Liquid-Liquid Extraction Efficiency Calculator (Multiple Extractions and Partition Coefficient)
Liquid-liquid extraction (LLE) separates target compounds by partitioning between two immiscible solvents. It is widely used in organic synthesis workup, natural product isolation, and HPLC sample preparation.
Core Equations
The distribution constant Kd describes a compound’s solvent preference:
Kd = c_organic / c_aqueous (neutral form, at equilibrium)
After n extractions (each with Vorg mL of fresh organic solvent), the aqueous-phase residual fraction and cumulative extraction efficiency are:
qₙ = (Vaq / (Vaq + D × Vorg))^n
Eₙ = (1 − qₙ) × 100%
where D is the distribution ratio; for neutral compounds D = Kd.
Worked example: D = 5, Vaq = 50 mL, Vorg = 10 mL, n = 3 (expressed as fractions to avoid rounding)
Residual fraction per step Vaq / (Vaq + D·Vorg) = 50 / (50 + 5×10) = 1/2
n = 1: residual 1/2 → cumulative efficiency 50.0%
n = 2: residual (1/2)² = 1/4 → cumulative efficiency 75.0%
n = 3: residual (1/2)³ = 1/8 → cumulative efficiency 87.5%
Comparison with a single extraction using the same total volume (30 mL): residual = 50 / (50 + 5×30) = 1/4, giving E = 75.0%. Three extractions reach 87.5%, recovering 12.5 percentage points more — these four numbers are what the tool displays.
Multiple Small-Volume Extractions Always Outperform a Single Large-Volume Extraction
For any D > 0, it can be shown that splitting into n equal-volume extractions always gives higher cumulative efficiency than a single extraction with n times the volume. The smaller D (the harder the extraction), the larger the difference; when D is very large (Kd ≫ Vaq/Vorg), a single extraction already approaches completion and the difference vanishes.
Ionizable Compounds: pH Correction of the Distribution Ratio
Only the neutral form of weak acids and bases partitions into the organic phase; the distribution ratio D replaces Kd:
Weak acid (enter the acidic pKa, i.e., pKa for HA ⇌ H⁺ + A⁻):
D = Kd / (1 + 10^(pH − pKa))
When pH ≪ pKa, D ≈ Kd (neutral HA predominates, extraction efficiency is high); when pH ≫ pKa, D → 0 (charged A⁻ predominates, almost no extraction).
Weak base (enter the pKa of the conjugate acid BH⁺):
D = Kd / (1 + 10^(pKa − pH))
When pH ≫ pKa, D ≈ Kd (neutral form B predominates, extraction efficiency is high).
Worked example: Kd = 10, weak acid pKa = 4.0, pH = 2.0, Vaq = 100 mL, Vorg = 50 mL, n = 2
pH is two units below pKa; ionization is small but not zero:
D = 10 / (1 + 10^(2.0 − 4.0)) = 10 / 1.01 = 9.901
n = 1: cumulative efficiency 83.2% (aqueous residual 16.8%)
n = 2: cumulative efficiency 97.2% (aqueous residual 2.8%)
Single extraction with 100 mL: cumulative efficiency 90.8%
Difference: 6.4 percentage points
Every number above is what the tool will display. Note that even with pH two units below pKa, D is only 9.901, not 10 — ionization always costs a small fraction of efficiency.
Applicability Limits
- Concentration: dilute solutions (≤ 0.01 mol/L); at high concentration, activity coefficients deviate and Kd shifts with concentration.
- Ionization model: single-protic weak acids/bases only; polyprotic acids or amphoteric molecules (e.g., amino acids) require step-by-step treatment and cannot be handled with a single pKa.
- Phase separation: a clean interface must form after thorough mixing; distribution ratio is meaningless when emulsification occurs.
- Equilibrium: each extraction step must be mixed thoroughly and allowed to separate completely before proceeding.
- Temperature: literature logP values are typically measured at 25°C; deviation at room temperature affects Kd by less than 10%.
Common Mistakes
Confusing logD with logP: logP is log₁₀(Kd) for the neutral compound with no ionization effect; logD is the apparent value at a specified pH (equivalent to D as calculated in this tool), and the two can differ by more than 3 orders of magnitude. When using literature data, confirm the pH at which logD was measured — otherwise you are effectively using logD = logP − 0.3, which corresponds to the pH = pKa condition.
Entering pKb instead of the conjugate acid pKa for a weak base: this tool requires the pKa of the conjugate acid BH⁺, not the base’s pKb. Conversion: pKa(BH⁺) = 14 − pKb (aqueous solution at 25°C). For example, morphine has pKb ≈ 6.0, giving pKa(BH⁺) ≈ 8.0; extraction efficiency only approaches its maximum when the aqueous phase pH > 10.
FAQ
Is logP the same as Kd?
logP specifically refers to log₁₀(Kd) for the neutral form in the octanol–water system at 25°C, and is the most widely used lipophilicity parameter. For this tool's 'partition coefficient Kd': enter 10^logP when using octanol; for other solvents (ethyl acetate, dichloromethane, etc.), enter the measured or literature Kd for that solvent–water system directly. Note that literature logD is an apparent value at a specified pH, equivalent to D as calculated internally by this tool; always note the measurement pH when citing logD.
Why do multiple small-volume extractions outperform a single large-volume extraction?
Let V mL be the organic phase volume per extraction over n steps: aqueous residual fraction = (Vaq/(Vaq+D·V))ⁿ; for a single extraction with nV mL: residual = Vaq/(Vaq+D·nV). It can be proven mathematically (AM–GM inequality) that for any D > 0 and n ≥ 2, the first residual is always lower — multiple small-volume extractions always recover more. The smaller D (the harder the extraction), the larger the advantage. Example: D = 5, Vaq = 50 mL — three 10 mL extractions give 87.5%; one 30 mL extraction gives only 75.0%, a difference of 12.5 percentage points.
How should I choose the aqueous phase pH when extracting a weak acid?
At low pH (pH ≪ pKa), a weak acid exists predominantly as the neutral form HA, so D ≈ Kd and extraction efficiency approaches its maximum. As pH rises, deprotonation to A⁻ reduces D toward zero and extraction becomes negligible. Practical rule: adjust the aqueous phase to pH ≤ pKa − 2, which keeps the neutral fraction above 99% and brings D close to Kd. For a weak base, the opposite applies: adjust pH to ≥ pKa + 2 (where pKa is the conjugate acid value) to approach maximum extraction efficiency.
When is this formula unreliable?
① High concentration (≥ 0.1 mol/L): activity coefficients deviate from 1 and Kd varies with concentration. ② Polyprotic acids or amphoteric molecules (e.g., amino acids): each ionization form must be treated separately; this tool supports single-protic systems only. ③ Emulsification: when the phase interface is unclear, the distribution ratio has no physical meaning. ④ Dimerization in the organic phase: some carboxylic acids form dimers in nonpolar solvents, raising the apparent Kd with concentration. ⑤ Literature logP values are typically measured at 25°C; room-temperature deviations affect Kd by less than 10%.
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