6. Liquid Liquid Extraction

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<p>TABLE OF CONTENTS</p> <p>Abstract</p> <p>2</p> <p>Introduction</p> <p>3</p> <p>Objectives</p> <p>3</p> <p>4.0 Theory5</p> <p>5.0 Materials and Apparatus6</p> <p>6.0 Procedures7</p> <p>7.0 Results8</p> <p>8.0 Calculation9</p> <p>9.0 Discussions13</p> <p>10.0 Conclusion14</p> <p>11.0 Recommendations15</p> <p>12.0 References15</p> <p>13.0 Appendices</p> <p>16</p> <p>ABSTRACT</p> <p>LIQUID LIQUID EXTRACTION</p> <p>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.</p> <p>3.0OBJECTIVES</p> <p>To separate two or more components of a solution from one another.</p> <p>Todetermine whether or not a certain compound is present in an unknown reagent.</p> <p>INTRODUCTION</p> <p>2.1Definitions</p> <p>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.</p> <p>2.2Properties of solvents</p> <p>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.</p> <p>2.3Properties of Solute</p> <p>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.</p> <p>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.</p> <p>2.4Different principle of Liquid-Liquid Extraction</p> <p>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</p> <p>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,</p> <p>Extractant= Extract</p> <p>= Solvent 2 + Solute</p> <p>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</p> <p>= Solvent 1 + Remaining Solute</p> <p>And if the extraction processis repeated, thenRaffinate= Feed phase</p> <p>= Solvent 1 + Remaining Solute</p> <p>4.0THEORY</p> <p>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.</p> <p>Feed</p> <p>Raffinate stream</p> <p>Extract stream</p> <p>Solvent stream</p> <p>Example Flowchart: Extraction of CH3COOHusing C6H13OH as solvent</p> <p>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:</p> <p>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:</p> <p>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.</p> <p>5.0MATERIALS/APPARATUS</p> <p>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</p> <p>6.0PROCEDURE</p> <p>6.1Determination of Distribution Coefficient</p> <p>A mixture of50 mL organic solvent and 50 mL of de-mineralised water is made up in a conical flask.</p> <p>5 mL of propionic acid is added to the mixture in the flask by using a pipette.</p> <p>A stopper is placed into the flask and the mixture is well shaken for 5 minutes.</p> <p>Themixture is then poured into a separating funnel and left for 5 minutes.</p> <p>After 5 minutes, the lower aqueous layer is removed.</p> <p>10 mL sample of this payer is taken and is titrated with 0.1 M NaOH solution, where phenolphthalein is used as indicator.</p> <p>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</p> <p>All data is recorded.</p> <p>6.2Determination of Mass Transfer Coefficient</p> <p>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.</p> <p>The level control is switched to the bottom of the column (electrode switch S2)</p> <p>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.</p> <p>The metering pump is started and is set at a flowrate of 0.2 L/min</p> <p>The system is left to operate for 15 to 20 minutes to achieve steady state conditions.</p> <p>Two 15mL samples of feed, raffinate and extract are taken, using 50 mL beakers.</p> <p>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.</p> <p>7.0RESULTS</p> <p>7.1Determination of Distribution Coefficient</p> <p>Table 7.0:Propionic acid added with sodium hydroxide titrationPropionic acid added (mL)Aqueous layer (Y)Organiclayer (X)K = Y/X</p> <p>Titre of M/10 NaOH (mL)Concentration (M)Titre of M/10 NaOH (mL)Concentration (M)</p> <p>5.022.52.2578.27.820.29</p> <p>3.020.52.0551.05.100.40</p> <p>1.016.31.6317.51.750.93</p> <p>7.2Determination of Mass Transfer Coefficient</p> <p>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</p> <p>Flow rate of organicphase (L/min)0.2</p> <p>Sodium hydroxide concentration (M)0.10.025Concentration of Propionic acid (kg/L)</p> <p>Feed (mL)35.054.10.3500.135</p> <p>Raffinite (mL)12.51.50.1250.004</p> <p>Extract (mL)20.023.20.2000.058</p> <p>Propionic acid extracted from the organic phase (mL)0.150.077</p> <p>Propionic acid extracted from the aqueous phase (mL)0.040.0116</p> <p>Mass transfercoefficient (kg/min)0.02230.024</p> <p>8.0CALCULATIONS</p> <p>8.1Determination of Distribution Coefficient</p> <p>Finding the distribution coefficient, K</p> <p>where ,</p> <p>K=YX</p> <p>Y = Aqueous layer concentrationX = Organic layer concentration</p> <p>8.1.1For 1.0 mL ofPropionic acid added</p> <p>Aqueous Layer , Y:Organic Layer , X:K=YX</p> <p>M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2= Concentration of Propionic acidV2= Volume of Propionic acid</p> <p>(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</p> <p>8.1.2For 3.0 mL of Propionic acid added</p> <p>Aqueous Layer , Y:Organic Layer , X:K=YX</p> <p>M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2=Concentration of Propionic acidV2= Volume of Propionic acid</p> <p>(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</p> <p>8.1.3For 5.0 mL of Propionic acid added</p> <p>Aqueous Layer , Y:OrganicLayer , X:K=YX</p> <p>M1V1= M2V2whereM1= Concentration of NaOHV1=Volume of NaOH addedM2= Concentration of Propionic acidV2= Volume of Propionic acid</p> <p>(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</p> <p>8.2Determination of Mass Transfer Coefficient</p> <p>Finding the mass transfer coefficient,</p> <p>where,Mass transfer coefficient = Rate of propionic acid transfer Volume of packing x mean driving force</p> <p>8.2.1For titration with 0.1 M of NaOH</p> <p>Feed:Raffinite:Extract:</p> <p>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)</p> <p>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 ,</p> <p>Rate of acid transfer =( 0.2 L/min ) (0.200 mol/L 0)=0.04 mol/min</p> <p>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 ,</p> <p>Rate of acid transfer =( 0.2 L/min ) (0.350 mol/L X2)By mass balance:</p> <p>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</p> <p>Next , the driving force ,Log mean driving force = X'1 -X2ln (X1/X2)</p> <p>WhereX1is the Driving Force at the top of column = (X2 0)X1 = 0.15 mol/L</p> <p>AndX2is the Driving Force at the bottom of column = (X1 X'1)</p> <p>Then by coefficient,K = Y1X'1</p> <p>By assuming the K is 0.93 ( fromExperiment A )The ,X'1 =0.215 mol/minSo,X2 =0.135 mol/min</p> <p>And log mean driving force is ,= (0.215 - 0.135)ln (0.15/0.135)= 0.759</p> <p>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</p> <p>8.2.2For titration with 0.025 M of NaOH</p> <p>Feed:Raffinite:Extract:</p> <p>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)</p> <p>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 ,</p> <p>Rate of acid transfer =( 0.2 L/min ) (0.058 mol/L 0)=0.0116 mol/min</p> <p>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 ,</p> <p>Rate of acid transfer =( 0.2 L/min ) (0.135 mol/L X2)</p> <p>By mass balance:</p> <p>Vo( X1 X2) = Vw( Y1 0)(0.2 L/min)(0.135 mol/L - X2) =(0.2 L/min)(0.058 m...</p>