Objective:-

The purpose of this experiment is to understand the concept of the cross current extraction and to

determine the efficiency of each stage in the multistage cross current extraction. This was done

by using mixture of toluene and acetic acid as a feed and distilled water as a solvent. Solute

acetic acid is extracted from toluene to the water.

Apparatus:-

Standard flash, conical flask, burette, pipette and funnel.

Components: -

Toluene as carrier (A), Acetic acid as solute (C), Distilled Water as (B), NaoH as titrating agent

and phenolphthalein as indicator.

Theory:-

Liquid-liquid extraction is a separation method and mass transfer operation in which liquid of

mixture (feed) is treated with an immiscible or nearly immiscible solvent to get the desirable

component which is soluble in solvent from the liquid of mixture. Separation of component can

be done by distillation which is most preferable method and gives higher purity components but

when relative volatility of component is very low and heating need to be avoided, distillation

method is not preferable, then liquid-liquid extraction method is used for separation. This

method separates compounds based on their relative solubility in two different immiscible

liquids.

There are three ways of extraction method. These are co-current, counter-current and cross-

current method.In this experiment cross current method was used. Cross current method is very

preferable in laboratory and for lab experiments because each stage is made up of an equilibrium

stage where the two liquid phases are mixed together for a period of time until equilibrium is

attained. Fresh solvent is added to each stage in this method while in co-current and cross-current

method solvent is added only once. This method is very rarely used in industry because of the

large volume of solvent required and low concentration of solute extracted by this method.

Efficiency of stage can be calculated by following formula:

% Efficiency = molesof acetic acid extractedmolesof acetic acid∈ feed x100

Figure 1:- Schematic of cross current method

Procedure:-

Initially in first stage a solution mixture is prepared by mixing 60 ml of toluene and 40 ml of

acetic acid in a flask. Then 60 ml of distilled water is mixed. Mixture is shaken for 10 minutes

and then it is left unattended until 2 layers of immiscible phases are not formed clearly in the

flask. 95 ml of the settled down phase is taken out and it is diluted to 285 ml thus the normality

of liquid taken out would be N/3, where N is actual normality. Then 10 ml of this solution is

titrated with NaOH solution which has normality of 0.93 N. In second and third stage again 60

ml of water is added in mixture and same procedure is followed but 63 ml of the settled down

phase in second stage and 60 ml of the settled down phase in third stage is extracted and solution

will not be diluted as in 1st stage. After that same procedure is repeated.

Observations & Readings:-

Obs no. Amount of

carrier

added

(ml)

Amount of

solute

added

(ml)

Amount of

solvent

added

(ml)

Amount of

extract

removed

(ml)

Amount of

NaoH

required

(ml)

1 Stage 1 60 40 60 95 21

2 Stage 2 - - 60 63 8.5

3 Stage 3 - - 60 60 0.5

Result:-

Mole of acetic acid extracted in stage 1 = 0.556

Mole of acetic acid extracted in stage 2 = 0.049

Mole of acetic acid extracted in stage 3 = 0.003

Efficiency of stage 1 = 79.58 %

Efficiency of stage 2 = 34.87 %

Efficiency of stage 3 = 2.99 %

Conclusion:-

It was observed that efficiency is decreasing stage by stage; this is because the amount of

extracted solute is decreasing stage by stage as solute is decreasing in carrier in each stage.

Amount of extracted solute was maximum in 1st stage and then decreases stage by stage due to

high driving force in 1st stage. Since no solute is added after 1st stage and volume of solvent is

added in each stage so concentration of solute decreases stage by stage thus driving force

decreases. Amount of extraction and efficiency depends on the amount of solvent added if

amount of solvent decreases, efficiency decreases. Selectivity, density, viscosity, relative

volatility and corrosivity are the major affecting factors. Selectivity factor determine the number

of stages required for extraction. Density difference should be high between the feed and the

solvent for better separation. High viscosity of solvent affects the mass transfer and separation

method.

Discussion:-

Initially in stage 1, reading of NaOH volume was not getting correctly measured in titration

procedure. Sometimes end point was coming very early and sometimes very late. It was may be

due to some human errors i.e. while making solution or putting indicator in solution.

Sample Calculation:

For stage3:

Volume of NaOH normalized = 21 ml

N1*V1= N2*V2

V1= Volume of NaOH

N1= Normality of NaOH

V2 = Volume of the extracted solution

N2= Normality of the extracted solution

N2= (N1*V1)/V2

= 0.93*8.5/10

= 0.7905

N = 0.7905 N

Moles extracted would be

N=n/V

n = N*V

= 0.7905*63/1000

= 0.04980 moles

Moles of acetic acid in total solution

Volume: 40 ml

Weight = 40*density

Density = 1.05

= 40*1.05

= 42 grams

Moles

= 42/60.05

= 0.7

Initial moles in raffinate of stage 1 = 0.7- 0.556=0.144 moles

Efficiency

= Moles extracted/moles in solution

= (0.049/0.144) *100

= 34.87 %

References:-

1. http://www.slideshare.net/GerardBHawkins/liquid-liquid-extraction-basic-principles?

next_slideshow=1

2. http://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction

3. http://www.powershow.com/view/3acbb3-ZmNkZ/Chapter_5_Liquid-

Liquid_Extraction_powerpoint_ppt_presentation