PRACTICAL 3: ADSORPTION
FROM SOLUTION
Objective:
This
experiment is carried out to study the adosorption of iodine from solution and
determine the surface area of activated charcoal via adsorption of iodine from
solution by using Langmuir equation.
Introduction:
Adsorption is totally different
with the absorption. Absorption is the process in which a fluid is dissolved by
a liquid or a solid (absorbent) while 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. Adsorption is nearly always an
exothermic process. 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 as it involves the transfer or sharing of electron ir
breakage of the adsorbate into atoms or radicals which are bound separately. Chemisorption
can also be very strong and not readily reversible. It takes place at high
temperature.
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. The individuality of the adsorbate and
the adsorbent are preserved. It usually takes place at lower temperature and
decreases with increasing temperature.
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. The amount of adsorbate on the adsorbent as a
function if its pressure or concentration at constant temperature. The quantity
adsorbed is nearly always normalized by the mass of the adsobent to allow
comparison of different materials.
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 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 thiosulphate
solution
Distilled water
Activated
charcoal
centrifudge machine
Procedure:
1.
Twelve conical flasks were prepared with 50ml mixtures of iodine solutions (A
and B) as stated in Table 1 by using 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
|
20
|
6
and 12
|
50
|
0
|
2.
1 to 2 drops of starch solution were added into flasks 1 to 6 as an indicator.
3.
0.1M sodium thiosulphate solution was triturated into the flasks 1 to 6 by
using burette until the colour of the solution changed from dark blue to
colourless.
4.
The volume of the sodium thiosulphate solution used was recorded and the actual
concentration of iodine in solution A (X) was calculated.
5.
0.1g activated charcoal was weighed by electronic balance and added into flasks
7 to 12 respectively.
6. The flasks were capped tightly and swirled or
shaked every 10 minutes for 2 hours.
7.
The solutions were transferred into centrifuge tubes after 2 hours and they
were labelled accordingly.
8.
The solutions were centrifuged at 3000 rom for 5 minutes and the resultant
supernatants were transferred into the
new conical flasks. Each conical flask was labelled accordingly.
9.
Steps 2, 3 and 4 were repeated by replacing flasks 1 to 6 with the resultant
supernatants.
10.
N was calculated for iodine in each flask according to the general notes given below.
11.
The graph of amount of iodine adsorbed (N) versus balance concentration of
solution (C) at equilibrium was plotted to obtain adsorption isotherm.
12.
The graph of C/N versus C was plotted to answer the questions given.
GENERAL
NOTES:
|
Titration
equation:
I2
+ 2Na2S2O3 → Na2S4O6
+ 2NaI
Na2S2O3
≡ 1/2 I2
|
Given:
(1 mole Na2S2O3
≡ 1/2 mole of I2
1 mole iodine
= 2 x 126.9 g = 253.8 g
1 ml 0.1M Na2S2O3
= 0.01269g I
|
If the amount
of activated charcoal used is Y
gram, therefore the total mole of iodine adsorbed by 1g of activated charcoal
(N) is given by the following equation:
N
= (X-C) x 50/1000 x 1/y
|
Result:
Solution
A: Iodine (0.05 M)
Solution
B: Potassium iodide (0.1 M)
Solution C: Sodium
thiosulphate solution (0.1M)
Flask
|
Volume of
solution A
(ml)
|
Volume of
solution B
(ml)
|
Volume of
Solution C (ml)
|
1
|
10
|
40
|
9.80
|
2
|
15
|
35
|
15.30
|
3
|
20
|
30
|
18.65
|
4
|
25
|
25
|
24.50
|
5
|
30
|
20
|
29.60
|
6
|
50
|
0
|
48.20
|
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
|
Flask
|
Number of
moles of Solution C used (mol)
n1 =
MV
|
Number of
moles of iodine (mol)
n2 =
n1/2
|
Actual
concentration of iodine in solution A (X)
X = n2/V
(M)
|
1
|
9.8x10-3 x 0.1 = 9.8x10-4
|
9.8x10-4/2 = 4.9x10-4
|
4.9x10-4/0.05L = 9.8x10-3
|
2
|
1.53x10-2 x 0.1 = 1.53x10-3
|
1.53x10-3/2 = 7.65x10-4
|
7.65x10-4/0.05L = 0.0153
|
3
|
1.865x10-2 x 0.1 = 1.865x10-3
|
1.865x10-3/ 2 = 9.325x10-4
|
9.325x10-4/0.05L = 0.01865
|
4
|
2.45x10-2 x 0.1 = 2.45x10-3
|
2.45x10-3/2 = 1.225x10-3
|
1.225x10-3/0.05L = 0.0245
|
5
|
2.96x10-2 x 0.1 = 2.96x10-3
|
2.96x10-3/2 = 1.48x10-3
|
1.48x10-3/0.05L = 0.0296
|
6
|
4.82x10-2 x 0.1 = 4.82x10-3
|
4.82x10-3/2 = 2.41x10-3
|
2.41x10-3/0.05L = 0.0482
|
Flask
|
Number of
moles of Solution C used (mol)
n3 =
MV
|
Number of
moles of iodine (mol)
n4 = n3/2
|
Concentration
of iodine in solution A at equilibrium (C)
C = n4/V
(M)
|
7
|
2.0
x 10-3 x 0.1 = 2.0 x 10-4
|
2.0 x 10-4 / 2 = 1.0 x 10-4
|
1.0 x 10-4 / 0.012L = 8.333
x 10-3
|
8
|
3.5
x 10-3 x 0.1 = 3.5 x 10-4
|
3.5 x 10-4 / 2 = 1.75 x 10-4
|
1.75 x10-4/0.012L = 0.0146
|
9
|
3.8
x 10-3 x 0.1 = 3.8 x 10-4
|
3.8 x 10-4/ 2 = 1.9 x 10-4
|
1.9 x 10-4/0.012L = 0.0158
|
10
|
4.5
x 10-3 x 0.1 = 4.5 x 10-4
|
4.5 x 10-4/ 2 = 2.25 x 10-4
|
2.25 x 10-4/0.012L = 0.01875
|
11
|
4.5
x 10-3 x 0.1 = 4.5 x 10-4
|
4.5 x 10-4/ 2 = 2.25 x 10-4
|
2.25 x 10-4/0.012L =
0.01875
|
12
|
8.0
x 10-3 x 0.1 = 8.0 x 10-4
|
8.0 x 10-4/ 2 = 4.0 x 10-4
|
4.0 x 10-4/0.012L = 0.03333
|
Questions:
1) Calculate N
for iodine in each flask.
N = (X-C) x
50/1000 x 1/y
a. For flask 1 and 7:
N = (9.8x10-3 - 8.333 x 10-3) x 50/1000 x 1/0.1
= 7.335 x 10-4
b. For flask 2 and 8:
N = (0.0153 - 0.0146) x 50/1000 x 1/0.1
= 3.5 x 10-4
c. For flask 3 and 9:
N = (0.01865 – 0.0158) x 50/1000 x 1/0.1
= 1.425 x 10-3
d. For flask 4 and 10:
N = (0.0245 –
0.01875) x 50/1000 x 1/0.1
= 2.875 x 10-3
e. For flask 5 and 11:
N = (0.0296 – 0.01875) x 50/1000 x 1/0.1
= 5.425 x 10-3
f. For flask 6 and 12:
N = (0.0482 – 0.03333) x 50/1000 x 1/0.1
= 7.435 x 10-3
2)
Plot amount of iodine adsorbed (N) versus balance concentration of solution (C)
at equilibrium to obtain adsorption isotherm.
Concentration
of iodine in solution A at equilibrium (C / M)
|
Amount of
iodine adsorbed (N / mol)
|
8.333 x 10-3
|
7.335 x 10-4
|
0.0146
|
3.5 x 10-4
|
0.0158
|
1.425 x 10-3
|
0.01875
|
2.875 x 10-3
|
0.01875
|
5.425 x 10-3
|
0.03333
|
7.435 x 10-3
|
Graph of amount of iodine adsorbed against
balance concentration of solution at equilibrium
Concentration
of iodine in solution A at equilibrium (C / x 10-2M)
|
Amount of
iodine adsorbed (N / x 10-3 mol)
|
C/N (Mg/mol)
|
0.8333
|
0.7335
|
11.357
|
0.146
|
0.35
|
41.714
|
1.58
|
1.425
|
11.088
|
1.875
|
2.875
|
6.522
|
1.875
|
5.425
|
3.456
|
3.333
|
7.435
|
4.479
|
Graph of C/N versus C
Gradient = (0-24)
------------
(2.9-0.5)×10-2
= -1000
Gradient = 1/Nm = -1000
Thus, Nm = 1 = -1.0×10-3molg-1
---------
-1000
No. of moles of charcoal =
-1×10-3 molg-1 × 0.1
g
= -1.0
x 10⁻⁴ mol
No. of charcoal molecules
= -1.0 x 10⁻⁴ mol x (6.023×1023) molecules/mol
= -6.023 x 1019 molecules
Area covered by charcoal
molecules = -6.023 x 1019 x (3.2 x 10-19m2)
= -19.2736
m2
Surface area of charcoal =
-19.2736 m2 / 0.1 g
= -19.2736 m2g-1
4) Discuss the results of the experiment. How do you determine experimentally that equilibrium has been reached after shaking for 2 hours?
The result of the experiment is unaccurate. The
slope of the original graph of C/N versus C must be in positive form but the
slope of the graph of our group is in negative form. After shaking for 2 hours,
the equilibrium
has been reached when the solution becomes homogenous and has no more colour
changes.
Discussion:
Adsorption is the sticking of molecules from the gas
or liquid phase onto the surface of a solid and it is different from absorption
which is the filling of pores in a solid. A molecule that undergoes adsorption
is referred to as the adsorbate, and the solid is the adsorbent. The theorectical result of
this experiment is not the same as the predicted result. This may be caused by
the errors made when carrying out this experiment.
From the first graph, it shows that
the solute concentration is one of the factors affecting adsorption. The amount
of adsorption occurring at equilibrium also will increase when the solute
concentration increases until a limiting value is reached. The shape of the
first graph shows that the higher the balance concentration of solution (C) at
equilibrium, the larger the amount of iodine adsorbed (N). The actual
concentration of iodine in solution A (X) is always smaller than the
concentration of iodine in solution A at equilibrium because of the addition of
the activated charcoal into flasks 7 to 12. This will produce the concentration
of iodine in solution A at equilibrium. The iodine is being adsorbed by the
activated charcoal when the charcoal is added into the solution. Therefore, the
total amount of iodine remains in the solution decreases and this can explain
that the actual concentration of iodine in solution A (X) is larger than the
concentration of iodine in solution A at equilibrium (C) as shown by the
results of the experiment. In conclusion, the total mole of iodine adsorbed by 1 g of
activated charcoal (N) is always positive as C is always smaller than X.
However, the second graph we
obtained is inaccurate as the slope of the graph is in negative form. It is
impossible to get a negative slope because there is no negative value for the
surface area of the charcoal. There are some possible errors that
occur while conducting this experiment. The amount of sodium thiosulphate
titrated recorded may be inaccurate because the position of eye s not
perpendicular to the solution. This will affect the overall changes result of
the experiment. Other than that, there are too much of charcoal added into the
solution and not distributed evenly over the solution. This causes the solution
cannot achieve the equilibrium. Next, the charcoal may react supernatant which
will affect the amount of sodium thiosulphate needed to change colour from dark
blue to colourless.
Conclusion:
Iodine can be adsorbed by activated charcoal. The
surface area of activated charcoal determined in this experiment is -19.2736 m2g-1. The experiment is not carried out successfully.
Reference:
1.
Martin’s Physical Pharmacy and Pharmaceutical
Sciences, 5th Edition, Patrick J. Sinko, Lippincott Williams and
Wilkins, page 39, 40
2.
E.A.Moelwyn-
Hughes. (1961). Physical Chemistry, 2nd Ed. Pergamon. New York.
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