Extraction: use partitioning and pH to separate a mixture

MCAT content library / Chemistry & physics / Extraction: use partitioning and pH to separate a mixture

Predict where a compound goes by tracking its chemical form, phase volumes, and the amount left after each extraction.

About 14 minutes, including the worked examples and checks.

Useful foundations

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Concept: identify the two phases before following the solute
Pattern: combine partitioning with conservation of amount
Rule: count all chemical forms when pH can change
Application: separate an acid from a neutral companion
Check your reasoning
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Concept: identify the two phases before following the solute

Liquid–liquid extraction separates compounds by allowing them to distribute between two immiscible liquids. A useful prediction names the target, its chemical form, and the phase to collect. Extraction is part of the MCAT separation-method scope; acid–base reasoning adds a way to change that distribution deliberately.¹

The aqueous phase contains water; the organic phase contains an organic solvent. Neither name determines which layer is on top. The denser liquid is the lower layer. A dissolved compound can be present in both layers even when one layer contains most of it. Molecular interactions influence its preference, but a supplied partition value is more precise than a general statement that a substance is “polar.”²

For one unchanged chemical form, define K = [solute]_org/[solute]_aq at the stated temperature and solvent composition. The numerator convention matters: some problems define the reciprocal. K compares equilibrium concentrations, not amounts. A large organic volume and a small organic volume can have the same concentration ratio while collecting different numbers of moles.

Pattern: combine partitioning with conservation of amount

Consider an original illustrative model: 90 μmol of neutral solute N begins in 30 mL of water. A fresh 15 mL organic portion is added. N has K = 4, does not ionize, and undergoes no other reaction. Assume equilibrium is reached, phase volumes remain unchanged, and both solutions remain below their solubility limits.

Let x μmol remain in water. Then the organic phase contains 90 − x μmol. Substituting amounts divided by their own volumes gives 4 = [(90 − x)/15]/[x/30]. Thus 4x = 180 − 2x, so x = 30 μmol. The organic extract contains 60 μmol, or two thirds of the starting amount. A fourfold concentration preference did not produce four fifths recovery because the phase volumes were unequal.

Quantity	Aqueous phase	Organic phase
Volume	30 mL	15 mL
Amount after equilibration	30 μmol	60 μmol
Concentration	1.0 μmol/mL	4.0 μmol/mL
Fraction of starting N	33.3%	66.7%

One extraction of neutral N: original illustrative values

The same conservation argument gives a reusable fraction: f_aq = V_aq/(V_aq + K V_org). The organic fraction is 1 − f_aq. Use matching volume units; the fraction is dimensionless. This expression assumes the same molecular form is being counted in both phases.²

Why two fresh portions can collect more

Now divide the same 15 mL of organic solvent into two fresh 7.5 mL portions. Each extraction leaves the fraction 30/(30 + 4 × 7.5) = 1/2 of the N present in water immediately before that extraction. The first leaves 45 μmol; the second leaves 22.5 μmol. Together the removed organic portions contain 67.5 μmol, or 75% of the original N.

The second fraction acts on the remaining amount, not the original 90 μmol. That is why the retained fractions multiply. Recombining the first organic portion with the same water would simply restore the already established equilibrium; it would not act like fresh solvent. Several smaller portions can improve recovery for a fixed total extracting volume under this model, although extra handling can introduce losses in an actual experiment.³

Rule: count all chemical forms when pH can change

Introduce a weak acid HA with pK_a = 4.5. Its neutral form has P = [HA]_org/[HA]_aq = 8. In this model A⁻ has negligible organic solubility, while aqueous pH is maintained by a sufficiently strong buffer. The acid’s total amount includes both HA and A⁻; proton transfer changes form without destroying the acid’s molecular skeleton.

In water, [A⁻]/[HA] = 10^{pH − pKa}. At pH 6.5 the ratio is 100; at pH 2.5 it is 0.01. This follows from the acid-dissociation expression. “Weak acid” describes its ionization equilibrium, not a rule that it must remain neutral at every pH.⁴

Define the total distribution ratio D = [acid pool]_org/[acid pool]_aq. Substituting the two aqueous forms gives D = P/(1 + 10^{pH − pKa}). P still describes neutral HA; D changes because the denominator now includes aqueous A⁻. With equal phase volumes, the fraction of the total acid pool in the organic layer is D/(1 + D).

Aqueous pH	[A−]/[HA] in water	Total D	Acid pool in organic phase
2.5	0.01	7.92	88.8%
6.5	100	0.0792	7.3%

Equal 20 mL phases: calculated distribution of the illustrative acid

Layer position follows the supplied densities. Percentages are fractions of each compound’s total amount, not percentages of the whole mixture.

Application: separate an acid from a neutral companion

Suppose N and the acid begin together, and the organic liquid has density 1.15 g/mL while the aqueous phase has density 1.00 g/mL. At pH 6.5, collecting the upper aqueous layer captures most of the acid pool. N, still governed by K = 4, has 80% of its amount in the lower organic layer when the volumes are equal. Each percentage refers to that compound’s own total.

Next, acidifying the collected aqueous portion to pH 2.5 favors neutral HA. Contact with a fresh organic portion then moves a larger fraction of that remaining acid pool into the organic phase. The switch exploits reversible proton transfer coupled to partitioning. A weak base can show the opposite pH preference: protonation can favor its aqueous ionic form, while deprotonation restores its neutral form.⁵

This sequence enriches the fractions; it does not guarantee perfect purity or complete recovery. Some N accompanied the acid into water, and some acid remained organic. Recoveries across successive steps must be applied to the amount entering each step. If pH drifts, a precipitate forms, the compound reacts, or the phases do not reach equilibrium, the stated two-phase calculation no longer describes the whole system.

For a new problem, work in this order: identify layer densities; determine the relevant chemical forms; read the direction of K or D; include phase volumes; and track the amount retained after each separation. This prevents a correct acid–base idea from being attached to the wrong physical layer.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A weak acid has pKₐ = 5.0 and neutral-form partition ratio [HA]org/[HA]aq = 6. The aqueous pH is maintained at 8.0, A− has negligible organic solubility, and equal phase volumes reach equilibrium. The organic and aqueous densities are 1.20 and 1.00 g/mL, respectively. Which layer contains the larger share of the total acid pool, and which form predominates there?

A. Lower layer; predominantly A−.

Review this choice. The lower layer is organic because its density is greater. The model gives A− negligible organic solubility, so this cannot be the main A− reservoir.

B. Lower layer; predominantly HA.

Review this choice. Neutral HA can enter the lower organic layer, but pH − pKₐ = 3 makes [A−]/[HA] = 1000 in water. D = 6/1001 ≈ 0.006, so the total acid pool strongly favors the aqueous layer.

C. Upper layer; predominantly A−.

Correct. The less dense aqueous phase is on top. Its [A−]/[HA] ratio is 1000, and D = 6/1001 ≈ 0.006. With equal volumes, about 99.4% of the acid pool is aqueous and A− predominates there.

D. Upper layer; predominantly HA.

Review this choice. The layer is correctly identified, but the form is not. A pH three units above pKₐ gives a 1000:1 aqueous ratio favoring A−.

The principle to carry forward

Layer position comes from density; the acid’s form comes from pH relative to pKₐ. Use both before choosing a fraction.

Check 2

A neutral solute has K = [solute]org/[solute]aq = 2 and begins in 40 mL of water. One extraction with 20 mL of organic solvent would recover 50%. Instead, two fresh 10 mL portions are used sequentially, removing each organic portion before adding the next. Assume unchanged volumes and complete equilibration. What percentage of the original solute is recovered in the combined organic extracts?

A. 44.4%

Review this choice. This is the fraction left in water: [40/(40 + 2 × 10)]² = (2/3)² = 4/9. The question asks for the complementary recovered fraction.

B. 50.0%

Review this choice. This is the one-portion recovery. Each fresh portion establishes a new equilibrium with the amount remaining, so dividing the total solvent changes the result.

C. 55.6%

Correct. Each extraction leaves 2/3 of the amount then present in water. The retained fraction is 4/9, so the combined extracts recover 1 − 4/9 = 5/9 ≈ 55.6%.

D. 66.7%

Review this choice. Adding the two one-third removal fractions treats both as fractions of the original amount. The second removes one third of the reduced remainder.

The principle to carry forward

Multiply sequential retained fractions, then subtract from one. Do not add removal fractions with different denominators.

Check 3

A different compound is a weak base B; its conjugate acid BH+ has pKₐ = 9.0. Only neutral B appreciably enters the organic phase. The aqueous phase initially has maintained pH 7.0. The solvent pair, phase volumes, temperature, and neutral-form partition ratio remain fixed. Which prediction best follows?

A. Lowering pH to 5.0 increases neutral B and raises the organic fraction.

Review this choice. Lower pH favors protonated BH+, not neutral B. Under the supplied partition model, the organic fraction decreases.

B. Raising pH to 11.0 increases neutral B and raises the organic fraction.

Correct. [B]/[BH+] = 10^(pH − pKₐ) rises from 0.01 to 100. More of the aqueous compound is neutral, increasing its total distribution into the organic phase under the stated model.

C. Lowering pH to 5.0 increases BH+ and raises the organic fraction.

Review this choice. The form change is correct, but its partition consequence is reversed. BH+ remains mainly aqueous in this model.

D. Raising pH to 11.0 increases BH+ and raises the organic fraction.

Review this choice. Higher pH favors deprotonation to B. The organic fraction rises because neutral B becomes more available, not because BH+ increases.

The principle to carry forward

For a base, higher pH can favor the neutral form; for an acid, lower pH can do so. Apply the supplied ionization and partition assumptions.