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Abstract Regarding to the objectives of the experiment, which are to conduct the simple experiments regarding liquid-liquid extraction and to determine the distribution of coefficient and mass transfer coefficient with the aqueous phase as the continuous medium through liquidliquid extraction. This experiment is based on the solubility. First experiment, we used separators funnel to separate two solutions of different solubility and densities, and then titrate with different of NaOH concentration (0.1M and 0.025M). The values for distribution coefficient by titration with 0.1M are 7.78 in 1.0 ml of propionic acid, 2.30 in 3.0 ml propionic acid and 1.80 in 5.0 ml propionic acid. Second experiment, we used liquid-liquid extraction column to obtain feed, raffinate and extract samples. The samples were titrated with different of NaOH concentration (0.1M and 0.025M). The value of mass transfer coefficient from liquid-liquid extraction are; -0.05 kg/min if titrated with 0.1M NaOH and -0.237 kg/min if titrated with 0.025M NaOH.

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Introduction A relatively cheap method to separate a two component mixture is through the process of liquid-liquid extraction. Liquid-liquid extraction, also known as solvent extraction and partitioning, is a method to separate compounds based on their relative solubility in two different immiscible liquids, usually water and an organic solvent ( propionic acid). It is an extraction of a substance from one liquid phase into another liquid phase. Liquid-liquid extraction is a basic technique in chemical laboratories, where it is performed using a separator funnel. This type of process is commonly performed after a chemical reaction as part of the work-up. In other words, this is the separation of a substance from a mixture by preferentially dissolving that substance in a suitable solvent. By this process a soluble compound is usually separated from an insoluble compound. The basic principle behind extraction involves the contacting of a solution with another solvent that is immiscible with the original. The solvent is also soluble with a specific solute contained in the solution. Two phases are formed after the addition of the solvent, due to the differences in densities. The solvent is chosen so that the solute in the solution has more affinity toward the added solvent. Therefore mass transfer of the solute from the solution to the solvent occurs. Further separation of the extracted solute and the solvent will be necessary. However, these separation costs may be desirable in contrast to distillation and other separation processes for situations where extraction is applicable. A general extraction column has two input stream and two output streams. The input streams consist of a solution feed at the top containing the solute to be extracted and a solvent feed at the bottom which extracts the solute from the solution. The solvent containing the extracted solute leaves the top of the column and is referred to as the extract stream. The solution exits the bottom of the column containing only small amounts of solute and is known as the raffinate. Further separation of the output streams may be required through other separation processes. The main advantage in using this technique is the lack of an energy expense, because these phases do not have to be boiled nor condensed. Also, the lack of heating allows us to perform this procedure on many components even if they are temperature sensitive. By examining current industry processes, it is quite evident that this method is the most attractive by its use in a wide variety of applications. For example, liquid – liquid extraction is used in the production of penicillin and for the separation of aromatic products from other mixed hydrocarbons that come out of the efflux of the reactor, etc. Another chic characteristic of this process is the generality of the equations used to design the unit. For example, in this experiment, we will study the extraction of acetic acid from toluene by water. The correlations and theories developed from this experiment will be applicable to the above mentioned extractions and many other extraction processes that may arise in the chemical engineer’s course of action. 2

Objectives 1. To conduct the simple experiments regarding liquid-liquid extraction. 2. To determine the distribution coefficient for the system organic solvent-propionic acid water and show its dependence on concentration. 3. To demonstrate how a mass balance is performed on the extraction column and to measure the mass transfer coefficient with the aqueous phase as the continuous medium.

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Theory In liquid-liquid extraction, two phases must be brought into contact to permit transfer of material and then be separated. Extraction equipment may be operated batch wise or continuous. The extract is the layer of solvent plus extracted solute and the raffinate is the layer from which solute has been removed. The extract may be lighter or heavier than the raffinate, and so the extract may be shown coming from top of the equipment in some cases and from the bottom in others. The operation may of course be repeated if more than one contact is required, but when the quantities involved are large and several contacts are needed, continuous flow becomes economical. In dilute solutions at equilibrium, the concentration of the solute in the two phases is called the distribution coefficient or distribution constant ‘K’. K=Y/X Where the Y and X are the concentrations of the solute in the extract and the raffinate phases respectively. The distribution coefficient can also be given as the weight fraction of the solute in the two phases in equilibrium contact: K’ =y*/x Where y* is the weight fraction of the solute in the extract and x is the weight fraction of the solute in the raffinate. The rate at which a soluble component is transferred from one solvent to another will be dependent, among other things, on the area of the interface between the two immiscible liquids. Therefore it is very advantageous for this interface to be formed by droplets and films, the situation being analogous to that existing in packed distillation columns. The theory for the system Trichloroethylene-Propanoic acid-Water is as follows: Let

Vw : Water flow rate, lt/s 4

Vo : Trichloroethylene flow rate, lt/s X : Propionic acid concentration in the organic phase, kg/lt Y : Propionic acid concentration in the aqueous phase, kg/lt Subscripts: 1: Top of column 2: Bottom of column Mass Balance: Propionic acid extracted from the organic phase (raffinate). = Vo (

X1

-

X2

) (8.1.1)

Propionic acid extracted by the aqueous phase (extract) = Vw (

Y1

-0) (8.1.2)

Therefore theoretically, Vo (

X1

-

X2

) = Vw (

Y1

-0) (8.1.3)

Mass transfer coefficient: MTC=Rate of acid transfer/volume of packing × mean driving force (8.1.4) Where Log mean driving force: (Δ

X1

-Δ

X2

) / ln (Δ

X1

/Δ

X2

-0)

Δ

X1

: Driving force at the top of the column = (

Δ

X2

: Driving force at the bottom of the column = (

5

X1

X2

-

)

X1

*)

Where

X1

* is the concentration in the organic phase which would be in

equilibrium with concentration

Y1

in the aqueous phase. The equilibrium

values can be found using the distribution coefficient found in the first experiment.

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Apparatus and Materials Apparatus:       

Liquid-liquid extraction column Water pump Separator funnel Burette Conical flask Volumetric flask Measuring cylinder

Materials:      

 0.1 M NaOH solution 0.025 M NaOH solution  Distillates product solution  Raffinate product solution Feed solution  Propionic acid

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Procedures Experiment A 1. 2. 3. 4. 5. 6. 7. 8.

Make sure valve V6 and V11 are closed. Valve S1, C3 and S3 are switched on. The feed flow rates (C1) are adjusted until maximum. S4 is switched on. When the water reaches the top column, C1 is set to 250 cc/min. Wait for 20 minutes. 150mL sample from raffinate, feed and extract is taken. 15 mL from each sample is taken in Experiment A and 3 drops of phenolphthalein is

added. 9. Each sample is titrated with 0.1M NaOH. 10. Step 8 and 9 is repeated with 0.025 M NaOH. 11. For each moles of NaOH is titrated twice. Experiment B 1. 50mL of water, 50mL of organic solvent and 1mL of propionic acid is added into a conical flask and is shakes well. 2. Step 1 is repeated with 3 and 5 mL of propionic acid. 3. Two layers of solution is formed (immiscible) and 10mL from each upper and bottom layer is taken. 4. 3 drops of phenolphthalein is added into each sample. 5. Each ample is titrated with 0.1 M of NaOH.

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Results Experiment A Concentration of NaOH (M)

Raffinate (mL)

Feed (mL)

Extract (mL)

0.1 M

0.6

37.6

6.9

0.025 M

1.0

72.6

22.5

Water flow rate

: 0.20 L/min

Organic flow rate

: 0.20 L/min

Packing dimension

: Length

= 1.2 m

Diameter = 50 mm Experiment B

Volume of NaOH(mL) Volume of propionic acid (mL)

Upper

Bottom

1.0

21.0

2.7

3.0

60.0

26.0

5.0

100.0

55.4

Calculations 9

Concentration of NaOH(M)

0.1 M

Formula for finding distribution coefficient (Experiment B): K = Y/X Where Y: Concentration of solute in exact phase. X: Concentration of solute in raffinate phase. For 0.1M of NaOH 1. 1.0mL of propionic acid Upper (Y) M1V1 = M2V2 (0.1M)(21.0 × 10-3L) = (M2)(1.0 × 10-3L) M2 = 2.10 M

Bottom (X) M1V1 = M2V2 (0.1M)(2.70 × 10-3L) = (M2)(1.0 × 10-3L) M2 = 0.27 M K = Y/X = 2.10M / 0.27M = 7.78

2. 3.0mL of propionic acid Upper (Y) M1V1 = M2V2 (0.1M)(60.0 × 10-3L) = (M2)(3.0 × 10-3L) M2 = 2.00 M

Bottom (X) M1V1 = M2V2 (0.1M)(26.0 × 10-3L) = (M2)(3.0 × 10-3L) M2 = 0.87 M K = Y/X = 2.00M / 0.87M = 2.30

3. 5.0mL of propionic acid Upper (Y) M1V1 = M2V2 (0.1M)(100 × 10-3L) = (M2)(5.0 × 10-3L) M2 = 2.00 M

Bottom (X) M1V1 = M2V2 (0.1M)(55.4 × 10-3L) = (M2)(5.0 × 10-3L) M2 = 1.11 M K = Y/X =2.00M / 1.11M = 1.80

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Formula for finding mass transfer coefficient (Experiment A): 

M1V1 = M2V2 Where:

   



M1 = concentration of NaOH M2 = concentration of propionic acid V1 = volume of NaOH V2 = volume of propionic acid Rate of acid transfer = Vw (Y1 – 0) Vo (X1 – X2) = Vw (Y1 – 0) K = Y1/X* Log mean driving force = (ΔX1-ΔX2) / ln (ΔX1/ΔX2) Where: ΔX1 : Driving force at the top of the column = (X2 – 0) ΔX2 : Driving force at the bottom of the column = (X1 - X1*) Packing dimension : Length = 1.2m Diameter = 50mm −3 50× 10 Radius = = 0.025m 2

(

)

Therefore, packing volume, V = πr2L = π ( 0.025m)2(1.2m) = 2.36 × 10-3 m3 = 2.36 L For 0.1M of NaOH Raffinate: M1V1 = M2V2 (0.1)(0.6×10-3) = M2 (0.015) M2 = 0.004M of propionic acid  X1 Feed: M1V1 = M2V2 (0.1)(37.6×10-3) = M2 (0.015) M2 = 0.251M of propoinic acid Extract: M1V1 = M2V2 (0.1)(6.9×10-3) = M2 (0.015) M2 = 0.046M of propoinic acid  Y1

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



Rate of acid transfer = Vw (Y1- 0) = (0.20 L/min)(0.046 mol/L) = 0.0092 mol/min Finding the value of X2:

Vo (X1-X2) = Vw (Y1 -0) (0.20 L/min)(0.004 mol/L - X2) = 0.0039 mol/min X2 = 0.0155 mol/L @ M 

Finding the log mean driving force: ΔX1 = (X2-0) = 0.0155 M K = Y1 / X* X* = Y1 / K

 At equilibrium, assume K = 7.78 (from experiment B)

= 0.046 / 7.78 = 0.0059 M ΔX2 = (X1-X1*) = 0.004 M – 0.0059M = -0.0019M Log mean driving force = (ΔX1-ΔX2) / ln (ΔX1/ΔX2) = ( 0.0155 +0.0019M / ln (0.0155 / -0.0019) = -0.0855 

Finding the mass transfer coefficient: Mass transfer coefficient =

=

Rate of acid transfer Volume of packing × Meandriving force 0.00 92mol /min 2.36 L x−0.0855

= -0.05 mol/L.min = -0.05 M/min = -0.05 kg/min 12

For 0.025M of NaOH Raffinate: M1V1 = M2V2 (0.025)(1.0×10-3) = M2 (0.015) M2 = 0.0017 M of propionic acid  X1 Feed: M1V1 = M2V2 (0.025)(72.6×10-3) = M2 (0.015) M2 = 0.1210 M of propoinic acid Extract: M1V1 = M2V2 (0.025)(22.5×10-3) = M2 (0.015) M2 = 0.0375 M of propoinic acid  Y1 



Rate of acid transfer = Vw (Y1- 0) = (0.20 L/min)(0.0375 mol/L) = 0.0075 mol/min Finding the value of X2:

Vo (X1-X2) = Vw (Y1 -0) (0.20 L/min)(0.0017 mol/L - X2) = 0.0075 mol/min X2 = - 0.0358 mol/L @ M 

Finding the log mean driving force: ΔX1 = (X2-0) = - 0.0358 M K = Y1 / X* X* = Y1 / K

 At equilibrium, assume K = 7.78 (from experiment B)

= 0.0375 / 7.78 = 0.0048 M 13

ΔX2 = (X1-X1*) = 0.0017 M – 0.0048 M = -0.0031 M Log mean driving force = (ΔX1-ΔX2) / ln (ΔX1/ΔX2) = ( -0.0358+ 0.0031)M / ln (-0.0358 / -0.0031) = -0.0134 

Finding the mass transfer coefficient: Mass transfer coefficient =

=

Rate of acid transfer Volume of packing × Meandriving force 0.0075 mol/min 2.36 L x−0.0134

= -0.237 mol/L.min = -0.237M/min = -0.237 kg/min

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Discussion The purpose of this experiment is to conduct the simple experiments regarding liquidliquid extraction ,to determine the distribution coefficient for the system organic solventpropionic acid water and show its dependence on concentration and to demonstrate how a mass balance is performed on the extraction column and to measure the mass transfer coefficient with the aqueous phase as the continuous medium. This experiment is referred to the solubility when the solvent is soluble with a specific solute contained in the solution and also about the separation of a substance from a mixture by preferentially dissolving the substance in a suitable solvent. For the first experiment, which is to determine the mass transfer coefficient, the feed, raffinate, and extract is got from liquid-liquid extraction column. The mass transfer coefficient depended on diameter. The samples then titrated with 0.1 M NaOH and 0.025 M NaOH. The value of mass transfer coefficient when titrated with 0.1 M NaOH is -0.05 kg/min and 0.237kg/min when titrated with 0.025M NaOH. The values of mass transfer coefficient for both concentrations were negative which is wrong from the theory. In order to get the successful results, we must get positive value for the mass transfer coefficient for both concentrations. It shows that our experiment is failed. Besides, we cannot show the relation between mass transfe rate with the concentration of NaOH. From theory, the mass transfer rate should be increase when the concentration of NaOH is decreasing. For the second experiment, which is to determine the distribution coefficient, titration method is used from upper (Y) and bottom (X) layer example. When titrated with 0.1M, the values are 7.78 in 1.0 ml of propionic acid. From the result of the experiment, it shows that the value of distribution coefficient decrease as the volume of propionic acid increase. The result is almost same with the theory which is the value of distribution coefficient must be decreased when the volume of propionic acid is increasing.

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Conclusion From the experiment we can conclude that our first experiment which is to determine the mass transfer coefficient for 0.1M NaOH and 0.025M NaOH. The mass transfer coefficient from liquid-liquid extraction are -0.05kg/min and -0.237kg/min. Both values are failed because the value that we get are negatives instead of positive. The outside mass transfer coefficient should decreases as the diameter increases. It was my conclusion that the increase in the volume to surface area ratio caused the internal oscillations (eddies) to become relatively slower. Hence less turbulence induced mass transfer was occurring and causing the additional surface area to become less effective. For the second experiment, the values for distribution coefficient by titration with 0.1 M are 7.78 in 1.0 ml of propionic acid, 2.30 in 3.0 ml of propionic acid and 1.80 in 5.0 ml of propionic acid.  

Recommendations These are some recommendations must be followed during the experiment. Before starting the experiment, we must make sure the apparatus is clean. However, if the apparatus is not clean, we must clean up those apparatus using distilled water. Furthermore, the eye position should be perpendicular to the meniscus and the scale. In addition, the experiment must repeat at least 2 or 3 times. Those ways is done to make sure the value is accurate. Besides that, titration process should be repeat several times to get the average values. Then, we must using magnetic stirrer during the titration process. Finally, we needed to make sure the change of color is constant in every titration.

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References 

W. McCabe, J. Smith, and P. Harriot, (2005). Unit Operations of Chemical Engineering,

7th ed. McGraw-Hill.  Liquid-liquid extraction, Formal Report, Thomas Salerno, (2008)  http://www.scribd.com/doc/5552507/3-Liquid-liquid-extraction

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Appendices

Liquid-liquid extraction column

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