EXPERIMENT 3: ADSORPTION FROM SOLUTION

Introduction

The process where the free moving molecules of a gaseous or solutes of a solution come close and attach themselves onto the surface of the solid is known as absorption.

The attachment or adsorption bonds can be strong and weak, depending on the nature of forces between adsorbent (solid surface) and adsorbate (gas or dissolved solutes). There are two types of adsorption. There are chemical adsorption and Physical adsorption.

Chemical adsorption or chemisorption is when adsorption involves only chemical bonds between absorbent and adsorbate. It acquires activation energy, can be very strong and not readily reversible. It generally produces adsorption of a layer of absorbate (monolayer adsorption)

Physical adsorption is also known as van der Waals adsorption. It happened when the adsorbate is bound to the surface through the weak van der Waals forces. It is a non - specific reaction which can be occurred at any condition and reversible, either by increasing the temperature or reducing the pressure of the gas or concentration of the solute. Physical adsorption can produce adsorption of more than one layer of adsorbate (multilayer adsorption). It is possible that chemical adsorption can be followed by physical adsorption on subsequent layers.

Adsorption isotherm is study of the relationship between the degree of adsorption and the partial pressure or concentration.. In some cases, the degree of adsorption at a specified temperature depends on the partial pressure of the gas or on the concentration of the adsorbate for adsorption from solution. 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.

Basically, physical adsorption is far more common than chemisorption

In this experiment, we use the adsorption from solution to determine the surface area of activated charcoal.

There are 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 adsorbent	An increase in surface area will increase the extent of adsorption.

Determination of Surface Area of Activated Charcoal via Adsorption from Solution

       It is important to determine the surface area of powder drug because it is one of the factors that govern the rate of dissolution and bioavailability of drugs that are absorbed through the gastrointestinal tract. Adsorption measurements is used to determine the surface area of a solid. For the rough surfaces and pores solid , 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

iodine solutions ( specified in Table 1 ), 1 % w/v starch solution, 0.1 M sodium thiosulphate solution, distilled water and activated charcoal.

Apparatus

12 conical flasks, 6 centrifuge tubes, measuring cylinders, analytical balance, Beckman J6M/E centrifuge, burettes, retort stand and clamps, Pasteur pipettes

Aim

To study the adsorption of iodine from solution and to determine the surface area of activated charcoal sample using Langmuir equation.

Procedure

12 conical flasks (labelled 1-12) is filled with 50ml mixture of iodine solutions (A and B) as stated in table 1.

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

Table 1

Solution A: iodine (0.05 M)

Solution B: potassium iodide (0.1 M)

Set 1: Actual concentration of iodine in solution A (X)

For flask 1-6:

1)     1-2 drops of starch solution is added as an indicator

2)    The solution is titrated using 0.1 M sodium thiosulphate solution until the colour of the solution changes from blue to colourless.

3)    The volume of sodium thiosulphate used is recorded.

Set 2: concentration of iodine in solution A at equilibrium (C)

For flask 7-12:

1)     0.1g of activated charcoal is added into the flasks.

2)    The flasks was capped tightly. The flask is swirled every 10 minutes for 1 hour.

3)    After an hour, the solutions is transferred into labelled centrifuge tubes.

4)    The solutions is centrifuged at 3000 rpm for 5 minutes

5)    The supernatant is transferred into new conical flasks and labelled accordingly.

6)    Step 1, 2, and 3 as carried out for flasks 1-6 in set 1 is repeated.

RESULTS  AND CALCULATIONS

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 (mL)
1	10	40	7.5
2	15	35	14.2
3	20	30	19.2
4	25	25	24.8
5	30	20	29.2
6	50	0	44.5

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 (mL)
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

Based on titration equation :

I₂ + 2Na₂S₄O₃ = Na₂S₄O₆ + 2NaI

Na₂S₄O₃             =  ½  I₂

Given :

1 mole iodine = 2 x 126.9 g

1 mL 0.1M Na₂S₄O₃ = 0.01269g of I

Flask 1

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

7.5 mL of 0.1M Na₂S₄O₃ = 0.095175 g I₂

2 mol Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n =  0.095175g / 253.8g/mol

   =  3.75 x 10 ^{– 4} mol I₂

[ X ] = 3.75 x 10 ^{– 4}mol / ( 50/1000 )

        = 7.5 x 10^{- 3} M

Flask 2

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

14.2 m L of 0.1 M Na₂S₄O₃ = 0.1801 g I₂

2 mol Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n= 0.1801g / 253.8g/mol

 = 7.1 x 10 ^{– 4} mol of I₂

[ X ] = 7.1 x 10 ^{– 4} mol / ( 50/1000 )

        = 0.0142 M

Flask 3

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

19.2 mL of 0.1 M Na₂S₄O₃ =  0.2436 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.2436g / 253.8g/mol

   = 9.6 x 10^{- 4} mol of I₂

[ X ] = 9.6 x 10 ^{– 4} mol / (50/1000)

        = 0.0192 M

Flask 4

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

24.8 mL of 0.1 M Na₂S₄O₃ = 0.3147 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.3147g / 253.8 g/mol

   = 1.240 x 10 ^-3 mol of I₂

[ X ] = 1.240 x 10 ^{– 4} mol / (50/1000)

        = 0.02480 M

Flask 5

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

29.2 mL of 0.1 M Na₂S₄O₃ = 0.3705 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.3705g / 253.8g/mol

   = 1.46 x 10 ^{– 3} mol I₂

[ X ] = 1.46 x 10 ^{– 3} mol / (50/1000)

        = 0.0292 M

Flask 6

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

44.5 mL of 0.1 M Na₂S₄O₃ = 0.5647 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.5647g / 253.8g/mol

   =  2.2250 x 10 ^{– 3} mol I₂

[ X ] =  2.2250 x 10 ^{– 3} mol / (50/1000)

        =  0.0445 M

 ____________________________________________________________________

Flask 7

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

2 mL of 0.1 M Na₂S₄O₃ = 0.02538 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.02538 g / 253.8g/mol

   = 1 x 10 ^{– 4} mol I₂

[ C ] = 1 x 10 ^{– 4} mol / (50/1000)

        = 2 x 10 ^{– 3} M

Flask 8

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

3.5 mL of 0.1 M Na₂S₄O₃ = 0.0444 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n= 0.0444g / 253.8g/mol

 = 1.7494 x 10 ^{– 4} mol I₂

[ C ] = 1.7494 x 10 ^{– 4} mol / (50/1000)

        = 3.4988 x 10 ^-3 M

Flask 9

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

4.5 mL of 0.1 M Na₂S₄O₃ = 0.0571 g I₂

2 mole Na₂S₄O₃  = 1 mole I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n= 0.0571g / 253.8g/mol

 = 2.2498 x 10 ^{– 4} mol I₂

[ C ] = 2.2498 x 10 ^{– 4} mol / (50/1000)

        = 4.4996 x 10 ^{– 3} M

Flask 10

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

4.5 mL of 0.1 M Na₂S₄O₃ = 0.0571 g I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.0571g / 253.8g/mol

   = 2.2498 x 10 ^{– 4} mol I₂

[ C ] = 2.2498 x 10 ^{– 4} mol / (50/1000)

        = 4.4996 x 10 ^{– 3} M

Flask 11

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

4.5 mL of 0.1 M Na₂S₄O₃ = 0.0571 g I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.0571g / 253.8g/mol

   = 2.2498 x 10 ^{– 4} mol I₂

[ C ] = 2.2498 x 10 ^{– 4} mol / (50/1000)

        = 4.4996 x 10 ^{– 3} M

Flask 12

1 mL of 0.1 M Na₂S₄O₃ = 0.01269 g I₂

8.5 mL of 0.1 M Na₂S₄O₃ = 0.1079 g I₂

1 mole I₂ = 2 x 126.9 g/mol = 253.8 g/mol

n = 0.1079g/ 253.8g/mol

   = 4.2514 x 10 ^{– 4} mol I₂

[ C ] = 4.2514 x 10 ^{– 4} mol / (50/1000)

        = 8.5028 x 10 ^{– 3} M

Graph of  C versus N

QUESTIONS

N = ( X – C ) x 50/1000 x 1/y y=  amount of activated charcoal in gram   = 0.1g N= Total mole of iodine adsorbed by 1g activated charcoal

 

1. Calculate N for iodine in each flask.

[ Flask 1 and 7 ]

X= 7.5 x 10 ^{– 3}M

C= 2 x 10^-3 M

N= (7.5 x 10 ^{– 3}M - 2 x 10^-3 M) x 50/1000 x 1/0.1g

   = 2.75 x 10 ^-3 mol/g

[ Flask 2 and 8 ]

X= 0.0142M

C= 3.4988 x 10 ^{– 3} M

N = (0.0142M - 3.4988 x 10 ^{– 3} M) x 50/1000 x 1/0.1g

   = 5.3506 x 10 ^{– 3} mol/g

[ Flask 3 and 9 ]

X= 0.0192M

C= 4.4996 x 10 ^-3 M

N= (0.0192M - 4.4996 x 10 ^-3 M) x 50/1000 x 1/0.1g

   = 7.3502 x 10 ^{– 3} mol/g

[ Flask 4 and 10 ]

X= 0.02480M

C= 4.4996 x 10 ^-3M

N= (0.02480M - 4.4996 x 10 ^-3M) x 50/1000 x 1/0.1g

  = 0.01015 mol/g

[ Flask 5 and 11 ]

X= 0.0292 M

C= 4.4996 x 10 ^{– 3} M

N= (0.0292 M – 4.4996 x 10 ^{– 3}M) x 50/1000 x 1/0.1g

   = 0.0124 mol/g

[ Flask 6 and 12 ]

X= 0.0445M

C= 8.5028 x 10 ^{– 3} M

N= (0.0445M – 8.5028 x 10 ^{– 3}M) x 50/1000 x 1/0.1g

  = 0.0180mol/g

2. Plot amount of iodine adsorbed (N) versus balanced concentration of solution (C) at equilibrium to obtain adsorption isotherm.

Flask	Amount of iodine adsorbed (N)	Balanced concentration of solution (C) at equilibrium
1 and 7	2.7500 x 10 -3 mol/g	2.0000 x 10-3 M
2 and 8	5.3506 x 10 – 3 mol/g	3.4988 x 10 – 3 M
3 and 9	7.3502 x 10 – 3 mol/g	4.4996 x 10 -3 M
4 and 10	0.01015 mol/g	4.4996 x 10 -3M
5 and 11	0.0124 mol/g	4.4996 x 10 – 3 M
6 and 12	0.0180mol/g	8.5028 x 10 – 3 M

Graph of C/N versus C

3. According to Langmuir theory, if there is no more than a monolayer of iodine adsorbed on the charcoal,

C/N = C/N_m  + 1/KN_m

Where C= concentration of solution at equilibrium

N_m = 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/N_m and intercept of 1/KN_m is obtained.

Obtained the value of N_m , then calculate the number of iodine molecule adsorbed on monomolecular layer. Assume that the area covered by one adsorbed molecule is 3.2x10^-19m² , Avogadro no= 6.023x10²³ molecule, calculate the surface area of charcoal in m²g ^{– 1}.

Flask	C/N	C
1 and 7	0.7273	2.0000 x 10-3 M
2 and 8	0.6539	3.4988 x 10 – 3 M
3 and 9	0.6122	4.4996 x 10 -3 M
4 and 10	0.4433	4.4996 x 10 -3M
5 and 11	0.3629	4.4996 x 10 – 3 M
6 and 12	0.4724	8.5028 x 10 – 3 M

m= (8.5028x10 ^-3 M – 2.0x10 ^-3M) / (0.4724g/l – 0.7273g/l)

   = - 0.0255

1 / N_m = 0.0255

     N_m  = 39.22 mole/g

Avogadro`s number = 6.023x10 ²³

Number of molecules = 6.023x10²³ x 39.22 mole/g

                                   = 2.3620x10 ²⁵ molecules/g

Area covered by one adsorbed molecule = 3.2x10 ^-19 m²

Surface area of charcoal = 3.2x10 ^-19 m² / molecules x 2.3620x10 ²⁵ molecules/ g

                                       =  7.558x10 ⁶ m²g^-1

Discussion

Absorption is a process that occurs when a gas or liquid solute accumulates on the surface of a solid or a liquid (absorbent), forming a molecular or atomic film (adsorbate). The driving force for adsorption is the reduction in interfacial tension between the fluid and the solid adsorbent as a result of the adsorption of the adsorbate on the surface of the solid.

Langmuir’s isotherm describing the Adsorption of Adsorbate onto the surface of the Adsorbent requires three assumptions:

·         The surface of the absorbent is in contact with a solution containing an adsorbate which is strongly attracted to the surface.

·         The surface has a specific number of sites where the solute molecules can be adsorbed.

·         The adsorption involves the attachment of only one layer of molecules to the surface, i.e. monolayer adsorption.

Adsorption by a solid is not a very important process unless the solid has a very large surface area compared to its mass. Consequently, charcoal is especially effective because it has a highly porous structure. Charcoal is activated by being heated to quite high temperatures in a vacuum or in a stream of dry air.

In pharmacy, activated charcoal is considered to be the most effective single agent available as an emergency decontaminant in the gastrointestinal tract. It is used after a person swallows or absorbs almost any toxic or chemical.

The surface area of activated charcoal can be obtained by the adsorption characteristics of iodine. The iodine number is defined as the milligrams of iodine absorbed by 1.0g of carbon when the iodine concentration of the filtrate is 0.02N (0.02 mol L^-1).

In this experiment, iodine is the absorbate while activated charcoal is the absorbent. In set 1, titration is used to calculate concentration of iodine because iodide ions and ion molecules are in equilibrium. The potassium iodide (KI) acts as a reservoir for the iodine by forming KI₃ (K⁺ and I₃^-). This is necessary to do this because of the limited solubility of iodine, as such, in water. Starch is added to act as an indication in the titration. The solution turns dark blue when starch is added as iodine molecules are present. When sodium thiosulfate is added, iodine molecules react with sodium thiosulfate to form sodium iodide.

I₂ + 2Na₂S₂O₃        Na₂S₄O₆ + 2NaI

When all the iodine molecules are totally reacted with sodium thiosulfate solution, the dark colour of solution change to colourless. From the equation, the number of moles of iodine can be calculated. In set 2, the activated charcoal acts as adsorbent to adsorb the iodine molecules.

The Langmuir isotherm shows that the amount of iodine adsorbed increases as the concentration increases up to a saturation point. Beyond this point, increasing the iodine concentration will not cause further increases. This behavior is typical of absorbents with a limited number of accessible sites. As long as there are available sites, adsorption will increase with increasing iodine concentration but as soon as all the sites are occupied, a further increase in quantity of iodine will result to a negative adsorption (desorption) at the monolayer.

A graph of amount of iodine adsorbed (N) versus balance concentration of solution(C) at equilibrium to obtain adsorption isotherm is plotted. The graph is hyperbolic-shaped. When the concentration of iodine increases from flasks 7 to 12, the total moles of iodine absorbed by 1g of activated charcoal (N) also increase. The actual concentration of iodine in solution A (X) is always lower than the concentration of iodine in solution A in equilibrium (C) because the iodine is being adsorbed by the activated charcoal. Hence, the amount of iodine left is decreased. We can determine that the equilibrium has been reached after shaking for 2 hours by observing there is no further change in the concentration of supernatant when titrated with sodium thiosulfate solution. We can also examine there is a gas produced from the solution in the conical flasks which is carbon dioxide gas due to the reaction.

A graph of C/N versus C is plotted and a straight line with the slope of 1/N_m and intercept of 1/KN_m should be obtained according to the equation C/N = C/N_m + 1/KN_m. However, we have obtained a negative slope. Hence, Langmuir theory is not obeyed that there is more than a monolayer of iodine adsorbed on the charcoal. In fact, the value of Nm (number of mole per gram charcoal required) can be calculated from the gradient of graph. Then, the number of iodine molecule adsorbed on the monomolecular layer is calculated. N_m is a measure of the adsorptive capacity of the activated charcoal for the iodine molecules. Finally, the surface area of charcoal can be calculated. The mass of charcoal is calculated by dividing the number of iodine molecule by Avogadro constant. Then, assumed value of area covered by one absorbed molecule (3.2 x 10^-19 m²) is divided by the mass of charcoal.

There are some errors made in this experiment. The volume of sodium thiosulfate solution titrated is recorded inaccurately as the position of eyes is not parallel to the lower meniscus of sodium thiosulfate solution. The concentration of solution may be affected when swirling as some solutions are accidentally poured out due to the conical flasks are not capped tightly. Hence, the conical flasks must be capped tightly before swirling. The conical flasks must be swirled constantly and vigorously to allow the charcoal molecules totally exposed to the solution in order for the adsorption to take place. The purity of the components of the solution and the avoidance of contamination during preparation and handling are essential. When working with organic media, contamination with water can have a drastic effect on the measurement. After equilibration, the adsorbent together with the adsorbate bound to it must be separated from the bulk equilibrium liquid by centrifugation. The separation must be carried out at the same temperature as that at which equilibrium was established.

Conclusion

As a conclusion, adsorption is a process where free moving molecules of a gaseous or solute of a solution come close and attach themselves onto the surface of the solid. The purpose of carrying out this experiment is to identify the surface area of the charcoal by studying the adsorption of iodine from solution process. The adsorption process is important in the field of pharmacy as it is a method to determine the surface area of powder drug. In this experiment, the adsorption process follows the Langmuir equation that is used to calculate the surface area of the charcoal and the adsorption process also undergo monolayer adsorption.

The Nm value is 1.4x10³ g mol-¹. The number of molecules of iodine adsorbed onto the monomolecular layer is 8.4325 x 10²⁰ molecules of iodine. The surface area of charcoal is 269.84 m²g-¹.

Reference

1.      http://www.fpharm.uniba.sk/fileadmin/user_upload/english/Physical_Chemistry/5-Adsorption.pdf

2.      http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0100-40422011000300020

3.      http://pac.iupac.org/publications/pac/pdf/1986/pdf/5807x0967.pdf