How to Calculate a Partition Coefficient

A partition coefficient describes how a solute distributes between two phases at equilibrium. In the common liquid–liquid case, the phases are immiscible solvents, such as water and an organic solvent. The coefficient is a concentration ratio, but you must specify which phase is in the numerator: different fields and references may use opposite conventions.

This article uses the convention

$$K = \frac{C_{\mathrm{org}}}{C_{\mathrm{aq}}},$$

where $C_{\mathrm{org}}$ and $C_{\mathrm{aq}}$ are the equilibrium concentrations in the organic and aqueous phases. If your source defines the inverse ratio, invert the value before comparing results.

Calculate it from concentrations

Measure the solute concentration in each phase after the system has reached equilibrium, then divide the concentration in the numerator phase by the concentration in the denominator phase. Use the same concentration units for both measurements, such as mol/L or mg/L. The units then cancel, leaving a dimensionless ratio in this practical concentration-based definition.

For example, suppose a solute has an equilibrium concentration of $0.80\ \mathrm{mol/L}$ in the organic phase and $0.20\ \mathrm{mol/L}$ in the aqueous phase. With the convention above,

$$K = \frac{0.80}{0.20} = 4.$$

The solute concentration is four times as high in the organic phase as in the aqueous phase. This does not mean that four times as much total solute is in the organic phase: total amount also depends on the volume of each phase.

Calculate it from measured amounts

If you measured the amount of solute in each layer, first calculate each concentration using $C=n/V$ (or mass divided by volume). Then form the ratio. For example, if the organic layer contains $12\ \mathrm{mg}$ in $30\ \mathrm{mL}$ and the aqueous layer contains $8\ \mathrm{mg}$ in $80\ \mathrm{mL}$, then

$$\begin{aligned} C_{\mathrm{org}} &= \frac{12\ \mathrm{mg}}{30\ \mathrm{mL}} = 0.40\ \mathrm{mg/mL},\\ C_{\mathrm{aq}} &= \frac{8\ \mathrm{mg}}{80\ \mathrm{mL}} = 0.10\ \mathrm{mg/mL},\\ K &= \frac{0.40}{0.10} = 4. \end{aligned}$$

The amount ratio is $12/8=1.5$, not 4, because the phase volumes differ. A partition coefficient compares concentrations, not total amounts.

Predict how much solute each phase contains

Once you know $K$ and the phase volumes, you can calculate the equilibrium distribution from the total amount of solute. Suppose $100\ \mathrm{mg}$ of solute is distributed between $100\ \mathrm{mL}$ of water and $50\ \mathrm{mL}$ of an organic solvent, and $K=C_{\mathrm{org}}/C_{\mathrm{aq}}=4$. Let the aqueous concentration be $C_{\mathrm{aq}}$. Then $C_{\mathrm{org}}=4C_{\mathrm{aq}}$. Assuming all the solute is in these two phases, conservation of mass gives

$$\begin{aligned} 100\ \mathrm{mg} &= C_{\mathrm{aq}}(0.100\ \mathrm{L}) + 4C_{\mathrm{aq}}(0.050\ \mathrm{L}),\\ C_{\mathrm{aq}} &= 333.3\ \mathrm{mg/L},\\ C_{\mathrm{org}} &= 1333.3\ \mathrm{mg/L}. \end{aligned}$$

The aqueous phase contains about $33.3\ \mathrm{mg}$ and the organic phase about $66.7\ \mathrm{mg}$. The organic concentration is four times larger, but the organic phase holds only twice as much solute because its volume is half as large.

More generally, for total solute amount $n_{\mathrm{tot}}$, aqueous volume $V_{\mathrm{aq}}$, and organic volume $V_{\mathrm{org}}$, the amount in the organic phase is

$$n_{\mathrm{org}} = n_{\mathrm{tot}}\frac{K V_{\mathrm{org}}}{V_{\mathrm{aq}}+K V_{\mathrm{org}}}.$$

This expression assumes the stated coefficient applies under the conditions used and that no significant solute is lost to a third phase, reaction, or container surface.

Partition coefficient, log P, and log D

For a neutral compound partitioning between octanol and water, the coefficient is often called $P$; its base-10 logarithm is reported as $\log P$. For example, if $P=4$, then $\log P=\log_{10}(4)\approx0.602$. Always check which phase is placed in the numerator when interpreting a reported value.

For an ionizable compound, the neutral form and charged forms may partition differently. The partition coefficient for the neutral species is not the same as the distribution ratio, commonly written $D$, which can account for the total concentration of all forms in each phase. Consequently, $D$ can vary with pH, while the neutral-species partition coefficient is defined for that species under specified conditions. In biological and pharmaceutical contexts, this pH-dependent quantity is often reported as $\log D$.

Common mistakes

• Using initial concentrations: The ratio must use concentrations after the phases have reached equilibrium.
• Comparing amounts instead of concentrations: Divide each measured amount by its phase volume first.
• Leaving the phase order unstated: Write the defining ratio, such as $C_{\mathrm{org}}/C_{\mathrm{aq}}$, alongside the answer.
• Mixing conditions or species: Temperature, solvent composition, pH, and whether you measure a neutral species or total solute can change the value being reported.
• Assuming the ratio predicts total amounts by itself: Phase volumes matter; use a mass balance to determine how much solute each phase contains.