Lipophilicity: LogP, LogD, and Octanol-Water Partitioning

Why Two Lipophilicity Measures Are Needed

A compound placed in a two-phase system containing water and octanol may distribute between the phases. The concentration ratio indicates which phase contains more compound per unit volume. For a neutral compound, the octanol-water partition coefficient provides a pH-independent description under specified conditions. For an ionizable compound, however, the total amount in each phase can depend on the aqueous pH because the compound may exist in more than one charge state. The pH-specific distribution coefficient describes that measured, overall distribution.

The distinction matters when interpreting physicochemical data for PBPK applications: a neutral-species partition measurement and a total-concentration distribution measurement are not interchangeable by default. Both are ratios based on octanol and water, but they refer to different concentration definitions.

Definition: The octanol-water partition coefficient \(P\) is the equilibrium concentration of a neutral solute in the octanol phase divided by its equilibrium concentration in the aqueous phase. Its base-10 logarithm is \(\log P=\log_{10}(P)\). The pH-specific distribution coefficient \(D_{\mathrm{pH}}\) is the ratio of the total equilibrium concentration of the compound in octanol to its total equilibrium concentration in the aqueous phase at a stated aqueous pH. Its logarithm is \(\log D_{\mathrm{pH}}=\log_{10}(D_{\mathrm{pH}})\). For an ionizable compound, the totals include all relevant forms of the compound in each phase.

Concentration Ratios and Their Logarithms

The ratio is always written octanol concentration divided by aqueous concentration. Reversing that order changes the sign of the logarithm. Concentrations in the two phases must use compatible units, such as mol/L in both. Since the units cancel, \(P\) and \(D_{\mathrm{pH}}\) are dimensionless; their logarithms are dimensionless as well.

P=\frac{C_{\mathrm{oct,neutral}}}{C_{\mathrm{aq,neutral}}} \qquad\text{and}\qquad \log P=\log_{10}\left(\frac{C_{\mathrm{oct,neutral}}}{C_{\mathrm{aq,neutral}}}\right)

For logD, the concentrations are totals rather than concentrations of the neutral form alone. The pH must be reported because it defines the aqueous condition under which that total distribution was measured.

D_{\mathrm{pH}}=\frac{C_{\mathrm{oct,total}}}{C_{\mathrm{aq,total}}} \qquad\text{and}\qquad \log D_{\mathrm{pH}}=\log_{10}\left(\frac{C_{\mathrm{oct,total}}}{C_{\mathrm{aq,total}}}\right)

A ratio greater than 1 gives a positive logarithm, a ratio equal to 1 gives a logarithm of 0, and a ratio between 0 and 1 gives a negative logarithm. For example, a ratio of 10 corresponds to a log value of 1, while a ratio of 0.1 corresponds to −1. A log value is therefore not itself a concentration or a percentage.

Worked Example: Calculating logP for a Neutral Compound

In a hypothetical equilibrium experiment, a neutral compound has an octanol concentration of \(3.20\times10^{-4}\) mol/L and an aqueous concentration of \(8.00\times10^{-5}\) mol/L. Calculate \(P\) and \(\log P\). Assume the reported concentrations refer to the same compound at equilibrium, and that the compound remains neutral under the stated experimental conditions.

First divide the octanol concentration by the aqueous concentration:

P=\frac{3.20\times10^{-4}\ \mathrm{mol/L}}{8.00\times10^{-5}\ \mathrm{mol/L}} =\frac{3.20}{8.00}\times10 =4.00

The units cancel, so \(P=4.00\) is dimensionless. Taking the base-10 logarithm gives:

\log P=\log_{10}(4.00)=0.6021

At equilibrium, the neutral compound concentration per unit volume is four times as high in octanol as in water. This concentration ratio does not, on its own, state what fraction of the total experimental amount is in either phase; that fraction would also depend on the phase volumes.

Neutral and Ionizable Compounds

For a compound that is neutral in both phases, the total concentration in each phase is the neutral concentration. If the measurements use the same compound and conditions, the logD ratio therefore reduces to the logP ratio. This is a consequence of the concentration definitions, not a separate pH adjustment.

For an ionizable compound, the conventional meaning of logP is the partitioning of the neutral form. LogD instead uses total concentrations in each phase at a specified aqueous pH. Changing pH can change the relative amounts of the compound’s forms and thereby change the total concentration ratio. The size and direction of the change depend on the compound and experimental conditions; a logD value must be tied to its stated pH.

LogP and logD can happen to have the same numerical value under particular conditions, but that numerical agreement does not make their definitions identical. Nor does a logD measurement alone identify the concentrations of the individual forms. The detailed relationship to ionization requires information beyond the concentration ratio itself.

Key result: If only the neutral form is present and the measurements refer to the same equilibrium system, then \(C_{\mathrm{oct,total}}=C_{\mathrm{oct,neutral}}\) and \(C_{\mathrm{aq,total}}=C_{\mathrm{aq,neutral}}\). Substitution into the definitions gives \(D_{\mathrm{pH}}=P\), and therefore \(\log D_{\mathrm{pH}}=\log P\). For an ionizable compound with multiple forms present, this equality is not guaranteed.

Worked Example: Calculating logD at a Stated pH

A hypothetical ionizable compound is equilibrated between octanol and buffered water at aqueous pH 7.4. The measured total concentrations are \(1.50\times10^{-5}\) mol/L in octanol and \(3.00\times10^{-4}\) mol/L in water. Assume equilibrium has been reached, the buffer maintains the stated pH, and both measurements include all relevant forms in their respective phases. Calculate \(D_{7.4}\) and \(\log D_{7.4}\).

Use the total concentrations and keep the ratio in the defined direction:

D_{7.4} =\frac{1.50\times10^{-5}\ \mathrm{mol/L}} {3.00\times10^{-4}\ \mathrm{mol/L}} =0.0500

Then take the base-10 logarithm:

\log D_{7.4}=\log_{10}(0.0500)=-1.301

The total concentration per unit volume is lower in octanol than in the aqueous phase at pH 7.4. The result is properly reported as \(\log D_{7.4}=-1.301\), not simply as an unqualified logD when the pH condition is important. It does not establish the compound’s logP, because the calculation used total rather than neutral-form concentrations.

A Short Derivation: Why a Neutral Compound Has Matching Values

The equality for a neutral-only system follows directly from the definitions. For clarity, take concentrations in compatible units and assume both ratios refer to the same equilibrium conditions. If the compound is neutral and no other form is present, the total concentration in either phase is just the concentration of that neutral form:

C_{\mathrm{oct,total}}=C_{\mathrm{oct,neutral}}, \qquad C_{\mathrm{aq,total}}=C_{\mathrm{aq,neutral}}

Substitute these equalities into the definition of the distribution coefficient:

D_{\mathrm{pH}} =\frac{C_{\mathrm{oct,total}}}{C_{\mathrm{aq,total}}} =\frac{C_{\mathrm{oct,neutral}}}{C_{\mathrm{aq,neutral}}} =P

Since equal positive ratios have equal base-10 logarithms, \(\log D_{\mathrm{pH}}=\log P\). The argument depends on the neutral-only condition. If total concentrations include additional forms, the substitution no longer follows merely from the definitions.

Worked Example: Comparing logD at Two pH Conditions

In a hypothetical set of measurements on the same ionizable compound, the total octanol concentration is \(2.40\times10^{-4}\) mol/L at both test conditions. The measured total aqueous concentration is \(3.00\times10^{-4}\) mol/L at pH 3.0 and \(6.00\times10^{-3}\) mol/L at pH 8.0. Assume each measurement reached equilibrium, the respective aqueous pH was maintained, and the same total-concentration basis was used in all phases. Calculate the two logD values and compare them.

At pH 3.0:

D_{3.0} =\frac{2.40\times10^{-4}}{3.00\times10^{-4}} =0.800 \qquad \log D_{3.0}=\log_{10}(0.800)=-0.09691

At pH 8.0:

D_{8.0} =\frac{2.40\times10^{-4}}{6.00\times10^{-3}} =0.0400 \qquad \log D_{8.0}=\log_{10}(0.0400)=-1.398

The measured logD is lower at pH 8.0 in this hypothetical data set. The calculation supports a comparison of the overall phase ratios under these two specified conditions. It does not, without further information, quantify the fraction of each ionization state or establish a value for logP.

Measurement Checks and Common Pitfalls

A calculated logarithm is only as interpretable as its concentration measurements. The ratio describes phase concentrations at equilibrium, not the ratio of amounts in the two vessels. If phase volumes differ, amounts are concentration multiplied by volume; using amounts directly in place of concentrations changes the quantity being calculated.

Before interpreting a reported value, check that the measurements are for octanol and water, that the numerator and denominator have not been reversed, and that both concentrations use the same compatible units and total-versus-neutral basis. For logD, confirm the aqueous pH and whether it was maintained during equilibration. Temperature and other experimental conditions should also be documented when comparing measurements.

Potential warning signs include poor recovery of the compound across the phases, failure to reach equilibrium, pH drift, precipitation, degradation, or loss to surfaces or an emulsion. These processes can make measured concentrations unrepresentative of the intended equilibrium ratio. A ratio can still be computed from such numbers, but the arithmetic alone cannot validate the experiment.

Reversing the ratio: using aqueous divided by octanol yields the reciprocal coefficient and reverses the sign of its logarithm.
Using amounts instead of concentrations: this ignores phase volumes and does not calculate the defined partition or distribution coefficient.
Dropping the pH from logD: an unqualified value can conceal the condition needed to interpret a pH-dependent measurement.
Calling total-concentration logD “logP”: logP conventionally refers to the neutral form; the total ratio at stated pH is logD.

These descriptors concern partitioning between octanol and water. They are not, by definition, tissue-to-plasma partition coefficients, and a log value is not itself a PBPK tissue-partition parameter. Moving from an octanol-water measurement to a model parameter requires its own definition and method.

Check Your Understanding

For each question, calculate or explain the result. Solutions follow the questions.

A neutral compound has equilibrium concentrations of \(6.00\times10^{-5}\) mol/L in octanol and \(2.00\times10^{-5}\) mol/L in water. Calculate \(P\) and \(\log P\).
At aqueous pH 6.5, an ionizable compound has total concentrations of \(4.00\times10^{-5}\) mol/L in octanol and \(8.00\times10^{-4}\) mol/L in water. Calculate \(D_{6.5}\) and \(\log D_{6.5}\).
Why must a logD value for an ionizable compound be reported with its aqueous pH?
An analyst calculates aqueous concentration divided by octanol concentration instead of octanol divided by aqueous concentration. What changes?

Solutions. (1) \(P=(6.00\times10^{-5})/(2.00\times10^{-5})=3.00\), so \(\log P=\log_{10}(3.00)=0.4771\). (2) \(D_{6.5}=(4.00\times10^{-5})/(8.00\times10^{-4})=0.0500\), so \(\log D_{6.5}=\log_{10}(0.0500)=-1.301\). (3) The total ratio for an ionizable compound can depend on aqueous pH; reporting pH identifies the condition for the measured value. (4) The analyst obtains the reciprocal ratio: the log changes sign. For example, a defined ratio of 4 has log 0.6021, while its reciprocal, 0.25, has log −0.6021.