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UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA ENGINEERING CHEMISTRY LAB II (CHE523) STUDENT NAME STUDENT ID

: WAHIDATON NISA’ BT AZMAN : 2011253798 GROUP : EH2203B EXPERIMENT : LIQUID-LIQUID EXTRACTION DATE PERFORMED : 01 OCTOBER 2012 SEMESTER :3 PROGRAMME / CODE : CHEMICAL ENGINEERING / EH220 SUBMIT TO : DR. ABDUL HADI

No . 1 2 3 4 5 6 7 8 9 10 11 12 13

Title Abstract Introduction Objectives Theory Apparatus Experimental Procedure Results Calculations Discussion Conclusion Recommendations Reference Appendix TOTAL MARKS

Remarks: Checked by : --------------------------Date :

Allocated Marks (%) 5 5 5 5 5 10 10 10 20 10 5 5 5 100

Marks

No.

Contents

Page

1.

Abstract

3

2.

Introduction

4-5

3.

Objectives

6

4.

Theory

7-9

5.

Apparatus

10

6.

Experimental Procedure

11

7.

Results

12

8.

Calculations

13-21

9.

Discussion

22-23

10.

Conclusion

24

11.

Recommendations

25

12.

References

26

13.

Appendix

27-29

Abstract

2

This liquid-liquid extraction experiment is conducted to find out the distribution coefficient and the mass transfer coefficient for the system organic solvent-propionic acid water and show its dependence on concentration. For this experiment, we conduct two experiments which are based on the solubility. First experiment conducted, we used separators funnel to separate two solutions of different solubility and densities, and then titrate with different of NaOH concentration which are 0.1 M and 0.0025M. The values for distribution coefficient by titration with 0.1M are 15.86 in 1.0 ml of propionic acid, 5.02 in 1.5 ml propionic acid and 2.74 in 2 ml propionic acid. In addition the value for distribution coefficients by titration with 0.025 M of NaOH are 5.88 in 1.0 ml of propionic acid, 5.85 in 1.5 ml propionic acid and 3.64 in 2 ml propionic acid. Next for the second experiment, we used liquid-liquid extraction column to obtain feed, raffinate and extract samples. The samples were titrated with different of NaOH concentration which are 0.1M and 0.025M. The values of mass transfer coefficient from liquid-liquid extraction are 0.0434 kg/min when titrated with 0.1 M NaOH and 0.1302 kg/min when titrated with 0.025 M NaOH. The experiment was successfully done completely.

Introduction

3

Liquid-liquid extraction is a separation technique based upon transfer of a solute between two immiscible liquids. The key principle underlying liquid-liquid extraction is that the solubility of the solute differs significantly between the liquids, providing a thermodynamic driving force for transfer from one phase to the other. Any two immiscible liquids may be used, but it is common to use water and an organic solvent. In a small-scale chemistry laboratory, batch liquid-liquid extraction is often performed using a separator funnel. Extraction is a process that separates components based upon chemical differences. The general principle behind extraction includes 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. For this experiment we use propionic acid as the 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 basic extraction column has two input stream and two output streams. The input streams contain 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.

4

Flow Sheet

Figure 1 : Extraction Flow Sheet for an Extraction Column

Figure 2 : Liquid-Liquid Extraction column schematic diagram

Objectives

5

The objectives of this experiment : 1.

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. 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.

Theory 6

Liquid-liquid extraction defined as 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 seen 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= X* / Y

Where X* is the weight fraction of the solute in the extract and Y is the weight fraction of the solute in the raffinate.

The theory for the system Trichloroethylene-Propionic acid-Water is as follows :

7

Let Vw : Water flow rate, mol/min Vo : Trichloroethylene flow rate , mol/min X : Propionic acid concentration in the organic phase, kg/mol Y : Propionic acid concentration in the aquous phase, kg/mol Subscripts 1 : Top of column 2 : Bottom of column Mass Balance Propionic acid extracted from the organic phase ( raffinate ) = Vo ( X1-X2 ) Propionic acid extacted by the aqueous phase ( extract ) = Vw ( Y1 – 0 ) Therefore, Vo( X1 – X2 ) = Vw ( Y1 – 0 ) Mass transfer coefficient : MTC = Rate of acid transfer / volume of packing x mean driving force

Where Log mean driving force : ( ∆X1 –∆X2 ) / ln (∆X1 / ∆X2 ) 8

∆X1 : Driving force at the top of the column = ( X2 – 0 ) ∆X2 : Driving force at the bottom of the column = ( X1 – 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.

Apparatus and Reagents

Apparatus 9

•

Liquid-liquid extraction column

•

Water pump

•

Separator funnel

•

Burette

•

Conical flask

•

Volumetric Flask

Reagents •

NaOH solution ( 0.1 M and 0.025 M )

•

Distillates product solution

•

Raffinate product solution

•

Feed solution

•

Propionic acid

10

Experimental Procedure

Experiment 1 1. Add 50ml of water, 50ml of organic solvent and 1ml of propionic acid into a conical flask

and shake it well. 2. Repeat step 1 with 1.5 and 2ml of propanoic acid. 3. It will form two layers and take 10ml from each upper and bottom layer. 4. Add 3 drops of phenophtalein into each sample. 5. Titrate each sample with 0.1M and 0.025M of NaOH. Experiment 2 1. Make sure valve V6 and V11 are closed. 2. Switch on valve S1, C3 and S3. 3. Adjust feed flow rate (C1) until maximum. 4. Switch on S4. 5. When the water reach the top column, set C1 to 300 cc/min. 6. Wait for 20 minutes. 7. Take 100ml sample from refinate, feed and extract. 8. Take 10ml from each sample in Experiment 1 and add 3 drops of phenolphthalein. 11

9. Titrate each sample with 0.1 M NaOH. 10. Repeat step 1 and 2 with 0.025 M NaOH. 11. Titrate twice for each moles of NaOH.

Result

Experiment 1 Volume of Propionic

Volume NaOH (ml) Upper Bottom

Acid (ml) 1.0 1.5 2.0 1.0 1.5 2.0

Concentration of NaoH (ml)

22.2 31.7 40.5 51.8 100.0 112.8

1.4 6.3 14.8 8.8 17.1 31.0

0.1 M 0.025 M

Experiment 2 Concentration of NaOH (M) 0.1 M NaOH 0.025 M NaOH

Raffinate (ml) 15.00 23.50

Feed (ml) 3.25 4.75

Extract (ml) 5.00 16.00

Water flow rate = 0.25 L/min Organic flow rate = 0.25 L/min Packing dimension : length, l = 1.2 m ; diameter, d = 0.05 m

12

Calculations

Formula : Finding distribution coefficient : 1. K = Y/X Where Y : concentration of solute in extract phase X : concentration of solute in raffinate phase Finding mass transfer coefficient : 1. M1 V1 = M2 V2

Where M1 : concentration of NaOH M2 : concentration of propanoic acid V1 : volume of NaoH V2 : volume of propionic acid

2. Rate of acid transfer = Vw ( Y1 -0 ) 3. Vo ( X1 –X2) = Vw (Y1 – 0) 4. K = Y1 / X* 13

5. Log mean driving force = ( ∆X1 –∆X2 ) / ln (∆X1 / ∆X2 )

Where : ∆X1 : Driving force at the top of the column = ( X2 -0 ) ∆X2 : Driving force at the bottom of the column = ( X1 – X*1 )

6. Packing Dimension : Length, l = 1.2 m Diameter, d = 0.05 m Packing Volume, V = πr2l = (π) (0.025 m)2 (1.2 m) = 2.36 x 10-3 m3 = 2.36 L

7. Mass transfer coefficient =

14

Experiment 1 Finding the distribution coefficient : For 0.1 M of NaOH 1. 1.0 ml of propionic acid

Upper ( Y ) M1 V1 = M2 V2 ( 0.1 )(0.0222) = M2 ( 0.001 ) M2 = 2.22 M

Bottom ( X ) M1 V1 = M2 V2 ( 0.1 )( 0.0014 ) = M2 ( 0.001 ) M2 = 0.14 M K = Y/X

= 2.22 M / 0.14 M = 15.86

2. 1.5 ml of propionic acid

Upper ( Y ) M1 V1 = M2 V2 ( 0.1 )(0.0317) = M2 ( 0.0015 ) M2 = 2.11 M

Bottom ( X ) M1 V1 = M2 V2 ( 0.1 )( 0.0063 ) = M2 ( 0.0015 ) M2 = 0.42 M 15

K = Y/X = 2.11 M / 0.42 M = 5.02

3. 2.0 ml of propionic acid

Upper ( Y ) M1 V1 = M2 V2 ( 0.1 )(0.0405) = M2 ( 0.002 ) M2 = 2.025 M

Bottom ( X ) M1 V1 = M2 V2 ( 0.1 )( 0.0148 ) = M2 ( 0.002 ) M2 = 0.74 M K = Y/X

= 2.025 M / 0.74 M = 2.74

For 0.025 M of NaOH 1. 1.0 ml of propionic acid Upper ( Y ) Bottom ( X ) M1 V1 = M2 V2 M1 V1 = M2 V2 ( 0.025 )(0.0518) = M2 ( 0.001 ) ( 0.025 )( 0.0088 ) = M2 ( 0.001 ) M2 = 1.295 M M2 = 0.22 M

K = Y/X = 1.295 M / 0.22 M = 5.88

16

2. 1.5 ml of propionic acid

Upper ( Y ) M1 V1 = M2 V2 ( 0.025 )(0.1) = M2 ( 0.0015 ) M2 = 1.67 M

Bottom ( X ) M1 V1 = M2 V2 ( 0.025 )( 0.0171 ) = M2 ( 0.0015 ) M2 = 0.285 M K = Y/X

= 1.67 M / 0.285 M = 5.85

3. 2.0 ml of propionic acid Upper ( Y ) Bottom ( X ) M1 V1 = M2 V2 M1 V1 = M2 V2 ( 0.025 )(0.1128) = M2 ( 0.002 ) ( 0.025 )( 0.031 ) = M2 ( 0.002 ) M2 = 1.41 M M2 = 0.3875

K = Y/X = 1.41 M / 0.3875 M = 3.64

Experiment 2 Finding the mass transfer coefficient : For 0.1 M of NaOH M1 V1 ( 0.1 )( 0.015 ) M2

= = =

M2 V2 M2 (0.015 ) 0.1 M of propionic acid

Raffinate : 17

X1

M1 V1 ( 0.1 )( 0.00325 ) M2

= = =

M2 V2 M2 (0.015 ) 0.0217 M of propionic acid

M1 V1 ( 0.1 )( 0.005 ) M2

= = =

M2 V2 M2 (0.015 ) 0.0333 M of propionic acid

Feed :

Extract :

Y1

Rate of acid transfer

= = =

Vw ( Y1 -0 ) 0.25 L/min ( 0.0333 mol/L ) 0.008325 mol/min

18

Vo ( X1 –X2) 0.25 L/min (0.1–X2 ) 0.025 – 0.25 X2 X2

= = = =

Vw (Y1 – 0) 0.008325 mol / min 0.008325 0.0667 M

Log mean driving force = ( ∆X1 –∆X2 ) / ln (∆X1 / ∆X2 ) ∆X1 = X2 - 0 = 0.0667 -0 = 0.0667 M K = Y1 / X* X* = Y1 / K At equilibrium, assume R = 15.86 ( From Experiment 1 ) X* = 0.0333 / 15.86 = 0.00209 M ∆X2 = ( X1 – X1* ) = 0.1 M - 0.00209 M = 0.09791 Log Mean Driving Force = ( 0.0667 – 0.09791 ) / ln ( 0.0667 / 0.09791 ) 19

= 0.0813

Mass transfer coefficient =

=

= 0.0434 M / min

= 0.0434 kg/min

For 0.025 M of NaOH M1 V1 ( 0.025 )( 0.0235 ) M2

= = =

M2 V2 M2 (0.015 ) 0.0392 M of propionic acid

Raffinate :

X1

M1 V1 ( 0.025 )( 0.00475 ) M2

= = =

M2 V2 M2 (0.015 ) 0.007916 M of propionic acid

Feed :

20

M1 V1 ( 0.025)( 0.016 ) M2

= = =

M2 V2 M2 (0.015 ) 0.0267 M of propionic acid

Extract :

Y1

Rate of acid transfer

Vo ( X1 –X2) 0.25 L/min (0.0392 –X2 ) 0.0098 – 0.25 X2 X2

= = =

= = = =

Vw ( Y1 -0 ) 0.25 L/min ( 0.0267 mol/L ) 0.006675 mol/min

Vw (Y1 – 0) 0.006675 mol / min 0.006675 0.0125 M

Log mean driving force = ( ∆X1 –∆X2 ) / ln (∆X1 / ∆X2 ) ∆X1 = X2 - 0 21

= 0.0125 -0 = 0.0125 M K = Y1 / X* X* = Y1 / K At equilibrium, assume R = 5.88 ( From Experiment 1 ) X* = 0.0267 / 5.88 = 0.00454 M ∆X2 = ( X1 – X1* ) = 0.0392 M - 0.00454 M = 0.03466 Log Mean Driving Force = (0.0125 – 0.03466) / ln (0.0125 / 0.03466) = 0.02173

Mass transfer coefficient =

=

= 0.1302 M / min

= 0.1302 kg/min

22

Discussion

We conduct this experiment to determine the distribution of coefficient and to determine the mass transfer coefficient. This experiment is based on the solubility, the solvent is also soluble with a specific solute contained in the solution. This is the separation of a substance from a mixture by preferentially dissolving the substance in a suitable solvent. The first experiment conducted which is to determine the distribution coefficient, we used titration method from the upper (Y) and bottom (X) layer sample. The values for distribution coefficient by titration with 0.1M are 15.86 in 1.0 ml of propionic acid, 5.02 in 1.5 ml propionic acid and 2.74 in 2 ml propionic acid. In addition the value for distribution coefficients by titration with 0.025 M of NaOH are 5.88 in 1.0 ml of propionic acid, 5.85 in 1.5 ml propionic acid and 3.64 in 2 ml propionic acid. For the second experiment, we used liquid-liquid extraction column to obtain feed, raffinate and extract samples. The samples were titrated with different of NaOH concentration which are 0.1M and 0.025M. The values of mass transfer coefficient from liquid-liquid extraction are 0.0434 kg/min when titrated with 0.1 M NaOH and 0.1302 kg/min when titrated with 0.025 M NaOH. In the first experiment, from the result if titrated with 0.1 M NaOH, it shows that the value of distribution coefficient decrease as the volume of propionic acid increase. Same as when titrated with 0.025 M NaOH, the value of distribution coefficient decrease as the volume of propionic acid increases.

23

For the second experiment, it shows that the increase of mass transfer rate as the concentration decrease. Thus we can summarize that the value of mass transfer rate is varies when the concentration of NaOH decrease. During titration we choose phenophtalein as the indicator. Its turn pinkish-purple colour during the titration, but to get all the same colour might be hard enough to classify. As the colour is not constant, the value of mass transfer coefficient and distribution coefficient will be different from the actual. The colour pinkish-purple indicates that the NaOH is at the equilibrium with the sample solution. For experiment 2, we can see from the result that the concentration of propionic acid in raffinate is more than in extract. The concentration of propionic acid should more in extract than in the raffinate. We can assume that the extraction does not occur efficiently and more time needed to extract the solution. The value from this experiment might be different from the actual. This might be because of several error occur during this experiment. The varies result might caused by the oil emission and impurities at the beaker, conical flask and burette. We can avoid this error by rinse and clean the apparatus before start the experiment. Other than that, the most common error that always occur during the experiment is the position of the eyes while taking reading at the burette. The eyes should be straight to the scale and must be perpendicular to the meniscus. Moreover to get the best and accurate result the experiment should be repeated at least 3 times and the error during the experiment progress can be determined.

24

Conclusion

Form the experiment 1, we can conclude that the value for distribution can be determine by titration with 0.1M of NaOH are 15.86 in 1.0 ml of propionic acid, 5.02 in 1.5 ml propionic acid and 2.74 in 2 ml propionic acid. In addition, the value for distribution coefficients by titration with 0.025 M of NaOH are 5.88 in 1.0 ml of propionic acid, 5.85 in 1.5 ml propionic acid and 3.64 in 2 ml propionic acid. For the Experiment 2, the values of mass transfer coefficient from liquid-liquid extraction are 0.0434 kg/min when titrated with 0.1 M NaOH and 0.1302 kg/min when titrated with 0.025 M NaOH..

25

Recommendation

1. Rinse the apparatus before use it on experiment. The apparatus contain impurities might effect the result of this experiment. 2. The experiment must be repeated at least 3 times per set. This is to achieve the accurate result as possible and also to determine the mistake that maybe done during the experiment. 3. The eyes should be straight and pependicular to the meniscus while reading the value at the burette. This common error should be avoid the most because it could effect the overall result. 4. Longer time is needed to extract the solution. This is because the concentration of propionic acid should more in extract than in the raffinate. We recommend the time is must be in 30 to 40 minutes instead of 20 minutes to extract the solution. 5. Wear mask and also the google while conduct the experiment. This is because we use the propanoic acid and tricholoroethylene that contain harmful substances and have stingy smell that could effect the respirotary system. 6. Used magnetic stirrer during titration to make sure the solution is mixed. 26

7. Make sure the changes of colour should be constant in every titration.

Reference

1. http://www.scribd.com/doc/29920099/Liquid-Liquid-Extraction 2. http://chemwiki.ucdavis.edu/VV_Lab_Techniques/Liquid-Liquid_Extraction 3. Liquid-Liquid Extraction: theory and laboratory practice, L. Alders, Elsevier Pub. Co.

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Appendix

28

1. Liquid-Liquid Extraction Unit

2. Liquid-Liquid Extraction Unit

29

30