Wednesday, 29 May 2013

Practical 3

TITLE: Adsorption from solution

DATE: 18 April 2013

OBJECTIVE
To study the adsorption process from solution that is important in determination of the surface area of powder drug, which is related to its particle size, is important in the field of Pharmacy.
INTRODUCTION:
            Adsorption is a process where free moving molecules of a gaseous or solutes of a solution come close and attach themselves onto the surface of the solid. The attachment or adsorption bonds can be strong or weak, depending on the nature of forces between adsorbent (solid surface) and adsorbate (gas or dissolved solutes). When adsorption involves only chemical bonds between adsorbent and adsorbate, it is recognized as chemical adsorption or chemisorption. Chemical adsorption or chemisorption acquires activation energy, can be very strong and not readily reversible.
            When the reaction between adsorbent and adsorbate is due solely to van der Waals forces, this type of adsorption is known as physical adsorption or van der Waals adsorption. This process is non-specific and can occur at any condition. This type of adsorption is reversible, either by increasing the temperature or reducing the pressure of the gas or concentration of the solute.
            Chemical adsorption generally produces adsorption of a layer of adsorbate (monolayer adsorption). On the other hand, physical adsorption can produce adsorption of more than one layer of adsorbate (multilayer adsorption). Nevertheless, it is possible that chemical adsorption can be followed by physical adsorption on subsequent layers. For a particular adsorbent/ adsorbate, the degree of adsorption at a specified temperature depends on the partial pressure of the gas or on concentration of the adsorbate for adsorption from solution. The relationship between the degree of adsorption and partial pressure or concentration is known as adsorption isomerism. The studies of types of isotherm and changes of isotherm with temperature can provide useful information on the characteristics of solid and the reactions involved when adsorption occurs.

            In adsorption from solution, physical adsorption is far more common than chemisorption. However, chemisorption is sometimes possible, for example, fatty acids are chemisorbed from benzene solutions on nickel and platinum catalysts.
PROCEDURE:
Material and apparatus:
12 conical flasks, 6 centrifuge tubes, measuring cylinders, analytical balance, Beckman J6M/E centrifuge, burettes, retort stand and clamps, Pasteur pipettes, iodine solutions (specified in Table 1), 1% w/v starch solution, 0.1 M sodium thiosulfate solution, distilled water and activated charcoal.
Experiment:
            12 conical flasks (labeled 1-12) were filled with 50ml mixtures of iodine solutions (A and B) as was stated in the Table 1 by using burettes or measuring cylinders.

Table 1: Solution A : Iodine (0.05 M)
            : 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 was added as an indicator.
2.       0.1 M sodium thiosulfate was titrated into the flasks until the colour of the solution had changed from dark blue to colourless.
3.      The volume of the sodium thiosulfate used was recorded.
 Set 2: Concentration of iodine in solution A at equilibrium (C)

For flask 7-12:
1.      0.1 g activated charcoal was added.
2.      The flasks were caped tightly. Every 10 minutes for 2 hours the flask was shaken and swirled.
3.      After 2 hours, the solutions were transferred into the centrifuge tubes and were labeled accordingly.
4.      The solutions were centrifuged at 3000rpm for 5 minutes and the resulting supernatants were transferred into new conical flasks. Each conical was labeled accordingly.
5.      The steps 1,2 and 3 were repeated as carried out for flasks 1-6 in Set 1.

RESULTS

Flasks
Volume of Na2S2O(mL)
Initial volume
Final volume
Total volume
1
0.0
8.0
8.0
2
8.0
21.0
13.0
3
21.0
38.0
17.0
4
0.0
21.2
21.2
5
0.0
28.0
28.0
6
0.0
43.5
43.5
7
0.0
9.9
0.8
8
6.0
19.0
1.2
9
0.0
16.1
2.2
10
19.0
37.0
3.0
11
10.0
31.0
4.5
12
0.0
38.0
8.0          



Flasks
X (M)
Flasks
C (M)
1
8.00x10-3
7
2.86 x10-3
2
13.00x10-3
8
4.39 x10-3
3
17.00x10-3
9
7.86 x10-3
4
21.20x10-3
10
10.70 x10-3
5
28.00x10-3
11
16.10 x10-3
6
43.50x10-3
12
28.60 x10-3

CALCULATION
Based on the titration equation,                                             
I2 + 2Na2S2O3 = Na2S4O6 + 2NaI
Na2S2O3 = ½ I2
Flask1:

From the results,
The volume of Na2S2O3 used = 8.0ml

The no. of moles of Na2S2O3 = 8.0ml x 0.1M
= 0.80 mol

The no. of moles of I2 = 0.80mol ÷ 2
                                     = 0.40mol

The concentration of I2 in solution A
 = 0.40mol ÷ 50ml
 =0.008 M

Thus X = 8.0 x10-3 M
For flask 7:
From the results,
The volume of Na2S2O3 used = 0.8ml

The no. of moles of Na2S2O3 = 0.8ml x 0.1M
= 0.08 mol

The no. of moles of I2 = 0.08mol ÷ 2
                                     = 0.04mol

The concentration of I2 in solution A
 = 0.04mol ÷ 14ml
 =0.00286M

Thus C = 2.86 x10-3 M
Flask 2:

From the results,
The volume of Na2S2O3 used = 13.0ml

The no. of moles of Na2S2O3 = 13.0ml x 0.1M
 = 1.30 mol

The no. of moles of I2 = 1.30mol ÷ 2
                                    = 0.65mol

The concentration of I2 in solution A
= 0.65mol ÷ 50ml
= 0.013 M

Thus X =  13.0 x10-3 M.
For flask 8:
From the results,
The volume of Na2S2O3 used = 1.2ml

The no. of moles of Na2S2O3 = 1.2ml x 0.1M
= 0.12 mol

The no. of moles of I2 = 0.12mol ÷ 2
                                     = 0.06mol

The concentration of I2 in solution A
 = 0.06mol ÷ 14ml
 =0.00439 M

Thus C = 4.39x10-3 M


Flask 3:

From the results,
The volume of Na2S2O3 used = 17.0 ml

The no. of moles of Na2S2O3 = 17.0ml x 0.1M
                                                = 1.70mol
The no. of moles of I2 = 1.70mol ÷ 2
                                     = 0.85mol

The concentration of I2 in solution A
 = 0.85mol ÷ 50ml
= 0.017 M

Thus X = 17.0 x10-3 M.
For flask 9:
From the results,
The volume of Na2S2O3 used = 2.2ml

The no. of moles of Na2S2O3 = 2.2ml x 0.1M
= 0.22 mol

The no. of moles of I2 = 0.22mol ÷ 2
                                     = 0.11mol

The concentration of I2 in solution A
 = 0.11mol ÷ 14ml
 =0.00786M

Thus C = 7.86 x10-3 M
Flask 4:

From the results,
The volume of Na2S2O3 used = 21.2ml

The no. of moles of Na2S2O= 21.2ml x 0.1M
                                                = 2.12 mol

The no. of moles of I2  = 2.12mol ÷ 2
                                     = 1.06mol

The concentration of I2 in solution A
= 1.06mol ÷ 50ml
= 0.0212 M
           
Thus X = 21.2 x10-3 M.
For flask 10:
From the results,
The volume of Na2S2O3 used = 3.0ml

The no. of moles of Na2S2O3 = 3.0ml x 0.1M
= 0.30 mol

The no. of moles of I2 = 0.30mol ÷ 2
                                     =0.15mol

The concentration of I2 in solution A
 = 0.15mol ÷ 14ml
 =0.0107 M

Thus C = 10.7 x10-3 M
Flask 5:

From the results,
The volume of Na2S2O3 used = 28.0ml

The no. of moles of Na2S2O3 = 28.0ml x 0.1M
                                                = 2.80 mol

The no. of moles of I2 = 2.80mol ÷ 2
                                    = 1.40mol

The concentration of I2 in solution A
= 1.40mol ÷ 50ml
= 0.028M
Thus X = 28.0 x10-3M
For flask 11:
From the results,
The volume of Na2S2O3 used = 4.5ml

The no. of moles of Na2S2O3 = 4.5ml x 0.1M
= 0.45mol

The no. of moles of I2 = 0.45mol ÷ 2
                                     = 0.225mol

The concentration of I2 in solution A
 = 0.225mol ÷ 14ml
 =0.0161M
Thus C = 16.1 x10-3 M
Flask 6:

From the results,
The volume of Na2S2O3 used = 43.5ml

The no. of moles of Na2S2O3 = 43.5ml x 0.1M
                                                = 4.35mol

The no. of moles of I2 = 4.35mol ÷ 2
                                    = 2.175mol

The concentration of I2 in solution A
= 2.175mol ÷ 50ml
= 0.0435 M

Thus X =43.5 x10-3M
Flask 12:

From the results,
The volume of Na2S2O3 used = 8.0ml

The no. of moles of Na2S2O3 = 8.0ml x 0.1M
                                                = 0.80mol

The no. of moles of I2 = 0.80mol ÷ 2
                                    = 0.40 mol

The concentration of I2 in solution A
= 0.40 mol ÷ 14ml
= 0.0286 M

Thus C =28.6 x10-3M
DISCUSSION
            Adsorption is the binding of molecules or particles onto a surface, it is different from absorption which is the filling of pores in a solid. It is also a process in which a gas, liquid, or solid adheres to the surface of a solid or (less frequently) a liquid but does not penetrate it, such as in adsorption of gases by activated carbon (charcoal). In comparison, a gas or liquid taken-in during absorption penetrates or mixes with the absorbent.Binding of molecules or particles to the surface is usually weak and reversible, but compounds with color and those that have taste or odor tend to bind strongly. The most common industrial adsorbents are activated carbon, silica gel, and alumina because they have enormous surface areas per unit weight.
      According to the theory, the amount of a substance that can be adsorbed onto activated charcoal depends on the nature of the substance and its concentration, the surface struc­ture of the activated carbon, the temperature and pH of the water. For a treatment system with a specific type of carbon and a known substance, there is a relationship between the amount of adsorbed matter per unit of weight of carbon and the equilibrium concentration in the water in which the temperature and pH are kept constant. This relationship is called an isotherm. The shape of the isotherm can be described in various mathematical ways. The most well-known is the Freundlich isotherm, the equation is as below:
where a and n are constants, the form 1/n being used to emphasize that c is raised to a power less than unity. 1/n is dimensionless parameter and is related to the intensity of drug adsorption. Method of B.E.T (Brunauer, Emmett and Teller) has been used to calculate the surface area of charcoal by using the adsorption of gas.
                  Throughout the experiment, the surface area of activated charcoal sample is determined by Langmuir equation and the interaction with iodine. Iodine is titrated with sodium thiosulphate solution to determine the amount of iodine present. The value of X, actual concentration of iodine in solution A and C, concentration of iodine in solution A at equilibrium of each flask can be calculated by using number of mole of iodine per volume of solution A.
      From the graph of amount of iodine adsorbed versus balance concentration of iodine, a curve graph is obtained which shows that the number of iodine adsorbed is gradually increasing in the solution. This phenomenon can be explained by solubility. For the iodine to be adsorbed onto the activated charcoal, solute-solvent bonds between the iodine and water must first be broken. Thus, it can be concluded that the greater the solubility, the stronger are the solute-solvent bonds and hence the smaller the extent of the adsorption of iodine onto the activated charcoal. For the second graph of C/N versus C, a straight line should be obtained in this graph according to Langmuir equation. The Langmuir isotherm was developed by Irving Langmuir in 1916 to describe the dependence of the surface the coverage or adsorption of molecules on a solid surface to gas pressure or concentration of a medium above the solid surface at a fixed temperature.
The surface area of charcoal calculated is deviated from the actual value is due to the there are some errors existed during the experiment. It may because of the inaccuracy titration of iodine with sodium thiosulphate solution. We may take out some charcoal accidentally from the centrifuged tube together with the solution that we needed in the titration. The sodium thiosulphate may be adsorbed onto the surface of the activated charcoal. This cause the activated charcoal does not achieved equilibrium with the solution after shaking for 2 hours and thus will require more sodium thiosulphate solution in the titration process compare to the actual volume that is needed. Besides, when the solution is shaking for two hours in every 10 minutes, the solution is not well swirled and the charcoal is not well distributed all over the solution, therefore adsorption could not happen completely. In addition, we may not weigh the 0.1g of activated charcoal accurately for every flask causing the result of adsorption is not accurate. The solution also may not be distributed evenly in each test tube which is supposed to be.
Equilibrium that has been reached after shaking for 2 hours can be determine experimentally by observing and recording the temperature of the solution. Since most of the adsorption is an exothermic reaction, temperature will increase as the adsorption of adsorbate onto the adsorbent is occurred. When the temperature remains the same, we can assume that it is at equilibrium and adsorption occur at monolayer. At equilibrium, no more iodine molecules can be adsorbed onto the activated charcoal since all the available surface areas of the activated charcoal have been used up for the adsorption of the activated charcoal. So, the equilibrium point is reached and we continue the titration process.
QUESTIONS

1.      Calculate N for iodine in each flask.
The value of N is calculated by using the formula
                        N = (X - C) x 50/1000 x 1/y
Where y = 0.1g                                                                                                                                  
Flasks 1 and 7:

N  = (X - C) x 50/1000 x 1/y
= (8.00x10-3- 2.86x10-3) x (50/1000) x   (1/0.1)
= 25.7 x 10-4mol
Flask 4 and 10:

N  = (X - C) x 50/1000 x 1/y
= (21.20x10-3-10.70x10-3) x (50/1000) x (1/0.1)
= 52.5 x 10-4mol

Flasks  2 and 8:

N  = (X - C) x 50/1000 x 1/y
= (13.00x10-3-4.39x10-3) x (50/1000) x (1/0.1)
= 43.1 x 10-4mol

Flask 5 and 11:

N  = (X - C) x 50/1000 x 1/y
= (28.00x10-3-16.10x10-3) x (50/1000) x (1/0.1)
= 59.5 x10-4 mol
Flasks 3 and 9:

N  = (X - C) x 50/1000 x 1/y
= (17.00x10-3-7.86x10-3) x (50/1000) x (1/0.1)
= 45.7 x 10-4mol
Flask 6 and 12:

N  = (X - C) x 50/1000 x 1/y
= (43.50x10-3-28.60x10-3) x (50/1000) x (1/0.1)
= 74.5 x 10-4mol

Flasks
X (x10-3M)
C(x10-3M)
N(x10-4)
1 and 7
8.00
2.86
25.7
2 and 8
13.00
4.39
43.1
3 and 9
17.00
7.86
45.7
4 and 10
21.20
10.70
52.5
5 and 11
28.00
16.10
59.5
6 and 12
43.50
28.60
74.5







































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

Graph N Versus C At Equilibrium

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

Graph N Versus C At Equilibrium



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 10-19 m2, Avogrado no. = 6.023 x 1023 molecule, calculate the surface area of charcoal in m2g-1.

CALCULATION


C (M)
N (mol)
C/N (M/mol)
2.86 x10-3
25.7 x10-4
1.113
4.39 x10-3
43.1 x10-4
1.019
7.86 x10-3
45.7 x10-4
1.720
10.70 x10-3
52.5 x10-4
2.038
16.10 x10-3
59.5 x10-4
2.706
28.60 x10-3
74.5 x10-4
3.839

C(x10-3M)
C/N (M/mol)
2.86
1.113
4.39
1.019
7.86
1.720
10.70
2.038
16.10
2.706
28.60
3.839


 From the graph obtained, the gradient of the graph
=
= 112
            Thus, 1/Nm = 112 and Nm = 8.93x10-3mol g-1
                No. of moles charcoal = 8.93x10-3mol g-1 x  0.1g
                                               = 8.93x10-4 mol
            No. of molecules = 8.93x10-4mol x (6.023 x 1023)
                                                = 5.378 x1020 molecules
            Area covered = (5.378 x1020  )x (3.2 x 10-19)
                                  = 172 m2
            The surface area of charcoal = 172m2 ÷ 0.1g
                                                          = 1720m2g-1


4.   Discuss the results of the experiment. How do you determine experimentally 
that equilibrium has been reached after shaking for 2 hours?
From the results of the experiment, the number of iodine adsorbed is gradually increasing in the solution due to its solubility as the concentration of iodine in solution at equilibrium keep on increasing. For the iodine to be adsorbed onto the activated charcoal, solute-solvent bonds between the iodine and water must first be broken. The greater the solubility, the stronger are the solute-solvent bonds and hence the smaller the extent of the adsorption of iodine onto the activated charcoal. The experiment is repeated and then it will be titrated with sodium thiosulfate. If the volume obtained stay constant then the equilibrium is reached.

CONCLUSION

The surface area of charcoal is 1720m2g-1
REFERENCES

1.      Florence A. T. and Attwood D., Physicochemical Principles of Pharmacy, 1998 3rd ed., Macmillan Press Ltd, London. Pages 204-236.
2.      Sinko P. J., 2011, Martin’s Physical Pharmacy and Pharmaceutical Sciences 6th ed., Lippincott Williams & Wilkins, a Wolters Kluwer Bussiness, Philadelphia. Pages 85-91.
3.      Conner C.W., 1998, Physical Adsorption : Experiment, Theory, and Application., Kluwer Academic Publisher, Netherlands. Pages 98-123.
4.      A. Zangwill, 1988, Physics at surfaces, Cambridge University Press. Pages 220-238.
5.      http://www.businessdictionary.com/definition/adsorption.html


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