Lesson 1 · 25 min

Extraction: partition equilibria and the separating funnel

Iodine is yellow-brown in water, but one shake with hexane turns the top layer violet. The partition coefficient tells you how much each extraction removes, why several small portions win, and how pH makes an acid or a base change layers.

In this lesson

Distinguish the partition coefficient $K_\mathrm{D}$ from the distribution ratio $D$ and write the fraction left in water after one extraction
Compute the result of repeated extractions and explain why several small portions beat one large one
Use acid–base speciation to predict how pH changes an extraction, and design the separation of acids, bases and neutral compounds
State the assumptions of the partition model: dilute solutions, immiscible phases, equilibrium reached

Pour a little hexane onto yellow-brown iodine water, stopper the tube, shake and let it stand: the top layer turns a vivid violet while the bottom layer nearly loses its colour. Nothing has reacted — the iodine has simply moved house, because $\ce{I2}$ is far more at home in non-polar hexane than in water. Brewing tea, pulling active compounds out of plants, recovering penicillin from fermentation broth and taking the caffeine out of coffee all rest on the same idea: solvent extraction.

Predict first

You have 30 mL of hexane to extract the iodine from 50 mL of iodine water. Which leaves less iodine behind?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

All 30 mL in one go Three portions of 10 mL The same — the total solvent is the same It depends on how hard you shake

Partition coefficient and distribution ratio

In 1891 Nernst found that when a solute is at equilibrium between two immiscible solvents at a fixed temperature, the ratio of its concentrations in the two phases is constant — the partition coefficient $K_\mathrm{D}$ . It is a true equilibrium constant and applies to one molecular form.

But solutes often exist in water in more than one form: a weak acid is both HA and $\ce{A-}$ , and iodine meeting $\ce{I-}$ becomes $\ce{I3-}$ . Usually only the neutral molecule crosses into the organic phase, while the analyst cares about how much has moved in total. That is the distribution ratio $D$ : the total concentration of all forms in the organic phase divided by the total in water. Without side reactions $D = K_\mathrm{D}$ ; with them, $D$ changes with pH or ligand concentration.

With an aqueous volume $V_\mathrm{aq}$ and an organic volume $V_\mathrm{org}$ , a mass balance gives the fraction $q$ left in water after one extraction.

K_\mathrm{D} = \frac{[\mathrm{S}]_\mathrm{org}}{[\mathrm{S}]_\mathrm{aq}}\qquad q = \frac{V_\mathrm{aq}}{V_\mathrm{aq} + D\,V_\mathrm{org}}\qquad q_n = \left(\frac{V_\mathrm{aq}}{V_\mathrm{aq} + D\,V_\mathrm{org}}\right)^{n}

Each fresh portion removes the same fraction, leaving q; after n portions qⁿ remains. The fraction extracted is E = 1 − q.

$K_\mathrm{D}$: partition coefficient: concentration ratio of one molecular form
$D$: distribution ratio: ratio of total concentrations of all forms
$V_\mathrm{aq}, V_\mathrm{org}$: volumes of water and of each organic portion
$q_n$: fraction left in water after n extractions

Virtual experiment · each partition equilibrium computed

Separating funnel: extracting iodine into hexane

KI in the water

Total hexane · 30 mL

Split into portions · 30.0 mL

No hexane added yet.

Water layer: 50 mL of 0.50 mM I₂

Distribution ratio D: 50
This portion took of what was left: —
Total in hexane so far: 0 %
Still in the water: 100 % · 0.5 mM

In 1 portions (computed)
Done
All 30 mL at once

q = V_aq/(V_aq + D·V_org) = 50/(50 + 50 × 30.0) = 0.032 · q¹ = 0.0323

Use all 30 mL in one go and note what is left; then compare five portions. Now redo it with 0.01 M KI in the water:

\ce{I3-}

holds the iodine back, D falls from 50 to about 6, and the hexane layer is visibly paler.

pH decides the layer

For a monoprotic weak acid HA only neutral HA enters the organic phase; charged $\ce{A-}$ is tightly wrapped in water. The organic phase holds $K_\mathrm{D}[\mathrm{HA}]_\mathrm{aq}$ while the water holds $[\mathrm{HA}] + [\ce{A-}]$ in total, so $D = K_\mathrm{D}\,\alpha_\mathrm{HA}$ — α being the species fraction from the acid–base unit.

Two units below the p $K_\mathrm{a}$ : 99 % is HA, $D \approx K_\mathrm{D}$ , and the acid goes into the organic phase;
at pH = p $K_\mathrm{a}$ : half is ionised and $D = K_\mathrm{D}/2$ ;
above the p $K_\mathrm{a}$ each pH unit cuts D about tenfold, and the acid stays in water as its anion.

Weak bases do the opposite: acidic water turns an amine into its ammonium ion and pulls it into the water; make the water basic and it returns to the organic phase. The “acid–base extraction” that separates benzoic acid, aniline and naphthalene in the organic lab is designed exactly this way.

D_\mathrm{HA} = K_\mathrm{D}\,\frac{[\ce{H+}]}{[\ce{H+}] + K_\mathrm{a}}\qquad D_\mathrm{B} = K_\mathrm{D}\,\frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}

Left: a monoprotic weak acid; right: a weak base B, with K_a that of its conjugate acid BH⁺. Only the neutral form is extracted.

$K_\mathrm{a}$: dissociation constant of the acid (or conjugate acid)
$[\ce{H+}]$: hydrogen-ion concentration of the water, 10⁻ᵖᴴ

Octanol–water partition · only neutral molecules cross

pH-controlled extraction

Solute A Solute B Aqueous pH · 7.0

V(organic) / V(aqueous)

A · Benzoic acid
B · Naphthalene

A Benzoic acid C₆H₅COOH ⇌ C₆H₅COO⁻

D: 0.118
Neutral fraction α: 0.0016
In the organic phase after one extraction: 10.6 %

B Naphthalene C₁₀H₈

D: 2000
Neutral fraction α: 1
In the organic phase after one extraction: 99.9 %

organic ←→ water

pH 7.0: A 10.6 % and B 99.9 % move into the organic phase — a partial separation; repeat or adjust the pH.

Separation factor D_A/D_B = 5.93 × 10⁻⁵

Drag the pH and find where benzoic acid and naphthalene separate in one step (one ≥ 99 % in the organic phase, the other ≤ 1 %). Then switch B to aniline: in water at pH 1 it stays behind as C₆H₅NH₃⁺.

Worked example

pH changes the extraction

100 mL of an aqueous benzoic acid solution is extracted once with 25 mL of 1-octanol. $K_\mathrm{D} = 10^{1.87} = 74.1$ and p $K_\mathrm{a} = 4.20$ . Find the fraction extracted with the water at pH 3.00 and at pH 6.00.

Concept check

An ether solution contains both benzoic acid (p $K_\mathrm{a}$ 4.20) and naphthalene. To move the benzoic acid into water and leave the naphthalene in the ether, which aqueous wash should you use?

Dilute hydrochloric acid (pH 1) Pure water (pH about 7) Saturated sodium bicarbonate (pH about 8.3) Another portion of hexane

Try the calculation

A solute has $D = 4.0$ between ether and water. 50 mL of its aqueous solution is extracted three times with 25 mL of ether each time. What percentage of the solute is left in the water?

Key takeaways

The partition coefficient $K_\mathrm{D}$ is an equilibrium constant for one form; the distribution ratio $D$ counts every form and changes with pH or ligands.
One extraction leaves $q = V_\mathrm{aq}/(V_\mathrm{aq} + DV_\mathrm{org})$ and n leave $q^n$ : splitting the same solvent is more thorough.
Weak acids extract below pH ≈ p $K_\mathrm{a}$ − 2 and stay in basic water as anions; weak bases do the opposite.
Side reactions such as $\ce{I2} + \ce{I-} \mathrel{\ce{<=>}} \ce{I3-}$ lower D even though $K_\mathrm{D}$ is unchanged.

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Data and references

D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Ch. 23 (solvent extraction; effect of pH).
D. A. Skoog, D. M. West, F. J. Holler, S. R. Crouch, Fundamentals of Analytical Chemistry, 9th ed. (2014), Ch. 31.
W. Nernst, “Verteilung eines Stoffes zwischen zwei Lösungsmitteln und zwischen Lösungsmittel und Dampfraum”, Z. Phys. Chem. 8, 110–139 (1891).
C. Hansch, A. Leo, D. Hoekman, Exploring QSAR: Hydrophobic, Electronic and Steric Constants (ACS, 1995); J. Sangster, Octanol–Water Partition Coefficients (Wiley, 1997).
C. A. Lipinski, F. Lombardo, B. W. Dominy, P. J. Feeney, Adv. Drug Deliv. Rev. 23, 3–25 (1997).
Triiodide constant K = 698: D. A. Palmer, R. W. Ramette, R. E. Mesmer, J. Solution Chem. 13, 673 (1984).