Questions: Liquid-Liquid Extraction

Question 1

An analyst needs to extract 90% of an analyte from 100 mL of aqueous solution. The distribution ratio D = 5. Which extraction scheme achieves better recovery?

• A. One extraction with 100 mL organic solvent — using the maximum total volume maximizes contact
• B. Two extractions with 50 mL organic solvent each — multiple smaller extractions recover more analyte than one large extraction of the same total volume
• C. Both schemes give identical recovery because the total volume of organic solvent is the same
• D. One extraction with 100 mL is better because each contact with fresh solvent is less efficient than a large single contact

Answer: (B)

The Craig equation shows that multiple extractions with smaller volumes always outperform one extraction with the same total volume, given D > 0. After the first extraction, the analyte concentration in the aqueous phase is reduced — a fresh portion of organic solvent contacts a depleted aqueous phase and extracts a fixed fraction of what remains. With D = 5 and 100 mL aqueous vs. 100 mL organic in one extraction: fraction remaining = 100/(100 + 5×100) = 100/600 = 1/6, so 83% extracted. With two extractions of 50 mL each: fraction remaining = [100/(100 + 5×50)]² = [100/350]² = 0.286² ≈ 8%, so 92% extracted. Multiple extractions win.

Question 2

A weak acid drug (pKa = 5) is being extracted from a biological sample into an organic solvent. The neutral form of the drug has a partition coefficient K = 100, but at pH 7.4 (blood pH), D = 0.03. Why is D so much lower than K?

• A. At pH 7.4, the drug binds to plasma proteins, reducing free drug concentration in the aqueous phase
• B. K and D are unrelated quantities — D at physiological pH reflects a different chemical equilibrium than K
• C. At pH 7.4, which is well above the pKa, the drug is predominantly ionized; the ionized form does not extract into organic solvent, greatly reducing D below K
• D. K is measured in different units than D, so direct comparison is not meaningful

Answer: (C)

The distribution ratio D accounts for all chemical forms of the analyte in each phase. At pH 7.4 (two units above pKa = 5), the Henderson-Hasselbalch equation predicts roughly 99.6% of the drug is in the ionized (conjugate base) form. The ionized form is polar and essentially insoluble in organic solvents, while only the tiny fraction of neutral form extracts. D therefore reflects the weighted average across all forms: mostly ionized drug that won't transfer, giving D ≪ K. This is why pH adjustment is the primary tool for LLE — lowering the pH toward or below pKa converts most of the drug to the neutral form, dramatically increasing D.

Question 3

The distribution ratio D is typically equal to or greater than the partition coefficient K for a given analyte and solvent pair, because D accounts for more chemical species.

• T. True
• F. False

Answer: False

D can be less than K, greater than K, or equal to K depending on the chemical conditions. D = K only when the analyte exists entirely in a single chemical form in both phases. If the analyte is partially ionized in the aqueous phase (as for a weak acid at high pH), D < K because the ionized form does not extract into the organic phase, reducing the effective concentration ratio. If conditions promote the formation of extractable complexes in the organic phase (e.g., metal-chelate complexes), D could exceed K. The relationship between D and K is not fixed — it depends on pH, complexing agents, and other equilibrium conditions.

Question 4

Performing three extractions with 10 mL of organic solvent each will recover more analyte than one extraction with 30 mL of the same solvent, assuming D > 0.

• T. True
• F. False

Answer: True

This follows directly from the Craig equation. Each fresh portion of organic solvent extracts a fixed fraction of the analyte remaining in the aqueous phase. After one extraction, the second portion contacts a depleted aqueous phase and still extracts that same fraction of the remainder. The cumulative effect is always greater recovery with multiple smaller extractions than with one large extraction of the same total volume, as long as D > 0. In practice, three to four extractions are sufficient to achieve >95% recovery for analytes with moderate D values, and adding more extractions beyond that yields diminishing returns.

Question 5

Why is the distribution ratio D a more useful quantity than the partition coefficient K for predicting and optimizing liquid-liquid extraction of real analytical samples?

Model answer: K describes the partitioning of a single, specific chemical species between two phases — it is a thermodynamic constant that applies only to one form of the analyte (e.g., the neutral molecule). In real samples, analytes often exist as multiple chemical species simultaneously: a weak acid can be neutral or ionized, a metal can be free or complexed, a drug can be protein-bound or free. D accounts for all of these forms: it is the ratio of total analytical concentration in the organic phase to total analytical concentration in the aqueous phase. Because extraction efficiency in practice depends on how much of the analyte actually transfers (in all its forms), D directly predicts extraction yield, whereas K alone would overestimate extraction if most of the analyte is in a form that cannot transfer.

The practical consequence is that D, not K, is what the analyst controls and optimizes. By adjusting pH to change the ionization state, or adding complexing agents to convert a poorly extractable form into an extractable complex, the analyst changes D without changing K. This is why pH is the most powerful LLE parameter: it shifts the equilibrium between forms with very different K values, dramatically changing the overall extraction efficiency. The key insight is that K is a property of a specific molecule under specific conditions, while D is a measurable property of the full chemical system that directly predicts experimental outcome.