Partition, Distribution Ratio, and the Distribution Constant

Solvent extraction built directly on $K_D$ and $D$ remains the workhorse of nuclear fuel reprocessing (the PUREX process separates uranium and plutonium as tributyl-phosphate complexes with carefully engineered distribution ratios), of rare-earth element purification, and of pharmaceutical downstream processing, where an active ingredient is repeatedly partitioned out of a fermentation broth into an organic solvent before crystallization. In drug discovery, the octanol–water partition coefficient $\log P$ — precisely a $K_D$ for the neutral drug species — is a core Lipinski "rule of five" descriptor predicting oral bioavailability and membrane permeability, while the pH-dependent distribution ratio $\log D_{7.4}$ (evaluated at physiological pH) is the quantity that actually predicts how much of an ionizable drug crosses a membrane at blood pH. Solid-phase extraction (SPE) cartridges used in environmental and clinical sample prep exploit the same equilibrium logic with a solid stationary phase replacing the second liquid. Every chromatographic separation — gas, liquid, ion, size-exclusion — is fundamentally the partitioning equilibrium of this article, replayed continuously as a solute travels down a column and repeatedly re-equilibrates between a mobile phase and a stationary phase, exactly the countercurrent-style repetition analyzed in Example 3, taken to the limit of an enormous number of infinitesimally small theoretical plates. Retention factors reported by chromatography software, $k' = K_D \cdot (V_s/V_m)$, are a direct restatement of the phase-ratio-weighted distribution constant developed here. Environmental chemists use the octanol–water $K_{ow}$ to model how pollutants partition between water, soil organic matter, and biological lipids, predicting bioaccumulation in fish and sediment. In every case, the analyst's actual lever is not the fixed thermodynamic $K_D$ but the tunable, chemistry-dependent $D$ — controlled through pH, ionic strength, temperature, and complexing or masking agents.

Ask the tutor: How do I choose the extracting solvent and pH to maximize D for a specific ionizable analyte in a real separation?

Consider two immiscible liquids in contact — water and an organic solvent, say — with a solute $A$ dissolved in the system. At the molecular scale, $A$ molecules are constantly crossing the interface in both directions. Initially, if all the $A$ starts in the aqueous phase, the net flux is from water into the organic phase, because that's the only direction available. As organic-phase $A$ builds up, molecules also start returning to water. Equilibrium is reached not when the crossing stops, but when the rate of $A$ leaving water equals the rate of $A$ returning to it — a dynamic balance identical in spirit to any other chemical equilibrium, whether it is a proton-transfer or a redox reaction. This is exactly the same kind of process treated in Chemical equilibrium: a reaction (here, a phase transfer rather than a bond-breaking transformation) proceeds until the chemical potential of $A$ is equal in both phases. Because chemical potential depends on concentration (strictly, activity — see Activity and concentration), equality of chemical potential translates into a fixed ratio of concentrations at a given temperature. That ratio is the partition constant. Crucially, $K_D$ applies to one specific chemical species. If $A$ can also protonate, dimerize, or complex with a metal ion, each of those derived species has its own equilibria and, in principle, its own partition behavior. The distribution ratio $D$ is the pragmatic, analytically important quantity: the ratio of the total analytical concentration of all $A$-containing species in phase 2 to the total in phase 1, regardless of chemical form. When $A$ exists as only one species in each phase, $D = K_D$ exactly. The moment side equilibria — acid dissociation, complexation, ion association — enter the picture, $D$ becomes a function of pH, ligand concentration, and ionic strength, while $K_D$ for the neutral species itself stays fixed.

Extraction efficiency follows directly from $D$ and the phase-volume ratio. Shake an aqueous sample with an organic solvent, let the phases separate, and some fraction of the solute leaves with the organic layer. Repeating the extraction with fresh solvent — rather than using one large volume — exploits the same equilibrium math multiple times, and (as the derivation below shows) strips out more total analyte for the same total solvent consumption. This "divide and re-equilibrate" logic, iterated many times over infinitesimally small stages, is precisely what a chromatographic column does as a solute repeatedly partitions between a stationary phase and a flowing mobile phase — the connection foreshadowed for Principles of Chromatographic Separation and developed fully for multistage extraction in Solvent Extraction and Countercurrent Distribution.

The thermodynamically rigorous partition constant, written in terms of activity,

where: $a_{A,2}$ and $a_{A,1}$ are the activities of the identical chemical species $A$ in phase 2 and phase 1 at equilibrium, and $K_D$ is dimensionless (a true equilibrium constant). In dilute solution, activity coefficients are close to unity and cancel approximately between the two expressions for each phase, giving the practical form

where: $[A]_2$, $[A]_1$ are molar equilibrium concentrations of the same species $A$ in each phase. Reading: $K_D$ is fixed by the chemistry of $A$ and the two solvents (and by $T$, $P$) — it is exactly analogous to any other equilibrium constant, e.g. $K_a$ or $K_{sp}$, and does not depend on how much $A$ you started with or on side reactions. The distribution ratio generalizes this to the whole analytical picture:

where: $C_{A,2}$ and $C_{A,1}$ are the total (analytical) concentrations of every chemical form of $A$ — protonated, deprotonated, complexed, dimerized — in phase 2 and phase 1 respectively. Reading: $D$ is what a balance or a spectrophotometer actually measures after extraction; it equals $K_D$ if and only if $A$ exists as a single, non-reacting species in both phases. The fraction of solute remaining unextracted after a single extraction, and after $n$ successive extractions, are given by

where: $V_1$ is the volume of the original (usually aqueous) phase, $V_2$ is the volume of extracting solvent used per stage, $D$ is the distribution ratio under the conditions of the extraction, and $q_n$ is the fraction of the original analyte left in phase 1 after $n$ repeated extractions each using fresh volume $V_2$. Reading: extraction efficiency improves as $D$ increases, as $V_2/V_1$ increases, and — dramatically — as $n$ increases at fixed total solvent volume, which is the mathematical root of why chromatography (many, many tiny equilibrations) vastly outperforms one giant separatory-funnel shake.

Example 1 — Foundational: iodine partitioning between water and CCl₄ (single species, so $D = K_D$). At 25 °C, the partition constant for I₂ between CCl₄ and water is reported as $K_D = [\text{I}_2]_{\text{CCl}_4}/[\text{I}_2]_{\text{H}_2\text{O}} = 85$ (a realistic literature value; iodine is not appreciably ionized or hydrolyzed at low concentration, so a single molecular species dominates in both phases and $D=K_D$). Suppose $25.0\text{ mL}$ of an aqueous solution containing $1.00\times10^{-3}\text{ M}$ I₂ ($n_0 = 25.0\times10^{-3}\text{ L} \times 1.00\times10^{-3}\text{ M} = 2.50\times10^{-5}\text{ mol}$) is shaken once with $10.0\text{ mL}$ of CCl₄. Using $q = V_1/(V_1+DV_2)$ with $V_1 = 25.0\text{ mL}$, $V_2 = 10.0\text{ mL}$, $D=85$:

So only $2.86\%$ of the iodine remains in the aqueous phase, and $97.1\%$ has moved into the CCl₄ layer after one equilibration. In moles, the aqueous phase retains

corresponding to a remaining aqueous concentration $[\text{I}_2]_1 = 7.14\times10^{-7}\text{ mol}/0.0250\text{ L} = 2.86\times10^{-5}\text{ M}$ — three significant figures throughout, consistent with a two-sig-fig $K_D$ value used as the limiting precision. Limiting-case check: if instead $D \to 1$ (a solute with no preference between phases), $q = 25.0/(25.0+10.0) = 0.714$ — the extraction merely dilutes the solute according to the volume ratio, exactly as expected when partitioning carries no thermodynamic driving force.

Example 2 — Applied: extracting a weak organic acid, where $D$ depends on pH ($D \neq K_D$). Benzoic acid ($\text{HA}$, $K_a = 6.3\times10^{-5}$, i.e. $\text{p}K_a = 4.20$) is to be extracted from water into toluene. Only the neutral form $\text{HA}$ partitions into the organic phase; the anion $\text{A}^-$ is far too polar to leave water in any appreciable amount, so its partition coefficient is taken as zero. The intrinsic partition constant of the neutral acid is $K_D = [\text{HA}]_{\text{org}}/[\text{HA}]_{\text{aq}} = 180$ (representative literature-range value for benzoic acid/toluene). The measured distribution ratio must account for both aqueous forms:

At pH 2.00 ($[\text{H}^+] = 1.00\times10^{-2}\text{ M}$), the ratio $K_a/[\text{H}^+] = 6.3\times10^{-5}/1.00\times10^{-2} = 6.3\times10^{-3}$, so

Essentially all the acid stays protonated at pH 2.00 (two pH units below p$K_a$), so $D \approx K_D$, as expected. At pH 6.00 ($[\text{H}^+] = 1.00\times10^{-6}\text{ M}$), $K_a/[\text{H}^+] = 6.3\times10^{-5}/1.00\times10^{-6} = 63.0$, giving

The distribution ratio has collapsed by a factor of $64$ simply by raising the pH four units above pH 2 — nothing about the intrinsic chemistry of benzoic acid ($K_D=180$, $K_a=6.3\times10^{-5}$) has changed. This is the operational meaning of "$D$ is tunable, $K_D$ is not": at pH 6.00 most of the analyte exists as $\text{A}^-$, which never leaves the water, so the apparent organic preference collapses even though $K_D$ for the neutral species alone is unchanged. Extraction efficiency at pH 6.00, using equal volumes $V_1=V_2=25.0\text{ mL}$ (ratio $V_2/V_1=1$):

so only $73.7\%$ of the benzoic acid is extracted per stage at pH 6, versus

i.e. $99.4\%$ extracted per stage at pH 2. Controlling pH — not changing solvent — is the standard analytical strategy for extracting or suppressing extraction of ionizable analytes. Limiting-case check: as $\text{pH}\to -\infty$ (very strongly acidic, $[\text{H}^+]\to\infty$), $K_a/[\text{H}^+]\to 0$ and $D\to K_D=180$ exactly, confirming that in the fully protonated limit the distribution ratio reduces to the true partition constant, as it must for a single dominant species.

Example 3 — Exam/competition level: how many extractions to reach 99.9% removal, and does splitting the solvent actually help? A contaminant has $D = 8.00$ between chloroform and water. You have $V_1 = 50.0\text{ mL}$ of aqueous sample and a total of $V_2^{\text{tot}} = 50.0\text{ mL}$ of chloroform available. Compare (a) one extraction using the full $50.0\text{ mL}$, (b) two extractions of $25.0\text{ mL}$ each, and (c) five extractions of $10.0\text{ mL}$ each. Then find the number of $10.0\text{ mL}$ extractions needed to leave less than $0.10\%$ of contaminant behind. (a) Single extraction, $V_2 = 50.0\text{ mL}$:

leaving $11.1\%$ behind — $88.9\%$ removed. (b) Two extractions of $25.0\text{ mL}$ each ($v = 25.0\text{ mL}$ per stage):

leaving $4.00\%$ behind — $96.0\%$ removed, better than the single large extraction with the same total solvent. (c) Five extractions of $10.0\text{ mL}$ each ($v=10.0\text{ mL}$):

Computing stepwise: $0.3846^2 = 0.1479$, $0.3846^3 = 0.05690$, $0.3846^4 = 0.02189$, $0.3846^5 = 8.42\times10^{-3}$. So $q_5 = 0.842\%$ remains — $99.16\%$ removed, using the same total $50.0\text{ mL}$ of chloroform as case (a), but distributed over five stages instead of one. This confirms the derivation: subdividing the extracting solvent and re-equilibrating repeatedly strips out substantially more analyte for the same total solvent budget. For the second part, find $n$ (using $10.0\text{ mL}$ portions, $q^{(1)}=0.3846$ per stage as above) such that $q_n < 0.00100$:

With $\ln(0.3846) = -0.9556$ and $\ln(10^{-3}) = -6.908$,

So $n=8$ extractions of $10.0\text{ mL}$ each (a total of $80.0\text{ mL}$ chloroform — more than the original $50.0\text{ mL}$ budget, illustrating the real trade-off: smaller stages need more total solvent to reach a stringent target, but the marginal return per milliliter is far higher than in a single giant extraction). Checking $n=8$: $q_8 = (0.3846)^8$. Continuing the power series, $0.3846^6=3.239\times10^{-3}$, $0.3846^7=1.246\times10^{-3}$, $0.3846^8=4.79\times10^{-4}$ — indeed below $0.00100$, while $n=7$ gives $1.246\times10^{-3}$, just above the target, confirming $n=8$ is the minimum integer number of stages required.

The modern theory of two-phase distribution equilibria traces to Walther Nernst, who in 1891 published the "distribution law" describing how a solute partitions between two immiscible solvents in contact — the direct ancestor of the $K_D$ used throughout this article. Nernst arrived at the law from strict thermodynamic reasoning about chemical potential equality across phases, treating the partitioning solute the same way he treated ions in his electrochemical work on the Nernst equation. He insisted on rigorous concentration/activity bookkeeping, in an era when many "distribution" measurements were reported with no correction for the dissociation or association of the solute in either phase — precisely the confusion between $K_D$ and $D$ that this article works to prevent. Nernst's distribution law became foundational to the physical chemistry of solutions and, decades later, to Archer Martin and Richard Synge's chromatographic partition theory, for which they won the 1952 Nobel Prize in Chemistry — explicitly citing Nernst's distribution law as the starting point for treating a chromatographic column as a very large number of discrete partitioning stages.