Objectives:
To study
the adsorption of iodine from the solution and its relationship with the
surface area of activated charcoal and the determination of the surface area of
activated charcoal by using Langmuir Isotherm Adsorption Theory.
Introduction:
Adsorption
is the
adhesion of free moving molecules of gas, liquid or dissolved solids to a
surface. It can be divided into two general types which are physical adsorption
and chemisorption. Physical adsorption is categorized by low heat adsorption in
which the adsorbate is bound to the surface through the weak van der Waals
forces while chemisorption involves only chemical bonds between adsorbent and
adsorbate. Chemisorption occurs at high adsorbent heat and it is not
reversible. Physical adsorption, however, is reversible. Physical adsorption
decreases when the temperature increases. This is because physical adsorption
is generally an exothermic process. Increase the pressure will increase the
physical adsorption. Both physical and chemical adsorption may be involved in a
particular adsorption process. Physical adsorption can produce adsorption of
more than one layer of adsorbate (multilayer adsorption) while chemical
adsorption generally produces adsorption of a layer of adsorbate (monolayer
adsorption). However, it is possible that chemical adsorption can be followed
by physical adsorption on subsequent layers.
Adsorption isotherm is the relationship between the
degree of adsorption and the partial pressure or concentration. Physical adsorption is far more common than chemisorption.
Chemisorption is more specific and usually involves an ion-exchange process.
There are several factors which will influence the extent of adsorption from
solution. When the solute concentration increases, the amount of adsorption
occurring at equilibrium until a limiting value is reached is also increases.
An increase in temperature will decrease the adsorption. The pH influences the
rate of ionization of the solute. Hence, the effect is dependent on the species
that is more strongly adsorbed. An
increase in surface area will also increase the extent of adsorption.
Material and Apparatus:
12 conical
flask, 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
thiosulpate solution, distilled water and activated charcoal.
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Procedure:
12 conical flasks
(labeled 1-12) are 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 (0.05 M)
Solution B: Potassium iodide
(0.1 M)
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
|
21
|
6 and 12
|
50
|
0
|
Set
1: Actual concentration of iodine in solution A (X)
For flasks 1-6:
1. 1-2 drops of starch
solution was added as an indicator.
2. The flasks were
titrated using 0.1 M sodium thiosulphate solution until the colour of the
solution changes
from dark blue to colourless.
3. The volume of the
sodium thiosulphate used was recorded.
Set 2:
Concentration of iodine in 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 swirled or shaked every 10 minutes for 2 hours.
3. After 2 hours, the
solution was transferred into centrifuge tubes and was labeled accordingly.
4. The solution was
centrifuged at 3000 rpm for 5 minutes and the resulting supernatant was transferred
into new conical flasks. Each conical flask was labeled accordingly.
5. Steps 1, 2, and 3 as
carried out for flasks 1-6 in Set 1 were repeated.
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RESULTS AND CALCULATION:
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 solution (mL)
|
1
|
10.0
|
40.0
|
8.1
|
2
|
15.0
|
35.0
|
13.7
|
3
|
20.0
|
30.0
|
20.8
|
4
|
25.0
|
25.0
|
26.8
|
5
|
30.0
|
20.0
|
33.4
|
6
|
50.0
|
0.0
|
49.5
|
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 solution (mL)
|
7
|
10.0
|
40.0
|
7.4
|
8
|
15.0
|
35.0
|
12.7
|
9
|
20.0
|
30.0
|
17.5
|
10
|
25.0
|
25.0
|
21.8
|
11
|
30.0
|
20.0
|
28.4
|
12
|
50.0
|
0.0
|
41.8
|
Titration equation: I2
+ 2Na2S2O3 = Na2S4O6
+ 2NaI
Based on the
equation:
2 mol Na2S2O3 ≈ 1 mol I2
2 mol Na2S2O3 ≈ 1 mol I2
1 mole I2 =
2 x 126.9 g/mol = 253.8 g/mol
1 ml 0.1M Na2S2O3
= 0.01269 g I2
0.0001 mol Na2S2O3
= 0.01269g I2
RESULTS:
For flasks 1-6: X =
Calculate the actual concentration of iodine in solution A
Flask 1:
1.0 ml 0.1M Na2S2O3
= 0.01269 g I2
8.1 ml 0.1M Na2S2O3
= 0.1028 g I2
1 mol I2
= 2 x 126.9 = 253.8g
n= 0.1028 / 253.8g
= 4.1 x 10-4 mole I2
[X] = No. of
mole (mol)/Volume (L)
= 4.1 x 10-4 mole /
50/1000L
= 8.2 x 10-3 M
Flask 2:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
13.7 ml 0.1M Na2S2O3
= 0.1739 g I2
1 mol I2
= 253.8 g
n =
0.1739/253.8 = 6.9 X10-4 mol
[C] = No. of
mole (mol)/Volume (L)
= 6.9 X10-4 mol/ 50/1000L
= 0.0138 M
Flask 3:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
20.8 ml 0.1M Na2S2O3
= 0.2640 g I2
1 mol I2
= 253.8 g
n =
0.2640 / 253.8 =1.04 x 10-3 mol
[C] = No. of
mole (mol)/Volume (L)
= 1.04 x 10-3 mol /
50/1000L
= 0.0208 M
Flask 4:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
26.8 ml 0.1M Na2S2O3
= 0.3401 g I2
1 mol I2
= 253.8 g
n =
0.3401/253.8 = 1.3 x 10-3 mol I2
[C] = No. of
mole (mol)/Volume (L)
= 1.3 x 10-3 mol /
50/1000L
= 0.0260 M
Flask 5:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
33.4 ml 0.1M Na2S2O3
= 0.4238 g I2
1 mol I2
= 253.8 g
n=0.4238/ 253.8
= 1.7 x 10-3 mol I2
[C] = No. of
mole (mol)/Volume (L)
= 1.7 x 10-3 mol /
50/1000L
= 0.0340 M
Flask 6:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
49.5 ml 0.1M Na2S2O3
= 0.6282 g I2
1 mol I2
= 253.8 g
n =0.6282/253.8
= 2.5 x 10-3 mol I2
[X] = No. of
mole (mol)/Volume (L)
= 2.5 x 10-3 mol / 0.05L
=0.0500 M
For flasks 7-12:
C =Calculate the concentration of iodine in solution A at equilibrium
Flask 7:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
7.4 ml 0.1M Na2S2O3
= 0.09391 g I2
1 mol I2
= 253.8 g
n= 0.09391 /
253.8 = 3.7 x 10-4 mol
[C]= No. of mole
(mol)/Volume (L)
= 5.5 x 10-5 mol /
50/1000L
= 7.4 x 10-3 M
Flask 8:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
12.7 ml 0.1M Na2S2O3
= 0.1611 g I2
1 mol I2
= 253.8g
n =
0.1611/253.8= 6.3 x 10-4 mol I2
[X]= No. of mole
(mol)/Volume (L)
= 6.3 x 10-4mol / 50/1000L
= 0.0126 M
Flask 9:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
17.5 ml 0.1M Na2S2O3
= 0.2221 g I2
1 mole I2
= 253.8 g
n=0.2221/253.8 g = 8.8 x 10-4 mol I2
X(M) = No. of
mole (mol)/Volume (L)
= 8.8 x 10-4 mol / 0.05L
= 0.0176 M
Flask 10:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
21.8 ml 0.1M Na2S2O3
= 0.2766 g I2
1 mole I2
= 253.8g
n= 0.2766/253.8
= 1.1 x 10-3 mol I2
[X]= No. of mole
(mol)/Volume (L)
= 1.1 x 10-3 mol / 0.05L
= 0.0220 M
Flask 11:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
28.4 ml 0.1M Na2S2O3
= 0.3604 g I2
1 mol I2
= 253.8 g
n=0.3604/253.8 =
1.4 x 10-3 mol I2
[X] = No. of
mole (mol)/Volume (L)
= 1.4 x 10-3 mole / 0.05L
= 0.0280 M
Flask 12:
1 ml 0.1M Na2S2O3
= 0.01269 g I2
41.8 ml 0.1M Na2S2O3
= 0.5304 g I2
1 mol I2
= 253.8 g
n = 0.5304/253.8
= 2.09 x10-3 mol I2
[C] = No. of
mole (mol)/Volume (L)
= 2.09 x10-3 mol / 0/1000L
= 0.0418 M
QUESTIONS:
1. Calculate N
for iodine in each flask.
N = (X – C) x 50/1000 x 1/y
Where y = Amount of activated charcoal used in gram
=
0.1g
N =
Total mole of iodine adsorbed by 1g of activated charcoal
Flask 1 and 7:
X = 8.2 x 10-3 M
C = 7.4 x 10-3 M
N = (X – C) x 50/1000 x 1/y
= (8.2 x 10-3 – 7.4 x 10-3) M x 50/1000 x 1/0.1g
= 4.0 x 10-4 mol/g
Flask 2 and 8:
X = 0.0138 M
C = 0.0126 M
N = (X – C) x 50/1000 x 1/y
= (0.0138 – 0.0126) M x 50/1000 x 1/0.1g
= 6.0 x 10-4 mol/g
Flask 3 and 9:
X = 0.0208 M
C = 0.0176 M
N = (X – C) x 50/1000 x 1/y
= (0.0208– 0.0176) M x 50/1000 x 1/0.1g
= 1.6 x 10-3 mol/g
Flask 4 and
10:
X = 0.0260 M
C = 0.0220 M
N = (X – C) x 50/1000 x 1/y
= (0.0260 – 0.0220) M x 50/1000 x 1/0.1g
= 2.0 x 10-3 mol/g
Flask 5 and
11:
X = 0.0340 M
C = 0.0280 M
N = (X – C) x 50/1000 x 1/y
= (0.0340 – 0.0280) M x 50/1000 x 1/0.1g
= 3.0 x 10-3 mol/g
Flask 6 and 12:
X = 0.0500 M
C = 0.0418 M
N = (X – C) x 50/1000 x 1/y
= (0.0500 – 0.0418) M x 50/1000 x 1/0.1g
= 4.1 x 10-3 mol/g
2. Plot amount of iodine adsorbed (N) versus
balance concentration of solution (C) at equilibrium to obtain adsorption
isotherm.
Flask
|
X
|
C
|
N
|
1 and 7
|
8.2 x 10-3
|
7.4 x 10-3
|
4.0 x 10-4
|
2 and 8
|
0.0138
|
0.0126
|
6.0 x 10-4
|
3 and 9
|
0.0208
|
0.0176
|
1.6 x 10-3
|
4 and 10
|
0.0260
|
0.0220
|
2.0 x 10-3
|
5 and 11
|
0.0340
|
0.0280
|
3.0 x 10-3
|
6 and 12
|
0.0500
|
0.0418
|
4.1 x 10-3
|
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.
C (M)
|
C/N (g/L)
|
7.4 x 10-3
|
18.5
|
0.0126
|
21
|
0.0176
|
11
|
0.0220
|
11
|
0.0280
|
9
|
0.0418
|
10
|
From the graph obtained, the
gradient of the graph
1/ Nm = Gradient of the graph
Thus,
1/Nm = 2.5
and Nm = 0.4 mole / g
Avogadro
number = 6.023 x 1023 molecule
Number
of molecules = Number of moles x Avogadro Number
= 0.4 moles
x 6.023 x 1023 molecule
= 2.4092 x
1023 molecules / g
Area
covered by one adsorbed molecule is 3.2 x 10-19 m2
Surface
area of charcoal = 2.4092 x 1023 molecules / g x 3.2 x 10-19
m2 / molecule
= 7.709 X 104 m2g-1
4. Discuss the results of the experiment. How do you
determine experimentally that equilibrium has been reached after shaking for 2
hours?
We
repeat the experiment and titrate with sodium thiosulphate. If the volume stays
constant then equilibrium is reached.
Discussion:
Common
charcoal is usually made from coal, wood or petroleum. Activated charcoal is
usually used in water filters; medicines which are selectively remove toxins,
and chemical purification processes. Activated charcoal is a carbon that has
been treated with oxygen. When activated is treated with oxygen, a highly
porous charcoal will be resulted. Tiny holes on these highly porous charcoals
give the charcoal a surface area of 300-2000 m2/g which are allowing liquids or
gases to pass through the charcoal and then interacting with the carbon that is
exposed. Carbon is known with its ability to adsorb a wide range of impurities
and contaminants, including chlorine and odours. Other substances which is
sodium, nitrates and fluorides are not attracted to the carbon and therefore
they will not be filtered out. As we know, adsorption works by chemically
binding the impurities to the carbon, so the active sites in the charcoal will
be filled. So, activated charcoal filters will become less effective and then
it have to be recharged and replaced after being used for so long.
There are several factors which
influence the effectiveness of activated charcoal. First factor is the pore
size and its distribution. This factor is depending on the source of the carbon
and the manufacturing process. It is known that large organic molecules are
adsorbed better than smaller ones. Adsorption will be increased when pH and
temperature decrease. Contaminants are also removed more effectively if they
are in contact with activated charcoal for a long time. Therefore, other factor
that influences the effectiveness of activated charcoal is flow rate through
the charcoal.
From the first graph that is drawn, the
relationship between the amount of iodine adsorbed and concentration of iodine
at equilibrium has been clearly shown. Based on the graph, the amount of iodine
adsorbed is proportional to the concentration of iodine at equilibrium. So,
when the amount of iodine at equilibrium is increase, it mean there will be
more collision between the iodine and the the adsorbent. Adsorption occurred
because of the collision between the adsorbent and the adsorbate. So, when, the
rate of collision higher, the adsorption will be increased.
Adsorption isotherm is used to
describe the equilibrium of the adsorption of particles at a surfaces at
constant temperature. Adsorption isotherm shows the amount of particles which
are attached at the surface as a function of the particles present in the
solution. In this experiment, adsorption of iodine from solution is studied and
Langmuir equation has been used to determine the surface area of activated
charcoal. Langmuir states that the rate of adsorption and the rate of adsorbate
evaporation were equal at constant when there is no change in temperature. The assumptions
of Langmuir Isotherm are adsorption cannot exceed monolayer coverage. Besides
that, Langmuir Isotherm also assume that all surface sites are uniform and
equivalent, and the ability of a particle to adsorb at specific site is
independent to the occupation of neighbouring sites.
Based on the results, the volume of
sodium thiosulphate that is used on experiment 7 to 12 which are contained with
activated charcoal are lesser than that is used for flask 1 to 6 (without
activated charcoal). In the beginning of the experiment, for each of the flask
7 to 12, the mixture of iodine and potassium iodide will be added. As we know,
when starch is added, iodine will tend to bind with a starch and then form a
starch iodide complex which make colour of solution become blue. For the flask
7 to 12, the activated charcoal is added first and then starch indicator is
added. Activated charcoal is useful in attracting non polar adsorbates.
Activated charcoal has enormous surface area because it is extremely porous. A
huge surface area of activated charcoal will be a target for the iodine
molecules to be attracted and holding it within its pores by a process called
adsorption. Therefore, this will reduce the amount of iodine present in the
solution which will react with added sodium thiosulphate. So, that’s why the
amount of sodium thiosulphate used in the experiment of flask 7 to 12 is
smaller than flask 1 to 6.
Based on the calculation, the
surface area of charcoal is 7.709 X 104 m2g-1 . This value is to big if we compared to the
general actual value of 1 g of charcoal which is 500 m2 This difference in value
may be occurred because of errors while conducting the experiment. The error
that looks significant in this experiment is over titration. This occurred when
the titration is not stopped directly after colour change occurred. Besides
that, other significant error in this experiment is the volume of the sodium
thiosulphate used for each titration is not accurate as the real end point of
the titration is not reached and then resulting an inaccurate calculation.
Other error is the shaking of the flask every 10 minute interval is not done
properly and cause the not constant value of iodine being adsorbed on the
surface of activated charcoal.
Starch solution is highly used in
detection of the end point of iodine- thiosulphate titration. This is because starch
gives a very definite colour change at the end point. Without starch indicator,
the colour of the iodine solution in the conical flask near the end point fades
slowly from pale yellow to colourless. When the starch indicator is added, the
colour of the solution in the conical flask at the end point changes suddenly
from blue black to colourless. The starch indicator should be added close to
the end point to give a sharp end point, while avoiding the formation of excess
starch-iodine complex, which would be difficult to decompose.
Centrifugation is done to these
flasks before being titrated with sodium thiosulphate. Fine particles suspended
in a liquid can be separated by centrifugation process. After the solution in
flasks has been centrifuged, the higher densities of activated charcoal with
bounded iodine will move down. Besides, lesser amount of free iodine present in
the solution will also reduce the volume of sodium thiosulphate used in
titration.
Conclusion:
The
surface area of the charcoal is 7.709 X 104 m2g-1 . It is proven that the surface area of the
charcoal can be calculated by using Langmuir theory.
References:
1)
Alfonso R. Gennaro al.1995. Remington:
The Science & Practice of Pharmacy.19th Edition. Easton,
Pennsylavania: Mack Publishing Company.
2) Alexander
T. Florence, David Attwood. 2006. Physiochemical
Principles of Pharmacy. Fourth Edition. London: Pharmaceutical Press.
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