7.3.1 Partition Coefficients

Definition

Definition: Partition Coefficient

• Ratio of the concentration of a solute in two immiscible solvents in contact with each other
• At equilibrium At a particular temperature

Expression

For a solute dissolved in two different immiscible solvents that are in contact with each other, the Partition Coefficient, is given by:

Units of Partition Coefficient

Since partition coefficient is a ratio of concentrations, it has no units.

Essential Conditions

For to be valid and constant:

1. Immiscible Solvents: The two solvents must form two distinct layers (e.g. Water and Ether).
2. Equilibrium: The rate of solute moving from Solvent A to B equals the rate from B to A.
3. Constant Temperature: Like all equilibrium constants, is temperature-dependent.
4. Same Molecular State: The solute must exist in the same physical state in both solvents (i.e. no dissociation or dimerization).

Warning: The Ratio Order

Unlike where it is always Products/Reactants, does not have a fixed universal rule for which solvent is on top, however:

• Cambridge Rule: The exam question will specify the ratio:

  • If it says “Calculate of between water and ether”, this implies that water is on top:

  • If it says “Calculate of between ether and water”, this implies that ether is on top:

In general, the solvent that comes first is the one on top

Factors Affecting Partition Coefficient

The value of is determined by the relative solubility of the solute in the two solvents.

The Role of Polarity

1. Polar Solutes (e.g. Ammonia, Short-chain alcohols/acids):
   • Prefer Polar Solvents (e.g. Water).
   • Reason: Formation of hydrogen bonds or strong permanent dipole-permanent dipole forces with water.
   • usually favours the aqueous layer.
2. Non-polar Solutes (e.g. Iodine, Halogenoalkanes, long-chain organic molecules):
   • Prefer Non-polar solvents (e.g. Hexane, Cyclohexane, Ether).
   • Reason:
     • Formation of instantaneous dipole-induced dipole forces with the organic solvent.
     • The energy cost to disrupt water’s H-bond network is too high for non-polar solutes to overcome.
   • favours the organic layer.

Example: Iodine ( ) between Hexane and Water

• Structure: is non-polar.
• Solvents: Hexane (non-polar), Water (polar)l
• Observation: is much more soluble in hexane.
• value: If defined as , will be very large.

Successive Extractions

Extracting a solute form a solvent using multiple small batches is more efficient than using a single large volume at the same time.

Scenario:

• You have grams of solute in of water.
• .
• You have of organic solvent available.

Method A: Single Extraction () Let be the grams extracted into the organic layer. Remaining in water = .

Method B: Double Extraction ()

Extraction 1: Let be grams extracted. Remaining in water: .

Extraction 2: Now we extract from the remaining . Let be grams extracted.

Total Extracted: ().

Conclusion: Method B (89%) is more efficient than Method A (80%).