Introduction
The
process where the free moving molecules of a gaseous or solutes of a solution
come close and attach themselves onto the surface of the solid is known as
absorption.
The
attachment or adsorption bonds can be strong and weak, depending on the nature
of forces between adsorbent (solid surface) and adsorbate (gas or dissolved
solutes). There are two types of adsorption. There are chemical
adsorption and Physical adsorption.
Chemical
adsorption
or chemisorption is when
adsorption involves only chemical bonds between absorbent and adsorbate. It acquires activation energy, can be very
strong and not readily reversible.
It generally
produces adsorption of a layer of absorbate (monolayer adsorption)
Physical
adsorption is also known
as van der Waals adsorption. It happened when the adsorbate is bound to the
surface through the weak van der Waals forces. It is a non - specific reaction which can be
occurred at any condition and reversible,
either by increasing the temperature or reducing the pressure of the gas or
concentration of the solute. Physical
adsorption can produce adsorption of more than one layer of adsorbate
(multilayer adsorption). It is possible that chemical adsorption can be
followed by physical adsorption on subsequent layers.
Adsorption
isotherm is study of the relationship between the degree of adsorption and the
partial pressure or concentration.. In
some cases,
the degree of adsorption at a specified temperature depends on the partial
pressure of the gas or on the concentration of the adsorbate for adsorption
from solution. 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.
Basically,
physical adsorption is far more common than chemisorption
In
this experiment, we use the adsorption from solution to determine the surface
area of activated charcoal.
There
are 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 adsorbent
|
An
increase in surface area will increase the extent of adsorption.
|
Determination
of Surface Area of Activated Charcoal via Adsorption from Solution
It is important to determine the
surface area of powder drug because
it is
one of the factors that govern the rate of dissolution and bioavailability of
drugs that are absorbed through the gastrointestinal tract. Adsorption
measurements is used to determine the surface area of a
solid. For the rough surfaces and pores solid , 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
iodine
solutions ( specified in Table 1 ), 1 % w/v starch solution, 0.1 M sodium
thiosulphate solution, distilled water and activated charcoal.
Apparatus
12
conical flasks, 6 centrifuge tubes, measuring cylinders, analytical balance,
Beckman J6M/E centrifuge, burettes, retort stand and clamps, Pasteur pipettes
Aim
To study the adsorption of iodine from solution and to
determine the surface area of activated charcoal sample using Langmuir
equation.
Procedure
12 conical flasks (labelled 1-12) is filled with 50ml
mixture of iodine solutions (A and B) as stated in table 1.
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 (0.05 M)
Solution B: potassium iodide (0.1 M)
Set 1: Actual
concentration of iodine in solution A (X)
For flask 1-6:
1)
1-2
drops of starch solution is added as an indicator
2)
The solution
is titrated using 0.1 M sodium thiosulphate solution until the colour of the
solution changes from blue to colourless.
3)
The volume
of sodium thiosulphate used is recorded.
Set 2: concentration
of iodine in solution A at equilibrium (C)
For flask 7-12:
1)
0.1g of activated
charcoal is added into the flasks.
2)
The flasks
was capped tightly. The flask is swirled every 10 minutes for 1 hour.
3)
After an
hour, the solutions is transferred into labelled centrifuge tubes.
4)
The solutions
is centrifuged at 3000 rpm for 5 minutes
5)
The supernatant
is transferred into new conical flasks and labelled accordingly.
6)
Step 1, 2,
and 3 as carried out for flasks 1-6 in set 1 is repeated.
RESULTS AND CALCULATIONS
Set 1 : Actual concentration of iodine in solution
A (X)
Flask
|
Volume
of solution A
(mL)
|
Volume
of solution B
(mL)
|
Volume
of sodium thiosulphate
(mL)
|
1
|
10
|
40
|
7.5
|
2
|
15
|
35
|
14.2
|
3
|
20
|
30
|
19.2
|
4
|
25
|
25
|
24.8
|
5
|
30
|
20
|
29.2
|
6
|
50
|
0
|
44.5
|
Set 2 : Concentration of iodine in solution A at
equilibrium (C)
Flask
|
Volume
of solution A
(mL)
|
Volume
of solution B
(mL)
|
Volume
of sodium thiosulphate
(mL)
|
7
|
10
|
40
|
2.0
|
8
|
15
|
35
|
3.5
|
9
|
20
|
30
|
4.5
|
10
|
25
|
25
|
4.5
|
11
|
30
|
20
|
4.5
|
12
|
50
|
0
|
28.5
|
Based on titration
equation :
I2 + 2Na2S4O3
= Na2S4O6 + 2NaI
Na2S4O3 = ½ I2
Given :
1 mole iodine = 2 x 126.9
g
1 mL 0.1M Na2S4O3
= 0.01269g of I
Flask 1
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
7.5 mL of 0.1M Na2S4O3
= 0.095175 g I2
2 mol Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n = 0.095175g / 253.8g/mol
=
3.75 x 10 – 4 mol I2
[ X ] = 3.75 x 10 – 4mol
/ ( 50/1000 )
= 7.5 x 10- 3 M
Flask 2
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
14.2
m L of 0.1 M Na2S4O3 = 0.1801 g I2
2 mol Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n=
0.1801g / 253.8g/mol
= 7.1 x 10 – 4 mol of I2
[ X ] = 7.1 x 10 – 4 mol
/ ( 50/1000 )
= 0.0142 M
Flask 3
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
19.2
mL of 0.1 M Na2S4O3 = 0.2436 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n
= 0.2436g / 253.8g/mol
= 9.6 x 10- 4 mol of I2
[
X ] = 9.6 x 10 – 4 mol / (50/1000)
= 0.0192 M
Flask 4
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
24.8 mL of 0.1 M Na2S4O3
= 0.3147 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n = 0.3147g / 253.8 g/mol
= 1.240 x 10 -3 mol of I2
[ X ] = 1.240 x 10 –
4 mol / (50/1000)
= 0.02480 M
Flask 5
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
29.2 mL of 0.1 M Na2S4O3
= 0.3705 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n = 0.3705g / 253.8g/mol
= 1.46 x 10 – 3 mol I2
[ X ] = 1.46 x 10 – 3
mol / (50/1000)
= 0.0292 M
Flask 6
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
44.5 mL of 0.1 M Na2S4O3
= 0.5647 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n = 0.5647g / 253.8g/mol
=
2.2250 x 10 – 3 mol I2
[ X ] = 2.2250 x 10 – 3 mol / (50/1000)
=
0.0445 M
____________________________________________________________________
Flask 7
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
2
mL of 0.1 M Na2S4O3 = 0.02538 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n
= 0.02538 g / 253.8g/mol
= 1 x 10 – 4 mol I2
[
C ] = 1 x 10 – 4 mol / (50/1000)
= 2 x 10 – 3 M
Flask 8
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
3.5
mL of 0.1 M Na2S4O3 = 0.0444 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n=
0.0444g / 253.8g/mol
= 1.7494 x 10 – 4 mol I2
[
C ] = 1.7494 x 10 – 4 mol / (50/1000)
= 3.4988 x 10 -3 M
Flask 9
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
4.5
mL of 0.1 M Na2S4O3 = 0.0571 g I2
2 mole Na2S4O3
= 1 mole I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n=
0.0571g / 253.8g/mol
= 2.2498 x 10 – 4 mol I2
[
C ] = 2.2498 x 10 – 4 mol / (50/1000)
= 4.4996 x 10 – 3 M
Flask 10
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
4.5
mL of 0.1 M Na2S4O3 = 0.0571 g I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n
= 0.0571g / 253.8g/mol
= 2.2498 x 10 – 4 mol I2
[
C ] = 2.2498 x 10 – 4 mol / (50/1000)
= 4.4996 x 10 – 3 M
Flask 11
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
4.5 mL of 0.1 M Na2S4O3
= 0.0571 g I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n
= 0.0571g / 253.8g/mol
= 2.2498 x 10 – 4 mol I2
[
C ] = 2.2498 x 10 – 4 mol / (50/1000)
= 4.4996 x 10 – 3 M
Flask 12
1 mL of 0.1 M Na2S4O3
= 0.01269 g I2
8.5 mL of 0.1 M Na2S4O3
= 0.1079 g I2
1 mole I2 = 2 x
126.9 g/mol = 253.8 g/mol
n
= 0.1079g/ 253.8g/mol
= 4.2514 x 10 – 4 mol I2
[
C ] = 4.2514 x 10 – 4 mol / (50/1000)
= 8.5028 x 10 – 3 M
 |
| Graph of C versus N |
QUESTIONS
|
1.
Calculate N for iodine in each flask.
[
Flask 1 and 7 ]
X=
7.5 x 10 – 3M
C=
2 x 10-3 M
N=
(7.5 x 10 – 3M - 2 x 10-3 M) x 50/1000 x 1/0.1g
= 2.75 x 10 -3 mol/g
[
Flask 2 and 8 ]
X=
0.0142M
C=
3.4988 x 10 – 3 M
N
= (0.0142M - 3.4988 x 10 – 3 M) x 50/1000 x 1/0.1g
= 5.3506 x 10 – 3 mol/g
[ Flask 3 and 9 ]
X=
0.0192M
C=
4.4996 x 10 -3 M
N=
(0.0192M - 4.4996 x 10 -3 M) x 50/1000 x 1/0.1g
= 7.3502 x 10 – 3 mol/g
[
Flask 4 and 10 ]
X=
0.02480M
C=
4.4996 x 10 -3M
N=
(0.02480M - 4.4996 x 10 -3M) x 50/1000 x 1/0.1g
= 0.01015 mol/g
[
Flask 5 and 11 ]
X=
0.0292 M
C=
4.4996 x 10 – 3 M
N=
(0.0292 M – 4.4996 x 10 – 3M) x 50/1000 x 1/0.1g
= 0.0124 mol/g
[
Flask 6 and 12 ]
X=
0.0445M
C=
8.5028 x 10 – 3 M
N=
(0.0445M – 8.5028 x 10 – 3M) x 50/1000 x 1/0.1g
= 0.0180mol/g
2.
Plot amount of iodine adsorbed (N) versus balanced concentration of solution
(C) at equilibrium to obtain adsorption isotherm.
Flask
|
Amount
of iodine adsorbed
(N)
|
Balanced
concentration of solution
(C) at equilibrium
|
1 and 7
|
2.7500 x 10 -3 mol/g
|
2.0000 x 10-3 M
|
2 and 8
|
5.3506 x 10 – 3 mol/g
|
3.4988 x 10 – 3 M
|
3 and 9
|
7.3502 x 10 – 3 mol/g
|
4.4996 x 10 -3 M
|
4 and 10
|
0.01015 mol/g
|
4.4996 x 10 -3M
|
5 and 11
|
0.0124 mol/g
|
4.4996 x 10 – 3 M
|
6 and 12
|
0.0180mol/g
|
8.5028 x 10 – 3 M
|
 |
| Graph of C/N versus C |
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.
Obtained
the value of Nm , then calculate the number of iodine molecule
adsorbed on 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.
Flask
|
C/N
|
C
|
1 and 7
|
0.7273
|
2.0000 x 10-3 M
|
2 and 8
|
0.6539
|
3.4988 x 10 – 3 M
|
3 and 9
|
0.6122
|
4.4996 x 10 -3 M
|
4 and 10
|
0.4433
|
4.4996 x 10 -3M
|
5 and 11
|
0.3629
|
4.4996 x 10 – 3 M
|
6 and 12
|
0.4724
|
8.5028 x 10 – 3 M
|
m=
(8.5028x10 -3 M – 2.0x10 -3M) / (0.4724g/l – 0.7273g/l)
= - 0.0255
1
/ Nm = 0.0255
Nm
= 39.22 mole/g
Avogadro`s
number = 6.023x10 23
Number
of molecules = 6.023x1023 x 39.22 mole/g
= 2.3620x10
25 molecules/g
Area covered by one
adsorbed molecule = 3.2x10 -19 m2
Surface area of charcoal =
3.2x10 -19 m2 / molecules x 2.3620x10 25 molecules/
g
= 7.558x10 6 m2g-1
Discussion
Absorption
is a process that occurs when a gas or liquid solute accumulates on the surface
of a solid or a liquid (absorbent), forming a molecular or atomic film
(adsorbate). The driving force for adsorption is the reduction in interfacial
tension between the fluid and the solid adsorbent as a result of the adsorption
of the adsorbate on the surface of the solid.
Langmuir’s isotherm describing
the Adsorption of Adsorbate onto the surface of the Adsorbent requires three
assumptions:
·
The surface of the absorbent is in contact with a solution containing an
adsorbate which is strongly attracted to the surface.
·
The surface has a specific number of sites where the solute molecules can
be adsorbed.
·
The adsorption involves the attachment of only one layer of molecules to
the surface, i.e. monolayer adsorption.
Adsorption by a solid is not a
very important process unless the solid has a very large surface area compared
to its mass. Consequently, charcoal is especially effective because it has a
highly porous structure. Charcoal is activated by being heated to quite high
temperatures in a vacuum or in a stream of dry air.
In pharmacy, activated charcoal
is considered to be the most effective single agent available as an emergency
decontaminant in the gastrointestinal tract. It is used after a person swallows
or absorbs almost any toxic or chemical.
The surface area of activated
charcoal can be obtained by the adsorption characteristics of iodine. The
iodine number is defined as the milligrams of iodine absorbed by 1.0g of carbon
when the iodine concentration of the filtrate is 0.02N (0.02 mol L-1).
In this experiment, iodine is
the absorbate while activated charcoal is the absorbent. In set 1, titration is
used to calculate concentration of iodine because iodide ions and ion molecules
are in equilibrium. The potassium iodide (KI) acts as a reservoir for the
iodine by forming KI3 (K+ and I3-).
This is necessary to do this because of the limited solubility of iodine, as
such, in water. Starch is added to act as an indication in the titration. The
solution turns dark blue when starch is added as iodine molecules are present.
When sodium thiosulfate is added, iodine molecules react with sodium thiosulfate
to form sodium iodide.
I2
+ 2Na2S2O3 Na2S4O6
+ 2NaI
When
all the iodine molecules are totally reacted with sodium thiosulfate solution,
the dark colour of solution change to colourless. From the equation, the number
of moles of iodine can be calculated. In set 2, the activated charcoal acts as
adsorbent to adsorb the iodine molecules.
The Langmuir isotherm shows that
the amount of iodine adsorbed increases as the concentration increases up to a
saturation point. Beyond this point, increasing the iodine concentration will
not cause further increases. This behavior is typical of absorbents with a
limited number of accessible sites. As long as there are available sites,
adsorption will increase with increasing iodine concentration but as soon as
all the sites are occupied, a further increase in quantity of iodine will
result to a negative adsorption (desorption) at the monolayer.
A graph of amount of iodine
adsorbed (N) versus balance concentration of solution(C) at equilibrium to
obtain adsorption isotherm is plotted. The graph is hyperbolic-shaped. When the
concentration of iodine increases from flasks 7 to 12, the total moles of
iodine absorbed by 1g of activated charcoal (N) also increase. The actual
concentration of iodine in solution A (X) is always lower than the
concentration of iodine in solution A in equilibrium (C) because the iodine is
being adsorbed by the activated charcoal. Hence, the amount of iodine left is
decreased. We can determine that the equilibrium has been reached after shaking
for 2 hours by observing there is no further change in the concentration of
supernatant when titrated with sodium thiosulfate solution. We can also examine
there is a gas produced from the solution in the conical flasks which is carbon
dioxide gas due to the reaction.
A graph of C/N versus C is
plotted and a straight line with the slope of 1/Nm and intercept of
1/KNm should be obtained according to the equation C/N = C/Nm
+ 1/KNm. However, we have obtained a negative slope. Hence, Langmuir
theory is not obeyed that there is more than a monolayer of iodine adsorbed on
the charcoal. In fact, the value of Nm (number of mole per gram charcoal
required) can be calculated from the gradient of graph. Then, the number of
iodine molecule adsorbed on the monomolecular layer is calculated. Nm
is a measure of the adsorptive capacity of the activated charcoal for the
iodine molecules. Finally, the surface area of charcoal can be calculated. The
mass of charcoal is calculated by dividing the number of iodine molecule by
Avogadro constant. Then, assumed value of area covered by one absorbed molecule
(3.2 x 10-19 m2) is divided by the mass of charcoal.
There are some errors made in
this experiment. The volume of sodium thiosulfate solution titrated is recorded
inaccurately as the position of eyes is not parallel to the lower meniscus of
sodium thiosulfate solution. The concentration of solution may be affected when
swirling as some solutions are accidentally poured out due to the conical
flasks are not capped tightly. Hence, the conical flasks must be capped tightly
before swirling. The conical flasks must be swirled constantly and vigorously
to allow the charcoal molecules totally exposed to the solution in order for
the adsorption to take place. The purity of the components of the solution and
the avoidance of contamination during preparation and handling are essential.
When working with organic media, contamination with water can have a drastic
effect on the measurement. After equilibration, the adsorbent together with the
adsorbate bound to it must be separated from the bulk equilibrium liquid by
centrifugation. The separation must be carried out at the same temperature as
that at which equilibrium was established.
Conclusion
As a conclusion, adsorption is a process where free
moving molecules of a gaseous or solute of a solution come close and attach
themselves onto the surface of the solid. The purpose of carrying out this
experiment is to identify the surface area of the charcoal by studying the
adsorption of iodine from solution process. The adsorption process is important
in the field of pharmacy as it is a method to determine the surface area of
powder drug. In this experiment, the adsorption process follows the Langmuir
equation that is used to calculate the surface area of the charcoal and the
adsorption process also undergo monolayer adsorption.
The Nm value is 1.4x103 g mol-1.
The number of molecules of iodine adsorbed onto the monomolecular layer is
8.4325 x 1020 molecules of iodine. The surface area of charcoal is
269.84 m2g-1.
Reference
1. http://www.fpharm.uniba.sk/fileadmin/user_upload/english/Physical_Chemistry/5-Adsorption.pdf
2. http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0100-40422011000300020
3. http://pac.iupac.org/publications/pac/pdf/1986/pdf/5807x0967.pdf
Reference
1. http://www.fpharm.uniba.sk/fileadmin/user_upload/english/Physical_Chemistry/5-Adsorption.pdf
2. http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0100-40422011000300020
3. http://pac.iupac.org/publications/pac/pdf/1986/pdf/5807x0967.pdf
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