To determine the surface area of the drug, related to its particle size.
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 isotherm. 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.
Several factors will influence the extent of
adsorption from solution and is summarized in the table below.
Factors
affecting adsorption
|
Effect
on adsorption
|
Solute concentration
|
Increased solute concentration will increase
the
|
amount of adsorption occurring at equilibrium
until
a limiting value is reached
|
|
Temperature
|
Process is usually exothermic, therefore, an
increase
in temperature will decrease adsorption
|
pH
|
pH influences the rate of ionization of the
solute,
hence, the effect is dependent on the species
that is
more strongly adsorbed
|
Surface area of absorbent
|
An increase in surface area will increase the
extent
Of adsorption.
|
Determination
of Surface Area of Activated Charcoal via Adsorption from Solution
Determination of the surface area of powder
drug, which is related to its particle size, is important in the field of
pharmacy. Surface area is one of the factors that govern the rate of
dissolution and bioavailability of drugs that are absorbed through the
gastrointestinal tract. It is also important in the field of colloidal science,
which is widely used in the pharmaceutical preparations.
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 the geometric apparent surface area. In the
method of B.E.T (Brunauer, Emmett and Teller), adsorption of gas was used to
measure the surface area. In this experiment, adsorption of iodine from
solution is studied and Langmuir equation is used to estimate the surface area
of activated charcoal sample.
Materials
 |
| Distilled water |
 |
| Activated charcoal |
 |
| 1% w/v Starch Solution and Iodine Solution (0.05M) |
0.1 M Sodium Thiosulphate Solution
Apparatus |
| Analytical Balance |
 |
| Centrifuge |
 |
Beckman J6M/E
centrifuge
|
 |
| Pasteur Pipette |
 |
| Burette, retort stands and clamps |
 |
| Conical Flask |
Experimental Procedures
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: Solution A: Iodine (0.05M), Solution B: Potassium iodide
(0.1M)
12 conical
flasks (labelled 1-12) were filled with 50 ml mixtures of iodine solutions (A
and B) as stated in Table 1.
Set 1: Actual Concentrations of iodine in solution A (X)
For flasks 1-6:
1. 1-2
drops of starch solution were added as an indicator
2. The
solution was titrated using 0.1 M
2
3. The
volume of the sodium thiosulphate used was recorded.
Set 2: Concentration of iodine solution A at
equilibrium (C)
For flasks 7-12:
1.
0.1 g activated charcoal was added.
2. The
flasks were capped tightly. The flask was shake or swirled every 10 minutes
for 2 hours.
3. After
2 hours, the solutions were transferred into centrifuge and labelled accordingly.
4. The solutions were centrifuged at 3000rpm for 5
minutes and the resulting supernatant was transferred to new conical flask and labelled accordingly.
5. Steps
1, 2 and 3 were repeated as carried out for flasks 1-6 in Set 1.
Compositions of mixtures
\
Flask
|
Volume of solution A (ml)
|
Volume of
solution B (ml)
|
1 & 7
|
10.00
|
40.00
|
2 & 8
|
15.00
|
35.00
|
3 & 9
|
20.00
|
30.00
|
4 & 10
|
25.00
|
25.00
|
5 & 11
|
30.00
|
20.00
|
6 & 12
|
50.00
|
0.00
|
Index
Solution
A : 0.05M Iodine
|
Solution
B : 0.10M Potassium iodide
|
Flask
|
VO
(mL)
|
V f
(mL)
|
Volume of Na2S2O3
(mL)
|
1
|
0.1
|
8.8
|
8.7
|
2
|
8.8
|
21.4
|
12.6
|
3
|
21.4
|
38.80
|
17.4
|
4
|
0.0
|
22.1
|
22.1
|
5
|
22.1
|
48.6
|
26.5
|
6
|
0.0
|
44.5
|
44.5
|
7
|
21.5
|
27.0
|
5.5
|
8
|
0.0
|
8.3
|
8.3
|
9
|
8.3
|
21.5
|
13.2
|
10
|
21.7
|
40.8
|
19.1
|
11
|
0.0
|
21.7
|
21.7
|
12
|
0.0
|
39.8
|
39.8
|
Discussion
Adsorption
is a process where molecules (adsorbate) such as substances in gas or liquid
forms attach themselves to solid surfaces (adsorbent). Adsorption from liquid
phase can take place at any of the three interfaces, which are liquid-solid, liquid-liquid
or liquid-vapor. In practice, more attention has been directed in liquid-solid
interfaces. This is due to the fact that purification of liquid such as water,
wine and oils and their decolourization and detoxification have been carried
out for centuries using charcoals and active carbons. In this experiment, the
charcoal is used as the adsorbent in order to identify how the adsorbent works
and to differentiate the reaction takes place in the presence and absence of
adsorbent.
In experiment 1, the solution is titrated with sodium
thiosulphate until the color of the solution change colorless This titration
is used to determine the actual concentration of the iodine in the solution A.
Then the volume of the solution is used to calculate the number of moles of
iodine present in the solution, hence, finding the mass of the iodine(gram) in
the solution.
In experiment 2, the activated charcoal is placed into the flask containing the adsorbate. Iodine number is the most fundamental used to characterize the performances of activated charcoal. The iodine number is referring to the amount (gram) of the iodine that being adsorbed on the pores of the charcoal. The higher the iodine number, the higher the performances of the charcoal as an adsorbent.
In experiment 2, the activated charcoal is placed into the flask containing the adsorbate. Iodine number is the most fundamental used to characterize the performances of activated charcoal. The iodine number is referring to the amount (gram) of the iodine that being adsorbed on the pores of the charcoal. The higher the iodine number, the higher the performances of the charcoal as an adsorbent.
Based on the results obtained, the amount of sodium thiosulphate used is much more when no charcoal is added compared to when charcoal is added. Charcoal is known as activated carbon and has a very large surface area available for adsorption. An increase in surface area will increase the extent of adsorption. Besides, charcoal can pack the molecules of iodine into its crevices and hold most of the iodine so that less iodine is present in the solution. That is why the amount of sodium thiosulphate used to decolourised the solution is less.
For safety,
laboratory coat and goggle should be worn by the students to avoid direct
contact with the chemicals that are harmful to body and eyes if they are
spilled. All the apparatus are rinsed with distilled water before use, to
remove any impurities as they will affect the final results.
Questions
Molecular weight of iodine, I2 : 253.8
gmol-1
1. Calculate N for iodine in each flask.
N=(X-C)x50/1000x1/y
(y=0.1g)
1ml 0.1M Na2S2O3
= 0.01269g I
No. mole = Mass / Molecular weight
Flask 1
Mole of iodine = 8.7 ml x 0.01269 gml-1/253.8
gmol-1
=
4.35 x 10-4 mol
X= 4.35 x 10-4 mol / (50 ml/1000 ml)
= 0.0087
M
Flask 7
Mole of iodine = 5.5 ml x 0.01269 gml-1/253.8
gmol-1
=
2.75 x 10-4 mol
C= 2.75 x 10-4 mol / (50 ml/1000 ml) = 0.0055
M
For
flask 1 and 7
N= (0.0087 -0.0055) x 50/1000 x 1/0.1
= 0.00595
mol g-1
Flask 2
Mole of iodine = 12.6 ml x 0.01269 gml-1/253.8
gmol-1
=
6.30 x 10-4 mol
X=6.30 x 10-4 mol / (50 ml/1000 ml)
= 0.0126
M
Flask 8
Mole of iodine = 8.3 ml x 0.01269 gml-1/253.8
gmol-1
=
4.15 x 10-4 mol
C= 4.15 x 10-4 mol / (50ml/1000 ml)
= 0.0083
M
For
flask 2 and 8
N= (0.0126-0.0083) x 50/1000 x 1/0.1
= 0.00845
mol g-1
Flask 3
Mole of iodine = 17.4 ml x 0.01269 gml-1/253.8
gmol-1
=
8.7 x 10-4 mol
X= 8.7 x 10-4 mol / (50 ml/1000 ml)
= 0.0174
M
Flask 9
Mole of iodine = 13.2 ml x 0.01269 gml-1/253.8
gmol-1
=
6.6 x 10-4 mol
C= 6.6 x 10-4 mol / (50 ml/1000 ml)
= 0.0132
M
For
flask 3 and 9
N= (0.0174-0.0132) x 50/1000 x 1/0.1
= 0.0108
mol g-1
Flask 4
Mole of iodine = 22.1 ml x 0.01269 gml-1/253.8
gmol-1
=
1.105 x 10-3 mol
X= 1.105x 10-3 mol / (50 ml/1000 ml)
= 0.0221
M
Flask 10
Mole of iodine = 19.1 ml x 0.01269 gml-1/253.8
gmol-1
=
9.55 x 10-4 mol
C= 9.55 x 10-4 mol / (50 ml/1000 ml)
= 0.0191
M
For
flask 4 and 10
N= (0.0221-0.0191) x 50/1000 x 1/0.1
= 0.01255
mol g-1
Flask 5
Mole of iodine = 26.5 ml x 0.01269 gml-1/253.8
gmol-1
=
1.625 x 10-3 mol
X= 1.325 x 10-3 mol / (50 ml/1000 ml)
= 0.0265 M
Flask 11
Mole of iodine = 21.7 ml x 0.01269 gml-1/253.8
gmol-1
=
1.058 x 10-3 mol
C= 1.058 x 10-3 mol / (50 ml/1000 ml)
= 0.0217
M
For
flask 5 and 11
N= (0.0265-0.0217) x 50/1000 x 1/0.1
= 0.01565
molg-1
Flask 6
Mole of iodine = 44.5 ml x 0.01269 gml-1/253.8
gmol-1
=
2.225 x 10-3 mol
X= 2.225 x 10-3 mol / (50 ml/1000 ml)
= 0.0445
M
Flask 12
Mole of iodine = 39.8 ml x 0.01269 gml-1/253.8
gmol-1
=1.99
x 10-3 mol
C= 1.99 x 10-3 mol / (50 ml/1000 ml)
= 0.0398
M
For
flask 6 and 12
N= (0.0445-0.0398) x 50/1000 x 1/0.1
= 0.0246
mol g-1
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
|
0.0087
|
0.0055
|
0.1
|
0.00595
|
2 and 8
|
0.0126
|
0.0083
|
0.1
|
0.00845
|
3 and 9
|
0.0174
|
0.0132
|
0.1
|
0.01080
|
4 and 10
|
0.0221
|
0.0191
|
0.1
|
0.01255
|
5 and 11
|
0.0265
|
0.0217
|
0.1
|
0.01565
|
6 and 12
|
0.0445
|
0.0398
|
0.1
|
0.02460
|
3. According to Langmuir theory, if there
is no more than a monolayer of iodine adsorbed on the charcoal,
C/N = C/Nm + 1/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 10-19 m2, Avogadro no. = 6.023
x 1023 molecule, calculate the surface area of charcoal in m2g-1.
Concentration Of
Solution, C (M)
|
Amount Of Iodine
Adsorbed, N (molg-1)
|
C/N
(M/molg-1 )
|
0.0055
|
0.00595
|
0.9244
|
0.0083
|
0.00845
|
0.9822
|
0.0132
|
0.01080
|
1.2222
|
0.0191
|
0.01255
|
1.5219
|
0.0217
|
0.01565
|
1.3866
|
0.0398
|
0.02460
|
1.6179
|
C/N (M/molg-1) 
From the graph the
gradient is used to determine the number of mole per gram charcoal required
(Nm)
Gradient,
m = (1.4-0.8) M/molg-1 ÷ (0.0225-0.00125) M
=
28.24 gmol-1
In general the gradient
is equal to 1/ Nm, therefore
1/Nm = 28.24
Nm = 1/28.24
= 0.03542molg-1
From the Nm, 0.03542 mol of iodine is adsorbed in 1 g of
charcoal. Therefore, the number of mol of iodine molecules that adsorbed on the
monomolecular layer is 0.03542 /1g = 0.03542mol .
The
number of molecules of the iodine molecules
= number
of mole of iodine molecules x Avogadro no.
=0.03542mol
x 6.023x 1023
=2.133x
1022 molecules of iodine
Assume the surface
covered by one adsorbed molecules is 3.2x 10-19 m2
1 molecules of iodine =
3.2x 10-19 m2 adsorbed on the charcoal layer.
Therefore,
2.133 x 1022 molecules of iodine
= (3.2x
10-19 m2) x (2.133
x 1022)
= 6825.6
m2g-1 adsorbed on the charcoal layer
4. Discuss the results of the experiment.
How do you determine experimentally that equilibrium has been reached after
shaking for 2 hours?
Adsorption occurs when
an adsorbent solid is mixed in a solution containing dissolved substances whicj
can act as an adsorbate. When iodine solution is shaken with activated
charcoal, part of the iodine is removed by the charcoal and the concentration
of iodine in the solution decreased. The amount of iodine adsorbed (N)
increases with the concentration of the adsorbate,C. The theoretical result should
be a rapid increase at first when the surface
of the adsorbent is relatively free. As the surface fills
with the adsorbate, the amount of iodine adsorbed increases, but the rate of adsorption decreases.
Eventually, the surface of the adsorbent becomes full
and further increases in the concentration cause no further increase in the
amount adsorbed. Therefore, the graph should obtained
have a parabolic pattern.
In our
experiment, the result shows a direct increase in the amount
of iodine adsorbed. Due to the difference in concentration of iodine between
the last two sets of flasks is large, resulting a drastic increase in N, we
could not obtained a constant line at the last part of the graph.
If the adsorption of the
adsorbate leads to a maximum single molecular 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 the
adsorbent equals the total area of the adsorbed molecules. An increase in
surface area will increase the extent of adsorption.
The graph of C/N versus
C has some points that deviate from the straight line. This may probably due to
error that occurred while carrying out the
experiment. The volume of sodium thiosulphate may not be accurate because we
might not stop the titration once the color of the solution changed from dark
blue to colorless This may also due to parallax error while we took the
reading.
The equilibrium can be
determined experimentally after shaking for 2 hours by titration
using 0.1M sodium thiosulphate When the volume
of sodium thiosulphate solution required to turn the solution from dark to colorless is constant for a few repeating titration, this indicate the
equilibrium has been reached.
Conclusions
Variation
of extend of adsorption with concentration of solute can be represented by the
Langmuir isotherm. The concentration of iodine (adsorbate) is the main factor
in this experiment.
Increased
iodine concentration will increase the amount of adsorption occurring at
equilibrium until a limiting value is reached.
The surface
area of activated charcoal as determined from the experimental result is
References
1.A.T.Florence
and D.Attwood. (1998). Physicochemical Principals of Pharmacy, 4th Edition.
Macmillan Press Ltd.page 194
2. en.wikipedia.org/wiki/Adsorption
3.
Brunauer,S. Pysical Adsorption.1971.
Pricetun University Press. Priceton. New Jersey.
4. U. B. Hadkar. 2007. Physical Pharmacy. 9th edition. Nirali Prakashan Publisher.
5. Mansoor M. Amiji. Beverly J. Sandmann. 2003. Applied physical pharmacy. 1st edition. McGraw-Hill
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