1.Abstract/summary Regarding to the experiment objectives, which is 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 liquid-liquid 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 7.53 in 1.0 ml of propionic acid, 4.29 in 1.5 ml propionic acid and 3.64 in 2 ml propionic acid. While the value for distribution coefficients by titration with 0.025M are 4.41 in 1.0 ml of propionic acid, 1.825 in 1.5 ml propionic acid and 4.71 in 2 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; 0.0102 kg/min if titrated with 0.1M NaOH and 0.068 kg/min if titrated with 0.025M NaOH. The experiment was completely and successfully done. 1 2. Introduction 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 2 Flow sheet Figure 1. Extraction Flow sheet for an Extraction Column Figure 2: Liquid-liquid extraction column schematic diagram 3 . 3. 4 . To determine the distribution coefficient for the system organic solvent-propionic acidwater and show its dependence on concentration. Aims/Objectives The objectives of this experiment: y y Conduct the simple experiments regarding liquid-liquid extraction. y 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. continuous flow becomes economical.4. the situation being analogous to that existing in packed distillation columns. The operation may of course be repeated if more than one contact is required. 5 . In dilute solutions at equilibrium. 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 extract is the layer of solvent plus extracted solute and the raffinate is the layer from which solute has been removed. but when the quantities involved are large and several contacts are needed. the concentration of the solute in the two phases is called the distribution coefficient or distribution constant µK¶. and so the extract may be shown coming from top of the equipment in some cases and from the bottom in others. K=Y/X Where the Y and X are the concentrations of the solute in the extract and the raffinate phases respectively. 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. Theory In liquid-liquid extraction. The rate at which a soluble component is transferred from one solvent to another will be dependent. Therefore it is very advantageous for this interface to be formed by droplets and films. among other things. The extract may be lighter or heavier than the raffinate. on the area of the interface between the two immiscible liquids. 2) Therefore theoretically. 6 .1.4) Where Log mean driving force: ( X . lt/s o X : Propionic acid concentration in the organic phase. = Vo (X -X ) 1 2 (8. lt/s w V : Trichloroethylene flow rate. kg/lt Y : Propionic acid concentration in the aqueous phase.3) Mass transfer coefficient: MTC=Rate of acid transfer/volume of packing × mean driving force (8. kg/lt Subscripts: 1 : Top of column 2 : Bottom of column Mass Balance: Propionic acid extracted from the organic phase (raffinate).1) Propionic acid extracted by the aqueous phase (extract) = Vw (Y -0) 1 (8.1. Vo (X -X ) = Vw (Y -0) 1 2 1 (8. The equilibrium values can be found using the distribution coefficient found in the first experiment.X ) / ln ( X / X ) 1 2 1 2 X1: Driving force at the top of the column = (X -0) 2 * X2: Driving force at the bottom of the column = (X -X ) 1 1 Where X1* is the concentration in the organic phase which would be in equilibrium with concentration Y1 in the aqueous phase.The theory for the system Trichloroethylene-Propionic acid-Water is as follows: Let V : Water flow rate.1.1. 025 M) Distillates product solution raffinate product solution Feed solution Propionic acid 7 . Apparatus Apparatus: y y y y y y Liquid-liquid extraction column Water pump Separator funnel Burette conical flask volumetric flask Reagents: y y y y y NaOH solution (0.1 M and 0.5. 5. Add 50mL of water. feed and extract.1 M NaOH. Repeat step 1 and 2 with 0.1 M and 0. Take 10mL from each sample in experiment A and add 3 drops of phenolphthalein. 50mL of organic solvent and 1mL of propanoic acid into a conical flask and shake it well. Titrate each sample with 0. It will form two layers and take 10mL from each upper and bottom layer. Procedure Experiment A 1. 11. Take 100mL sample from refinate. 2. Adjust feed flow rate (C1) until maximum.025 M of naOH. C3 and S3. 8 . 8. 4. 4. Make sure valve V6 and V11 are closed. 2. 9. Repeat step 1 with 1.6. Wait for 20 minutes. Switch on valve S1. 3. 5. 6. 10. When the water reach the top column. Switch on S4. Experiment B 1. 3. Add 3 drops of phenophtalein into each sample. Titrate each sample with 0.025 M NaOH. 7. set C1 to 300 cc/min. Titrate twice for each moles of NaOH.5 and 2 mL of propanoic acid. 6 8.7 6.5 29.3 0.5 2.0 1.2 m Diameter = 50 mm 9 .1 mL 0.0 1.3 mL 1.8 27.1 M Upper Bottom Concentration of NaOH (M) Water flow rate = 0.4 mL Raffinate Feed Extract EXPERIMENT B Volume of NaOH (mL) Volume of Propionic acid (mL) 1.025 M NaOH 1.5 2.0 1.025 M 0.7.0 12.7 37.1 mL 0.6 26.6 mL 0. Result EXPERIMENT A Concentration of NaOH (M) 0.9 48.0 mL 0.8 1.3 L/min Packing dimension: length = 1.3 7.1 M NaOH 0.3 L/min Organic flow rate = 0.2 137.0 27. 4.8. Rate of acid transfer = Vw (Y1. Calculations Formula: Finding distribution coefficient: K = Y/X Where Y: concentration of solute in extract phase.0) Vo (X -X ) = Vw (Y -0) 1 2 1 K = Y1/X* Log mean driving force= ( X . Finding mass transfer coefficient: 1. M1V1 = M2V2 Where. X1: Driving force at the top of the column = (X -0) 2 X2: Driving force at the bottom of the column = (X -X* ) 1 1 10 . 5. 3.X ) / ln ( X / X ) 1 2 1 2 Where. X: concentration of solute in raffinate phase. M1 ± concentration of NaOH M2 ± concentration of propionic acid V1 ± volume of NaOH V2 ± volume of propionic acid 2. 2m) = 2.36 L 7. Packing dimension: length = 1.2 m Diameter = 50 mm Therefore.36×10-3 m3 = 2.025 m)2 (1.6. Mass transfer coefficient = rate of acid transfer Volumes of packing ×mean driving force 11 . packing volume. V= r2L = ( ) (0. 1.29 3.385 M/0.1) (0.28 M/0.1) (0.027) = M2 (0.0017) = M2 (0.0063) = M2 (0.17 M K = Y/X = 1.0015) M2 = 0.0128) = M2 (0.1) (0.53 2.1) (0.17 M = 7.0277) = M2 (0.1) (0.1M of NaOH 1.28 M K = Y/X = 1.5 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.80 M/0.Finding the distribution coefficient: For 0.64 12 .80 M K = Y/X = 1.002) M2 = 1.0015) M2 = 1.002) M2 = 0.42 M = 4.0 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.385 M bottom (X) M1V1 = M2V2 (0.001) M2 = 0.0076) = M2 (0.1) (0.38 M = 3.42 M bottom (X) M1V1 = M2V2 (0.0 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.38 M bottom (X) M1V1 = M2V2 (0. 2.001) M2 = 1. 1. 002) M2 = 1.825 3.0293) = M2 (0. 1.1378) = M2 (0.723 M bottom (X) M1V1 = M2V2 (0.0482) = M2 (0.215 M K = Y/X = 0.0 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.025) (0.44 M bottom (X) M1V1 = M2V2 (0.803 M/0.9475 M/0.025) (0.00265) = M2 (0.0015) M2 = 0.215 M = 4.1) (0.002) M2 = 0.723 M/0. 2.025 M of NaOH 1.0015) M2 = 0.44 M = 1.803 M bottom (X) M1V1 = M2V2 (0.5 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.1) (0.366 M = 4.0379) = M2 (0.1) (0.001) M2 = 0. 1.For 0.025) (0.0086) = M2 (0.0 ml of propionic acid Upper (Y): M1V1 = M2V2 (0.001) M2 = 0.41 2.9475 M K = Y/X = 0.71 13 .366 M K = Y/X = 1. 012 M 14 .01) M2 = 0.0003 mol/min Vo (X -X ) = Vw (Y -0) 1 2 1 0.001M of propionic acid Extract: M1V1 = M2V2 (0.0001) = M2 (0.1)(0.1 M of NaOH Raffinate: M1V1 = M2V2 (0.01) M2 = 0.0) = 0.001 mol/L) = 0.013 mol/min ± X2) = 0.0001) = M2 (0.1)(0.013M of propionic acid Feed: M1V1 = M2V2 (0.1)(0.3 L/min (0.0013) = M2 (0.01) M2 = 0.001M of propionic acid Y1 X1 Rate of acid transfer = Vw (Y1.003 mol /min X2 = 0.Finding the mass transfer coefficient: For 0.3 L/min (0. 0129) / ln (0.0003 mol/min 2. assume R = 7.0129 Log mean driving force= (0.53 = 0.X ) / ln ( X / X ) 1 2 1 2 X1 = (X -0) 2 = 0.000133M) = 0.001/7.36 L × 0.012/0.0124 Mass transfer coefficient = rate of acid transfer Volumes of packing ×mean driving force = 0.012 M K = Y1/X* X* = Y1/K = 0.Log mean driving force= ( X .0102M/min = 0.0102 kg/min 15 .012 ± 0.0.53 (from experiment A) = ( 0.0129) = 0.013 M .0124 = 0.0102 mol/L.min = 0.000133 M X2= (X -X* ) 1 1 at equilibrium. 3 L/min (0.003 mol /min X2 = 0.01) M2 = 0.0) = 0.0003 mol/min Vo (X -X ) = Vw (Y -0) 1 2 1 0.025)(0.001) = M2 (0.025 M of NaOH Raffinate: M1V1 = M2V2 (0.0015M of propionic acid Extract: M1V1 = M2V2 (0.001 mol/L) = 0.0006) = M2 (0.025)(0.025)(0.0004) = M2 (0.01) M2 = 0.For 0.0025mol/min ± X2) = 0.001M of propionic acid Y1 X1 Rate of acid transfer = Vw (Y1.3 L/min (= 0.0025M of propionic acid Feed: M1V1 = M2V2 (0.0015 M 16 .01) M2 = 0. X ) / ln ( X / X ) 1 2 1 2 X1 = (X -0) 2 = 0.0015/0.0003 mol/min 2.002273 Log mean driving force= (0.36 L × 0.00186 = 0.min = 0.41 (from experiment A) = (0.41 = 0.068 mol/L.0025M .002273) / ln (0.000227M) = 0.002273) = 0.0.Log mean driving force= ( X .001/4.068 M/min = 0.068 kg/min 17 .0015± 0.000227 M X2= (X -X* ) 1 1 at equilibrium. assume R = 4.0015 M K = Y1/X* X* = Y1/K = 0.00186 Mass transfer coefficient = rate of acid transfer Volumes of packing ×mean driving force = 0. This experiment is based on the solubility.025M NaOH is 0.64 in 2 ml propionic acid. the solvent is also soluble with a specific solute contained in the solution.9. Discussions In this experiment. The value from this experiment might be different with the actual. if titrated with 0.068 kg/min.29 in 1. The most common error is the position of the eye during taking the volume value at the burette.024M NaOH. 18 . we used the liquid-liquid extraction column to get the feed.0102 kg/min while the mass transfer coefficient value when titrated with 0. if titrated with 0. So. By titration with 0. For the second experiment. While the value for distribution coefficients by titration with 0. The samples then were titrated with 0. 4. here we say that the mass transfer rate will differ as the concentration of NaOH increase. 1. The value of distribution coefficient should decrease as the volume of propionic acid increase.41 in 1.0 ml of propionic acid.825 in 1. From the result in the first experiment. we want to determine the distribution of coefficient and to determine the mass transfer coefficient.5 ml propionic acid and 3.025M are 4.53 in 1. The mass transfer coefficient value when titrated with 0. raffinate and extract solution. it shows that the value of distribution coefficient decrease as the volume of propionic acid increase.1M NaOH. which is to determine the mass transfer coefficient. the values are 7.1M NaOH and 0. which is to determine the distribution coefficient.0 ml of propionic acid. For the first experiment. the value of distribution coefficient firstly decrease then increase back. Meanwhile. Result for the second experiment shows the increase of mass transfer rate as the concentration of NaOH decrease.1M NaOH is 0. this might be of the errors during titration.1M. This maybe because of several error occur during the experiment progress.71 in 2 ml propionic acid.5 ml propionic acid and 4. the eye position should be straight to the scale and must be perpendicular to the meniscus.025M NaOH. we used titration method from the upper (Y) and bottom (X) layer sample. and this is the separation of a substance from a mixture by preferentially dissolving the substance in a suitable solvent. the equipment or apparatus we use must be in clean condition. More time needed to extract the solution. The differences of the experiment result from the actual result might also caused by the oil emission and impurities at the beaker. The color indicates that the NaOH is at equilibrium with the sample solution. but to get all the same color might be hard enough to identify. this might be the extraction process does not occur efficiently. As the color is not constant. To avoid the error. the value of mass transfer coefficient and distribution coefficient will be different from the actual. The concentration of propionic acid in raffinate is more than in extract. The concentration of propionic acid should more in extract than in raffinate. Moreover. the experiment should be repeated at least 3 times to get the accurate values and the mistake during the experiment progress can be identified. 19 . conical flask or burette.We choose pinkish-purple color for the color indicator during titration. 20 .0102 kg/min if titrated with 0. While the value for distribution coefficients by titration with 0.025M are 4.64 in 2ml propionic acid.1M and the values are.025M NaOH.825 in 1.0 ml of propionic acid. 0. Conclusion As a conclusion: y The value for distribution coefficient can be determine by titration with 0.068 kg/min if titrated with 0. 1.53 in 1.71 in 2 ml propionic acid.29 in 1.5 ml propionic acid and 4.0ml of propionic acid.41 in 1. y The value of mass transfer coefficient from liquid-liquid extraction are.5ml propionic acid and 3.1M NaOH and 0.10. 4. 7. Recommendations y y y y y y Use clean apparatus. F. McGraw-Hill. 1983.. 12.com/doc/7009722/Liquid-Liquid-Extraction Perry.com/doc/5552507/3-Liquid-Liquid-Extraction http://www. 6th edition. R. Make sure the changes of color should be constant in every titration.scribd. J.wikipedia.scribd. Used magnetic stirrer during titration. 1st edition.org/wiki/Liquid-liquid_extraction http://rothfus. and J. 3rd edition. References http://en. Myers. and J.cheme. y y y y y 21 .K. 1984. Heat and Mass Transfer. Momentum. M. Titration should be repeat several times to get the average values. 1983.pdf http://www. y Sinnott. C. Repeat the experiment at least 3 times to get accurate values and to make comparisons. if not clean the apparatus using distilled water The eye position should be perpendicular to the meniscus and the scale.. y Bennett. Coulson. Perry¶s Chemical Engineering Handbook. Green. and D.11. Richardson. McGraw-Hill.H.cmu. Pergamon Press.edu/uolab/lle/projects/t8_f00/t8_f00.. R. An Introduction to Chemical Engineering Design. E. O. 13. Appendices 22 .