EXPERIMENT 5 TITLE Adsorption OBJECTIVES 1. To determine the absorption isotherm of acetic acid on activated carbon. 2. To determine the surface area of activated carbon. THEORY Adsorption is a process that occurs when a gas or liquid solute accumulates on the surface of a solid or a liquid (adsorbent), forming a film of molecules or atoms (the adsorbate). The matter that is attracting to the surface is call adsorbate and the matter that attract or where others matters adhere on is called adsorbent. It is different from absorption, in which a substance diffuses into a liquid or solid to form a solution. The term sorption encompasses both processes, while desorption is the reverse process. Similar to surface tension, adsorption is a consequence of surface energy. In a bulk material, all the bonding requirements (be they ionic, covalent, or metallic) of the constituent atoms of the material are filled by other atoms in the material. However, atoms on the surface of the adsorbent are not wholly surrounded by other adsorbent atoms and therefore can attract adsorbates. The exact nature of the bonding depends on the details of the species involved, but the adsorption process is generally classified as physisorption (characteristic of weak van der Waals forces) or chemisorption (characteristic of covalent bonding). Freundlich stated that the equation for adsorption isotherm can be written as : N = KCa Where; N = total number of mol adsorbate adsorbed by 1 g of activated carbon C = concentration at equilibrium K = constant a = constant (range between 0 – 1) However, Langmuir introduced an equation for adsorption isotherm as follows :
Where, : surface fraction that is covered by the adsorbate molecules
Where, : total number of mole adsorbate adsorbed by 1 g of the adsorbent to form a monolayer All of the equations on the aboce can be rearranged in the form of :-
A plot of
versus C will give a straight line with a slope of
.
If the cross section area (σ) of the absorbed molecule is known ( which area occupied by each adsorbate molecule on the surface of adsorbent). The specific surface area, A (m2g-1) can be calculated using the following equation: σ x 10-20 m2g-1
A= Where,
A = specific surface area of 1 g of solid (m2) = Avogrado number =
from the graph
versus C
σ = cross sectional area of the adsorbate molecule
APPARATUS Conical flask and stopper, beaker, pipette, glass rod, weighing bottle, volumetric flask 250 cm3, burette 50 Ml CHEMICALS Activated carbon, Acetic acid (2M), pellet NaOH, standard solution of HCl (0.10 M), phenolphthalein
RESULTS 1. We prepared Standard solution of NaOH by using the formula; n = MV/ 1000 = 0.1 M x 500 mL 1000 = 0.05 mol Mass of NaOH = n x molar mass
, Mr of NaOH = 40g mol-1
=0.05 mol x 40 g mol-1 = 2.0 g 2. We used to prepare 250 mL of each of the following acetic solutions ; 0.01M, 0.05M, 0.08M, 0.10M, 0.12M, and 0.15M. The concentration of Acetic Acid given is 2M. Therefore we can prepared each of the solution by dilution process : M1V1 = M2V2 1. 0.01M of Acetic Acid
M1V1 = M2V2 (2M) (V1) = (0.01M) (250mL) V1 =1.25 mL 2. 0.05M of Acetic Acid
M1V1 = M2V2 (2M) (V1) = (0.05M) (250mL) V1 =6.25 mL 3. 0.08M of Acetic Acid M1V1 = M2V2 (2M) (V1) = (0.08M) (250mL) V1 =10 mL 4. 0.10M of Acetic Acid
M1V1 = M2V2 (2M) (V1) = (0.10M) (250mL) V1 =12.5 mL
5. 0.12M of Acetic Acid M1V1 = M2V2 (2M) (V1) = (0.12M) (250mL) V1 =15 mL 6. 0.15M of Acetic Acid
M1V1 = M2V2 (2M) (V1) = (0.15M) (250mL) V1 =18.75 mL First Titration ( Day 1) Acetic acid 0.01 0.05 0.08 0.1 0.12 0.15
Initial Volume of NaOH (mL) 0.0 0.0 0.0 0.0 0.0 0.0
Final Volume of NaOH (mL) 2.5 12 19.5 25.3 28.3 35.3
Volume NaOH used
Initial Volume of NaOH (mL) 0.0 0.0 0.0 0.0 0.0 0.0
Final Volume of NaOH (mL) 1.2 8.9 13.8 25.3 22.5 33.0
Volume NaOH used
2.5 12 19.5 25.3 28.3 35.3
Second Titration ( Day 6) Acetic acid 0.01 0.05 0.08 0.1 (control) 0.12 0.15 Mass of activated carbon used
1.2 8.9 13.8 25.3 22.5 33.0
Concentration of Acetic acid(M) 0.01 0.05 0.08 0.1 (control) 0.12 0.15
Mass of activated carbon(g) 1.000 1.002 1.001 1.003 1.000 1.001
CALCULATIONS 1. The final (equilibrium) concentration of each acetic solution after titrated with NaOH
1. 0.01M of Acetic Acid
M1V1 = M2V2 (M2) (25mL) = (0.10M) (1.2mL) M2 =4.8 x 10-3M 2. 0.05M of Acetic Acid
M1V1 = M2V2 (M2) (25mL) = (0.10M) (8.9mL) M2 =0.0356 M 3. 0.08M of Acetic Acid M1V1 = M2V2 (M2) (25mL)= (0.10M) (13.8mL) M2 =0.0552M 4. 0.10M of Acetic Acid
M1V1 = M2V2 (M2) (25mL)= (0.10M) (25.3mL) M2 =0.1012M 5. 0.12M of Acetic Acid
M1V1 = M2V2 (M2) (25mL) = (0.10M) (22.5mL) M2 =0.09M 6. 0.15M of Acetic Acid
M1V1 = M2V2 (M2) (25mL) = (0.10M) (33mL) M2 =0.132 M
2. The number of moles of acetic acid adsorbed by 1 g of activated solution of each acetic acid
1. 0.01M of Acetic Acid
a) Initial concentration M1V1 = M2V2 (M1) (25mL) = (0.10M) (2.5mL) M1 =0.01M Initial mol = n = MV/ 1000 = 0.01 M x 100 mL 1000
= 1.0 x 10-3 mol
b) Final concentration = 4.8 x 10-3M
Final mol = n = MV/ 1000 = 4.8 x 10-3M x 100 mL 1000
= 4.8 x 10-4 mol
c) Number of mole of CH3COOH adsorb
Initial mol – final mol =1.0 x 10-3 mol – 4.8 x 10-4 mol =5.2 x 10-4 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 5.2 x 10-4mol = 5.2 x 10-4 mol/g 1.000g 2. 0.05M of Acetic Acid
a) Initial concentration M1V1 = M2V2 (M1) (25mL) = (0.10M) (12mL) M1 =0.048M Initial mol = n = MV/ 1000 = 0.048M x 100 mL 1000
= 4.8 x 10-3mol
b) Final concentration = 0.0356 M
Final mol = n = MV/ 1000 = 0.0356 M x 100 mL 1000
= 3.56 x 10-3 mol
c) Number of mole of CH3COOH adsorb
Initial mol – final mol =4.8 x 10-3mol - 3.56 x 10-3 mol = 1.24 x 10-3mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 1.24 x 10-3mol = 1.238 x 10-3mol/g 1.002 g 3. 0.08M of Acetic Acid a) Initial concentration M1V1 = M2V2 (M1) (25mL) = (0.10M) (19.5mL) M1 =0.078M Initial mol = n = MV/ 1000
= 0.078M x 100 mL 1000
= 7.8 x 10 -3 mol
b) Final concentration = 0.0552M
Final mol = n = MV/ 1000 = 0.0552M x 100 mL = 5.52 x 10-3 mol 1000 c) Number of mole of CH3COOH adsorb
Initial mol – final mol = 7.8 x 10-3mol – 5.52 x 10-3mol = 2.28 x 10-3mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 2.28 x 10-3 mol 1.001 g = 2.278 x 10-3mol/g 4. 0.10M of Acetic Acid
a) Initial concentration M1V1 = M2V2 (M1) (25mL) = (0.10M) (25.3mL) M1 =0.1012M Initial mol = n = MV/ 1000 = 0.1012M x 100 mL 1000
= 0.01012 mol
b) Final concentration = 0.1012M
Final mol = n = MV/ 1000 = 0.1012 M x 100 mL 1000
= 0.01012 mol
c) Number of mole of CH3COOH adsorb
Initial mol- final mol = 0.01012 mol- 0.01012 mol = 0.0 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.0 mol 1.003 g =0
5. 0.12M of Acetic Acid a) Initial concentration M1V1 = M2V2 (M1) (25mL) = (0.10M) (28.3mL) M1 =0.1132M Initial mol = n = MV/ 1000 = 0.1132 M x 100 mL 1000
= 0.01132mol
b) Final concentration = 0.09M
Final mol = n = MV/ 1000
= 0.09M x 100 mL 1000 = 9.0 x 10 -3mol c) Number of mole of CH3COOH adsorb
Initial mol- Final mol = 0.01132- 9.0 x 10 -3mol = 2.32 x 10-3 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 2.32 x 10-3 mol 1.000g = 2.32 x 10-3mol/g 6. 0.15M of Acetic Acid
a) Initial concentration
M1V1 = M2V2 (M1) (25mL) = (0.10M) (35.3mL) M1 = 0.1412M Initial mol = n = MV/ 1000 = 0.1412M x 100 mL 1000 = 0.01412 mol b) Final concentration = 0.132 M
Final mol = n = MV/ 1000 = 0.132 M x 100 mL 1000 = 0.0132 mol c) Number of mole of CH3COOH adsorb
Initial mol – Final mol = 0.01412 mol – 0.0132 mol = 9.2 x 10-4mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon
N = number of mole of CH3COOH adsorb Mass of activated carbon = 9.2 x 10-4mol 1.001 g = 9.19 x 10 -4mol/g
3. Plot also the Langmuir isotherm, C/N versus C. Calculate Nm from the slope of the
graph. C 4.8 x 10-3 M
0.036 M
0.055M
0.090M
C/N = 4.8 x 10-3 M 5.2 x 10 -4 molg-1 = 9.23 g/ L = 0.036M 1.238 x 10-3molg-1 = 29.08 g/L = 0.055M 2.278 x 10-3 molg-1 =24.14 g/L = 0.090M 2.32 x 10-3molg-1
= 38.79 g/L = 0.132M 9.19 x 10-4molg-1 =143.63g/L
0.132 M
From equation
We know that
= gradient from the graph C/N versus C
From the graph =
(62-20) (0.16 – 0.034)
= 333.33 Nm = 1 / 333.33 = 3 x 10-3 4. By assuming that the cross sectional area of acetic acid is 21 A2, calculate the specific
surface area of activated carbon used in this experiment
A=
σ x 10-20 m2g-1
= 3 x 10-3 x 6.02x 1023 x 21 x 10-20 m2g-1 = 379.26 m2g-1
DISCUSSION
1. Absorption involves a substance being taken into a bulk of a phase, while absorption
involves a substance being taken onto a surface. The adsorption process is represented by A + surface site A ( adsorbed ) 2. It is assumed that the surface contains a set of site at each of which a molecule of A
can be adsorbed, and that only a single layer of molecules of A ( a monolayer ) can be adsorbed on the surface. The sites might include all of the atoms of the solid surface, or might be special locations such as a “step” between two layers of atoms. It is
assumed that the total number of surface sites is fixed for a fixed amount of catalyst. The fraction of the surface sites occupied by adsorbed A molecules is denoted by θ. 3. The activity level of adsorption is based on the concentration of substance in the
water, the temperature and the polarity of the substance. A polar substance (a substance which is good soluble in water) cannot or is badly removed by active carbon, a non-polar substance can be removed totally by active carbon. Every kind of carbon has its own adsorption isotherm and in the water treatment business this isotherm is definite by the function of Freundlich. 4. For this experiment, it is influenced by the concentration of the solution. The type of
adsorption involve is chemisorptions. Concentration of solution will affect the adsorption. As the concentration increase, the rate of adsorption will increase too. 5. The solutions that we left for six days to ensure the adsorption process occur
efficiently. But for certain time, the adsorption will stop and remain constant as the layers of charcoals are already full with acetic acid molecules. 6. Activated carbon is used as an adsorbent. Adsorbents are used usually in the form of spherical pellets, rods, moldings, or monoliths with hydrodynamic diameters between 0.5 and 10 mm. They must have high abrasion resistance, high thermal stability and small pore diameters, which results in higher exposed surface area and hence high surface capacity for adsorption. The adsorbents must also have a distinct pore structure which enables fast transport of the gaseous vapors. 7. While doing the experiment, some precautions step should be taken: • •
Shake all the solution properly Filter the solution before proceeding for titration and discard the initial small volume of the filtrate • Do not use wet filter paper in filtration as it may dilute the solution CONCLUSION From the experiment, we can conclude that 1. Adsorption is usually described through isotherms, that is, the amount of adsorbate on
the adsorbent as a function of its pressure (if gas) or concentration (if liquid) at constant temperature. Activated carbon is used for adsorption of organic substances and non-polar adsorbates. It is the most widely used adsorbent. Its usefulness derives
mainly from its large micropore and mesopore volumes and the resulting high surface area.
2. The surface area of activated carbon can be calculated by using the equation A=
σ x 10-20 m2g-1
= 3 x 10-3 x 6.02x 1023 x 21 x 10-20 m2g-1 = 379.26 m2g-1
REFERENCES 1. Robert G. Mortimer (1993) Physical Chemistry. California, The Benjamin/ Cummings Publishing Company 2. O.P Pandey, D.N. Bajpai,S.Giri(1972) Practical Chemistry( for B.Sc. I.II &III year
students, India, S.Chand 3. http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Adsorb/langmuir.htm 4. http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Adsorb/equation.htm 5. http://en.wikipedia.org/wiki/Adsorption 6. http://www.lenntech.com/adsorption.htm
