LAB LLE

CHE 523 CHEMICAL ENGINEERING LABORATORY 2

MEMBERS NAME :

ABSTRACT Based

on

experiment objectives, which is to conduct the simple experiments

regarding liquid-liquid extraction and to determine the distribution of coefficient and the distribution for the system organic solvent-propanoic acid-water. 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 4.16 in 1.0 ml of propionic acid, 0.56 in 3.0 ml propionic acid and 0.35in 5.0 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; 0.0025 kg/min if titrated with 0.1M NaOH and 0.00193 kg/min if titrated with 0.025M NaOH. The experiment was completely conducted and successfully done.

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 anorganic solvent(propionic acid) as shown at the diamgram above. It is an

extraction of a substance from one liquid phase into another liquid phase. Liquidliquid 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 thework-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 solventoccurs. 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 solventfeed 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

OBJECTIVE There are several objectives for this liquid-liquid extraction experiment which are:  To determine the distribution coefficient for the system organic solvent-Propionic acidWater and to show its dependence on concentration.  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. THEORY Extraction is a process that used to separate component. This process is based on the chemical differences rather than differences in the properties. Liquid-liquid extraction is Liquid-liquid extraction is based on the transfer of a solute substance from one liquid phase into another liquid phase according to the solubility. Extraction becomes a very useful tool if you choose a suitable extraction solvent.You can use extraction to separate a substance selectively from a mixture, or to remove unwanted impurities from a solution Liquid-liquid extraction process has it own basic principle which involve the contracting of solution with another solvent that is immiscible with the original. The theory is well developed for the prediction of liquid-liquid extraction column operations. When liquid-liquid extrection is performed, different phases form the continuous phase and the dispersed phase. When an experiment is performed, the column will be first filled with water. This is the continuous phase. The success of this method depends upon the difference in solubility of a compound in various solvents. For a given compound, solubility differences between solvents is quantified as the "distribution coefficient. When a compound is shaken in a separatory funnel with two immiscible solvents, the compound will distribute itself between the two solvents. Normally one solvent is water and the other solvent is a water-immiscible organic solvent. Most organic compounds are more soluble in organic solvents, while some organic compounds are more soluble in water.

In this process when solvent (water) is mix together with a solution (organic solvent/propanoic acid) and then allow them to separate into the extracted phace and raffinate phase. The extract phase wil be the water and propanoic acid and the raffinate, organic solvent with a trace of propanoic acid. In dilute solutions at equilibrium, the concentration of the solute in the two phases is called the distribution coefficient or distribution constant, K, which can be defined as the ratio of :

K=

concentration of solute ∈the extract phase ,Y concentrationof solute∈the raffinate phase , X

The theory for the system Trichloroethylene-Propionic acid-Water is as follows: Let

Vw : Water flow rate ( L/s ) Vo : Organic solvent, Trichloroethylene flow rate ( L/s ) X : Propionic acid concentration in the organic phase ( kg/L ) Y : Propionic acid concentration in the aqueous phase ( kg/L )

Subscripts:

1 2

= Top of column

= Bottom of column

1. Mass Balance: Propionic acid extracted from the organic phase (raffinate). = V0 (X1-X2) Propionic acid extracted by the aqueous phase (extract) = Vw (Y1-0)

Therefore, V0 (X1-X2) = Vw (Y1-0)

2. Extraction Efficiency Mass transfer coefficient ( based on the raffinate phase ) =

Rate of acid transfer Volume of packing × Meandriving force

Log mean driving force =

ΔX 1−ΔX 2 ΔX 1 ln ΔX 2

Where ΔX1 : Driving force at the top of the column = (X2-0) ΔX2 : Driving force at the bottom of the column = (X1-X1*)

Where X1* is the concentration in the organic phase which would be in equilibrium with concentration Y1 in the aqueous phase. The equilibrium values can be found using the distribution coefficient found in the first experiment.

MATERIALS AND APPARATUS  Liquid- liquid extraction unit  Conical flask with stopper  Measuring cylinder  Separating funnel  Burette  Conical flask  Beaker  0.1 M sodium hydroxide solution  0.025 M sodium hydroxide solution  Phenolphthalein  Propionic acid  Trichloroethylene  De-mineralised water

PROCEDURE PART 1 1.

50 ml organic solvent and 50 ml of de-mineralised water was mixed in a conical flask.

2.

5 ml of propionic acid is added into the flask by using dropper.

3.

A stopper was placed on the flask and the flask was shaked for 5 minutes.

4.

The mixture was poured into a separating funnel,the mixture were leaved for 5 minutes.

5.

10 ml sample of the bottom layer was taken and 3 drop phenolphthalein was added in the sample and the sample was titrated against 0.1M sodium hydroxide solution.

6.

The remaining bottom layer was removed until we can take the upper layer sample.

7.

10 ml sample of the upper layer was taken and 3 drop of phenolphthalein was added in the sample.

8.

Then, the sample was titrated against 0.1M sodium hydroxide until light pink solution form.

9.

The experiment is repeated for two further concentrations of propionic acid which are 3 ml and 1 ml of propionic acid.

PART 2 1.

100 ml of propionic acid is added to 10 litres of the organic phase. The mixture is mixed well to ensure an even concentration then organic phase feed tank (bottom tank) is filled with the mixture.

2.

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

3.

The water feed tank is filled with 15 litres of clean de-mineralised water, the water feed pump is started and the column is filled with water at a high flow rate.

4.

As soon as the water is above the top of the packing, the flow rate is reduced to 0.21/min.

5.

The metering pump is started and the flow rate is set at 0.21/min.

6.

The experiment is run for 20 minutes until steady conditions are achieved, the flow rates are monitored during this period to ensure that they remain constant.

7.

Sample from feed, raffinate and extract stream was taken 100ml each and the liquid-

8.

liquid extraction column was shut down. Then, 15ml of each sample was taken and 3 drop phenolphthalein was added in each

9.

sample. Each sample was titrated against 0.1M sodium hydroxide solution until light pink

solution formed. 10. Step 8 was repeated and the sample was titrated against 0.025M sodium hydroxide solution until light pink solution formed.

RESULTS

1. Experiment A Aqueous layer (Y) Propionic acid added (ml)

Organic layer (X) K=

Y X

Titre of M/10 Naoh (mL)

Concentration, M (mol/L)

Titre of M/10 Naoh (mL)

Concentration, M (mol/L)

1

10.4

1.04

25.1

0.25

4.16

3

24.5

0.82

43.8

1.46

0.56

5

31.5

0.63

89.0

1.78

0.35

2. Experiment B Flowrate of aqueous phase (L/min) Flowrate of organic phase (L/min) Sodium hydroxide concentration (M) Feed (mL)

0.2 0.2 0.1

0.025

20

72.6

Raffinate (mL)

0.05

1.0

0.005

0.0025

Extract (mL)

6.3

22.5

0.063

0.056

Propionic acid extracted form aqueous phase for 0.1 M NaOH (mol/min) Mass transfer coefficient for 0.1 M NaOH (kg/min)

0.0126

0.0025

Concentration of propionic acid (kg/L) 0.20 0.0182

Propionic acid extracted form aqueous phase for 0.025 M NaOH (mol/min) Mass transfer coefficient for 0.025 M NaOH (kg/min)

0.0112

0.00193

CALCULATIONS Experiment A calculation Finding the distribution coefficient, K : - Titration with 0.1 M sodium hydroxide solution. 1. 1.0 mL of propionic acid added Aqueous layer (Y) : M 1 V 1=M 2 V 2 (0.1)(0.0104) = (M2)(0.001) M2 = 1.04 M

Organic layer (X) : M 1 V 1=M 2 V 2 (0.1)(0.0025) = (M2)(0.001) M2 = 0.25 M

K = Y/X = 1.04 M / 0.25 M = 4.16

2. 3.0 mL of propionic acid added Aqueous layer (Y) : M 1 V 1=M 2 V 2 (0.1)(0.0245) = (M2)(0.003) M2 = 0.82 M

K = Y/X = 0.82 M / 1.46 M = 0.56

Organic layer (X) : M 1 V 1=M 2 V 2 (0.1)(0.0438) = (M2)(0.003) M2 = 1.46 M

3. 5.0 mL of propionic acid added Aqueous layer (Y) : M 1 V 1=M 2 V 2 (0.1)(0.0315) = (M2)(0.005) M2 = 0.63 M

K = Y/X = 0.63 M / 1.78M = 0.35

Experiment B Calculation Finding the mass transfer coefficient : I.

0.1 M of NaOH Feed :

M 1 V 1=M 2 V 2 (0.1)(0.02) = (M2)(0.01) M2 = 0.20 M X1

Raffinate :

Organic layer (X) : M 1 V 1=M 2 V 2 (0.1)(0.089) = (M2)(0.005) M2 = 1.4 M

M 1 V 1=M 2 V 2 (0.1)(0.005) = (M2)(0.01) M2 = 0.005 M

Extract :

M 1 V 1=M 2 V 2 (0.1)(0.0063) = (M2)(0.01) M2 = 0.063 M Y1

Propionic acid extracted from the organic phase (raffinate) = Vo (X1 – X2) = 0.2 L/min (0.20 mol/L - X2) Propionic acid extracted by the aqueous phase (extract) : Rate acid transfer = Vw (Y1 – 0 ) = 0.2 L/min (0.063 mol/L) = 0.0126 mol/min Since Vo (X1 – X2) = Vw (Y1 – 0 ) 0.2 L/min (0.20 mol/L - X2) = 0.0126 mol/min X2 = 0.137 mol/L

Log mean driving force =

∆ x ₁−∆ x ₂ ∆x₁ ln ∆x₂

∆ x ₁ = Driving force at the top of the column = (X2 – 0) ∆ x₂

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

∆ x ₁ = 0.137 mol/L

By using the K from experiment A, assume the value is 4.16 X1* = Y1/K = 0.063 / 3.45 = 0.018 Thus , ∆ x₂

= (X1 – X1*)

= 0.20 – 0.018 = 0.182

Log mean driving force

= (0.137 – 0.182) / ln (0.137/0.182) = 0.158

Mass transfer coefficient

II.

0.025 M of NaOH Feed :

=

rate of acid transfer Volumes of packing ×mean driving force

= =

0.0126 mol/min 2.36 L × 0.158 0.0338 mol/L.min

=

0.0338 M/min x 0.07408 kg/mol

=

0.0025 kg/min

M 1 V 1=M 2 V 2 (0.025)(0.0726) = (M2)(0.01) M2 = 0.182 M X1

Raffinate :

M 1 V 1=M 2 V 2 (0.025)(0.001) = (M2)(0.01) M2 = 0.0025 M Extract :

M 1 V 1=M 2 V 2 (0.025)(0.0225) = (M2)(0.01) M2 = 0.056 M Y1

Propionic acid extracted from the organic phase (raffinate) = Vo (X1 – X2) = 0.2 L/min (0.182 mol/L - X2)

Propionic acid extracted by the aqueous phase (extract) : Rate acid transfer = Vw (Y1 – 0 ) = 0.2 L/min (0.056mol/L) = 0.0112 mol/min

Since Vo (X1 – X2) = Vw (Y1 – 0 ) 0.2 L/min (0.182 mol/L - X2) = 0.0112 mol/min X2 = 0.126 mol/L

Log mean driving force =

∆ x ₁−∆ x ₂ ∆x₁ ln ∆x₂

∆ x ₁ = Driving force at the top of the column = (X2 – 0) ∆ x₂

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

∆ x ₁ = 0.188 mol/L

By using the K from experiment A, assume the value is 4.16 X1* = Y1/K = 0.056 / 4.16 = 0.0135 Thus , ∆ x₂

= (X1 – X1*)

= 0.186 – 0.0135 = 0.173

Log mean driving force

= (0.186 – 0.173) / ln (0.186/0.173 = 0.18

Mass transfer coefficient

=

rate of acid transfer Volumes of packing ×mean driving force

= =

0.0112 mol/min 2.36 L × 0.18 0.026 mol/L.min

=

0.026 M/min x 0.07408 g/mol

=

0.00193 kg/min

DISCUSSION Extraction in chemistry is a separation process consisting in the separation of substance from a matrix. It may include liquid-liquid extraction or solid phase extraction. For our experiment, we just perform the Liquid-liquid extraction. Extraction process also separates components based upon chemical differences rather than differences in physical properties. The basic principle extraction involve the contacting of a solution with another solvent that is immiscible with the original. The two liquid phases must be either immiscible, or partially miscible. • usually isothermal and isobaric • can be done at low temperature (good for thermally fragile solutes, such as large organic molecules or biomolecules) • can be very difficult to achieve good contact between poorly miscible liquids (low stage efficiency) • extracting solvent is usually recycled, often by distillation (expensive and energy-intensive) • can be single stage (mixer-settler) or multistage (cascade) 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.

In our experiment have two parts, 1st finding the distribution coefficient,K . 2nd finding the mass transfer coefficient.

We are able to extract propionic acid (organic solvent) with water until separation occur into extract phase and the raffinate phase is produce then the distribution coefficient is determine. For the first part, to finding the distribution coefficient, the ratio of concentrations of a compound in a mixture of two immiscible phases at equilibrium is must be determine. These coefficients are measure of the difference in solubility/density of the compound in these two phases. K = y/x We are supplied with 1.0mL ,3.0mL and 5.0mL of volume of propionic acid and then titrate with 0.1 M sodium hydroxide. Three drops of phenolphthalein is added to the propionic acid before titration process is carry out. Then, the mixture must be shaken for the five minutes. Next, the mixture was poured into a separating funnel for titration process, two layers of solution can be observed. The two layers seen are aqueous on the lower part and organic on the top side. All the changes of the colour while titration process is observe. Noted that the colourless colour of phenolphthalein turns to light pink. At this moment, the product of reaction propionic acid and NaOH with adding phenolphthalein as indicator completely produce salt and water. Then, the water is used to further extraction. Basically, phenolphthalein is used as the common indicator in neutralization reaction. Water is then used as aqueous layer to propionic acid to perform extraction process. For the other tests results, it can be seen in the calculation and results section.

The formula;

M 1 V 1=M 2 V 2

Is used to identify the aqueous (y) and organic (x)concentration .Thus, the distribution coefficient can be determined.

For the second part, we are provided with 0.1 M NaOH and 0.025 M NaOH. The second part of the experiment was also a titration experiment to determine the mass transfer coefficient. This time the concentration of sodium hydroxide (NaOH) used is 0.1 M and 0.025 M. The solutions titrated in this experiment are taken from the feed, raffinate and extract. The volume

needed to titrate the feed with 0.1 M of NaOH was 3 mL. As for raffinate 0.2 mL and extract 7 mL. With the concentration of NaOH at 0.025 M, the volume needed to titrate the 15 mL feed was 79.5 mL. And as for the raffinate and extract, 40 mL and 25.5 mL was needed to titrate the NaOH solution respectively. The calculation of the mass transfer coefficient can be calculated with all the data obtained. The calculation of the mass transfer coefficient can be seen in the results and calculation section.

There is some partial list of physical properties is needed in liquid-liquid extraction separations. It is by no means complete, other properties will be needed for some of the calculations, and especially those needed to size the diameter of the column. It is however complete as it relates to the described theory. 

 



Temperature plays a smaller role in extraction than in other separation processes. It is only dependent upon the temperatures of the streams fed into the column. There is not a heating requirement for the process and DH of mixing is generally insignificant. For these reasons, extraction can be regarded as an isothermal process. Activity coefficients are the most important physical property in the extraction process. The reason for this is that these are used to determine the miscibility of the solute in both of the solvents involved. While there are many different equations available to determine a particular activity some are better than others for our purposes. When working with liquid-liquid systems the NRTL and the UNIFAC models are the most accurate in predicting the activities of the liquids involved. Although better than such predictive models such as Van Laar or Margules they still fall short of perfection. Once a predictive model has been plotted on a diagram it will most likely be necessary to fix the exact equilibrium line experimentally for the most accurate data. The activity coefficients also determine the partition factor which will determine whether or not a good separation is possible. Viscosity is a property that cannot be overlooked, its presence appears in two different areas, flooding and choice of equipment. Flooding is a phenomenon that can

occur in extraction just as it can for other unit operations we will encounter. Viscosity is also valuable in the determination of what type of system to use for extraction. Components having a high viscosity cannot be used in spray or packed columns. In a scientific research, there are always unknown variables that could disrupt us from obtaining the best result possible. In terms of making a reading of the measurements, there may have occurred an error. The error might have come from parallax error where the observer’s eyes were not perpendicular to the reading which caused a slight alteration in the reading by a few millilitres. The most important part of conducting an experiment is to understand the contents of the experimental procedure so that throughout applying the research

CONCLUSION For the conclusion in this experiment, we can conclude that the experiment is success generally as all the data needed for this experiment are obtain. For the first experiment we have to find the distribution coefficient, K and in this first experiment we have two type of layer which is aqueous layer (Y) and organic layer (X) and we need to titre both of the layer with M/10 Naoh (mL) in oder to get the concentration, M (mol/L) of each layer. Propanoic acid will be add both the layer. There are 1, 3, 5 ml of propanoic acid added respectively. For the 1 ml propanoic acid add for titration the distribution coefficient, K get is 4.16,. for the 3 ml propanoic acid add the distribution coefficient, K get is 0.56 and for 5 ml added the the distribution coefficient, K get is 0.35. For experiment 2 we have to find the mass transfer coefficient. The propionic acid extracted form aqueous phase for 0.1 M NaOH (mol/min) is 0.0126 and the mass transfer coefficient obtain is 0.0025. For the propionic acid extracted form aqueous phase for 0.025 M NaOH (mol/min) the value obtain is 0.0112 and the mass transfer coefficient for 0.025 M NaOH (kg/min) obtain is 0.00193.

RECOMMENDATION When conducting this experiment, there are several recommendations that will produce better observation which will not differ much from the theoretical observations.

Firstly, safety is very important when doing experiment. Thus, we need to wear laboratory coat, helmet and fully cover shoes to avoid any danger for safety precaution. Titration must take place in fume chamber and must be stop when the solution turns to light pink. Before conducting the experiment, we must ensure that all the apparatus are in good condition and follow all the procedures in order to get more accurate result. Next, when taking the reading of volume of sodium hydroxide solution, make sure that eyes is directly perpendicular with the level of sodium hydroxide solution inside the burette to avoid any parallax error.

REFERENCES   

http://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction –the concept of liquid-liquid extraction. Liquid-Liquid Extraction Equipment, Jack D. Law and Terry A. Todd, Idaho National Laboratory. UiTM’s chemical engineering lab manual.

APPENDIX