Introduction:
Adsorption is the process when free moving molecules of a gaseous or solutes of a solution come close and attach themselves onto the surface of the solid. Physical adsorption is the condition where adsorbate is bound to the surface of adsorbent through weak Van der Waals forces. It is reversible. Whereas chemical adsorption or chemisorption is the condition when involved stronger valence forces and not readily reversible. Chemical adsorption generally produces monolayer adsorption while physical adsorption can produce multilayer adsorption.
For a particular adsorbent or 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 know as adsorption isotherm.
The factors will influence the extent of adsorption from solution are solute concentration, tempeature, pH and surface area of adsorbent.
Surface area of activated charcoal can be determined via adsorption from solution. In the method of B.E.T. ( Burnauer, Emmett and Teller), adsorption of gas was used to measure the surface area.
Materials:
iodine solutions ( specified in Table 1 ), 1 % w/v starch solution, 0.1 M sodium thiosulphate solution, distilled water and activated charcoal.Apparatus:
Procedures:
1. The 12 conical flasks are labeled 1 – 12.
2. The conical flasks is filled with 50 ml mixtures of iodine solutions ( A and B ) by using measuring cylinders as stated in the Table 1.
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
|
Set 1 : Actual concentration of iodine in solution A (X)For flasks 1 – 6 :
1. 1 – 2 drops of the starch solution was added as an indicator.
2. 0.1 M sodium thiosulphate solution was used to titrate until the colour of the solution changed 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 of activated charcoal was added
2. The flasks were capped tightly. The flask was shaken and swirled every 10 minutes for 1 ½ hours.
1. 0.1 g of activated charcoal was added
3. After 1 ½ hours, the solution was transferred into the centrifuge tubes and labeled accordingly.
4. The solutions were centrifuged at 3000rpm for 5 minutes and the resulting supernatant was transferred into the new conical flasks. Each conical flask was labeled accordingly.
5. Steps 1, 2, and 3 were repeated as carried out for flasks 1 – 6 in Set 1.
GENERAL NOTES:
|
Titration equation:
I2 + 2Na2S2O3 ↔ Na2S4O6 + 2NaI
Na2S2O3 = ½ I2
Given:
(1 mole Na2S2O3 = ½ mole I2)
1 mole iodine = 2 x 126.9g
1 ml 0.1M Na2S2O3 = 0.01269g I
If the amount of the 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
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
|
10.90
|
2
|
15.0
|
35.0
|
13.50
|
3
|
20.0
|
30.0
|
19.30
|
4
|
25.0
|
25.0
|
23.00
|
5
|
30.0
|
20.0
|
28.50
|
6
|
50.0
|
0.0
|
46.40
|
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
|
1.70
|
8
|
15.0
|
35.0
|
3.10
|
9
|
20.0
|
30.0
|
5.30
|
10
|
25.0
|
25.0
|
7.50
|
11
|
30.0
|
20.0
|
8.90
|
12
|
50.0
|
0.0
|
10.00
|
Titration equation:
I2 + 2Na2S2O3 = Na2S4O6
+ 2NaI
2 mol Na2S2O3 ≡ 1 mol I2
1 mole iodine = 2 x 126.9 g
1 ml 0.1M Na2S2O3 = 0.01269 g I2
Results:
For flasks 1-6: X = Calculate the actual concentration of iodine in
solution A
Flask 1:
1 ml 0.1M Na2S2O3 = 0.01269 g I
10.90ml 0.1M Na2S2O3 = 0.13832 g I
1 mole iodine = 2 x 126.9 g
5.45x 10-4 mole I2 =0.13832 g I2
X(M) = No. of mole (mol)/Volume (L)
= 5.45 x 10-4 mole /
50/1000L
= 1.09x 10-2
M
Flask 2:
1 ml 0.1M Na2S2O3 = 0.01269 g I
13.50 ml 0.1M Na2S2O3 = 0.17132 g I
1 mole iodine = 2 x 126.9 g
6.75 x 10-4 mole I2 = 0.17132 g I
X(M) = No. of mole (mol)/Volume (L)
= 6.75 x 10-4mole
/ 50/1000L
= 0.0135 M
Flask 3:
1 ml 0.1M Na2S2O3 = 0.01269 g I
19.30 ml 0.1M Na2S2O3 = 0.24492g I
1 mole iodine = 2 x 126.9 g
9.65 x 10-4 mole I2= 0.24492g I2
X(M) = No. of mole (mol)/Volume (L)
= 9.65 x 10-4
mole / 50/1000L
= 0.0193M
Flask 4:
1 ml 0.1M Na2S2O3 = 0.01269 g I
23.0ml 0.1M Na2S2O3 = 0.29187g I
1 mole iodine = 2 x 126.9 g
1.15 x 10-3 mole I2
= 0.29187 g I2
X(M) = No. of mole (mol)/Volume (L)
= 1.15 x 10-3 mole
/ 50/1000L
= 0.023 M
Flask 5:
Flask 5:
1 ml 0.1M Na2S2O3 = 0.01269 g I
28.50ml 0.1M Na2S2O3 = 0.36167 g I2
1 mole iodine = 2 x 126.9 g
1.425 x 10-3 mole I2
=0.36167 g I2
X(M) = No. of mole (mol)/Volume (L)
= 1.425 x 10-3 mole
/ 50/1000L
= 0.0285 M
Flask 6:
1 ml 0.1M Na2S2O3 = 0.01269 g I2
46.40 ml 0.1M Na2S2O3 = 0.58882g I2
1 mole iodine = 2 x 126.9 g
2.32 x 10-3 mole I2. = 0.58882 g I2
X(M) = No. of mole (mol)/Volume (L)
= 2.32x 10-3
mole / 50/1000
=0.0464 M
For flasks 7-12: C = Calculate the concentration of iodine in solution
A at equilibrium
1 ml 0.1M Na2S2O3 = 0.01269 g I
1.70 ml 0.1M Na2S2O3 = 0.021573 g I
1 mole iodine = 2 x 126.9 g
8.5x 10-5 mole = 0.021573 g I2
C(M) = No. of mole (mol)/Volume (L)
= 8.5 x 10-5
mole / 10/1000L
= 8.5 x 10-3
M
Flask 8:
1 ml 0.1M Na2S2O3
= 0.01269 g I
3.10 ml 0.1M Na2S2O3
= 0.03934 g I
1 mole iodine = 2 x 126.9 g
1.55 X10-4mole= 0.03934 g I2
= 1.55 x 10-4 mole / 10/1000L
= 1.55 x 10-2
M
Flask 9:
1 ml 0.1M Na2S2O3 = 0.01269 g I
5.30 ml 0.1M Na2S2O3 = 0.06726g I
1 mole iodine = 2 x 126.9 g
2.65 x 10-4 mole= 0.06726 g I2
C(M) = No. of mole (mol)/Volume (L)
= 2.65 x 10-4
mole / 10/1000L
= 0.0265 M
Flask 10:
1 ml 0.1M Na2S2O3 = 0.01269 g I
7.50 ml 0.1M Na2S2O3 = 0.095175g I
1 mole iodine = 2 x 126.9 g
3.75 x 10-4 mole =0.095175 g I2
C(M) = No. of mole (mol)/Volume (L)
= 3.75 x 10-4 mole / 10/1000L
= 0.0375M
Flask 11:
1 ml 0.1M Na2S2O3 = 0.01269 g I
8.90 ml 0.1M Na2S2O3 = 0.112941 g I
1 mole iodine = 2 x 126.9 g
0.112941 g I2 = 4.45 x 10-4 mole I2
C(M) = No. of mole (mol)/Volume (L)
= 4.45 x 10-4 mole / 10/1000L
= 0.0445 M
Flask 12:
1 ml 0.1M Na2S2O3 = 0.01269 g I
10.0 ml 0.1M Na2S2O3 = 0.1269 g I
1 mole iodine = 2 x 126.9 g
5 x10-4 mole I2= 0.1269g I2
C(M) = No. of mole (mol)/Volume (L)
= 5.00x 10-4 mole / 10/1000L
= 0.05M
Questions:
2） Plot
amount of iodine adsorbed (N) versus balance concentration of solution (C) at
equilibrium to obtain adsorption isotherm.
4 a)
Discuss the result of the experiment
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 = 1.09 x 10-2
M
C = 8.5 x 10-3 M
N = (X – C) x
50/1000 x 1/y
=
(1.09 x 10-2 – 8.5x 10-3)M x
50/1000 x 1/0.1g
= 1.2 x 10-3
mol/g
Flask 2 and 8:
Flask 2 and 8:
X = 0.0135 M
C = 0.0155 M
N = (X – C) x
50/1000 x 1/y
=
(0.0135- 0.0155) M x 50/1000 x 1/0.1g
= -1
x 10-3 mol/g
Flask 3 and 9:
X = 0.0193 M
C = 0.0265 M
N = (X – C) x
50/1000 x 1/y
= (0.0193 – 0.0265) M x 50/1000 x
1/0.1g
=
-3.6x 10-3 mol/g
Flask 4 and 10:
X = 0.023 M
C = 0.0375 M
N = (X – C) x
50/1000 x 1/y
=
(0.023 –
0.0375)
M x 50/1000 x 1/0.1g
=
-7.25 x 10-3 mol/g
Flask 5 and 11:
X = 0.0285 M
C = 0.0445 M
N = (X – C) x
50/1000 x 1/y
=
(0.0285 –
0.0445)
M x 50/1000 x 1/0.1g
= -8
x 10-3 mol/g
Flask 6 and 12:
X = 0.0464
C = 0.05M
N = (X – C) x
50/1000 x 1/y
=
(0.0464– 0.0500)
M x 50/1000 x 1/0.1g
=
-1.8x 10-3 mol/g
C (M)
|
N x
10-3 (mol/g)
|
0.0085
|
1.20
|
0.0155
|
-1.00
|
0.0265
|
-3.60
|
0.0375
|
-7.25
|
0.0445
|
-8.00
|
0.05
|
-1.80
|
Graph of
Amount of Iodine Adsorbed (N) against
Balance Concentration of Solution (C) at Equilibrium
3.
According to Langmuir theory , if there is no more than a monolayer of iodine
adsorbed on the charcoal,
C/N = C/Nm
+ I/KNm
where C :
Balance concentration of iodine in solution A at equilibrium (moles)
N: Total
mole of iodine adsorbed by 1 gram of activated charcoal (moles)
Nm: number
of mole per gram charcoal required
I : Number
of mole of iodine absorbed on the monomolecular lay er (moles)
K :
constant to complete a monolayer
a)
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)
|
0.0085
|
7.08
|
0.0155
|
-15.5
|
0.0265
|
-7.36
|
0.0375
|
-5.17
|
0.0445
|
-5.56
|
0.05
|
-0.028
|
b) Calculate the number of iodine molecule absorbed on the
monomolecular layer.
From equation C/N = C/Nm + I/KNm, the number of iodine
molecule absorbed on the monomolecular layer can be derive;
I = KNm (C/N-C/Nm)
I=(-20x0.3091) (C/N-C/Nm)
I = -0.162 x (C/N-C/Nm)
I x Av ogadro number
Avogadro number (NA) = 6.023 x 1 023 molecule.
c) Calculate the surface area of charcoal in m 2g-1.
Assume that the area covered by one adsorbed molecule is 3.2
x 1 0-1 9m2
Molar mass of iodine is 253.8 g/mol
Slope (1 /Nm) = (-15.5-(-5.17)) / (0.0155-0.0375)
Nm = 2.13×10-3
y -ax is intercept (1 /KNm) = -20
K = 1 / -20Nm
= 1 / (-20x 2.13x10-3 )
= -23.48
No of moles of charcoal = 2.13х10-3mol g-1
х 0.1g
= 2.13x10-4mol
No. of molecules =2.13x10-4mol x (6.023x1023)
=1.28x1020
molecules
Area covered =1.28 x1020 molecules x 3.2x10-19
= 41.05m2
The surface area of charcoal= 41.05m2÷ 0.1g
= 410.5m2g-1
The colour changes in the flask was observed.
The initial colour of iodine is dark brown colour, the colour become light
(light brown) when the reaction proceed and remain the unchanged until a long
period of time because some of the iodine is being adsorbed onto the charcoal.
This is when the reaction is at equilibrium.
The results
and readings obtained from the result are not accurate compare to the
theoretical one as we studied. This could be due to several reasons. Firstly,
the time for swirling the flask with charcoal is reduced from 2 hours to 1
hour. This will cause some of the iodine did not completely adsorbed onto the
charcoal. Thus, balance concentration C were affected.Then,the different person
shaking the flask might affected the results because different people uses
different strength and ways to shake the flask. This caused the reaction did
not achieve equilibrium.
Besides, some
of the charcoal might split when it was added into the small mouth of the
conical flask.This will greatly affecting the amount of charcoal to adsorb the
iodine. More iodine will be presented in the analyte during titration.
In this
experiment, flasks 1-6 use two solutions. Solution A uses iodine while solution
uses potassium iodide. Iodine
produces a deep blue colour when it forms a complex with starch solution. Thus,
a starch solution is used as an indicator for free elemental iodine in solution
during titration. Iodine has low water solubility. Solutions for analytical
purposes are made up in moderately concentrated potassium iodide solution.
The iodine is
taken into solution as the I3- ion according to the equation:
I2 +
I- I3-
The I3- ion is much more
soluble in water than iodine is.
Iodine
solutions lack stability. Iodine is volatile and loss of iodine occurs when a
vessel is open even for relatively short periods. Iodine slowly attacks organic
materials - thus cork or rubber stoppers should not be used. Iodide is also susceptible to air oxidation
in an acidic medium:
4I- +
4H+ + O2 → 2I2 + 2H2O
For flasks 7-12, activated charcoal is added
to assay tubes to terminate the incubation. It is an approach to estimate the
amount of radio ligand-receptor complex that forms in a detergent-solubilized
binding assay. The equilibrium state can be determined shaking iodine solution
with 0.1 g of active charcoal for every 10 minutes in one and half hours. Often it is found that radio ligands,
particularly those of small molecular weight, are rapidly adsorbed to the
charcoal (i.e. within 0-5 minutes). After centrifugation, to sediment the
charcoal and adsorbed ligand, the receptor-bound ligand can be estimated by
determining the radioactivity in the supernatant. Centrifugation is done to
these flasks before being titrated with sodium thiosulphate. Fine particles
suspended in a liquid can be separated by centrifugation.
From this
experiment, we know that the higher the actual concentration of iodine in
solution A, the higher the mole of iodine adsorbed by 1 g of activated
charcoal. The higher the solute concentration, the higher the amount of the adsorption
occurring at equilibrium until a limiting value is reached. In fact, the
adsorption of a solute is inversely proportional to its solubility in the
solvent .In this experiment, when the concentration of solution increases, the
amount of iodine adsorbed is also increased. This can be proven from the graph
above which show the amount of iodine adsorbed is proportional to the balance
concentration of solution. Hence, we can conclude that the higher the
solubility, the higher the degree of adsorption.
(b) How do you determine experimentally that
equilibrium has been reach after shaking for 2hours?
We can determine the equilibrium when
the solution in the test tube starting producing gas which is carbon dioxide gas due to their
reaction.
Conclusion:
As a conclusion, the adsorption process is
important in the field of pharmacy as it is a method to determined the surface
area of powder drug. In this experiment, the adsorption
process follows the Langmuir equation and undergo monolayer adsorption.
Precaution:
There are a few precautions should be taken
during the experiment.
- Careful handling and good laboratory technique should always be used when working with chemicals.
- The conical flask has to be labeled properly to prevent the confusion occur.
- They should not be any leakage from the burette during titration.
- Keep your eye in level with the liquid surface while taking the burette reading or while reading the pipette or measuring flask.
- Keep
away from fire when handling and storing the products.
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