LLE

1. Abstract

There are two different parts in this experiment but related to each other. The first objectives was to determine the distribution coefficient for the system organic solvent (Propionic Acid-Water) and to show its dependence on concentration. Another objective is to 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. In the first experiment, we use separating funnel to separate two solutions of different solubility and density. By titrating with 0.1M NaOH, the values of K are 3.58 for 1.0mL of propionic acid, 3.66 for 2mL of propionic acid and 2.95 for 3.0mL of propionic acid. The value of K for 0.025 NaOH are 4.97 for 1.0mL of propionic acid, 3.06 for 2mL of propionic acid and 2.99 for 3.0mL of propionic acid. In the second experiment, we want to perform a mass balance and calculate the mass transfer coefficient. This experiment must be taken seriously and some recommendation must be applied to get the best result. This is because from the results, a long calculation must be performed and this calculation is related to each other.

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2. Introduction Extraction is a process that separates components based upon chemical differences rather than differences in physical properties. 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.

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3. Aims/Objectives

In Experiment A, the objectives are to determine the distribution coefficient for the system organic solvent-Propionic acid-water and to show its dependence on concentration. For the Experiment B, it is conducted to 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.

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4. Theory Experiment A: Determination of distribution coefficient of solution (organic solvent + Propionic acid) with water and its dependence on concentration.

For this experiment, by mixing two immiscible liquid of solution (organic solvent + propionic acid) and water and then it is allowed to separate into extract and raffinate phase. The extract phase would be the water plus the propionic acid, while the raffinate phase would be the solution of organic solvent and residual of propionic acid.

Figure 4.1 250mL Separating funnel showing upper layer and lower layer for raffinate and extract phase.

The distribution coefficient, K, can be calculated using the following equation: (Equation 4.1)

It can be assume that the equilibrium state exist between these two phases.

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At a low concentrations, the distribution coefficient, K, is dependent on the concentration and thus Y = KX (equation 4.2)

Experiment B: Demonstrating a mass balance in the extraction column and measuring the mass coefficient, Kc with aqueous phase as the continuous medium.

For this experiment, we used the extraction column to extract the acid from an organic solvent and aqueous phase as continuous medium.

Figure 4.2 Liquid-liquid extraction equipment set-up. Let

Subscripts:

Vw

=

Water flow rate (L/s)

V0

=

organic solvent flow rate (L/s)

X

=

Propionic acid concentration in the organic phase (Kg/L)

Y

=

Propionic acid concentration in the aqueous phase (Kg/L)

1

=

Top of the column

2

=

Bottom of the column

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So, the Mass balance for this experiment can be shown as: (Equation 4.3) (Equation 4.4)

Therefore,

=

(Equation 4.5)

In calculation of the Extraction efficiency: Mass transfer coefficient, Kc (based on the raffinate phase) =

(equation 4.5)

Log mean driving force =

Where,

(Equation 4.6)

∆x1

= Driving force at the top of the column

= (X2 – 0)

∆x2

= Driving the at the bottom of the column = (X1 – X1*)

Where X1* is the concentration in the organic phase. The equilibrium values can be found using the distribution coefficient found in the first experiment.

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5. Apparatus



250 ml Conical stoppered flask



250 ml measuring cylinder



250 ml Separating funnel



Pipette with rubber bulb



Sodium Hydroxide Solution (0.1M)



Phenolphthalein



Propionic acid



Liquid/liquid extraction unit

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6. Procedure

EXPERIMENT A 1. A mixture of 50ml organic solvent and 50ml of de-mineralized water is made up in a conical flask. 2. 5ml of propionic acid was added. 3. A stopper was placed into the flask and was shaked for a minimum of 5 minutes. 4. Mixture was poured into a separating funnel and left for 5 minutes and the lower aqueous layer was removed. 5. 10 ml of sample of this layer was taken and titrated against 0.1M sodium hydroxide solution using phenolphthalein as indicator. 6. The experiment was repeated for two different concentrations of propionic acid.

EXPERIMENT B 1. 100ml of propionic acid was added to 10 liters of the organic phase. It was mixed well to ensure an even concentration then the organic phase tank was filled with the mixture. 2. The level control was switched to the bottom of the column. (electrode switch S2) 3. The water tank was filled with 15 liters of clean de-mineralized water, the water feed pump was stared and the column was filled with water at a high flow rate. 4. As soon as the water was above the top of the packing, the flow rate was reduced to 0.21 per min. 5. The metering pump was started and the flow rate was set at 0.21 per min. 6. It was runned for 15-20 minutes until steady conditions are achieved; flow rates are monitored during this period to ensure that they remained constant. 7. 15ml samples were taken from the feed, raffinate and extract streams. 8. 10ml of each sample was titrated against 0.1M NaOH using phenolphthalein as the indicator.

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7. Results

Experiment A: Determination of distribution coefficient of solution (organic solvent + Propionic acid) with water and its dependence on concentration.

Propionic acid added (mL)

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

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

K=

5

44

0.44

77

0.77

0.571

8

82

0.82

140

1.40

0.587

11

107

1.07

196

1.96

0.550

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Experiment B: Demonstrating a mass balance in the extraction column and measuring the mass coefficient, Kc with aqueous phase as the continuous medium.

Flow rate of aqueous phase (L/min) Flow rate of organic phase (L/min)

0.2

0.2

0.1

0.025

Concentration of Propionic acid (kg/L)

(mL)

14

51

0.14

Raffinate (mL)

6

13

0.06

Extract (mL) Propionic acid extracted from the organic phase (mol/min) Propionic acid extracted from the aqueous phase (mol/min) Mass transfer coefficient (kg/min)

5.5

20

0.055

Sodium hydroxide concentration (M) Feed

0.0165

0.0165

0.444

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8. Sample of Calculations

Finding the distribution coefficient

Where Y: concentration of solute in extract phase. X: concentration of solute in raffinate phase. For 0.1 M NaOH Volume of propionic acid (mL)

Upper

bottom

5.0 (0.1)(0.044)=

(0.005)

=0.88 M

(0.1)(0.077)=

(0.005)

0.57

=1.54 M

8.0 (0.1)(0.082)=

(0.008)

=1.025 M

(0.1)(0.14)=

(0.008)

0.59

=1.75 M

11.0 (0.1)(0.107)= =0.973 M

(0.011)

(0.1)(0.196)=

(0.011)

0.55

=1.78 M

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Finding the mass transfer coefficient For 0.1 M of NaOH Raffinate (0.1)(0.006)=

(0.01)

=0.06 M Feed (0.1)(0.014)=

(0.01)

=0.14 M Extract (0.1)(0.0055)=

(0.01)

=0.055 M

Rate of acid transfer = Vw (Y1- 0) = 0.3 L/min (0.055mol/L) = 0.0165 mol/min

Vo (X1-X2) = Vw (Y1 -0) 0.3 L/min (0.06 mol/min – X2) = 0.0165 mol /min X2 = 0.005 M

Log mean driving force= (ΔX1-ΔX2) / ln (ΔX1/ΔX2) ΔX1 = (X2-0) = 0.005 M

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K = Y1/X* at equilibrium, assume R = 0.57 (from experiment B) X* = Y1/K = 0.055/0.57 =0.09649 M

ΔX2= (X1-X*1) = (0.06 M -0.09649 M) =- 0.03649M

Log mean driving force= 0.005 - 0.03649 / ln (0.005 /0.03649) = 0.01575

Mass transfer coefficient =

=

= 0.444mol/L.min = 0.444 M/min = 0.444 kg/min

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For 0.025 M of NaOH Raffinate (0.025)(0.013)=

(0.01)

=0.0325 M Feed (0.025)(0.051)=

(0.01)

=0.1122 M Extract (0.025)(0.02)=

(0.01)

=0.05 M

Rate of acid transfer = Vw (Y1- 0) = 0.3 L/min (0.05mol/L) = 0.015 mol/min

Vo (X1-X2) = Vw (Y1 -0) 0.3 L/min (0.0325 mol/min – X2) = 0.015 mol /min X2 = 0.0175 M

Log mean driving force= (ΔX1-ΔX2) / ln (ΔX1/ΔX2) ΔX1 = (X2-0) = 0.0175 M

K = Y1/X* at equilibrium, assume R = 4.97 (from experiment B) X* = Y1/K = 0.05/4.97 =0.01006 M

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ΔX2= (X1-X*1) = (0.05 M -0.01006 M) = 0.03994M

Log mean driving force= (0.0175 – 0.03994) / ln (0.0175 /0.03994) = 0.0176

Mass transfer coefficient =

=

= 0.357 mol/L.min = 0.357 M/min = 0.357 kg/min

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9. Discussions In this experiment, we want to identify the distribution of coefficient and to identify the mass transfer coefficient. This experiment is referred to the solubility which is the solvent is also soluble with a specific solute contained in the solution and also about the separation of a substance from a mixture by preferentially dissolving the substance in a suitable solvent.

For the first experiment, which is to determine the distribution coefficient, we used the titration method from the upper (Y) and bottom (X) layer sample. By titration with 0.1 M, the values are 0.571 in 5.0 ml of propionic acid, 0.587 in 8.0 ml of propionic acid and 0.550 in 11.0 ml of propionic acid. From the result in this experiment, if titrated with 0.1 M NaOH, it shows that the value of distribution coefficient decrease as the volume of propionic acid increase. This experiment is totally failed because there were increment in distribution coefficient at 8mL of propanoic acid. This is because of the error that maybe occur during the experiment and this will discussed later.

For the second experiment, which is to determine the mass transfer coefficient, we used the liquid-liquid extraction column to get the feed, raffinate and extract solution to be the samples. The samples then are titrated with 0.1 M NaOH and 0.025 M NaOH. The value of mass transfer coefficient when titrated with 0.1 M NaOH is 0.444 kg/min while titrated with 0.025 M, the value of mass transfer coefficient is 0.357 kg/min. For the actual result, this experiment can shows the increment of mass transfer rate when the concentration of NaOH is decreasing. But from our result,the mass transfer coefficient is decrease as the concentration of NaOH decreasing. These part of experiment is conclude to be failed because does not follows the actual result.

Several error occurred during this experiment progress that totally effect the result of first experiment which is for finding the mass transfer coefficient. The most common error is the position of the eye during taking the volume value at the burette. As the solution, the eye position should be straight to the scale and must be perpendicular to the meniscus. Besides that, the error also occurred while using the apparatus that is not properly clean. We should used the clean apparatus to avoid the oil emission and impurities at the beaker, conical flask or burette. If the apparatus not clean, we should used the distilled water to clean up those apparatus.

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10. Conclusions

In the first experiment, when the propionic acid added is 5ml, the concentration of propionic acid in aqueous layer is 0.44 M whereas the concentration of propionic acid in organic layer is 0.77M. Its distribution coefficient calculated is 0.571. At 3ml propionic acid, the propionic acid concentration in aqueous layer is 0.82M while the propionic acid concentration in organic layer is 1.4M. When 1ml of propionic acid is added, after been titrate, the concentration in aqueous layer is 1.07M and the concentration in organic layer is 1.96M which results to 0.550 of distribution coefficient. For the second experiment, the flow rate of aqueous and organic phase is the same,0.2 L/min. The concentration of propionic acid from the feed is 0.14 kg/L. Sample from the raffinate that has been titrated gives the concentration of propionic acid with 0.06 kg/L. Beside that, the concentration of propionic acid from the extract sample is 0.055 kg/L. Based on the result that we get, the propionic acid extracted from the organic phase is 0.0165M in concentration whereas the propionic acid extracted from the aqueous phase is 0.0165M. After being calculated, the mass transfer coefficient is 0.444. Therefore, by successfully determining the distribution coefficient and the mass transfer coefficient, it can be safely said that the objectives had been fulfilled.

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11. Recommendations

There are several recommendations and precautions that should be taken seriously to get the best result for this experiment. One of the recommendations is that all the apparatus that want to be used must be cleaned and dry properly because it will affect the rate of titration. Another precaution is the experiment must be repeated at least three times to get the best result. Furthermore, time taken must be taken seriously because it will affect the result. On top of that, eye must be parallax to the reading meter to avoid parallax error. Last but not least, during titration, the change of colour must be taken seriously.

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12. Reference  



Finchsigmate K. (2010), ‘Liquid-Liquid Extraction’, retrieved 17th April 2013 from http://chemwiki.ucdavis.edu/VV_Lab_Techniques/Liquid-Liquid_Extraction Wikipedia, (n.d.) , ‘ Types of aqueous two phase extractions’, retrieved 17th April 2013 from http://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction#Types_of_aqueous _two-phase_extractions Perry, R.H., and D. Green, Perry’s Chemical Engineering Handbook, 6th edition, McGraw-Hill, 1984.



Bennett, C.O., and J.E. Myers, Momentum, Heat and Mass Transfer, 3rd edition, McGraw-Hill, 1983.



n.d, retrieved 16th April 2013 from http://rothfus.cheme.cmu.edu/uolab/lle/projects/t8_f00/t8_f00.pdf

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13. Appendices

Figure 13.1: Liquid-Liquid Extraction Continous column equipment.

Figure 13.2: bubble effect when the extraction started in the column. 20

Figure 13.3: Results for the experiment.

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