Group members

CHAN SII NING   A140292

LILIAN LAU MEI MEI   A139695

NABILAH BINTI ISMAIL    A140108

LIM MING CHIANG   A140272

MOHD HANIF BIN YUSUF    A139505

Lab Time

PRACTICAL 4: DETERMINATION OF DIFFUSION COEFFICIENT

OBJECTIVE :

To determine the value of diffusion coefficient, D.

THEORY :

Diffusion, which is the spontaneous movement of solutes from an area of high concentration to an area of low concentration can be explained by Fick’s law which states that the flux of material (amount dm in time dt) across a given plane (area A) is proportional to the concentration gradient dc/dx.

                                  dc

            dm  =   - DA        dt      ----------- (i)

                                  dx

D is the diffusion coefficient or diffusivity for the solute, in unit m²sֿ¹.

If a solution containing neutral particles with the concentration  Mₒ, is placed within a cylindrical tube next to a water column, diffusion can be stated as

              M = M_o eksp ^{(-x2/ 4Dt)}      ---------- (ii)

where M is the concentration at distance x from the intersection between water and solution that is measured at time t.

By changing equation (ii) to its logarithm form, we get

               ln M  =  ln M₀ – x²/ 4Dt

or  2.303 x 4 D ( log₁₀ M₀ – log₁₀ M) t = x² ---------- (iii)

Thus aplot of x² against t can produce a straight line that passes through the origin with the slope 2.303 x 4D (log ₁₀ M₀–log ₁₀ M). From here, D can be counted.

If the particles in the solution are assumed to be spherical, their size and molecular weight can be calculated by the Stokes-Einstein equation.

                D = kT/ 6πηa

It is known that molecular weight M=mN (N is Avogadro’s number 6.02 x 10²³ mol^-1 ).

∴M = 4/3πa³N ρ ---------- (v)

Diffusion for charged particles, equation (iii) needs to be modified to include potential gradient effect that exists between the solution and solvent. However, this can be prevented by adding a little sodium chloride into the solvent to avoid the formation of this potential gradient.

Agar gels contain a partially strong network of molecules that is penetrated by water. The water molecules form a continuous phase around the gel. Thus, the molecules of solutes can diffuse freely in the water if chemical interactions and adsorption effects do not exist entirely. Therefore, the gel forms an appropriate support system to be used in diffusion studies for molecules in a medium of water.

MATERIALS:                                                               APPARATUS:

Agar powder                                                                    500 mL beaker

Ringer’s solution                                                               5 mL pipette

1 : 500000 crystal violet solution                                       Glass rod

1 : 200 crystal violet solution                                             14 test tubes with covers

1 : 400 crystal violet solution                                             Hot plate

1 : 600 crystal violet solution

1 : 500000 bromothymol blue solution

1 : 200 bromothymol blue solution

1 : 400 bromothymol blue solution

1 : 600 bromothymol blue solution

EXPERIMENTAL PROCEDURES :

1)      7g of agar powder was weighed and mixed with 420ml of Ringer solution in the 500mL beaker..

2)      The mixture in the beaker was stirred and boiled on a hot plate until a transparent yellowish solution was obtained.

3)      About 20ml of the agar solution was pour into each 6 test tubes. The test tubes were then put in the fridge to let them cool.

4)      An agar test tube which contained 5ml of 1:500,000 crystal violet was being prepared and it was used as a standard system to measure the distance of the colour as a result of the diffusion of crystal violet.

5)      After the agar solutions in the test tubes solidifying, 5ml of each 1:200, 1:400, 1:600 crystal violet solution were pour into each test tubes.

6)      The test tubes were closed immediately to prevent the vaporization of the solutions.

7)      Three test tubes were put in room temperature,28 ºC while another three were put in 37ºC water bath.

8)      The distance between the agar surface and the end of crystal violet where that area has the same color as in the indicator was measured accurately.

9)      Average of the readings were obtained, this value is x in meter.

10)  The x values were recorded after 2 hours and at appropriate intervals for 1 weeks.

11)  Procedures 3 to 10 were repeated for bromothymol blue solutions.

12)  Graph of x² values ( in m²) versus time ( in hours) was potted.

The diffusion coefficient , D was determined from the graph gradient for both   28 ºC and 37 ºC ; the molecular mass of crystal violet and bromothymol blue  were also determined by using N and V equation.

RESULTS

Crystal violets at room temperature (28°C)

Crystal violets in water bath (37°C)

Bromothymol blue at room temperature (28°C)

Bromothymol blue in water bath (37°C)

At concentration 1:200

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(14-6)x10^-4         =2.437x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6.2-2.4)(86400)

2.437x10^-9               = 2.303x4D[log₁₀(1/200)-log₁₀(1/500 000)]

D                     = 7.785 x10^-11 m²s^-1

At concentration 1:400

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(12-6)x10^-4         =1.929x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6.6-3)(86400)

1.929x10^-9               = 2.303x4D[log₁₀(1/400)-log₁₀(1/500 000)]

D                     = 6.762x10^-11 m²s^-1

At concentration 1:600

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(5.2-1.2)x10^-4         =1.157x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6-2)(86400)

1.157x10^-9               = 2.303x4D[log₁₀(1/600)-log₁₀(1/500 000)]

D                     = 4.3x10^-11 m²s^-1

Average diffusion coefficient: 6.282x10^-11 m²s^-1 

At concentration 1:200

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(25-10)x10^-4         =4.34x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(5.8-1.8)(86400)

4.34x10^-9                   = 2.303x4D[log₁₀(1/200)-log₁₀(1/500 000)]

D                     = 13.865x10^-11 m²s^-1

At concentration 1:400

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(12-5)x10^-4         =2.7x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(5-2)(86400)

2.7x10^-9                       = 2.303x4D[log₁₀(1/400)-log₁₀(1/500 000)]

D                     =9.464 x10^-11 m²s^-1

At concentration 1:600

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(10-1)x10^-4         =2.604x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(5-1)(86400)

2.604x10^-9               = 2.303x4D[log₁₀(1/600)-log₁₀(1/500 000)]

D                     = 9.678x10^-11 m²s^-1

Average diffusion coefficient: 11.002x10^-11 m²s^-1 

At concentration 1:200

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(12-4)x10^-4         =2.205x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(5.8-1.6)(86400)

2.205x10^-9               = 2.303x4D[log₁₀(1/200)-log₁₀(1/500 000)]

D                     = 7.044x10^-11 m²s^-1

At concentration 1:400

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(8-4)x10^-4         =1.781x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(4.4-1.8)(86400)

1.781x10^-9               = 2.303x4D[log₁₀(1/400)-log₁₀(1/500 000)]

D                     =6.243 x10^-11 m²s^-1

At concentration 1:600

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(7.6-2)x10^-4         =1.62x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6.4-2.4)(86400)

1.62x10^-9                   = 2.303x4D[log₁₀(1/600)-log₁₀(1/500 000)]

D                     = 6.021x10^-11 m²s^-1

Average diffusion coefficient: 6.436x10^-11 m²s^-1 

At concentration 1:200

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(16-6)x10^-4         =2.894x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(5.4-1.4)(86400)

2.894x10^-9               = 2.303x4D[log₁₀(1/200)-log₁₀(1/500 000)]

D                     =9.245 x10^-11 m²s^-1

At concentration 1:400

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(15-5)x10^-4         =2.63x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6-1.6)(86400)

2.63x10^-9                   = 2.303x4D[log₁₀(1/400)-log₁₀(1/500 000)]

D                     =9.219x10^-11 m²s^-1

At concentration 1:600

Gradient          = 2.303x4D(log₁₀Ma-log₁₀ M)

(11-3)x10^-4         =2.205x10^-9 = 2.303x4D(log₁₀Ma-log₁₀ M)

(6-1.8)(86400)

2.205x10^-9               = 2.303x4D[log₁₀(1/600)-log₁₀(1/500 000)]

D                     = 8.195x10^-11 m²s^-1

Average diffusion coefficient: 8.886x10^-11 m²s^-1 

Discussion

Agar diffusion refers to the movement of molecules through the matrix that is formed by the gelling of agar. When performed under controlled conditions, the degree of the molecule's movement can be related to the concentration of the molecule. Agar will be the inert medium that we are using to investigate diffusion through. Agar is extracted from seaweed and after dissolving it in hot water it cools to form a 'solid' jelly. The agar that will be used will be made alkaline by adding small amounts of sodium hydroxide

In the experiment, crystal violet diffuse faster than bromothymol blue solution.  Crystal violet with molecular formula C₂₅N₃H₃₀Cl has molecular weight of 407.979 g mol^-1 while bromothymol blue solution with molecular formula C₂₇H₂₈Br₂O₅S has molecular weight of 624.38 g mol⁻¹.  Molecular weight is how much mass each particle has or how heavy it is. The heavier the particle, the slower it is going to move ii solidified agar solution, assuming energy of the system remains constant.

For a given concentration gradient, a molecule's rate of diffusion is inversely proportional to its frictional coefficient, which depends on both size and shape. Assuming that the particle has a constant shape,  a sphere, then rate of diffusion is inversely proportional to the radius or diameter of the diffusing molecule. Assuming particle density doesn’t change, particle mass is proportional to the cube of particle radius. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight

Rate₁ / Rate₂ = square root of (Mass₂ / Mass ₁)

As the experiment is done in two different temperature of 28^oC water bath and 37^oC room temperature, the higher the temperature gives higher rate of diffusion. As the temperature increases, the amount of energy available for diffusion is increased. There would be increase in molecules' speed (kinetic energy). So the molecules move faster and there will be more spontaneous spreading of the material which means that diffusion occurs quicker. Thus the rate of diffusion will be faster as the temperature increases. From Stokes-Einstein equation:

D = kT/6пŋa

                       (D = kT/9 and 9 =6пŋa)

Where D is the diffusion constant, K is Boltzmann's constant, T is the absolute temperature.

k_B is Boltzmann's constant, η is viscosity, a is the radius of the spherical particle.

The concentration also varies for each crystal violet and bromothymol blue solution. For higher concentration of solution gives higher rate of diffusion. When a substance is diffusing between two compartments, the greater the concentration difference between the two compartments, the faster the substance will diffuse (faster rate of diffusion). diffusion will occur from areas of high concentration to low concentration. Fick's First Law states that the flux, J, of a component of concentration, C, across a membrane of unit area, in a predefined plane, is proportional to the concentration differential across that plane), and is expressed by:

Where J – Flux, D – diffusion coefficient, δC/δx – concentration gradient

(C is concentration and x is distance of movement perpendicular to the membrane surface )

Crystal violet diffuse faster than bromothymol blue solution. The diffusion coefficient value calculated from the experiment for crystal violet has higher value compared  to the value of diffusion coefficient for bromothymol blue solution. 

CONCLUSION

The diffusion coefficient of crystal violet at 28^oC is 6.282x10^-11 m²s^-1 while at 37^oC is 11.002x10^-11 m²s^-1. diffusion coefficient of bromothymol blue at 28^oC is 6.436x10^-11 m²s^-1  while at 37^oC is 8.886x10^-11 m²s^-1 the factor that influence the ratof diffusion is concentration, temperature and molecular weight since the surface area, Permeability is kept constant. diffusion rate is faster in the concentration of diffusing molecules 1:200> 1:400> 1:600.

REFERENCE

http://www.pojman.com/mg_materials/Diffusion/Diffusion.html

http://www.bookrags.com/research/agar-diffusion-wmi/

http://www.coursework.info/GCSE/Chemistry/Aqueous_Chemistry/Factors_affecting_the_rate_of_Diffusion_L38515.html

http://district.bluegrass.kctcs.edu/rmccane0001/shared_files/137labexercise3.pdf

http://answers.yahoo.com/question/index?qid=20090120064728AA2wNLH

http://wiki.answers.com/Q/How_does_temperature_affect_diffusion_rate

Practical 3: Adsorption from solution

Objective:

To study the adsorption of iodine from solution anduse  Langmuir equation to estimate the surface area of activated charcoal sample.

Theory:

            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. 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 chemisorptions acquires activation energy, can be very strong and not readily reversible.

            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.

            Chemical adsorption generally produces adsorption of a layer of adsorbate (monolayer adsorption). On the other, physical adsorption can produce adsorption of more than one layer adsorbate (multilayer adsorption). Nevertheless, it is possible that chemical adsorption can be followed by physical adsorption on subsequent layers. For a particular adsorbate/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 with temperature can provide useful information on the characteristics of solid and the reactions involved when adsorption occurs.

            In adsorption from solution, physical adsorption is far more common than chemisorption. However, chemisorptions 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.

Determination of Surface Area of Activated Charcoal via Adsorption from Solution

Determination of surface area of powder drug is important in the field of pharmacy and colloidal science as surface area is one of the factors that govern the rate dissolution and bioavailability of drugs that are absorbed through the gastrointestinal tract.

Adsorption measurement can be used to determine the surface area of solid. There are two methods to measure the surface area which are Langmuir and B.E.T (Brunauer, Emmett and Teller). 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, paster pipettes, iodine solutions, 1%w/v starch solution, 0.1M sodium thiosulphate solution, distilled water and activated charcoal.

Bottom of Form

Procedure:

Using burettes or measuring cylinders, 12 conical flasks （labeled 1-12）are filled with 50ml mixtures of iodine solutions (A and B) as stated in the table 1.

Table 1: Solution A: Iodine (0.05M)

              Solution B: Potassium Iodide (0.1M)

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

For flask 1 - 6:

1.      1- 2 drops of starch solution are added as an indicator.

2.      The solution is then titrated using 0.1 M sodium thiosulphate solution until the colour of the solution change from dark blue to colourless.

3.      The volume of the thiosulphate used was recorded.

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

For flask 7-12:

1.      0.1g of activated charcoal is added.

2.      The flask is capped tightly. The flask is swirled every 10 minutes for 2 hours.

3.      After 2 hours, the solutions are transferred into centrifuge tubes and they are labeled accordingly.

4.      The solutions are centrifuged at 3000rpm for 5 minutes and the resulting supernatant is transferred into the new conical flask. Each conical flask is labeled accordingly.

5.      Steps 1, 2 are 3 were repeated as carried out for flask 1-6 in set 1.

Results:

Questions:

1.      Calculate N for iodine in each flask.

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

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

Calculation:

From the graph C/N versus C, 1/KN_mobtained is 4.

Number of iodine molecule adsorbed on the monomolecular layer

= N_m × 0.1g charcoal × Avogadro no.

= 4.375×10^-3 × 0.1 × 6.023 × 10²³

= 2.635 × 10²⁰molecules

N_m = 4.375×10^-3 mole/gram ; 1 mole iodine = 2×126.9g

Weight of iodine = N_m × 0.1g × (2×126.9g)

                            = 4.375×10^-3 × 0.1 × (2×126.9)

                            = 0.111g

Thus, surface area of charcoal can be calculated. Since surface area covered by one adsorbed molecule is  3.2 × 10^-19 m²,

4.     Discuss the results of the experiment. How do you determine experimentally that equilibrium has been reached after shaking for 2 hours?

Equilibrium has been reached when the solution becomes homogenous and has no more colour changed.

Discussions:

            Adsorption is usually described using isotherms which are the amount of adsorbate on the adsorbent as a function of its pressure for gases or concentration for liquid at constant temperature. The quantity adsorbed is always normalized by the mass of the adsorbent in order to allow comparison of different substances.

            Langmuir isotherm adsorption is based on the theoretical equation on adsorption based on the short distance forces that present between molecules. It is a semi-empirical isotherm derived from a proposed kinetic mechanism. The equation got four assumptions as below :

(a) The surface of the adsorbent is uniform which all adsorption sites are equivalent. The gas molecules are adsorbed at fixed site on the surface.

(b)  The adsorbed molecules do not react with each other.

(c)  The thickness layer of gas molecules being adsorbed is one molecule thick, the adsorbate molecule do not deposit on other already adsorbed surface, only on the free surface of the adsorbent.

(d) All adsorption occurs through the same mechanism.

In this experiment, iodine is the adsorbate while charcoal is the adsorbent. In the set 1 experiment, titration method was used to calculate the concentration of iodine. This is because the iodide ion and iodine molecule are in equilibrium in the conical flask. Starch is used as an indicator in the titration. The solution turn dark blue colour when starch is added as iodine molecules are present. Then, when sodium thiosulfate is added, the iodine molecule react with sodium thiosulfate to form sodium iodide.

            I₂ + 2Na₂S₂O₃ → Na₂S₄O₆ + 2 NaI

When there is totally noiodine molecule in the solution, the dark colour change to colourless. From the equation, the moles of iodine can be calculated.

      In the set 2 experiment, 0.1 g of activated charcoal was added into flask 7-12 and capped tightly. The activated charcoal added is act as adsorbent to adsorp the iodine molecule. Adsorption of iodine molecule on the activated charcoal is a result from Van der Waal’s forces which exists between molecules. The forces are extremely short ranged and therefore sensitive to the distance between the carbon surface and the adsorbate molecule. They are also addictive which the adsorption force is the total of all interactions between all the atoms.

      A large specific surface area is preferable for providing large adsorption capacity, however, large internal surface area in a limited volume will give rise to large numbers of small pores in between the adsorption surface. The size of the pores will determine the accessibility of the adsorbate molecules into the internal adsorption surface area of adsorbent. Hence, the distribution of the micro pores size is an important property for characterized the adsorptivity of adsorbents.

C/N = C/N_m + 1/KN_m.

      In this experiment, there are some errors and precautions that need to be notified. There may be more than one layer of iodine molecules adsorbed on the surface of the activated charcoal. Besides, the concentration of solution may affected when swirling if the conical flask is not capped tightly. So the conical flask should be capped tightly. Moreover, the volume measured should also be accurate and avoid from parallax error as it may affect the results. The conical flasks also have to be swirled constantly to allow all charcoal molecules exposed to the mixture solution in order for adsorption to take place.

Conclusion :

The surface area of the activated charcoal is 759.64m²g^-1.

References :

1.     1.  A.T.Florence and D.Attwood.Physicochemical Principles of Pharmacy, 3^rd edition,                                             1998. Macmillan Press LTD.

2.      http://en.wikipedia.org/wiki/Adsorption

3.      infohost.nmt.edu/~jaltig/Langmuir.pdf‎