Reporters: Jonathan Jamito & Paula TayoGroup 3, AB2

It is a method of separating a component from a liquid solution by mixing it with a liquid solvent.

Ratio of concentration = Ratio of solubility

The added solvent must be more volatile than the desired component.

It must also specifically dissolve the desired component.

The component must have a greater tendency to dissolve in the added solvent than in the solution.

KD= CO = SO =

CH20 SH20

The added solvent must be immiscible with the solution.

The added solvent and the solution must be of different densities.

If possible, the added solvent must not be toxic.

To understand the theory and to develop the skills needed in liquid-liquid extraction.

To identify the characteristics of an ideal solvent in such a process.

To study the different types of solvents. To determine the best possible solvent

based on the outcomes of both solubility and miscibility tests.

To learn how to compute for the percent recovery [%rec] for each extraction.

Figure 1.1 The actual solubility test.L-R: Acetone+benzoic acid, Ether+benzoic acid, Hexane+benzoic acid

Figure 1.2 The idealized solubility test. L-R: Acetone+benzoic acid, Ether+benzoic acid, Hexane+benzoic acid

Figure 1.1 The actual miscibility test.L-R: Acetone+water, Ether+water, Hexane+water

Figure 1.2 The idealized miscibility test. L-R: Acetone+water, Ether+water, Hexane+water

Solvent Water Benzoic Acid

Acetone Miscible Soluble

Ether Immiscible Soluble

Hexane Immiscible Insoluble

The best extracting solvent: Ether Benzoic acid in saturated solution: 0.47 g Amount of undissolved benzoic acid: 0.03 g %rec : 10.64%

In choosing the ideal solvent for the extraction, there are parameters measured in the solubility and miscibility tests.

1.The component must have a greater tendency to dissolve in the added solvent than in the solution. ◦ Measured by the solubility test

[Why? To allow solute transfer].

◦ Greater solubility means greater rate of recovery.

Figure 2.1 A diagramatic representation of a separatory funnel showing layer formation.

2. The added solvent must be immiscible with the solution.

◦ Measured by the miscibility test [Why? Layer formation].

Figure 2.2 Layer separation showing ether forming above the denser aqueous layer.

3. The added solvent and the solution must be of different densities.

◦ Why? Layer separation◦ The less dense ether

layer forms above the aqueous layer.

Figure 2.3 The ether component is separated from the benzoic acid by evaporation.

4. The added solvent must be more volatile than the desired component.

◦ Why? Benzoic acid is retrieved from the ether layer by evaporation.

5. The added solvent must specifically dissolve the desired component and nothing else from the solution.

◦ Why? To prevent the extraction of undesired and unnecessary solutes.

6. If possible, the added solvent should not be toxic.

◦ Why? For reasons of good health.

The result of both miscibility and solubility tests can be deduced from the polarity of the substances:

Benzoic acid is polar and its IMF forms two H-bonds.

Hexane is non-polar and its IMF comprises of London Dispersion Forces.

Ether is slightly polar and its IMF forms a dipole-dipole bond.

Acetone is polar and its IMF forms an H-bond.

Water is polar is polar and its IMF forms H-bonds

The principle behind the solubility test?“Like dissolves like”

◦ Benzoic acid [polar] + Acetone [polar] = soluble◦ Benzoic acid [polar] + Ether [slightly polar] = soluble◦ Benzoic acid [polar] + Hexane [non-polar] = insoluble

◦ Benzoic acid [polar] + Water [polar] = slightly soluble

Why?Because water is too polar to completely dissolve benzoic

acid.

On the miscibility test…

o Water [polar] + Acetone [polar] = miscibleo Water [polar] + Ether [slightly polar] = immiscibleo Water [polar] + Hexane [non-polar] = immiscible

Solvent Water Benzoic Acid

Acetone Miscible Soluble

Ether Immiscible Soluble

Hexane Immiscible Insoluble

This based on the results of the two tests, the best extraction solvent is ether.o It can dissolve the desired solute [benzoic acid]o It is immiscible in the solute [water] of the solution o It is of a different density than the solute [water] of the

solutiono It is more volatile than the desired solute [benzoic acid]

So solving for the %recovery:1. Determine the amount of benzoic acid dissolved in the solution.

0.5g initial amount – 0.03g amount recovered from filter paper = 0.47g

Only 20mL of the 100mL was used in the solution.

2. Determine the amount of benzoic acid recovered from the ether layer.Amount = 0.01g

3. Determine the %recovery . %recovery = Xpure x 100% = 0.01g x 100% = 10.64% benzoic acid

Ximpure 0.094 g benzoic acid

⇓* 20mL x 0.47g = 0.094 g benzoic acid

100mL

1. Suppose 20mL of the saturated benzoic acid solution was treated with 10% NaOH, which solvent [acetone, ether, hexane] could extract most of the benzoic acid from the aqueous solution?

The presence of sodium hydroxide makes the benzoic acid more polar as the hydroxide ions ionize the benzoic acid, forming an organic salt. Despite the addition of sodium hydroxide, ether is still the best extracting solvent because it dissolves more sodium benzoate than hexane and unlike acetone, it is immiscible with the sodium hydroxide solution.

2. What effect does partial miscibility of the two solvents have on the efficiency of extraction?

◦ Partial miscibility of the two solvents makes the extraction less efficient. This makes the two layers more difficult to separate as compared to complete immiscibility between the two solvents.

◦ Partial miscibility also implies that the relative affinity of the solute to the extracting solvent is not as great as to efficiently extract the solute from the solution.

3. A substance X can be isolated from its plant source by solvent extraction. However, a minor component Y has an appreciable solubility in the solvents that may be used. Given below are the solubilities of X and Y in different solvents:

Solvent T, ˚C Solubility in 100g solvent at 28˚C

X Y

Ethyl methyl ketone 80 6 5

Cyclohexane 81 8 2

Benzene 80 5 1.8

CCl4 78 8.75 1.25

Water 100 2 1

a. Which is the best extracting solvent? CCl4 is the best extracting solvent

it dissolves only a little amount of Y relative to the X component needed.

It also dissolves the greatest amount of X among the choices.

It is immiscible with water, which happens to be the solvent in the solution.

◦ Mathematically speaking, the choice relies on the KD and the best result is the one with the highest KD. The KD values with respect to water as calculated are the following:

◦ From this table, it is evident that CCl4 is the yields the highest ratio, ergo the best possible choice as the extracting solvent.

Solvent KD

Ethyl methyl ketone

3

Cyclohexane 4

Benzene 2.5

CCl4 4.375

b. Given a saturated aqueous solution of X and Y and using 100mL of solvent in (1), determine the percent recovery of X in a single extraction.

KD = SCCl4 = 8.75g/100g = 4.375

SH20 2.00g/100g

KD = CCCl4 = Xg/100ml = 4.375

CH20 (2.00g-Xg)/100ml

X = 1.63g

%rec = Xpure x 100% = 1.63g x 100% = 81.4%

Ximpure 2.00g

c. Repeat (b) using 50mL of solvent in each of the two successive extractions. Determine the percent recovery and compare this with (b).

KD = SCCl4 = 8.75g/100g = 4.375 SH20 2.00g/100g First extraction:KD = CCCl4 = Xg/50ml = 4.375 CH20 (2.00g-Xg)/100ml

X = 1.37g 

Second extraction:KD = CCCl4 = Xg/50ml = 4.375

CH20 (0.63g-Xg)/100ml

X = 0.43g %rec = Xpure x 100%= (1.37+0.43)g x 100%= 90.0%

Ximpure 2.00g

 In the double extraction, the percent recovery is comparably

higher, considering the same amount of the solvent added.

d. What is the percent recovery of the minor component in a single extraction using 100mL of the solvent in (a)?

 KD = SCCl4 = 1.25g/100g = 1.25 SH20 1.00g/100g

KD = CCCl4 = Yg/100g = 1.25 CH20 (1.00g-Yg)/100g Y = 0.56

%rec = Ypure x 100% = 0.56g x 100% = 56% Yimpure 1.00g 

That these are the characteristics of an ideal solvent for a liquid-liquid extraction:

1. It can dissolve the solute to be extracted better than the original solvent of the solution.

2. It is immiscible with the solvent of the solution like how ether is immiscible with water.

3. It is of a different density with the solvent of the solution in order for layering to occur.

4. It can also be separated from the solute by evaporation like how ether is more volatile than its solute benzoic acid.

The aforementioned criteria can be used in determining the best extracting solvent which is of utmost importance in the liquid-liquid extraction.

Solubility and miscibility tests are effective means by which one can determine the best extracting solvent.

To avoid experimental errors, laboratory apparatuses should be in good shape, Laboratory techniques should be properly executed, The filter papers used should be thoroughly dried and the

organic layer of the ether-benzoic acid solution, completely evaporated.

In extracting benzoic acid from an aqueous solution, ether proved to be the best extracting solvent:◦ Benzoic acid is insoluble in hexane and…◦ Acetone is miscible in water…making both solvents inefficient in extracting benzoic acid from water.

So the best choice for the extraction: ether.

The more appropriate the solvent used, and the more extractions done, the more solute can be recovered from the solution.