Shake a solute with two immiscible solvents and it shares itself between them in a fixed ratio. That ratio, the partition coefficient, tells you how much you can extract and why several small extractions beat one big one.
When a solute X is shaken with two immiscible solvents, such as hexane and water, it distributes itself between them and reaches a dynamic equilibrium, X(aq) ⇌ X(organic). The partition coefficient is the ratio of the equilibrium concentrations, Kpc = [X]organic ÷ [X]aqueous, which is constant at a fixed temperature and has no units. A non-polar solute such as iodine is more soluble in a non-polar solvent, so Kpc (organic over water) is greater than 1; a polar solute such as ammonia, which hydrogen bonds to water, has Kpc less than 1. Because the amount extracted depends on the volumes as well as Kpc, several small portions of solvent extract more than one large portion.
Partition is an equilibrium like any other: molecules move across the boundary in both directions until the rates are equal, and the ratio of concentrations is then fixed.
| Solute | Solvents | Polarity match | Kpc (organic / aqueous) |
|---|---|---|---|
| iodine, I2 | hexane / water | non-polar solute, non-polar solvent | much greater than 1 |
| ammonia, NH3 | trichloromethane / water | polar solute that hydrogen bonds with water | less than 1 |
| butanedioic acid | ether / water | polar, two –COOH groups | less than 1 |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Which is the correct definition of the partition coefficient of X between hexane and water?
What are the units of Kpc?
Ammonia is shaken with trichloromethane and water. Which statement about Kpc = [NH3]organic ÷ [NH3]aqueous is correct?
Kpc for X between hexane and water is 4.00. The water layer contains 0.0500 mol dm−3 X. What is [X] in the hexane?
Let x be the amount that moves into the organic layer. Write each concentration as amount ÷ volume, substitute into Kpc and solve for x.
| Case (1.00 g of X in 100 cm3 water, Kpc = 4.00) | Equation | Extracted |
|---|---|---|
| one 50.0 cm3 portion of ether | (x ÷ 50.0) ÷ ((1.00 − x) ÷ 100) = 4.00 | x = 0.667 g |
| first 25.0 cm3 portion | (x ÷ 25.0) ÷ ((1.00 − x) ÷ 100) = 4.00 | x = 0.500 g |
| second 25.0 cm3 portion (0.500 g left) | (y ÷ 25.0) ÷ ((0.500 − y) ÷ 100) = 4.00 | y = 0.250 g |
| two 25.0 cm3 portions in total | 0.750 g (more than 0.667 g) |
Kpc is a ratio of concentrations, not masses: divide each amount by its own layer's volume. Keep the order as stated (organic over aqueous here); inverting it gives 1 ÷ Kpc. The amount left in water after extraction is the starting amount minus x, and each new portion starts from what is left. A large Kpc means the solute prefers the organic layer, which needs a non-polar solute.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
1.00 g of X is dissolved in 100 cm3 of water and shaken with 50.0 cm3 of ether. Kpc (ether / water) = 4.00. What mass of X is extracted?
The same 1.00 g of X in 100 cm3 of water (Kpc = 4.00) is extracted with two successive 25.0 cm3 portions of ether. What total mass is extracted?
1.50 g of a solute in 100 cm3 of water is shaken with 20.0 cm3 of an organic solvent, which removes 1.20 g. What is Kpc (organic / aqueous)?
0.0100 mol of Y in 50.0 cm3 of water is shaken with 10.0 cm3 of hexane (Kpc = 20.0). What percentage of Y is extracted?
This skill is now recorded on your profile. Keep the chain going.