25.2 Partition coefficients

For a solute distributed between two immiscible solvents, Kpc = concentration in one named solvent divided by concentration in the other at equilibrium and fixed temperature.

State which solvent is numerator and keep phases/units consistent. Kpc describes equilibrium distribution, not the total amount extracted in one operation.

If Kpc(organic/aqueous)=4, the equilibrium concentration in the organic phase is four times that in the aqueous phase under the stated conditions.

Do not invert Kpc without changing the definition, and do not assume a large Kpc means one extraction removes all solute.

For a solute in the same physical state in two immiscible solvents, Kpc = concentration in solvent 1 divided by concentration in solvent 2 at equilibrium.

State the numerator solvent, use matched units and measure after the two phases have equilibrated. If the solute associates or reacts in one phase, the simple expression may not apply.

If [solute]organic=0.80 mol dm⁻³ and [solute]aqueous=0.20 mol dm⁻³, Kpc(organic/aqueous)=4.0.

Do not invert the ratio without changing the label, and do not calculate from initial concentrations before equilibrium.

A solute partitions according to its relative affinity for the two phases. Similar polarity and intermolecular forces favour dissolution in a solvent, while a polarity mismatch favours the other phase.

Hydrogen bonding, ionisation and temperature can alter Kpc. Predict direction qualitatively, but do not treat polarity as the only possible factor.

A non-polar hydrocarbon generally partitions more into an organic solvent than water; an ionised acid may remain preferentially in the aqueous phase.

A larger Kpc is not an intrinsic label independent of solvent order; reversing numerator and denominator gives the reciprocal.