6.  burettes
7.  retort stand and clamps
8. Pasteur pipettes

PROCEDURE:

12 conical flasks (labeled 1-12) were filled with 50ml mixtures of iodine solutions (A and B) as stated in the table 1 by using burettes or measuring cylinders.

Table 1: 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

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

For flask 1 – 6:

1. 1- 2 drops of starch solution were added as an indicator.

2. The solution was then titrated using 0.1 M sodium thiosulphate solution until the colour of the solution change from dark blue to colourless.

3. The volume of the thiosulphate used was recorded.

Set 2: Concentration of iodine in solution A at equilibrium (C)

For flasks 7-12

1.  0.1g of activated charcoal was added

2. The flask was capped tightly. The flask was swirled every 10 minutes for 2 hours.

3. After 2 hours, the solutions are transferred into centrifuge tubes and they are labeled accordingly.

4. The solutions are centrifuged at 3000rpm for 5 minutes and the resulting supernatant was transferred into the new conical flask. Each conical flask was labeled accordingly.

5. Steps 1,2 and 3 were repeated as carried out for flask 1-6 in set 1.

RESULT :

Flask	VO	V f	Volume of Na2S2O3 (10ml)	Volume of Na2S2O3   (50ml)
1	0.0	7.1	-	7.1
2	7.1	20.3	-	13.2
3	20.3	38.8	-	18.5
4	0.0	21.9	-	21.9
5	21.9	47.8	-	25.9
6	0.0	46.6	-	46.6
7	0.7	1.8	1.1	5.5
8	1.8	3.9	2.1	10.5
9	3.9	6.9	3.0	15.0
10	6.9	10.5	3.6	18.0
11	10.5	15.6	5.1	25.5
12	15.9	24.1	8.2	41.0

QUESTIONS:

1. Calculate N for iodine in each flask.

N=(X-C)x50/1000x1/y

1ml 0.1M Na₂S₂O₃ =  0.01269g I

1mol iodine = 2 x 126.9= 253.8gmol^-1

Flask 1

Mole of iodine = 7.1 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 3.55x10^-4 mol

X= 3.55 x10^-4 mol / (50ml/1000ml)

   = 0.0071M

Flask 7

Mole of iodine = 5.5 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 2.75 x10^-4 mol

C= 2.75x10^-4 mol / (50ml/1000ml)

   = 0.0055 M

For flask 1 and 7

N= (0.0071 -0.0055)x50/1000x1/0.1

   = 0.0008 mol

Flask 2

Mole of iodine = 13.2 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 6.6 x10^-4 mol

X= 6.6 x10^-4 mol / (50ml/1000ml)

   = 0.0132 M

Flask 8

Mole of iodine = 10.5 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 5.25x10^-4 mol

C= 5.25x10^-4 mol / (50ml/1000ml)

   = 0.0105 M

For flask 2 and 8

N= (0.0132-0.0105)x50/1000x1/0.1

   = 0.00135 mol

Flask 3

Mole of iodine = 18.5 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 9.25x10^-4 mol

X= 9.25x10^-4 mol / (50ml/1000ml)

   = 0.0185 M

Flask 9

Mole of iodine = 15.0 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 7.5x10^-4 mol

C= 7.5x10^-4 mol / (50ml/1000ml)

   = 0.015 M

For flask 3 and 9

N= (0.0185-0.015)x50/1000x1/0.1

   = 0.00175 mol

Flask 4

Mole of iodine = 21.9 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 1.095x10^-3 mol

X= 1.095x10^-3 mol / (50ml/1000ml)

   = 0.0219M

Flask 10

Mole of iodine = 18.0ml x (0.01269gml^-1/253.8gmol^-1)

                        = 9x10^-4 mol

C= 9x10^-4 mol / (50ml/1000ml)

   = 0.0180 M

For flask 4 and 10

N= (0.0219-0.0180)x50/1000x1/0.1

   = 0.00195 mol

Flask 5

Mole of iodine = 25.9 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 1.5x10^-3 mol

X= 1.5x10^-3 mol / (50ml/1000ml)

   = 0.03M

Flask 11

Mole of iodine = 25.5 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 1.275x10^-3 mol

C= 1.275x10^-3 mol / (50ml/1000ml)

   = 0.0255M

For flask 5 and 11

N= (0.0300-0.0255)x50/1000x1/0.1

   = 0.00225 mol

Flask 6

Mole of iodine = 46.6ml x (0.01269gml^-1/253.8gmol^-1)

                        = 2.33x10^-3 mol

X = 2.33x10^-3 mol / (50ml/1000ml)

    = 0.0466 M

Flask 12

Mole of iodine = 41.0 ml x (0.01269gml^-1/253.8gmol^-1)

                        = 2.05 x10^-3 mol

C= 2.05x10^-3 mol / (50ml/1000ml)

   = 0.041 M

For flask 6 and 12

N= (0.0466-0.0410)x50/1000x1/0.1

   = 0.0028 mol

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

Flasks	X (M)	C (M)	Y (g)	N (mol)
1 and 7	0.0071	0.0055	0.1	0.00080
2 and 8	0.0132	0.0105	0.1	0.00135
3 and 9	0.0185	0.0150	0.1	0.00175
4 and 10	0.0219	0.0180	0.1	0.00195
5 and 11	0.0300	0.0255	0.1	0.00225
6 and 12	0.0466	0.0410	0.1	0.00280

Graph of iodine adsorbed (N) versus balance concentration of solution (C)

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

C/N=C/Nm+I/KN_m

Where C= concentration of solution at equilibrium

N_m=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 1/N_m and intercept of 1/KN_m is obtained.

Obtain the value of N_m and then calculate the number of iodine molecule adsorbed on the monomolecular layer. Assume that the area covered by one adsorbed molecule is    3.2x10^-19m², Avogadro no. = 6.023x10²³ molecule, calculate the surface area of charcoal in m²g^-1

C (M)	C/N (1/L)
0.0055	6.875
0.0105	7.778
0.0150	8.571
0.0180	9.231
0.0255	11.333
0.0410	14.642

From the equation C/N = C/Nm + 1/KN_m.

From the graph, it is shown that the intercept of 1/KN_m is 13.

From the graph, the calculated slope, 1/N_m  = (14.642-6.875) ÷ (0.0410-0.0055)

                                                                                  = 7.767 ÷ 0.0355

                                                                                  = 218.789

            1/N_m  = 218.789

            N_m  = 1/218.789

                   = 0.00457 mol g^-1 charcoal

No. of molecules of charcoal

= N_m   x Avogadro no.

= (0.00457 mol g^-1) x (6.023 x 10²³ molecules per mole)

= 2.75 x 10²¹ molecules g^-1

Surface area of charcoal = (3.2 x 10^-19 m² molecules^-1) x (2.75 x 10²¹ molecules g^-1)

                                        = 880 m² g^-1

DISCUSSION:

Adsorption, the binding of molecules or particles to a surface, must be distinguished from absorption, the filling of pores in a solid. The binding to the surface is usually weak and reversible. Just about anything including the fluid that dissolves or suspends the material of interest is bound, but compounds with color and those that have taste or odor tend to bind strongly. Compounds that contain chromogenic groups (atomic arrangements that vibrate at frequencies in the visible spectrum) very often are strongly adsorbed on activated carbon. Decolorization can be wonderfully efficient by adsorption and with negligible loss of other materials. The most common industrial adsorbents are activated carbon. Activated carbon is produced specifically so as to achieve a very big internal surface (between 500 - 1500 m²/g). This big internal surface makes active carbon ideal for adsorption. Active carbon comes in two variations: Powder Activated Carbon (PAC) and Granular Activated Carbon (GAC). Due to its high degree of microporosity, just 1 gram of activated carbon has a surface area in excess of 500 m2, as determined typically by nitrogen gas adsorption. Sufficient activation for useful applications may come solely from the high surface area, though further chemical treatment often enhances the adsorbing properties of the material. Activated carbon is usually derived from charcoal.

         Several factors influence the effectiveness of activated charcoal. The pore size and distribution varies depending on the source of the carbon and the manufacturing process. Large organic molecules are absorbed better than smaller ones. Adsorption tends to increase as pH and temperature decrease. Contaminants are also removed more effectively if they are in contact with the activated charcoal for a longer time, so flow rate through the charcoal affects filtration.

         In this experiment, each flask is filled with different concentration of iodine solution. For flask 1 to 6, it is filled with the actual concentration of iodine in the solution A, while for the flask 7 to 12, it is filled with the concentration of iodine in solution A at equilibrium. Based on the result obtained, we know that the increasing actual concentration of iodine in the solution A, the increasing the concentration of iodine being absorbed (N). This means that the increased solute concentration will increase the amount adsorption occurring at equilibrium until a limiting value is reached. As the charcoal achieved saturated level, the amount of solute being adsorbed also will not further increase. So when the equilibrium is achieved, there is no further adsorption occur.

Solubility is the important factor that affecting adsorption as the solute concentration highly depends on the solubility of the adsorbate. In fact , the adsorption of a solute is inversely proportional to its solubility in the solvent .In this experiment, when  the concentration of solution increases, the amount of iodine adsorbed is also increased. This can be proven from the graph above which show the amount of iodine adsorbed is proportional to the balance concentration of solution. Hence ,we can conclude that the higher the solubility, the higher the degree of adsorption.

         Basically, the equilibrium state can be determined shaking iodine solution with 0.1 g of active charcoal for different intervals of time ranging from 2 to 120 minutes. It was observed that the adsorption process is instantaneous and attained equilibrium within five minutes. Therefore, a shaking time of 10 minutes was selected for all further studies. The quick establishment of equilibrium indicates the high adsorption capacity of the active carbon for iodine. So, after the shaking and ensure the solute has been absorbed, the solution is titrated again to determine the balance of concentration of solute in solution A.

       In this experiment, when we plot C/N versus C, a straight line with slope 1/N_m and intercept of 1/KN_m is obtained, this means that the Langmuir equation is followed and the surface area of charcoal can be calculated. The gradient of the graph is 1/ N_m is 218.789 . N_m (number of mole per gram charcoal required) can be calculated from the gradient of the graph as it is equal to the 0.00457 mole / g . The surface area of charcoal obtained from this experiment is 880m²g^-1. Surface area of the adsorbent is the one of the factor that affect the absorption . The greater the the surface area, the greater for the rate of the absorption.

        Iodine number is a measure of the iodine adsorbed in the pores and, as such, is an indication of the pore volume available in the activated carbon of interest. The use of iodine number as a measure of the degree of exhaustion of a carbon bed can only be recommended if it has been shown to be free of chemical interactions with adsorbates and if an experimental correlation between iodine number and the degree of exhaustion has been determined for the particular application.

      If the adsorption of the adsorbate leads to a maximum of a single monomolecular layer when the adsorption is complete, it is possible to calculate the area of the adsorbent. When a monomolecular layer is adsorbed, it may be assumed that the area of an adsorbent equals the total area of the adsorbed molecules. Solid surfaces can adsorb dissolved substances from solution. When a solution of iodine is shaken with activated charcoal, part of the iodine is removed by the charcoal and the concentration of the solution decreased. From the results gathered, it is realized that K increases as the concentration of iodine is decreased with respect to time. Hence, the degree to which a solid will adsorb material depends on a number of things including temperature, nature of molecule being adsorbed, degree of surface pore structure, and, solute concentration & solvent. Other factors are important factors dealing with the process of adsorption of solutes from aqueous solution by highly porous solids.

CONCLUSION:

Adsorption from solution follows generally the principles laid down for the adsorption of gases. The surface area of charcoal as determined from the experimental result is 880 m² g^-1.

REFERENCES:

1.      Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5^th Edition, Patrick J. Sinko, Lippincott
      Williams and Wilkins, page 39, 40

2.      www.wikipedia.org/wiki/Activated_carbon

3.      E.A.Moelwyn- Hughes. (1961). Physical Chemistry, 2^nd Ed. Pergamon. New York.