PRACTICAL 3 : ADSORPTION
THEORY
Adsorption is a phenomenon occurs when the
ion, atom, or particles on the surface of solids are capable of attracting
other molecules due to instability of energies such as electrostatic force,
valency force or van der waals force around the particles. There are two
general types of adsorption which is physical adsorption and chemical
adsorption.
Physical adsorption which is known as van
der waals adsorption, is low heat adsorption. The equilibrium of physical
adsorption is basically depend on properties of the adsorbent substance,
properties of the adsorbate substance, surface area of the adsorbed substance,
temperature and pressure. Since physical adsorption is low heat adsorption, the
higher the temperature, the lesser the adsorption. Therefore by lowering the
temperature and at the same time increasing the pressure will result in
increase of adsorption.
Chemical adsorption or chemisorption
involve the stronger valence forces. Compared to physical adsorption, chemical
adsorption is more specific, and usually involve an ion – exchange process.
Chemisorption occurs at high adsorbent heat and it is reversible.
Chemical adsorption generally produce a
layer of adsorbate while physical adsorption involve more than one layer adsorbate.
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 at a specific temperature depends on the
partial pressure of the gas or on the concentration of the adsorbate for
adsorption from solution.
There are several factors that influence
the extent of adsorption from solution. Firstly, solute concentration. The
higher the solute concentration the higher the rate of adsorption until reach
the limiting value. Another factor is temperature, the higher the temperature,
the higher the rate of adsorption. Then, pH influences the rate of ionization
of the solute. Lastly, the surface area of the adsorbent, an increase in
surface area will extent of adsorption.
In the field of pharmacy, it is important
to determine the surface area of powder drug since it effect the dissolution
and bioavailability of drugs. It is also important in the pharmaceutical
preparations. Therefore, adsorptiom measurement can be used in other to
identify the surface area of a solid. With rough surfaces and pores, the actual
surface can be large than geometric apparent surface area. In method of B.E.T
(Brunauer, Emmett and Teller ) ,adsorption of gas is used to determine the
surface area. In the experiment, Longmuir Equation is used to estimate the
surface area of actvated 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, 1% w/v starch solution, 0.1 M sodium thiosulphate solution, distilled water and activated charcoal.
12 conical flasks, 6 centrifuge tubes, measuring cylinders, analytical balance, Beckman J6M/E centrifuge, burettes, retort stand and clamps, Pasteur pipettes, iodine solutions, 1% w/v starch solution, 0.1 M sodium thiosulphate solution, distilled water and activated charcoal.
PROCEDURE
12 conical flasks is filled with 50 ml
mixtures of iodine solutions ( A and B ) as stated in Table 1 by using
measuring cylinders.
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
|
Solution A : Iodine ( 0.05 )
Solution B : Potassium iodide ( 0.1M )
Table
1
Set 1 which is for flasks 1 – 6 is prepared
for the actual concentration of iodine
in solution A ( X ).
1 ) 1 – 2 drops of starch solution is added
as an indicator.
2 ) The flasks is 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 is recorded.
Set 2 which is for flasks 7 – 12 is
prepared for the concentration of iodine
in solution A at equilibrium ( C )
1 ) 0.1 g activated charcoal is added to
the flasks.
2 ) The cap is placed on the flasks
tightly. The flask is swirled and shaked every 10 minutes for 2 hours.
3 ) The solution is transferred into
centrifuge tubes and labelled them accordingly.
4 ) The solutions is centrifuged at 300 rpm
for 5 minutes and the resulting supernatant is transerred into new conical
flasks. Each conical flasks is labelled accordingly.
5 ) Steps 1, 2 and 3 is repeated as carried
out for flasks 1 -6 in Set 1.
RESULT
SET
1
flask
|
Volume of Na2S2O3 (ml)
|
1
|
10.8
|
2
|
14.1
|
3
|
18.7
|
4
|
23.8
|
5
|
28.4
|
6
|
46.1
|
SET
2
Flask
|
Volume of Na2S2O3 (ml)
|
7
|
0.9
|
8
|
1.4
|
9
|
2.3
|
10
|
2.9
|
11
|
4.0
|
12
|
7.5
|
QUESTION
1)
Calculate N for iodine in each flask.
Titration equation:
I2 + 2Na2S203 = Na2S4O6 + 2NaI
Na2S2O3 = ½ I2
Total mole of iodine adsorbed by 1 g of
charcoal (N ) is calculated by using the
formula given :
N
= (X –C) X 50 / 1000 X 1/y
Where
N: Total mole of iodine adsorbed by 1 gram
of activated charcoal (moles)
X: Actual concentration of iodine in
solution A (moles)
C: Balance concentration of iodine in
solution A at equilibrium (moles)
y: Amount of activated charcoal used (gram)
= 0.1g
Therefore,
Flask 1 :
From the results,
The volume of Na2S2O3 used = 10.8 ml
No. of moles of Na2S2O3 =,1.08x10-3
No. of moles of I2
=
=,1.08x10-3 / 2
= 5.4
X 10 – 4
The concentration of I2 in solution A
=
5.4 X 10 – 4 / 0.05 L
=
0.011 M
FLASK 7
From the results,
The volume of Na2S2O3 used = 0.9 ml
No. of moles of Na2S2O3 =0.09 x 10-3
No. of moles of I2
= 0.09
x 10-3 / 2
= 0.045
X 10 – 3
The concentration of I2 in solution A
= 0.045 x 10 – 3 / 0.012 L
= 3.75 X 10 – 3 M
N =
(X-C) x 50/1000 x 1/0.1
N = (0.011 M - 3.75 X 10 – 3 M) X
50/1000 X 1 / 0.1
N = 3.63 X 10-3 mol
FLASK 2 :
From the results,
The volume of Na2S2O3 used = 14.1 ml
No. of moles of Na2S2O3 =,1.41x10-3
No. of moles of I2
= 1.41
x 10-3 / 2
= 7.05
X 10 – 4
The concentration of I2 in solution A
= 7.05 X 10 – 4 / 0.05 L
= 0.012 M
FLASK 8
From the results,
The volume of Na2S2O3 used = 1.4 ml
No. of moles of Na2S2O3 = 0.14 x10-3 mol
No. of moles of I2
= 0.14
x 10-3 / 2
= 0.7 X
10 – 4
The concentration of I2 in solution A
= 0.7 X 10 – 4 / 0.012L
= 5.83 X 10 – 3 M
N =
(X-C) x 50/1000 x 1/0.1
N = ( 0.012 – 5.82 x 10-3 ) ) x 50/1000 x
1/0.1
N = 3.09 X 10 -3 mol.
FLASK 3 :
From the results,
The volume of Na2S2O3 used = 18.7 ml
No. of moles of Na2S2O3 = 1.87 x 10-3
No. of moles of I2
= 1.87
x 10-3 / 2
= 9.35
X 10 – 4
The concentration of I2 in solution A
= 9.35 X 10 – 4 / 0.05 L
= 0.019M
FLASK
9
From the results,
The volume of Na2S2O3 used = 2.3 ml
No. of moles of Na2S2O3 = 0.23 x10-3 mol
No. of moles of I2
= 0.23
x 10-3 / 2
= 1.15
X 10 – 4 mol
The concentration of I2 in solution A
= 1.15 X 10 – 4 / 0.012L
= 9.58 x 10 – 3 M
N = (X-C) x 50/1000 x 1/0.1
N = ( 0.019 – 9.58 x 10 – 3 M) x 50/1000 x
1/0.1
N = 4.71 X 10 -3 mol.
FLASK 4
From the results,
The volume of Na2S2O3 used =23.8 ml
No. of moles of Na2S2O3 = 2.38 x10-3
No. of moles of I2
= 2.38
x 10-3 / 2
= 1.19
X 10 – 3
The concentration of I2 in solution A
= 1.19 X 10 -3 / 0.05 L
= 0.024 M
FLASK 10
The volume of Na2S2O3 used = 30.0 ml
No. of moles of Na2S2O3 = 0.30 x10-3 mol
No. of moles of I2
= 0.30
x 10-3 / 2
= 1.5 X
10 – 4 mol
The concentration of I2 in solution A
= 1.5 X 10 – 4 / 0.012L
= 1.3 X 10 – 2 M
N =
(X-C) x 50/1000 x 1/0.1
N = ( 0.024 – 1.2 x 10 -2) x 50 / 1000 x 1
/ 0.1
N = 5.96 x 10 -3 mol
FLASK 5
From the results,
The volume of Na2S2O3 used = 28.4 ml
No. of moles of Na2S2O3 = 2.84 x10-3
No. of moles of I2
= 2.84
x 10-3 / 2
= 1.42
X 10 – 3
The concentration of I2 in solution A
= 1.42 X 10 – 3 / 0.05 L
= 0.028 M
FLASK 11
The volume of Na2S2O3 used = 4.0 ml
No. of moles of Na2S2O3 = 0.4 x10-3 mol
No. of moles of I2
= 0.4
x 10-3 / 2
= 0.2 X
10 – 3 mol
The concentration of I2 in solution A
= 0.2X 10 – 3 / 0.012L
= 1.67 X 10 – 2 M
N =
(X-C) x 50/1000 x 1/0.1
N = ( 0.028 – 1.67 X 10 -2 ) x 50/1000 x
1/0.1
N = 5.65 X 10 -3 mol
FLASK 6
From the results,
The volume of Na2S2O3 used = 46.1 ml
No. of moles of Na2S2O3 = 4.61 x10-3
No. of moles of I2
= 4.61
x 10-3 / 2
= 2.31
X 10 – 3
The concentration of I2 in solution A
= 2.31 X 10-3 / 0.05 L
= 0.046 M
FLASK 12
From the results,
The volume of Na2S2O3 used = 7.5 ml
No. of moles of Na2S2O3 = 0.75 x10-3 mol
No. of moles of I2
= 0.75
x 10-3 / 2
= 3.75
X 10 – 4 mol
The concentration of I2 in solution A
= 3.75 X 10 – 4 / 0.012L
= 3.13 X 10 – 2 M.
N =
(X-C) x 50/1000 x 1/0.1
N = (0.046 X 10 -3 – 3.13 X 10-2) x 50/1000
x 1/0.1
N = 7.35 X 10-3 mol
2 ) Plot amount of iodine adsorbed (N)
versus balance concentration of solution ( C)
FLASK
|
C (M)
|
N (mol)
|
1 & 7
|
3.75 X 10 – 3 M
|
3.63 X 10-3
|
2 & 8
|
5.83 X 10 – 3 M
|
3.09 X 10 -3
|
3 & 9
|
9.58 X 10 – 3 M
|
4.71 X 10 -3
|
4 & 10
|
13.0 x 10 -3 M
|
5.65 X 10 -3
|
5 & 11
|
16.7 X 10 – 3 M
|
5.96 x 10 -3
|
6 & 12
|
31.3 X 10 – 3 M
|
7.35 X 10-3
|
Gradient , m = ( 7.35 – 4.00 ) M
( 31.3 – 3.5 ) mol
= 0.12 M/mol
Therefore, the adsorption isotherm is 0.12 M/mol
3) Plot C/N versus C
FLASK
|
C ( M )
|
C/N ( M/mol)
|
1 & 7
|
3.75 X 10 – 3 M
|
1.03
|
2 & 8
|
5.83 X 10 – 3 M
|
1.89
|
3 & 9
|
9.58 X 10 – 3 M
|
2.03
|
4 & 10
|
13.0 x 10 -3 M
|
2.10
|
6 & 11
|
16.7 X 10 – 3 M
|
2.96
|
7 & 12
|
31.3 X 10 – 3 M.
|
4.26
|
Gradient,
m = ( 4.26 – 0.65 ) M/mol
( 31.3 – 0 ) x
10-3 mol
= 115.33 mol -1
m = 1 / Nm
Nm = 1 / 115.33
= 8.67 x 10 -3 mol
The number of iodine molecule adsorbed on the 0.1g of charcoal :
= 8.67 x 10^-3 mol x 0.1 x 6.023 x 10^23
= 5.22 x 10^ 20 molecules
The surface area of the charcoal :
=5.22 x 10^ 20 X 3.2x10-19 ÷ 0.1
= 1670.4
m^2 / g
DISCUSSION
The aim of this experiment is to obtain the
surface area of activated charcoal sample by using Langmuir equation.
For flask 1 until 6 is the titration of
iodine with sodium thiosulphate is in order to standardize the concentration of
Iodine in solution. Then, for the flask 7 until 12, the charcoal is been added
in order to determine the number of molecules of Iodine is adsorbed by the
charcoal. This can be done by calculating the concentration of Iodine after
titration with charcoal. Generally, the concentration of iodine has decrease.
Graph 1 which is the amount of iodine
adsorb (N) versus balance concentration of solution (C)., shows that number of
iodine adsorbed by the charcoal gradually increase in the solution. Graph 2, C/N
versus C which follow Langmuir equation. Langmuir equation explained the
relationship between the number of active sites of the surface undergoing
adsorption and pressure. This equation is called Langmuir isotherm equation.
Cf/N = (1/Nm) Cf + (1/kNm)
Where (1/Nm) is the slope, and (1/kNm) is
the intercept, when Cf/N is plotted versus the concentration C. Nm represents
the moles adsorbed at monolayer coverage which can be used to determine the
specific surface area of a solid.
We can determine the equilibrium has been
reached after two hours by observing the changes in the flasks. At first, the
flask which containing iodine solution is in dark brown colour. However, after
the reaction, when equilibrium has been reached, the solution will turn to
light brown. This is due to some iodine adsorbed by the charcoal.
There are few mistake that occur, therefore
here are the precaution steps that must be taken during handle the experiment
in order to obtain more accurate results. Firstly, during recording the volume
of sodium thiosulphate, make sure eyes is always directly proportional to the
scale meter. Next, during transferring charcoal to the iodine solution, make
sure that all of the amount of charcoal is insert to the solution. However, do
not put extra charcoal because this will be result in extra adsorbtion iodine
to the charcoal.
Conclusion
From this experiment, the adsorption of iodine solution in charcoal follows the Langmuir theory of adsorption isotherm. The result shows that the adsorption decrease as the concentration of the iodine solution decrease. From the experimental result, the surface area of charcoal is 1670.4 m^2 / g
From this experiment, the adsorption of iodine solution in charcoal follows the Langmuir theory of adsorption isotherm. The result shows that the adsorption decrease as the concentration of the iodine solution decrease. From the experimental result, the surface area of charcoal is 1670.4 m^2 / g
References
1. Alexander T Florence and David Attwood. (2006). Physiocochemical
Principles of Pharmacy (page 194-197). 4th Ed. Palgrave. USA. 3.
http://www.slideshare.net/ionizer_88/adsorption-from-solutions-acetic-acid-on-charcoal-14522884
2. Patrick J. Sinko, Lippincott Williams and Wilkins. Martin’s
Physical Pharmacy and Pharmaceutical Sciences (page 39,40), 5th Edition.
No comments:
Post a Comment