Partition Coefficients

This lesson covers: 

1. What the partition coefficient (K_pc) is
2. How to calculate partition coefficients
3. Factors affecting partition coefficients
4. Partition coefficients and paper chromatography

Partition coefficients (K_pc) describe solute distribution

A partition coefficient (K_pc) is the equilibrium constant that describes how a solute is distributed between two immiscible solvents at a given temperature.

When we dissolve a solute in two immiscible solvents (which don't mix), a dynamic equilibrium is reached where the solute molecules move between the two layers at equal rates:

X_{(solvent A)} ⇌ X_{(solvent B)}

The formula to express the partition coefficient (K_pc) is:

Kpc​=[X(solvent A)​][X(solvent B)​]​

Example

For ammonia distributed between water and an organic solvent, the equilibrium is:

NH_3(aq) ⇌ NH_{3(organic solvent)}

And the equilibrium expression for the partition coefficient of ammonia is:

Kpc​=[NH3(aq)​][NH3(organic solvent)​]​

Partition coefficients don't have units because they are simply ratios of two concentrations.

To calculate partition coefficients, we use the equilibrium concentrations of the solute in each solvent layer.

Worked example 1 - Calculating the partition coefficient of ammonia

100 cm³ of a 0.100 mol dm^-3 aqueous ammonia solution is shaken with 50 cm³ of trichloromethane (CHCl₃). After reaching equilibrium, a 20.0 cm³ portion of the aqueous layer requires 18.8 cm³ of 0.100 mol dm^-3 HCl for neutralisation.

Calculate the partition coefficient of ammonia between water and trichloromethane.

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

100 cm³ = 0.100 dm³

50 cm³ = 0.050 dm³

18.8 cm³ = 0.0188 dm³

Step 2: Calculate number of moles of HCl used in titration

n = c× V=0.100×0.0188=1.88×10−3 mol

Step 3: Calculate number moles of NH₃ in the 20.0 cm³ portion

NH_3(aq) + HCl_(aq) ➔ NH₄Cl_(aq)

NH₃ : HCl mole ratio = 1:1

Moles of NH₃ = 1.88 x 10^-3 mol

Step 4: Calculate total number moles of NH₃ in the 100 cm³ aqueous layer

Moles of NH₃ = (1.88 x 10^-3) x 5 = 9.40 x 10^-3 mol

Step 5: Calculate initial number of moles of NH₃ in the solution

n = c× V=0.100×0.100=1.00×10−2 mol

Step 6: Calculate number of moles of NH₃ in the organic layer

Moles of NH3​ in organic layer = Initial moles of NH3​− Moles of NH3​ in aqueous layer

=(1.00×10−2)−(9.40×10−3)=6.00×10−4 mol

Step 7: Calculate [NH_3(aq)] and [NH_3(organic)]

[NH3(aq)​]=Vn​=0.100(9.40×10−3)​=0.094 mol dm−3

[NH3(organic)​]=Vn​=0.050(6.00×10−4)​=0.012 mol dm−3

Step 8: Calculate partition coefficient (K_pc)

Kpc​=[NH3(aq)​][NH3(organic)​]​=0.0940.012​=0.13

The partition coefficient of ammonia between water and trichloromethane is 0.13. This indicates that ammonia is more soluble in water than in trichloromethane.

Worked example 2 - Calculating the partition coefficient of iodine

200 cm³ of a 0.050 mol dm^-3 aqueous iodine solution is shaken with 100 cm³ of cyclohexane. After reaching equilibrium, a 25.0 cm³ portion of the cyclohexane layer requires 15.0 cm³ of 0.200 mol dm^-3 Na₂S₂O₃ for titration to the endpoint.

Calculate the partition coefficient of iodine between water and cyclohexane.

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

200 cm³ = 0.200 dm³

100 cm³ = 0.100 dm³

15.0 cm³ = 0.0150 dm³

Step 2: Calculate number of moles of Na₂S₂O₃ used in titration

n = c× V=0.200×0.0150=3.00×10−3 mol

Step 3: Calculate number moles of I₂ in the 25.0 cm³ portion

I_2(aq) + 2Na₂S₂O_3(aq) ➔ 2NaI_(aq) + Na₂S₄O_6(aq)

I₂ : Na₂S₂O₃ mole ratio = 1:2

Moles of I₂ = 2(3.00×10−3)​=1.50×10−3 mol

Step 4: Calculate total number moles of I₂ in the 100 cm³ organic layer

Moles of I₂ = (1.50 x 10^-3) x 4 = 6.00 x 10^-3 mol

Step 5: Calculate initial number moles of I₂ in the solution

n = c× V=0.050×0.200=1.00×10−2 mol

Step 6: Calculate number of moles of I₂ in the aqueous layer

Moles of NH3​ in aqueous layer = Initial moles of I2​− Moles of I2​ in organic layer

=(1.00×10−2)−(6.00×10−3)=4.00×10−3 mol

Step 7: Calculate [I_2(aq)] and [I_2(organic)]

[I2(aq)​]=Vn​=0.200(4.00×10−3)​=0.020 mol dm−3

[I2(organic)​]=Vn​=0.100(6.00×10−3)​=0.060 mol dm−3

Step 8: Calculate partition coefficient (K_pc)

Kpc​=[I2(aq)​][I2(organic)​]​=0.0200.060​=3.0

The partition coefficient of iodine between water and cyclohexane is 3.0, showing that iodine is more soluble in cyclohexane than in water.

Factors affecting partition coefficients

The value of K_pc is influenced by the solute's relative solubilities in the two solvents, which depends on the strength of intermolecular forces between the solute and solvent molecules. This strength is determined by the polarity of the solute and solvent molecules.

Examples:

• Ammonia (NH₃) is more soluble in water than in organic solvents like trichloromethane (CHCl₃) due to its ability to form stronger hydrogen bonds with water compared to the weaker permanent dipole-dipole forces it would form with CHCl₃.
• Iodine (I₂) is more soluble in organic solvents like cyclohexane (C₆H₁₂) than in water because both iodine and cyclohexane are non-polar, leading to favourable induced dipole-dipole forces. Iodine cannot disrupt the strong hydrogen-bonded structure of water.

Partition coefficients and paper chromatography

In paper chromatography, the partition coefficient determines the solutes' relative solubility in the mobile phase (the solvent moving over the paper) and the stationary phase (the water trapped within the paper's structure).

Solutes with higher solubility in the mobile phase will have larger K_pc values and move faster up the paper. Those more soluble in the stationary phase will have smaller K_pc values and move slower.

This differential solubility, reflected in varying partition coefficients, allows for the separation of mixture components during paper chromatography.