UNIVERSITI TEKNOLOGI MARA FACULTY OF CHEMICAL ENGINEERING PROCESS ENGINEERING LABORATORY II (CPE 554) NAME
:
GROUP EXPERIMENT DATE PERFORMED SEMESTER PROGRAMME/CODE SUBMIT TO
: 04 (GROUP REPORT) : EXPERIMENT 8 (LIQUID-LIQUID EXTRACTION) : 4/10/2016 : 05 : EH 242 5B :
NO. 1 2 3 4 5 6 7 8 9 10 11 12 13
TITLE Abstract/Summary Introduction Aim Theory Apparatus Procedure Results Calculations Discussion Conclusion Recommendations Reference Appendices TOTAL MARKS
ALLOCATED MARKS (%)
MARKS
5 5 5 5 5 10 10 10 20 10 5 5 5 100
Remarks: Checked by:
Date:
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TABLE CONTENT
Number
Title
1.
Abstract/Summary
2.
Introduction
3.
Aims
4.
Theory
5.
Apparatus
6.
Methodology/Procedure
7.
Results
8.
Calculations
9.
Discussion
10.
Conclusion
11.
Recommendations
12.
Reference
13.
Appendix
1.0 Abstract/Summary
This experiment was conducted in order to study how to perform a simple experiment with applying principle of liquid-liquid extraction and also to obtain the distribution coefficient for organic solvent-propionic acid-water. This experiment was begin with flowrate was set to 200 l/hr and the feed was collected for about 100ml directly. Next, the system was run for 20 minutes to stabilize the operation. After 20 minutes, the sample from raffinate and extract column was collected same volume like feed which is 100ml for each. Then, all the sample must be titrated with two different concentration of sodium hydroxide (NaOH) which is 0.1M and 0.025M for three times each sample. 15ml of sample will be taken for each titration process. Before the titration process begin, the indicator which is phenolphthalein must be added 3 drops. All the sample must be titrated until light pink colour appeared. There are few calculation involve in this experiment in order to find the distribution coefficient and mass transfer coefficient.
2.0 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. (Stevens, Lo, & Baird, 2007). 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.
Figure 2.1: General Extraction Column. (Google image).
A general and the simplest extraction column as shown in the Figure 2.1 above has two input stream and two output streams. At the inlet, the feed containing the solute to be extracted will be fed in through the input streams at the top of the column. Then, the solvent which extracted the solute from the solution feed at the second inlet which is at the bottom of column. The solvent containing extracted solute leaves through extract stream at the top of the column. Hence, the solution that leaves through the raffinate streams at the bottom of the column containing only small amounts of solute.
3.0 Aim The objectives of the experiment are:1. To performed the simple experiment with applying principle of liquid -liquid extraction. 2. To obtain the distribution coefficient for organic solvent-propionic acid-water. 3. To determine the mass transfer coefficient with the aqueous phase as continuous medium.
4.0 Theory Liquid-liquid extraction (LLE) or known as solvent extraction is applied whenever the method of distillation for separation of mixtures of liquid is no used. This is perhaps due to the method of distillation is less economical and impossible for separation process. Generally, the principle of liquid-liquid extraction (LLE) process is based on varies solubility of components of components in partially miscible or immiscible liquids. (Dutta, 2007). The idea is that the components need to be recovered or purified must be extracted from the feed stream with the aid of extractant which is organic solvent. In fact, both liquids need to be contacted intimately and consequently they will be separated again. It is said that the usage of multiple stages in counter-current flow is very crucial in order to gain high purity and yields. (Sinnott, 2009). Few standards that must be fulfilled when aiming high quality of separation is the extractant must have encouraging partition coefficient and high selectivity. Besides that, both liquids must have differences in density and low common solubility. This conditions act as driving force for the motion of droplets. In addition almost all solvent extraction processes, one of the liquid is distributed into the other liquid in the form of droplets. (Frank, 2008). As far to concern, high performance of separation process by this method is an improved size of droplets and a constant liquid hold-up profile inside the column. Transfer of mass occur when there is relation between droplets of dispersed phase and surrounding liquid of continuous phase. (Weatherley, 2013). Difference in density is very crucial in order for the liquid to separate. Liquid relative densities influences the droplets to accumulate above or below the continuous phase.
Figure 4.1: Mechanism of liquid-liquid extraction (LLE)
Mass balance is a method that uses the conservation of mass where the entering mass and leaving mass is equal. On the other hand, random or ordered packings such as reaching rings helps in increasing the efficiency of extraction of the column. The packings helps in decreasing gross back mixing, control the size of droplet dispersion and induce some additional turbulence inside and outside of droplets in order for diffusion to occur from one phase to the other happen more swiftly. (Rydberg, 2004). Let; Vw
: Flow rate of water (L/s)
V o : Flow rate of solvent (L/s) X: Concentration of propanoic acid in organic phase (kg/L) Y: Concentration of propanoic acid in an aqueous phase (kg/L) Subscripts; : Top stream
1
: Bottom stream
2
Mass balance raffinate (the propanoic acid that being extracted from the organic phase) =
V o ( X 1− X 2 )
Mass balance extract (the propanoic acid extracted by aqueous phase) =
V w ( Y 1−0 )
Hence, V o ( X 1− X 2 )=V w ( Y 1−0 )
Efficiency of extraction
Mass
transfer
coefficient
(refer
to
raffinate
phase)
Rate of Acid Transfer Volume Packing × Mean Driving Force
Log mean driving force =
∆ x1 : ∆ x2
∆ x1 −∆ x 2 ∆ x1 ln ∆ x2
X −0 Driving force at top stream = ( 2 )
: Driving force at bottom stream =
¿
( X 1 −X 1 )
X ¿1 : Concentration in organic phase that would be balance (equilibrium) with concentration
Y1
in aqueous phase.
5.0 Apparatus and Materials:
0.1M Sodium hydroxide (NaOH)
0.025M Sodium hydroxide (NaOH)
Liquid-liquid extraction column
Phenolphthalein
Funnel
Pipette
250ml Conical flask
Water pump
Propionic acid
=
Distillates product solution
Raffinate product solution
Feed solution
Volumetric flask
250ml Measuring cylinder
Retort stand
Stopwatch
6.0 Procedure Experiment A: Distribution coefficient, k 1) 50 ml organic solvent and 50ml of de-mineralised water was mixed in the conical flask. 2) 5ml of propionic acid was added where 5ml can be pipetted into the flask using a pipette with a rubber bulb. 3) A stopper was placed into the flask and shake for a minimum 5minutes. 4) Then, it poured into separating funnel and left for 5 minutes. The lower aqueous layer was removed. 5) 10ml sample of this layer was taken and titrated against 0.1M sodium hydroxide solution using phenolphthalein as indicator. 6) The experiment was repeated for other concentrations of propionic acid. Experiment B: The mass transfer coefficient 1) 100ml of propionic acid was added to 10 litres of organic phase. Then, it has been mixed well to ensure even concentration fill the organic phase feed tank (bottom tank) with the mixture. 2) The level control was switched to bottom of column (electrode switch S2).
3) The water feed tank was filled with 15 litres of clean de-mineralised water. Then, the water feed pump was started and the column was filled with water at high flow rate. 4) The flow rate was reduced to 0.2 l/min as soon as water is above the top of the packing. 5) The metering pump was then started and set at flow rate of 0.2 l/min. 6) It has been run for 20 minutes until steady conditions achieved. Flow rate has been monitored by our group during this period to make sure they remain constant. 7) 200ml samples from the feed (V13), raffinate (V11), and extract (V1) streams were taken. 8) 15ml of each sample was titrating against 0.1M NaOH using 3 drops of phenolphthalein as indicator. Time for sample to turn light pink was taken by using stopwatch. Titration was done three times so that we can take the average time. 9) Step 8 was repeated by using 0.025M NaOH
7.0 Results Table 7.1: Result value for Concentration of 0.1 M NaOH and 0.025 M NaOH Concentration of NaOH (M)
No of Titration
Raffinate (ml)
Feed (ml)
Extract (ml)
1.0
11.3
4.2
1.0
12.5
4.4
0.6
10.9
4.3
0.87
11.57
4.3
4.0
38.9
16.1
3.0
39.3
16.1
0.8
41.4
16.0
2.6
39.87
16.03
1st 2nd 3rd 0.1 M NaOH Average 1st 2nd 3rd 0.025 M NaOH Average
8.0 Calculations Formula; I.
To find the Distibution Coefficient, K
K=
Concentration of solute ∈ Extract ,Y Concentration of solute ∈Raffinate , X
II.
To find the Mass Transfer Coefficient,
M 1 V 1=M 2 V 2 Where; M1 – Concentration of NaOH M2 – Concentration of propionic acid V1 – Volume of NaOH V2 – Volume of propionic acid Rate of Acid Transfer =V W (Y 1−0) V 0 ( X 1− X 2 )=V W (Y 1−0) 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-X*1) Packing dimension; Length = 1.2 Diameter = 50 mm Volume of Packing, V= πr2L = (π) (0.025 m)2(1.2m) = 2.36×10-3 m3 = 2.36 L
MassTransfer Coefficient=
Rate of Acid Transfer Volume of Packing× Mean Driving Force
A. To calculate the Distribution Coefficient, K For 0.1M of NaOH at Extract phase, Y; M 1 V 1=M 2 V 2
(0.1
mol ) ( 0.0043 L ) =(M 2)( 0.015 L) L
M 1=0.02867
mol L
For 0.1M of NaOH at Raffinate phase, X; M 1 V 1=M 2 V 2
(0.1
mol ) ( 0.00087 L )=( M 2 )(0.015 L) L
M 1=0.0058
Kd=
¿
mol L
X Y
0.02867 =4.94 0.0058
For 0.025M NaOH at Extract phase, Y; M 1 V 1=M 2 V 2
(0.025
mol ) ( 0.0163 L )=(M 2 )(0.015 L) L
M 1=0.02717
mol L
For 0.025M NaOH at Raffinate phase, X; M 1 V 1=M 2 V 2
(0.025
mol ) ( 0.0026 L )=( M 2 )(0.015 L) L
M 1=0.00433
Kd=
¿
mol L
Y X
0.02717 =6.28 0.00433
B. To calculate the Mass transfer coefficient, For 0.1M NaOH Raffinate; M 1 V 1=M 2 V 2
(0.1
mol ) ( 0.00087 L )=( M 2 )(0.015 L) L
M 1=0.0058
mol L ……………………………….X
Feed; M 1 V 1=M 2 V 2
(0.1
mol ) ( 0.01157 L )=( M 2)(0.015 L) L
M 2=0.07713
mol L
Extract; M 1 V 1=M 2 V 2
(0.1
mol ) ( 0.0043 L ) =(M 2)( 0.015 L) L
M 1=0.02867
mol L …………………………………Y
Rate of Acid Transfer =V W (Y 1−0) ¿(0.3
L mol )(0.02867 ) min L
¿ 0.00860
mol min
V 0 ( X 1− X 2 )=V W (Y 1−0) ) (0.3 minL )(0.0058 molL − X )=(0.00860 mol min 2
X 2=0.02287
mol L Log mean driving force= (ΔX1-ΔX2) / ln (ΔX1/ΔX2)
∆ X 1=( X 2 −0) = 0.02287 M K=
Y1 ¿ X
¿ The value of K ∈experiment A=4.94 X ¿=
0.02287 4.94
¿ 0.004630 M
∆ X 2=( X 1 −X ¿1 ) = (0.0058 – 0.004630) M = 0.00117 M log Mean Driving Force=
0.02287−0.00117 0.02287 ¿ 0.00117
¿ 0.00730
MassTransfer Coefficient=
mol min ¿ 2.36 L ×0.00730 0.00860
¿ 0.499
M min
¿ 0.499
kg min
For 0.025 M NaOH Raffinate;
Rate of Acid Transfer Volume of Packing× Mean Driving Force
M 1 V 1=M 2 V 2
(0.025
mol ) ( 0.0026 L )=( M 2 )(0.015 L) L
M 1=0.00433
mol L ………………………………X
Feed;
M 1 V 1=M 2 V 2
(0.025
mol ) ( 0.03987 L )=( M 2 )(0.015 L) L
M 1=0.06645
mol L
Extract; M 1 V 1=M 2 V 2
(0.025
mol ) ( 0.0163 L )=(M 2 )(0.015 L) L
M 1=0.02717
Rate of Acid Transfer =V W (Y 1−0) ¿(0.3
L mol )(0.02717 ) min L
¿ 0.008151
mol min
V 0 ( X 1− X 2 )=V W (Y 1−0)
mol L …………………………….Y
) (0.3 minL )(0.00433 molL − X )=(0.008151 mol min 2
X 2=0.15937
mol L Log mean driving force= (ΔX1-ΔX2) / ln (ΔX1/ΔX2)
∆ X 1=( X 2 −0) = 0. 15937M K=
Y1 X¿ ¿ The value of K ∈experiment A=6.28
¿
X=
0.02717 6.28
¿ 0.004326 M ∆ X 2=( X 1 −X ¿1 ) = (0.00433 – 0.004326) M = 0.000006 M log Mean Driving Force=
0.15937−0.000006 0.15937 ¿ 0.000006
¿ 0.01564 MassTransfer Coefficient=
mol min ¿ 2.36 L ×0.01564 0.008151
¿ 0.221
M min
Rate of Acid Transfer Volume of Packing× Mean Driving Force
¿ 0.221
kg min
9.0 Discussion The experiment liquid-liquid extraction was conducted in the purpose of determining the coefficient distribution, K as well as the mass transfer coefficient. For that, the experiment was divided into two parts with each carrying a different objective. The extraction method that was used is based on the relative principle of solubility where a solution can be separated and extracted into its individual component.
The first experiment was carried out to determine the distribution coefficient of the upper layers (Y) and bottom layers (X). We run the experiment by using LLE equipment for 20 minutes, and the sample is taken out from the valve V11 and V1. The samples that we used are propionic acid and water. From the sample we take 15 ml of propionic acid-water as used to titration method with 0.1 M of NaOH and 0.025 M of NaOH. The calculation of the distribution coefficient for 1.0 M of NaOH is 4.94 while the value of distribution coefficient for 0.025 M of NaOH is 6.28. From the results we can conclude that the value of the distribution coefficient increases when concentration of NaOH increase due to the volume of the titration small because high concentration will give the faster effect on the solution to change pinkish-purple colour. The second experiment was conducted in order to determine the mass transfer coefficient. We run the experiment of liquid-liquid extraction to get the feed, raffinate and extract solution. The sample then we titrate with NaOH solution with 0.1 M and 0.025 M of concentration. For 0.1 M of NaOH, we get the value of the mass transfer coefficient is 0.499 kg/min while for 0.025 M concentration of NaOH is 0.221 kg/min. The results that we get shown that increases the mass transfer coefficient as the concentration of NaOH increase. The results that we get is correct due to actual data that the increase the mass transfer coefficient when concentration of NaOH increase. In this experiment, we used different concentration of Sodium hydroxide (NaOH) to titrate the sample. Concentrations used were 0.1M and 0.025M. For experiment A, the distribution coefficient for 0.1M NaOH is
4.94
while for 0.025M NaOH is 6.28 . From
the result, we can see that it is differ. If we used 0.1M NaOH, average reading for raffinate is 0.87ml while if we used 0.025M the average reading for raffinate is 2.6ml. This same goes to feed and extract where reading for concentration 0.025M is greater. The mass transfer coefficient value when titrated with 0.1M NaOH is 0.499 M/min while the mass transfer coefficient value when titrated with 0.025M NaOH is 0.22083 M/min. Therefore, we can say that as concentration of NaOH decrease, mass transfer rate will decrease because mass transfer rate will definitely different as NaOH concentration increase. During the titration, it was repeated at least for three times for each sample because, to obtain the most accurate readings among the three results. If the titration was done two times, we cannot determine which of these results are outlier meanings that the one that is extremely
high or low reading. By repeating it three times and consider the mean as the final result would be probably gives the most accurate reading and this was done in order to minimize the systematic error while taking the reading. However, there are some precautions that we have to be careful in order to get the accurate results. Firstly, the eye position must be straight to the scale and should be perpendicular to the meniscus during taking the readings of the volume at the burettes. Secondly, we have to performed the titration process in lamina hood to prevent the oil emission and impurities might be enters into the conical flask, beakers or burettes. Then, the readings the titration should be taken at least 3 times to get the average values therefore we can get the accurate results and avoid the mistake during taking the readings of the volume at burette apparatus. For experiment to determine the distribution coefficient for the system of organic solvent propionic acid-water, 100 ml from the feed, extract and raffinate was obtained. From each sample, 15 ml was titrated with 0.1M NaOH and 0.025M NaOH using a burette. Any changes in the samples were determined by a change in colour due to adding an indicator known as phenolphthalein. When the samples were titrated with 0.1M NaOH, for the feed, it took 11.57 ml of NaOH with a total concentration of
0.07713 kg/L to cause a change of colour in the
sample. The same can be said when 0.1M NaOH was titrated into 15 ml of extract and raffinate. For the extract the volume needed was 4.3 ml with a concentration of
0.02867
kg/L. While for the raffinate, it took a volume of 0.87 ml with a concentration
0.0058
kg/L. Although the values are correct in a sense that the volume and concentration needed to cause a colour change in the extract being much higher than needed by the raffinate, theoretically, the numbers are still wrong due to the time taken before titrating each sample with NaOH was more than 20 minutes. When the samples were titrated with 0.025M NaOH, for the feed, it took 39.87 ml of NaOH with a total concentration of
0.06645 kg/L to cause
a change of colour in the sample. For the extract the volume needed was 16.03 ml with a concentration of concentration
0.02717 kg/L. While for the raffinate, it took a volume of 2.6 ml with a
0.00433 kg/L. The values are correct due to the volume and concentration
needed to cause a colour change in the extract being much higher than needed by the raffinate. From the data collected, when titrating the feed, extract and raffinate with 0.1M NaOH, it was determined that the volume of propanoic acid extracted from the aqueous phase to be
0.00860 mole/min and the mass transfer coefficient was determined to be a value of
0.499 M/min. When titrating the feed, extract and raffinate with 0.025M NaOH, it was
determined that the volume of propanoic acid extracted from the aqueous phase to be 0.008151 mole/min and the mass transfer coefficient was determined to be a value of 0.22083 M/min. A quantitative measure of the how an organic compound will distribute
between aqueous and organic phases is called the distribution or partition coefficient. It is the ratio, K, of the solubility of solute dissolved in the organic layer to the solubility of material dissolved in the aqueous layer. The constant K, is essentially the ratio of the concentrations of the solute in the two different solvents once the system reaches equilibrium. At equilibrium the molecules naturally distribute themselves in the solvent where they are more soluble. Inorganic and water soluble materials will stay in the water layer and more organic molecules will remain in the organic layer. By using thecorrect solvent system, a molecule can be specifically selected and extracted from another solvent. Solvent extraction is unlikely recrystallization and distillation because it does not give the outcome of a pure product. Basically, it is based on the principle of equilibrium dispersion of solute in between two immiscible liquid in which one of it is known as solvent. Nevertheless, the solvent does not need to be in pure state but it can be a mixture in order to form soluble substance in the solution. Most commonly, the solvent is a volatile organic liquid that can be eliminated by the means of evaporation after extraction of desired product. In fact, if the substance is insoluble in two immiscible liquids where it can be homogenized even though have difference in density but this can be overcome by shaking method. Propionic acid is under carboxylic acid group hence it soluble in water. Propionic acid is way less volatile than diethyl ether due to higher boiling point. As mentioned earlier, high volatility means low boiling point hence diethyl ether is very suitable for liquid-liquid extraction rather than propionic acid. However, propionic acid is also flammable hence safety precaution need to be alerted. It also can cause irritation to skin, eyes and respiratory tract if there is direct contact
or overexposed occur. Besides that, it also harmful to aquatic life hence it is not allowed to dispose waste propionic acid in sink but instead must be properly dispose in waste bottle.
10.0
Conclusion
As a conclusion, there are few factor that affect the result which is the concentration of the titrant that indicate the composition of the substances at raffinate, feed and extract. Based on the result, for the concentration of NaOH 0.1M, the highest reading goes to feed sample and the lowest reading was from raffinate sample. Meanwhile, for concentration of NaOH 0.025M, the highest and lowest reading same as the previous concentration but higher which is for feed sample the average reading is 39.87ml and the reading for raffinate is 2.6ml. It might be some error during taking the readings and calculation that may be affect the result. 11.0 Recommendations 1 Make sure to withdraw samples for the feed and extract after 20 minutes of withdrawing 2
sample from the feed. This is to ensure that the volume of titrate used to be accurate. Avoid parallax error to ensure accurate reading by making sure the eyes are parallel to
3
the scale of the instruments. Make sure to wear gloves and a pair of goggles when handling chemical substances. This
4
is to avoid injury. Make sure to use an Erlenmeyer flask during titration so that we can properly mixed the
5
two solutions by shaking the flask. Read and understanding the procedure written in laboratory manual in order to prevent any error while handling the experiment and equipment.
12.0 Reference 1. Google image.
General
Extraction
Column.
Retrieved
from
http://www.aiche.org/resources/publications/cep/2015/november/design principles-liquid-liquid-extraction 2. G.W. Stevens, T.C., Lo, & M. H. I. Baird, 2007, "Extraction, liquid–liquid", Retrieved from
http://chem.libretexts.org/Core/Analytical_Chemistry/Lab_Techniques/Liqui
-Liquid_Extraction 3. Laboratory manual, (n.d). Laboratory manual: Liquid-Liquid Extraction. Faculty of Chemical Engineering and Bioprocess. UiTM Shah Alam.
4. Dutta, B. K. (2007). Principles of mass transfer and seperation processes. . PHI Learning Pvt. Ltd.. 5. Frank, T. C. (2008). Frank, T. C., Dahuron, L., Holden, B. S., Prince, W. D., Seibert, A. F., & Wilson, L. C. (2008). Liquid-liquid extraction and other liquid-liquid operations and equipment. Perry’s Chemical Engineering Handbook. 6. Müller, E. B. (1985). Liquid–liquid extraction. . Ullmann's Encyclopedia of Industrial Chemistry. 7. Rydberg, J. (. (2004). Solvent extraction principles and practice, revised and expanded. . CRC Press. 8. Sinnott, R. K. (2009). Sinnott, R. K. (2009). Chemical engineering design: SI Edition. . Elsevier. 9. Weatherley, L. R. (2013). Weatherley, L. R. (Ed.). (2013). Engineering processes for . . Elsevier.
13.0
Appendices
Appendix A
Appendix B
Figure 13.1: Liquid-liquid extraction column
