Partition Coefficient Calculator
Enter the equilibrium concentrations of a solute in the organic and aqueous phases to get the partition coefficient P, log P, and the percentage of solute in each layer.
Find How a Solute Divides Between Two Immiscible Liquids
Partition Coefficient Formula
Where:
- \(P\) — partition coefficient (dimensionless)
- \(C_{org}\) — equilibrium concentration of the solute in the organic phase (mol/L)
- \(C_{aq}\) — equilibrium concentration of the solute in the aqueous phase (mol/L)
- \(\log P\) — base-10 logarithm of \(P\), the standard way of reporting lipophilicity
The formula expresses the equilibrium distribution of a solute between two immiscible liquids: the bigger the ratio, the stronger the solute's preference for the organic (non-polar) layer over water. Because both concentrations are converted to the same unit before dividing, \(P\) itself has no units.
When equal volumes of the two phases are equilibrated, the fraction of solute ending up in each layer follows directly from \(P\):
\[ f_{org} = \frac{P}{1+P} \qquad f_{aq} = \frac{1}{1+P} \]
Partition Coefficient – Calculation Example
Example: after shaking a water sample with an equal volume of an organic solvent, analysis shows the solute at 12 mmol/L in the organic layer and 3 mmol/L in the water layer.
\[ C_{org} = 12\ \text{mM} = 0.012\ \text{mol/L}, \quad C_{aq} = 3\ \text{mM} = 0.003\ \text{mol/L} \]
\[ P = \frac{0.012}{0.003} = 4 \]
\[ \log P = \log_{10} 4 \approx 0.602 \]
\[ f_{org} = \frac{4}{1+4} = 0.80 = 80\%, \quad f_{aq} = 20\% \]
Result: \(P = 4\), \(\log P \approx 0.60\). With equal phase volumes, 80 % of the solute partitions into the organic layer and 20 % remains in water — the compound is mildly lipophilic.
How to Use the Calculator
- Enter the equilibrium Concentration in Organic Phase and pick its unit (mol/L, mmol/L or µmol/L) from the dropdown.
- Enter the equilibrium Concentration in Aqueous Phase with its unit. Both values must describe the same solute at the same temperature after the layers have equilibrated.
- Read the results: Partition Coefficient, log P, the percentage of solute in each phase (for equal volumes), plus an automatic interpretation and a step-by-step calculation breakdown.
What Is the Partition Coefficient?
When a solute is shaken with two immiscible liquids — typically water and an organic solvent such as octanol, ether or dichloromethane — it distributes between the two layers until equilibrium is reached. The partition coefficient \(P\) (also written \(K_D\), the distribution coefficient) is the ratio of the solute's equilibrium concentration in the organic phase to that in the aqueous phase. It is a thermodynamic property of the solute–solvent pair at a given temperature: it does not depend on how much solute or solvent you use, only on their chemical identities.
A \(P\) greater than 1 means the solute prefers the organic layer; a \(P\) less than 1 means it prefers water. Because partition coefficients of real compounds span many orders of magnitude, chemists almost always report \(\log P\). A drug candidate with \(\log P\approx 2\), for instance, is a thousand times more concentrated in octanol than in water at equilibrium — a useful proxy for how easily it crosses lipid membranes. The classic reference system is 1-octanol/water, and tabulated \(\log P\) values usually refer to it unless stated otherwise.
The percentages shown by the calculator describe a single equilibration of equal phase volumes: 80 % organic / 20 % aqueous in the example above. If the phase volumes differ, the mass balance changes — with organic volume \(V_{org}\) and aqueous volume \(V_{aq}\), the fraction remaining in water is \(1/(1 + P V_{org}/V_{aq})\). The same logic underlies liquid–liquid extraction: a solute with a high \(P\) can be pulled nearly quantitatively into a small volume of organic solvent.
Interpreting log P Values
| log P range | Typical behaviour |
|---|---|
| < −3 | Very hydrophilic; essentially all solute stays in water |
| −3 to 0 | Hydrophilic; favours the aqueous phase |
| 0 to 1 | Mildly lipophilic; roughly balanced |
| 1 to 3 | Moderately lipophilic; many oral drugs fall here |
| 3 to 5 | Strongly lipophilic; accumulates in fatty tissue and membranes |
| ≥ 5 | Very lipophilic; poorly absorbed, bioaccumulation risk |
Assumptions, Limitations and Practical Tips
- Dilute, ideal solutions. The simple ratio is constant only at low concentrations; at high concentrations activity effects make the apparent \(P\) drift.
- One chemical form only. \(P\) assumes the solute does not ionise, associate or react in either phase. For ionisable compounds measured at a fixed pH, the ratio is called the distribution coefficient \(D\) (log D); it depends on pH through the Henderson–Hasselbalch relationship.
- Equilibrium. The input concentrations must be measured after thorough mixing and phase separation, at a fixed temperature — \(P\) changes with temperature just like any other equilibrium constant.
- Several small extractions beat one large one. Shaking the aqueous layer with \(n\) equal portions of fresh organic solvent leaves \((1/(1+P))^n\) of the solute in water — the classic liquid–liquid extraction trick.
- Equal volumes assumed for the percentages. The reported organic/aqueous percentages assume \(V_{org} = V_{aq}\); adjust manually for other volume ratios using the formula above.
Related Calculators
You may also find the Chromatography Resolution Calculator and the TLC Rf Calculator useful — both techniques separate compounds partly on the basis of the same partitioning behaviour described by \(P\).
Frequently Asked Questions
How do I calculate the partition coefficient by hand?
Shake the solute with the two immiscible solvents until equilibrium, separate the layers, and measure the solute concentration in each. Then divide: \( P = C_{org}/C_{aq} \). Make sure both concentrations are expressed in the same unit (e.g. mol/L) before dividing — 12 mmol/L over 3 mmol/L gives \( P = 4 \), just like 0.012 mol/L over 0.003 mol/L. Take \( \log_{10} P \) if you want log P.
What is a good or typical log P value?
There is no universal "good" value — it depends on the purpose. For oral drugs, Lipinski's rule of thumb suggests log P below 5; values between 1 and 3 often balance membrane penetration with solubility. Environmental chemists get concerned above roughly 4–5 because such compounds bioaccumulate in fatty tissue. For extraction work, a higher log P simply means the solute is easier to pull into the organic solvent.
What is the difference between log P and log D?
log P describes a single, unionised chemical form distributing between the phases. log D is measured at a fixed pH and includes all forms of an ionisable compound — ionised plus neutral. Since charged species stay in water, log D is usually lower than log P at pH values where the compound ionises, and it changes with pH. If your solute is an acid or base, decide which quantity you actually need before comparing values.
Do the two concentrations have to be in the same unit?
Yes and no — the calculator converts mol/L, mmol/L and µmol/L automatically, so you can enter them in different units and still get the correct ratio. But if you compute P manually, convert both to the same unit first. A common mistake is entering, say, 12 mM and 3 µmol/L and dividing the raw numbers, which understates or overstates P by a factor of 1000.
Why does the calculator refuse a zero aqueous concentration?
If nothing detectable dissolves in water, \( P = C_{org}/C_{aq} \) would divide by zero and tend to infinity. In practice this means the solute is so lipophilic (or the water layer so free of it) that the ratio cannot be measured this way — you would need a more sensitive analytical method or a different experimental design. At least one non-zero concentration is required for any result.
How do I handle phase volumes that are not equal?
The percentages from the calculator assume \( V_{org} = V_{aq} \). For unequal volumes, the fraction of solute remaining in water is \( 1/(1 + P \cdot V_{org}/V_{aq}) \). Doubling the organic volume has the same effect as squaring P for a single extraction, which is why using more solvent — or several smaller portions — extracts more solute from water.
What common mistakes distort a measured partition coefficient?
Frequent pitfalls include measuring before equilibrium is reached (shake thoroughly and give it time), letting temperature drift between measurements, using concentrations so high that activity effects bend the ratio, ignoring ionisation of the solute, and not correcting for the solute's own solubility in the mutual saturation of the two solvents. Also double-check which layer is which — dense solvents like dichloromethane form the bottom layer, while ether or octanol float on water.