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TABLE OF CONTENTS

Abstract

2

Introduction

3

Objectives

3

4.0 Theory5

5.0 Materials and Apparatus6

6.0 Procedures7

7.0 Results8

8.0 Calculation9

9.0 Discussions13

10.0 Conclusion14

11.0 Recommendations15

12.0 References15

13.0 Appendices

16

ABSTRACT

LIQUID LIQUID EXTRACTION

The purposes of the liquid-liquid extraction experiment is to separate two or more componentsof a solution from one another and then to determine whether a certain compound is present in an unknown reagent or not. There are three liquids used which are propionic acid, oil and water. Oil and propionic acid are mixed together in the feed tank, which is then mixed through the recycle system of the pump before it is sent to be mixed with the water in the first stage. Out of each stage samples of propionic acid was taken, and titrated with sodium hydroxide to determine the amount of propionic acid contained in the feed, extract and raffinite. For the first experiment, the data shows that the propionic acid added lower will gives the concentration of sodium hydroxide lower. The higher value is for 5.0 mL propionic acid added, which is 22.5 mL of sodium hydroxide was titrated and have the concentration of 2.25 M. Next, the data illustrates that the majority of the Propionic Acid was extracted from oil to the water in the first stage, which is 0.350 M and 0.135 M after titrated with 35.0 mL of 0.1 M and 54.1 mL of 0.025 M of sodium hydroxide respectively. The amount of Propionic Acid in water decreased in the next stage, which is at raffinite and then increase thereafter. This means that the amount of Acetic Acid extracted into water not constant with each stage. It was reliably assumed that the longer the mixing of the liquids the better the molecules are able to partition into the preferred solvent.

3.0OBJECTIVES

To separate two or more components of a solution from one another.

Todetermine whether or not a certain compound is present in an unknown reagent.

INTRODUCTION

2.1Definitions

Liquid-liquid Extraction is a method of separation where an aqueous solution is usually brought into contact with another immiscible organic solvent, so as to affect a transfer of one or more solute from the aqueous solution into the organic solvent. This is a fast, easy and convenient method of separation. It is usually performed by moderately shaking the liquids inside a separating funnel for a few minute and can be used for either large volume or trace level.

2.2Properties of solvents

Solvent are selected based on, make sure both solvents must be immiscible to each other. Then, solvent 1 can be aqueous / H2O and solvent 2 can be Organic in nature. It is because most targeted compound (or solute in solvent 1 ) are soluble in Organic Solvents (if the solutes are organic ) and most organic solvent has lower boiling point (Eg: Methanol, Chloroform, Carbon Tetrachloride has boiling point 50C, 61.2C and 76.8C respectively), and since, after extraction the solvent 2 is evaporated to collect the targeted solute, there will be a less chance for the targeted solute to be degraded at high temperature. Make sure, solvent 2 must be chosen accordingly. Sothat, the Target Compound must have higher affinity toward Solvent 2, so that, during extraction maximum amount of solute is transferred to Solvent 2. Density of both of the solvents must be known. So, that in the separating funnel it is possible to know which solvent is at the lower phase and which one is at the upper (Usually, the solvent with higher density is at the lower phase). Then, make sure that density difference between Solvent 1 and Solvent 2 should be high, so that there will be a less chance of emulsion formation and even if emulsion forms, it will easy to breakdown. Then, Kpvalue / Partition Coefficient of immiscible solvent pair must be higher, since it is important for the effectiveness of extraction.

2.3Properties of Solute

Solute / Target Compound can be organic solute and inorganic solute.Organic Solute :Most Organic Solutes are soluble in Organic Solvents, which can be extracted with any organic solvents ( Eg: Extraction Caffeine by Chloroform ). However, for both pharmaceuticaland therapeutic purpose, some organic solutes are introduced as salts in order to increase their solubility in Water ( Eg : Promethazine hydrochloride, a hydrochloric salts of Promethazine). So, before performing liquid-liquid extraction of such organic solutes, they are converted into their original organic forms by means of a chemical reaction, so that they can be extracted by any organic solvents ( E.g. Promethazine Hydrochloride is converted into Promethazine before they are extracted with organic solvents like n-hexane or Chloroform ). However, it is to say that upon chemical reaction the concentration of original compound does not changes.

Inorganic Solute :Inorganic solutes are frequently encountered in aqueous solvents either as impurities or as Pharmaceutical Ingredients. And before their extraction, it is absolutely necessary to form ion-association complexes or metal-chelates (by using organic-ligands), so that they may be extracted by an appropriate organic solvent. For example, Cu (2) can be extracted by acetyl acetone by forming Cu(2) - acetyl acetone complex.

2.4Different principle of Liquid-Liquid Extraction

Feed phase:Feed phase is the initial solution containing the target compound / Solute and Solvent 1. In the feed phase, thesolvent 2 added to perform the extraction. So,Feed Phase = Solute + Solvent 1

Extractant and Extract :Extractant is the Solvent 2 that is added to the Feed Phase for extraction. After extraction, the Extractant is called Extract which contains the target compound. The Extract is evaporated to collect and measure the amount of the solute. So before extraction,Extractant = Solvent 2And after extraction,

Extractant= Extract

= Solvent 2 + Solute

Raffinate :Raffinate is generally, termed for theFeed Phase after extraction. And if the extraction process is repeated, the Raffinate becomes the feed phase to which the Solvent 2 is added again. So,Raffinate= Feed phase after Extraction

= Solvent 1 + Remaining Solute

And if the extraction processis repeated, thenRaffinate= Feed phase

= Solvent 1 + Remaining Solute

4.0THEORY

Liquid-liquid extraction is a separation process in which substances or compounds are separated based on their properties, usually their solubility or polarity. In general, two streams, one consisting of solution+solute and the other consisting of the extracting solvent shall come in contact, and the solvent shall extract the solutes from the solution. This solubility and polarity characteristic is the most importantfor the extraction process, as this characteristic determines the solvent that needs to be used, and the effectiveness of the solvent. The feed solution is the solution that contains solute plus the solution, where this is the solution that needs to be separated from the solutes. The raffinate solution is the product stream, where this stream is the one that contains low composition of solutes, as the solutes have been extracted out from the feed solution stream by the solvent. The extract solution isthe second product stream, where this stream is the one that contains a high amount of solute. The extract stream is basically solvent+solute.

Feed

Raffinate stream

Extract stream

Solvent stream

Example Flowchart: Extraction of CH3COOHusing C6H13OH as solvent

There are a few ways to determine the effectiveness of the extraction process. The term distribution coefficient is used as a quantitative measure for the effectiveness of the solvent extraction. This coefficient is usually denoted with K. The formula is given as below:

In other words, the distribution coefficient K is simply the ratio of amount of solute in solvent (extract) to the amount of solute in aqueous solution (raffinate) once the process has achieved equilibrium.In certain cases, there are multiple solvents used in the extraction process. Some solvents are more effective than others. This defines in multiple solvent extraction, certain solvents will extract more solute than the other solvent. Thus, to determine the total amount of solutes that has been extracted by a certain solvent in multiple solvent system, the following formula can be used:

Where VAand VBare volumes of solvents, K is the distribution coefficient and n is the number of extractionsperformed. However, in this experiment, only one solvent is used, therefore this formula shall not be applied in the calculations.

5.0MATERIALS/APPARATUS

The apparatus used in this experiment:50 mL burette-for titration process250 mL Conicalflask-for titration process50 mL beakers-to measure the feed, raffinate and extract solutionsMeasuring cylindersPhenolphthalein as indicator0.025 M and 0.1 M NaOH as titrate

6.0PROCEDURE

6.1Determination of Distribution Coefficient

A mixture of50 mL organic solvent and 50 mL of de-mineralised water is made up in a conical flask.

5 mL of propionic acid is added to the mixture in the flask by using a pipette.

A stopper is placed into the flask and the mixture is well shaken for 5 minutes.

Themixture is then poured into a separating funnel and left for 5 minutes.

After 5 minutes, the lower aqueous layer is removed.

10 mL sample of this payer is taken and is titrated with 0.1 M NaOH solution, where phenolphthalein is used as indicator.

Steps 2 to 6 are repeated, by manipulating the amount of propionic acid added to the mixture from 5 mL to 3 mL and 1 mL

All data is recorded.

6.2Determination of Mass Transfer Coefficient

100 mL of propionic acid is added to 10 L of the organic phase. The mixture is mixed well and the organic feed tank is filled with the organic mixture.

The level control is switched to the bottom of the column (electrode switch S2)

The water feed tank is filled with 15 L of de-mineralised water, by starting the water feed pump and is operated at high flowrate. Once the water is above the packing, the flowrate is reduced.

The metering pump is started and is set at a flowrate of 0.2 L/min

The system is left to operate for 15 to 20 minutes to achieve steady state conditions.

Two 15mL samples of feed, raffinate and extract are taken, using 50 mL beakers.

One set of the three samples are titrated using 0.1 M NaOH solution, while the other set of three samples are titrated using 0.025 M NaOH. Phenolphthalein is used as indicator for all six titration processes.

7.0RESULTS

7.1Determination of Distribution Coefficient

Table 7.0:Propionic acid added with sodium hydroxide titrationPropionic acid added (mL)Aqueous layer (Y)Organiclayer (X)K = Y/X

Titre of M/10 NaOH (mL)Concentration (M)Titre of M/10 NaOH (mL)Concentration (M)

5.022.52.2578.27.820.29

3.020.52.0551.05.100.40

1.016.31.6317.51.750.93

7.2Determination of Mass Transfer Coefficient

Table7.1: Titration of 0.1 M and 0.025 M of sodium hydroxide in propionic acidFlow rate of aqueous phase (L/min)0.2

Flow rate of organicphase (L/min)0.2

Sodium hydroxide concentration (M)0.10.025Concentration of Propionic acid (kg/L)

Feed (mL)35.054.10.3500.135

Raffinite (mL)12.51.50.1250.004

Extract (mL)20.023.20.2000.058

Propionic acid extracted from the organic phase (mL)0.150.077

Propionic acid extracted from the aqueous phase (mL)0.040.0116

Mass transfercoefficient (kg/min)0.02230.024

8.0CALCULATIONS

8.1Determination of Distribution Coefficient

Finding the distribution coefficient, K

where ,

K=YX

Y = Aqueous layer concentrationX = Organic layer concentration

8.1.1For 1.0 mL ofPropionic acid added

Aqueous Layer , Y:Organic Layer , X:K=YX

M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2= Concentration of Propionic acidV2= Volume of Propionic acid

(0.1M) (16.3 mL) = M2(1 mL)Y = 1.63 MM1V1= M2V2(0.1M) (17.5mL) = M2(1 mL)X = 1.75 MK =16.31.75= 0.93

8.1.2For 3.0 mL of Propionic acid added

Aqueous Layer , Y:Organic Layer , X:K=YX

M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2=Concentration of Propionic acidV2= Volume of Propionic acid

(0.1M) (20.5 mL) = M2(1 mL)Y = 2.05 MM1V1= M2V2(0.1M) (51.0mL) = M2(1 mL)X = 5.10 MK =20.551.0= 0.40

8.1.3For 5.0 mL of Propionic acid added

Aqueous Layer , Y:OrganicLayer , X:K=YX

M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2= Concentration of Propionic acidV2= Volume of Propionic acid

(0.1M) (22.5 mL) = M2(1 mL)Y = 2.25 MM1V1= M2V2(0.1M) (78.2mL) = M2(1 mL)X = 7.82 MK =22.578.2= 0.29

8.2Determination of Mass Transfer Coefficient

Finding the mass transfer coefficient,

where,Mass transfer coefficient = Rate of propionic acid transfer Volume of packing x mean driving force

8.2.1For titration with 0.1 M of NaOH

Feed:Raffinite:Extract:

M1V1= M2V2(0.1M) (35.0mL) = M2(10 mL)M2= 0.350 M(X1)M1V1= M2V2(0.1M) (12.5mL) = M2(10 mL)M2= 0.125 MM1V1= M2V2(0.1M) (20.0mL) = M2(10 mL)M2= 0.200 M (Y1)

i. Propionic acid extracted from aqueous phaseRate of acid transfer = Vw(Y1 0)Where Vwis flowrate of aqueous phase is equalto 0.2 L/minSo ,

Rate of acid transfer =( 0.2 L/min ) (0.200 mol/L 0)=0.04 mol/min

ii. Propionic acid extracted from the organic phaseRate of acid transfer = Vo(X1 X2)Where Vois flowrate of aqueous phaseis equal to 0.2 L/minSo ,

Rate of acid transfer =( 0.2 L/min ) (0.350 mol/L X2)By mass balance:

Vo( X1 X2) = Vw( Y1 0)(0.2 L/min)(0.350 mol/L - X2) = (0.2 L/min)(0.200 mol/L - 0)X2= 0.15 mol/L

Next , the driving force ,Log mean driving force = X'1 -X2ln (X1/X2)

WhereX1is the Driving Force at the top of column = (X2 0)X1 = 0.15 mol/L

AndX2is the Driving Force at the bottom of column = (X1 X'1)

Then by coefficient,K = Y1X'1

By assuming the K is 0.93 ( fromExperiment A )The ,X'1 =0.215 mol/minSo,X2 =0.135 mol/min

And log mean driving force is ,= (0.215 - 0.135)ln (0.15/0.135)= 0.759

Lastly to find mass transfercoefficient =Rate of propionic acid transfer Volume of packing x mean driving force= 0.042.36 x 0.759= 0.0223

8.2.2For titration with 0.025 M of NaOH

Feed:Raffinite:Extract:

M1V1= M2V2(0.025M)(54.1mL)=M2(10 mL)M2= 0.135 M (X1)M1V1= M2V2(0.025M)(1.5mL) = M2(10 mL)M2=0.004 MM1V1= M2V2(0.025M)(23.2mL) = M2(10 mL)M2= 0.058 M (Y1)

i. Propionic acid extracted from aqueous phaseRate of acid transfer = Vw( Y1 0)Where Vw is flowrate of aqueousphase is equal to 0.2 L/minSo ,

Rate of acid transfer =( 0.2 L/min ) (0.058 mol/L 0)=0.0116 mol/min

ii.Propionic acid extracted from the organic phaseRate of acid transfer = Vo( X1 X2)Where Vois flowrateof aqueous phase is equal to 0.2 L/minSo ,

Rate of acid transfer =( 0.2 L/min ) (0.135 mol/L X2)

By mass balance:

Vo( X1 X2) = Vw( Y1 0)(0.2 L/min)(0.135 mol/L - X2) =(0.2 L/min)(0.058 mol/L - 0)X2= 0.077 mol/L

Next, the driving force ,Log mean driving force = X'1 -X2ln (X1/X2)

WhereX1is the Driving Force at the top of column = (X2 0)X1 = 0.077mol/L

AndX2 is the Driving Force at the bottom of column = (X1 X'1)

Then by coefficient,K = Y1X'1

By assuming the K is 0.93 ( from Experiment A )Then ,X'1 =0.062 mol/minSo,X2 =0.073 mol/min

And log mean driving force is= (0.062- 0.073)ln (0.077 / 0.073)

= 0.206

Lastly to find mass transfer coefficient =Rate of propionic acid transfer Volume of packing x mean driving force= 0.01162.36 x 0.206=0.024

9.0DISCUSSIONS

The purposes of the liquid-liquid extraction experiment is to separate two or more components of a solution from one another andthen to determine whether a certain compound is present in an unknown reagent or not.The system works off the premise that the components involved are immiscible with one another. In this case of this lab there are three liquids used which are propionicacid, oil and water. Oil and propionic acid are mixed together in the feed tank, which is then mixed through the recycle system of the pump before it is sent to be mixed with the water in the first stage. When it is properly mixed it is then sent to the first stage and allowed to interact with the water which is flowing countercurrent to the feed of the acid-oil mix. Throughout the three stages the acetic acid will transfer from theoil, which is the raffinate, to the extract, water in this case. Out of eachstage samples of propionic acid was taken, and titrated with sodium hydroxide to determine the amount of propionic acid contained in the feed, extract and raffinite. For the first experiment, the data was collected in Table 7.0. it shows that the propionic acid added lower will gives the concentration of sodium hydroxide lower. The higher value is for 5.0 mL propionic acid added, which is 22.5 mL of sodium hydroxide was titrated and have the concentration of 2.25 M.

Next, from Table 7.1, the dataillustrates that the majority of the Propionic Acid was extracted from oil to the water in the first stage, which is 0.350 M and 0.135 M after titrated with 35.0 mL of 0.1 M and 54.1 mL of 0.025 M of sodium hydroxide respectively. The amount of Propionic Acid in water decreased in the next stage, which is at raffinite and then increase thereafter. This means that the amount of Acetic Acid extracted into water not constant with each stage. It was reliably assumed that the longer the mixing of the liquids thebetter the molecules are able to partition into the preferred solvent. As stated, the data does not agree with this for the simple fact that the majority of Propionic Acid used is evaporated. It should be noted that almost 95% of the Propionic Acid used in this experiment evaporated to the air. This explains the trend seen in the data, because the majority of the Propionic Acid that was extracted into the water occurred in the first stage shortly after the Propionic Acid Oil mixture was added to the system. Once the liquids were allowed to mix and settle throughout the other two stages all the mixing and time to settle out allowed the Propionic Acid to evaporate out due to uncovered tanks. With so much Propionic Acid evaporating out it is clear that withtime the amount of Acetic Acid extracting into the water decreased with increasing evaporation as seen in the data.

Furthermore, the data should show that the amount of Propionic Acid in the oil is decreasing between each stage (meaning that the Propionic is partitioning into the water), which it does for the most part but there are a few discrepancies most likely due to the fact that the oil was not standardized and that the tank covers were removed for a significant portion of both trials. There are a few causes for this. One major complication of this experiment was the fact that the system kept clogging. This not only affected themixing between the two liquids but also required the tank covers to be removed in order to manually fix the clogging as stated before. The drain valves of the settling compartments were very tiny and prone to clogging. This could have been avoided if the drain valves were increased in size to avoid clogging from all the debris that enters in from the dirty holding tanks. Also,the rotameter for the Propionic Acid Oil mixture did not indicate any oil flow rate, and it also had a large clog in it. Without knowing the flowrate of the Oil-Acid mixture, or whether its flow was steady, it was difficult to determine the water flowrate required for the 2:1 countercurrent flow necessary to achieve steady state. The Water Rotameter was constantly being changed by operators of the apparatus (leading to a nonsteady state operation) and it is likely that these complications lead to someof the discrepancies observed.

There were many errors in the performance of the lab which yielded improper data. Therefore it is very difficult to make any definite conclusions on how the speeds of the agitators affected the overall efficiency of removingPropionic acid from the oil. The most significant error was that the Propionic acid was lost, and because of this fact the rest of the data cannot definitively draw any conclusions. From data collected for the concentrations of the Propionic acid found ineach stage it can be seen that most of the acid was lost between the first and third stage. Concentration of acetic acid stayed relatively constant in the oil samples indicating that it was indeed transferred from the oil to the water in the mixing stages. It is recommended that covers for all of the tanks should be kept on at all times while the process is in operation to ensure there is no loss of the acetic acid through evaporation. Another recommendation is to understand early on how exactly the systemworks because it is difficult to get the countercurrent flows to perform properly and to ensure that the mixing is successful in transferring the acetic acid.

10.0CONCLUSION

The experiment was conducted to separate the mixture using two immiscible solvent. From experiment A, table 7.0 shows that in aqueous and organic layer, both are decreasing in concentration due to decreasing volume of propanoic acid titrate with sodium hydroxide. The differences in degree of extraction of propanoic acid by thesediluents were explained in terms of relative permittivity and dipole moment. The higher the volume of propanoic acid much stronger the dipole moment hence the concentration will increase. From table 7.1 in experiment B, the volume of feed titrate with 0.025M of sodium hydroxide is higher compare to 0.1M of sodium hydroxide similar with extract but different for raffinite. From calculation, the concentration of propanoic acid in each sample was higher when the sample was titrated with 0.1M sodium hydroxide.The volume of propanoic acid extracted from the organic and aqueous phase, both higher in 0.1M sodium hydroxide. This shows that, the higher the concentration of sodium hydroxide titrated with the sample, much easier for propanoic acid to be extracted.11.0RECOMMENDATIONS

The component mixture must dissolve in a suitable solvent.

The mixture is thoroughly mixed (shaking) to make sure that the entire components are well mixed.

Gloves and goggles are used in order to prevent any unexpected contact with thereagents as well as lab coat is used when conducting an experiment.

Do not use any crack glassware as well as used glove when handling sodium hydroxide solid or solution as it can cause corrosive effect.

When transferring the solutions, make sure all of the solutions was transferred completely to avoid inaccurate volume of the solution transferred resulting in slightly different value in the molarity.

12.0REFERENCES

Dr. Pahlavan, Extraction Determination of Distribution Coefficient

Liquid extraction isan operation used to separate the component. (n.d.). Retrieved from Solution Inn:http://www.solutioninn.com/engineering/chemical-engineering/chemical-processes/liquid-extraction-is-an-operation-used-to-separate-the-components

Liquid-Liquid Extraction. (n.d.). Retrieved from Wikipedia:http://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction

Liquid-Liquid Extraction. (n.d.). Retrieved from Chemistry Courses:http://courses.chem.psu.edu/chem36/Experiments/PDF's_for_techniques/Liquid_Liquid.pdf

Techniques in Organic Chemistry 2nd ed pages 75-99, 3rd ed pages 113-140.

13.0APPENDICES

C:\Users\umira\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\P_20141103_094223.jpgC:\Users\umira\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\P_20141103_094339.jpg

Figure 13.0:sample from raffiniteFigure 13.1: Sample for feed

overview

Figure 13.2: Liquid Liquid extraction unit

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