1. The solute should not react chemically with either solvent.
2. The temperature should remain constant.
3. Both solvents must be immiscible (do not mix).
4. The solute must dissolve in both solvents.

1. Extraction Processes: Used in separating components of a mixture using solvents (e.g., extracting caffeine from tea using chloroform).
2. Drug Absorption: Helps in understanding how drugs distribute between aqueous body fluids and lipid layers of cells.
3. Chromatography: Principle of distribution between two phases is fundamental in chromatographic techniques.

One of the application is the extraction of solute from one component by mixing the solution with the second liquid that has no solute at all.

The liquid component which is removing the solute is called Extracting component and that in which the solute is removed from is called Extracted component.

In terms of extractions:

$K_d = \frac{x \text{ in } \frac{g}{cc} \text{ in extracting}}{x \text{ in } \frac{g}{cc} \text{ in extracted}}$

Note: Concentration is the amount of substance per unit volume.

$K_d = \frac{\text{Conc. of } x \frac{g}{l} \text{ in extracting}}{\text{Conc. of } x \frac{g}{l} \text{ in extracted}}$

$K_d = \frac{\text{Conc. of } x \frac{m}{dm^3} \text{ in extracting}}{\text{Conc. of } x \frac{mol}{dm^3} \text{ in extracted}}$

Also:

$K_d = \frac{x \text{ in } \frac{g}{cc} \text{ in extracting}}{x \text{ in } \frac{g}{cc} \text{ in extracted}}$

Note: During extraction of solute the amount of solute in extracted component will be decreasing while the amount of extracting component will be increasing.

An aqueous solution containing 10g per litre of solute X. This solution was shake with 100cc of ether, on shaking 6g of X was extracted. Calculate the amount of X extracted from aqueous so residue after shaking with 100cc of ether.

Solution

Given:

Extracted component = 10g/ l

Volume of extracting component = 100cc

Mass of X in water (aqueous solution) = 10g

The concentration calculation in the original seems problematic. Let me work through this carefully.

Given data suggests:

• Initial aqueous solution: 10g/L
• After extraction with 100cc ether: 6g extracted
• Therefore residue has 4g in aqueous phase

Using the distribution law:

$Kd = \frac{[x]_{\text{in ether}}}{[x]_{\text{in water}}}$

Given that 6g was extracted into ether (100cc) and 4g remained in water (1000cc):

$Kd = \frac{6g/100cc}{4g/1000cc}$

$Kd = \frac{0.06}{0.004}$

Kd = 15

Let the amount extracted be 'a'

From Kd = (a/100) / ((4-a)/1000)

$15 = \frac{a}{100} \times \frac{1000}{4-a}$

$15 = \frac{10a}{4-a}$

15(4 - a) = 10a

60 - 15a = 10a

60 = 25a

a = 60/25

a = 2.4g

∴ The amount of X extracted from the aqueous residue is 2.4g.

Note: If layers are not specified then the word "between" shows the numerator and denominator of the formula.

A solid X is added to a mixture of benzene and water after shaking well and allowing to hand, 10ml of benzene layer was found to certain 0.13g of X and 100ml of water layer contained 0.22g of X.

Calculate volume of distribution coefficient of solute X between benzene and water layer

Solution

Mass of solute X in benzene = 0.13g

Volume of benzene = 10ml

$\text{Conc. of } x \text{ in benzene} = \frac{0.13}{10}$

$= 0.013\text{g/ml}$

$\text{Mass of solute } x \text{ in water} = 0.22\text{g}$

$\text{Volume of water} = 100\text{ml}$

$\text{Conc. of } x \text{ in water} = \frac{0.22}{100}$

$= 0.0022\text{g/ml}$

$K_d = \frac{\text{Conc. of } x \text{ in benzene}}{\text{Conc. of } x \text{ in water}}$

$= \frac{0.013}{0.0022}$

$\therefore \text{The value of distribution coefficient is } 5.9$

In the distribution of succinic acid between ether and water at 15°C, 20ml of the ethereal layer contains 0.092g of the acid. Find out the weight of the acid present in 50ml of the aqueous solution in equilibrium with it. If the coefficient Kb between water and ether is 1.196g and Kd is 5.2

Solution

Mass of succinic acid in ethereal = 0.092g

Volume of ethereal = 20ml

Conc. of succinic acid in ethereal = 0.092/20

= 4.6 × 10⁻³ g/ml

Coefficient of succinic acid between water and ethereal = 5.2

Volume of water = 50ml

Let w be the weight of succinic acid in water

Conc. of succinic in water = w/50

From:

$Kd = \frac{[\text{Succinic}] \text{ in water}}{[\text{Succinic}] \text{ in ether}}$

$5.2 = \frac{w/50}{4.6 \times 10^{-3}}$

$w = 5.2 \times 50 \times 4.6 \times 10^{-3}$

$w = 1.196\text{g}$

∴ The weight of the acid present in aqueous solution is 1.196g

An aqueous solution of succinic acid at 15°C containing 0.07g in 10ml is in equilibrium with an ethereal solution which has 0.013g in 10ml. The acid has its normal molecular weight in both solvents. What is the concentration of the ethereal solution which is in equilibrium with aqueous solution containing 0.024g in 10ml?

(Ans: 0.00044g/ml)

Solution

Mass of succinic acid in aqueous solution = 0.07g

Volume of aqueous solution = 10ml

Concentration of succinic in ethereal = 0.013g/10ml = 0.0013g/ml

Concentration of succinic in aqueous solution = 0.07/10

= 7 × 10⁻³ g/ml

Now

For second extraction:

Mass of aqueous solution = 0.024g

Volume of aqueous solution = 10ml

Conc. of succinic in aqueous solution = 0.024/10

= 2.4 × 10⁻³ g/ml

Volume of ethereal = 10ml

Let x be the concentration of succinic acid in ethereal

$k_d = \frac{[\text{succinic}] \text{ in } H_2O}{[\text{succinic}] \text{ in ethereal}}$

$5.38 = \frac{2.4 \times 10^{-3}}{x}$

$x = \frac{2.4 \times 10^{-3}}{5.38}$

$x = 4.46 \times 10^{-3}\text{g/ml}$

∴ The concentration of succinic acid in the ethereal solution is 0.0044g/ml.

For more than one extraction we normally use the following formula:

$A_r e = \left( \frac{V_a}{V_a + k + V_b} \right)^n w_o$

Whereby:

$A_re$ is the amount of solute remaining

$V_a$ is the volume of extracted solution

$V_b$ is the volume of extracting solution

$k$ is the constant of distribution

$w_o$ is the original weight of the solute

From the formula, the amount extracted can be calculated as ($A_{ex}$):

$A_{ex} = w_o - A_r e$

$A_{ex} = w_o - \left( \frac{V_a}{V_a + V_b k} \right)^n w_o$

If the amount extracted and amount remained are known, then their respective percentages can be calculated.

Percentage remaining:

Percentage extracted: