Unit 8: Solvent Extraction: Partition Thermodynamics & Metal Chelation
Comprehensive physical and analytical treatise on liquid-liquid extraction: Nernst distribution law, activity coefficient effects, pH-dependent conditional distribution ratios (D), mathematical induction proof of multiple batch depletion factors (q_n), liquid-liquid extraction of iron(III) as tetrachloroferrate into MIBK, colorimetric copper estimation via diethyldithiocarbamate in carbon tetrachloride, Craig countercurrent distribution theory, and modern solid-phase extraction (SPE).
§8.1 Thermodynamic Basis of Liquid-Liquid Partition: Nernst Distribution Law & Activity Coefficients
Solvent extraction (liquid-liquid partition) is a fundamental separation process based on the unequal distribution of a chemical solute between two immiscible liquid phases (typically an aqueous phase and an organic solvent such as diethyl ether, methyl isobutyl ketone, chloroform, or hexane).
``` Phase Partition Equilibrium in a Separatory Funnel +-------------------------------------+ | | | ORGANIC PHASE (org) | Analyte A(org) | Volume V_org, Activity a_org | [A]_org, γ_org | | +=====================================+ <--- Liquid-Liquid Interface | | | AQUEOUS PHASE (aq) | Analyte A(aq) | Volume V_aq, Activity a_aq | [A]_aq, γ_aq | | +-------------------------------------+ ```
Thermodynamic Derivation of the Nernst Distribution Law
Consider a solute $A$ partitioned between an aqueous phase ($aq$) and an organic phase ($org$) at constant temperature and pressure:
At thermodynamic equilibrium, the chemical potentials of solute $A$ in the two contacting phases must be identical:
Expressing chemical potentials in terms of standard chemical potentials and thermodynamic activities:
Rearranging terms:
Exponentiating both sides yields the thermodynamic partition constant:
where $\Delta G_{\text{transfer}}^\circ$ is the standard Gibbs free energy of transfer of one mole of solute from the aqueous phase to the organic phase.
Practical Distribution Constant ($K_D$) and Activity Coefficients
Expressing thermodynamic activities as the product of molar concentration and activity coefficient ($a = \gamma [A]$):
The practical distribution constant (or partition coefficient) is defined as:
Under conditions of infinite dilution (or constant high ionic strength in the aqueous phase and ideal dilute behavior in the organic phase), the activity coefficient ratio $\gamma_{A,aq} / \gamma_{A,org}$ remains constant, and the classical Nernst Distribution Law holds:
The magnitude of $K_D$ is governed by the relative intermolecular forces: nonpolar hydrophobic solutes preferentially dissolve into organic solvents with low dielectric constants ($\Delta G_{\text{transfer}}^\circ < 0$, $K_D \gg 1$), whereas hydrated ionic species remain overwhelmingly in the polar aqueous phase ($K_D \ll 1$).
Synergistic Liquid-Liquid Extraction Thermodynamics
Synergistic extraction occurs when a mixture of two extractants extracts a metal cation with an efficiency vastly exceeding the sum of the individual extraction yields:
The classic mechanism involves the combination of an acidic chelating agent (e.g., thenoyltrifluoroacetone, HTTA) and a neutral donor ligand (e.g., tributyl phosphate, TBP, or trioctylphosphine oxide, TOPO):
- The chelating ligand ($\text{TTA}^-$) neutralizes the formal positive charge of the metal cation.
- The neutral organophosphorus ligand ($\text{TBP}$) displaces residual coordinated water molecules from the inner coordination sphere, forming a coordinatively saturated, lipophilic mixed complex that partitions into nonpolar organic diluents with a distribution ratio up to $10^5$ times greater than with HTTA alone.
§8.2 Distribution Ratio (D), pH-Dependent Extraction of Weak Acids/Bases & Chelate Complexes
While the thermodynamic distribution constant $K_D$ describes the partition of a single, specific chemical species, real analytical solutes frequently participate in chemical equilibria such as ionization, association, dimerization, or coordination in either phase. To treat the total mass transfer of the element or compound, we define the operational Distribution Ratio ($D$).
The Distribution Ratio ($D$)
The distribution ratio $D$ is the ratio of the total analytical concentration of all chemical forms of the solute in the organic phase to its total analytical concentration in the aqueous phase:
``` pH-Dependent Extraction of a Weak Monoprotic Acid (HA) ORGANIC PHASE: [HA]_org <=======> [(HA)₂]_org (Dimerization) ^ | K_D v AQUEOUS PHASE: [HA]_aq <=======> [A⁻]_aq + [H⁺]_aq (Ionization) | (Ionic: non-extractable) +---------------> Ka = [H⁺][A⁻] / [HA] ```
Extraction of a Weak Monoprotic Acid ($HA$)
Consider a weak organic acid $HA$ that ionizes in water but exists only as neutral $HA$ in the organic phase:
Because charged ions ($A^-$) do not partition into nonpolar organic solvents to any appreciable extent ($[A^-]_{org} \approx 0$):
where $\alpha_{HA} = \frac{[H^+]}{[H^+] + K_a}$ is the fraction of acid in the neutral, undissociated form.
- In strongly acidic media ($\text{pH} \ll pK_a$): $[H^+] \gg K_a \implies \alpha_{HA} \approx 1 \implies D \approx K_D$ (maximum extraction).
- In strongly basic media ($\text{pH} \gg pK_a$): $[H^+] \ll K_a \implies D \approx \frac{K_D [H^+]}{K_a} \implies \log D = \log K_D + pK_a - \text{pH}$ (extraction drops by 1 order of magnitude per pH unit).
- At $\text{pH} = pK_a$: $\alpha_{HA} = 0.5 \implies D = K_D / 2$.
Metal Chelation Extraction Thermodynamics
Metal cations ($M^{n+}$) cannot be extracted directly into organic solvents because of their large positive hydration enthalpies. However, by reacting with an organic chelating extractant ($HL$) dissolved in the organic phase, neutral metal chelates ($ML_n$) are formed:
The overall extraction equilibrium constant is:
Assuming no significant intermediate complexes or side reactions:
Taking the common logarithm:
The pH of 50% extraction ($\text{pH}_{1/2}$) is the pH at which $D = 1$ ($\log D = 0$) when equal volumes of organic and aqueous phases are used ($50\%$ of the metal is extracted):
Because $\text{pH}_{1/2}$ depends inversely on the stability of the metal chelate, metals with different $K_{\text{ex}}$ values can be separated cleanly by selective pH buffering.
§8.3 Efficiency of Multiple Batch Extractions: Exact Mathematical Proof of Depletion Factor q_n
A central principle of separation science is that multiple successive extractions using small portions of organic solvent achieve a vastly superior recovery compared to a single extraction using the entire volume of solvent.
``` Batch Multi-Stage Extraction Sequence & Depletion Proof Initial Aqueous: V_aq, Mass w₀ | +---> 1st Extraction (V_org) ===> Extracted: w₀(1 - q₁) | Remaining in aq: w₁ = w₀ · q₁ | +---> 2nd Extraction (V_org) ===> Extracted: w₁(1 - q₁) | Remaining in aq: w₂ = w₁ · q₁ = w₀ · q₁² | +---> nth Extraction (V_org) ===> Remaining in aq: w_n = w₀ · q₁ⁿ ```
Derivation of the Single Extraction Fraction ($q_1$)
Let an aqueous sample of volume $V_{aq}$ contain an initial mass $w_0$ of solute. Suppose this solution is equilibrated with a volume $V_{org}$ of organic solvent. Let $w_1$ be the mass of solute remaining in the aqueous phase at equilibrium. The mass extracted into the organic phase is $(w_0 - w_1)$. The equilibrium concentrations in the two phases are:
By definition of the distribution ratio $D$:
Multiplying by $w_1 V_{org}$ and solving for $w_1$:
We define the single-stage unextracted fraction $q_1$ as:
Mathematical Induction Proof for $n$ Successive Extractions
Suppose the aqueous phase (containing remaining mass $w_1$) is separated and extracted with a second identical portion of fresh organic solvent of volume $V_{org}$. The remaining mass $w_2$ satisfies:
Theorem: After $n$ successive extractions with fresh aliquots of organic solvent of volume $V_{org}$, the mass remaining unextracted in the aqueous phase is:
Proof by Induction:
- Base Case ($n = 1$): Holds by the single-stage derivation above.
- Inductive Step: Assume $w_k = w_0 q_1^k$ holds for some integer $k \ge 1$.
The $(k+1)$-th extraction uses remaining aqueous mass $w_k$ as the starting mass with fresh solvent $V_{org}$:
Thus, the formula holds for all integers $n \ge 1$. $\blacksquare$
The Percent Extraction ($\%E$)
The total percentage of solute extracted into the combined organic phases after $n$ cycles is:
Limiting Continuous Extraction Efficiency
If a fixed total solvent volume $V_{\text{total}}$ is divided into $n$ equal portions ($V_{org} = V_{\text{total}} / n$):
Taking the mathematical limit as $n \to \infty$ (continuous countercurrent partition):
This proves that splitting solvent into infinitesimal aliquots achieves the theoretical maximum possible recovery.
§8.4 Liquid-Liquid Extraction of Iron(III) as Tetrachloroferrate(III) into Methyl Isobutyl Ketone (MIBK)
The quantitative extraction of iron(III) from concentrated hydrochloric acid into methyl isobutyl ketone (MIBK, 4-methylpentan-2-one) is one of the classic, highly selective separation methods in analytical chemistry. It is widely used to remove bulk iron matrix prior to trace element determination in alloy and ore analysis.
``` Mechanism of Iron(III) Ion-Pair Extraction into MIBK Aqueous Phase (6-8 M HCl): Fe³⁺ + 4 Cl⁻ <===================> [FeCl₄]⁻ (Tetrachloroferrate anion) | H⁺ + n MIBK <===================> [H(MIBK)ₙ]⁺ (Solvated Oxonium Cation) | v Ion-Pair Formation: [H(MIBK)ₙ]⁺ + [FeCl₄]⁻ <===> {[H(MIBK)ₙ]⁺[FeCl₄]⁻} | v (Partitions!) Organic Phase (MIBK Layer): {[H(MIBK)ₙ]⁺[FeCl₄]⁻}(org) ```
Chemical Mechanism: Ion-Pair Extraction
In aqueous solutions of high hydrochloric acid concentration ($6\text{ to }8\text{ M HCl}$), iron(III) forms successive chloro complexes:
Although the stepwise stability constants are modest, the massive mass-action driving force of $6\text{--}8\text{ M Cl}^-$ shifts the equilibrium predominantly into the yellow tetrachloroferrate(III) complex anion, $[\text{FeCl}_4]^-$.
Because $[\text{FeCl}_4]^-$ is negatively charged, it cannot partition into MIBK as an isolated ion. Instead, the basic oxygen atom of MIBK acts as a Lewis base, solvating hydronium ions to produce large, bulky, lipophilic oxonium cations:
The oxonium cation pairs electrostatically with the tetrachloroferrate anion to form an overall electrically neutral ion-association complex:
This neutral ion-pair dissolves readily in the organic phase, yielding a distribution ratio $D_{\text{Fe}} > 1000$ (corresponding to $> 99.9\%$ single-stage extraction).
Selectivity Over Interfering Divalent Cations
Under $6\text{ M HCl}$ conditions:
- Divalent cations ($\text{Ni}^{2+}, \text{Co}^{2+}, \text{Mn}^{2+}, \text{Cr}^{3+}, \text{Al}^{3+}$) do not form stable, singly charged tetrahedral tetrachloro-complexes and exhibit virtually zero extraction ($D < 0.001$).
- Cobalt(II) forms blue $[\text{CoCl}_4]^{2-}$, but because it carries a $-2$ charge, it requires two oxonium cations to pair, resulting in negligible extraction into MIBK unless the $\text{HCl}$ concentration exceeds $9\text{ M}$.
- Stripping (Back-Extraction): Once iron is separated in the organic layer, it can be stripped back into an aqueous phase quantitatively by shaking with pure water or dilute acid ($0.1\text{ M HCl}$). In the absence of high chloride concentration, $[\text{FeCl}_4]^-$ immediately dissociates into hydrated $\text{Fe}^{3+}$ and free $\text{Cl}^-$, driving $D_{\text{Fe}} \to 0$.
§8.5 Extraction and Colorimetric Estimation of Copper as Copper(II) Diethyldithiocarbamate (Cu(DDTC)2) in Carbon Tetrachloride
The extraction of copper(II) as copper diethyldithiocarbamate, $\text{Cu(DDTC)}_2$, is an internationally recognized standard method for the trace estimation of copper in copper-base alloys, food matrices, and environmental samples.
``` Molecular Structure of Copper(II) Diethyldithiocarbamate Et₂N - C = S S = C - NEt₂ \ \ / / S - Cu(II) - S [Neutral Square-Planar Bis-Chelate Complex] Intense Golden-Brown Color in CCl₄ / CHCl₃ λ_max = 436 nm, ε ≈ 13,000 L·mol⁻¹·cm⁻¹ ```
Synthesis and Chelation Chemistry
Sodium diethyldithiocarbamate ($\text{Na-DDTC}$) is the water-soluble sodium salt of diethyldithiocarbamic acid:
The diethyldithiocarbamate anion is a powerful bidentate dithio-ligand that coordinates through both sulfur atoms. When introduced into a solution containing copper(II), it forms a neutral, water-insoluble, square-planar coordination complex:
Because the complex is neutral and possesses four hydrophobic ethyl substituents, it is readily extracted into nonpolar organic solvents such as carbon tetrachloride ($\text{CCl}_4$) or chloroform ($\text{CHCl}_3$):
The extracted organic phase exhibits an intense golden-yellow to brown coloration with an absorption maximum at $\lambda_{\text{max}} = 436\text{ nm}$ and a molar absorptivity $\varepsilon \approx 13,000\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$.
Masking Interferences with Citrate and EDTA
Diethyldithiocarbamate is a versatile chelating agent that also forms extractable complexes with many other transition and heavy metals ($\text{Fe}^{3+}, \text{Ni}^{2+}, \text{Co}^{2+}, \text{Bi}^{3+}, \text{Zn}^{2+}, \text{Pb}^{2+}$). To render the extraction completely specific for copper:
1. EDTA ($\text{Na}_2\text{H}_2\text{EDTA}$) is added as an auxiliary masking agent:
EDTA forms exceptionally stable, water-soluble, highly charged hexadentate complexes with $\text{Fe}^{3+}$, $\text{Ni}^{2+}$, $\text{Co}^{2+}$, $\text{Zn}^{2+}$, and $\text{Pb}^{2+}$. Because these EDTA complexes carry formal negative charges (e.g., $[\text{Fe(EDTA)}]^-$), they cannot be extracted into carbon tetrachloride.
2. Selectivity Thermodynamic Driving Force: Although EDTA also coordinates with copper ($\log K_f = 18.8$), the copper-diethyldithiocarbamate complex has an even higher conditional stability constant ($\log \beta_2 \approx 28$), so $\text{DDTC}^-$ selectively displaces EDTA from copper:
None of the other metal ions can be displaced from their EDTA complexes by $\text{DDTC}^-$.
3. Citrate Buffer ($\text{pH } 8.5\text{--}9.5$): Ammonium citrate prevents the precipitation of metal hydroxides and buffers the aqueous medium in the optimal range where DDTC does not undergo acid-catalyzed decomposition into diethylamine and carbon disulfide.
§8.6 Continuous Countercurrent Liquid-Liquid Extraction & Craig Distribution Theory
When two solutes have very similar distribution ratios ($D_A \approx D_B$), a single batch extraction or even a few repeated batch extractions cannot achieve complete separation. In such cases, multi-stage continuous countercurrent extraction is required. The theoretical foundation was formulated by Lyman C. Craig through the Craig Countercurrent Distribution (CCD) apparatus.
``` Craig Countercurrent Distribution Architecture Stage 0 Stage 1 Stage 2 Stage 3 ... Stage n +-----------+ +-----------+ +-----------+ +-----------+ | V_org (0) | | V_org (1) | | V_org (2) | | V_org (3) | Mobile Phase (shifts ->) +===========+ +===========+ +===========+ +===========+ | V_aq (0) | | V_aq (1) | | V_aq (2) | | V_aq (3) | Stationary Phase (fixed) +-----------+ +-----------+ +-----------+ +-----------+ ```
Mathematical Model of Craig Distribution
Consider a battery of $n$ extraction tubes ($r = 0, 1, 2, \dots, n$), each containing an identical volume of stationary lower aqueous phase ($V_{aq}$). A mobile upper organic phase of volume $V_{org}$ moves sequentially from tube $r$ to tube $r+1$ after each equilibrium shaking cycle. At each equilibration step, solute partitions according to its distribution ratio $D$:
- Fraction in the upper organic phase:
- Fraction in the lower aqueous phase:
The Binomial Distribution Formalism
Initially ($n = 0$), sample mass $w_0$ is placed in tube $0$. After equilibration, fraction $p$ is in the upper phase and $q$ in the lower phase. The upper phase is transferred to tube $1$, while fresh upper phase is added to tube $0$. Both tubes are equilibrated. By mathematical induction, the fraction of solute $T_{n,r}$ present in tube $r$ after $n$ complete transfers corresponds precisely to the $(r+1)$-th term of the binomial expansion:
Gaussian Approximation for Large Numbers of Transfers
When the number of transfers $n$ is large ($n > 25$), the discrete binomial distribution converges asymptotically to a continuous Gaussian normal distribution:
where:
- Tube of maximum concentration ($r_{\text{max}}$):
- Standard deviation ($\sigma$):
Resolution Between Two Components
If two solutes $A$ and $B$ have distribution ratios $D_A > D_B$, their peak maxima will migrate to different tube numbers $r_{\text{max},A} = n p_A$ and $r_{\text{max},B} = n p_B$. Complete baseline separation ($R_s \ge 1.5$) requires:
Craig distribution directly bridges classical liquid-liquid extraction and modern partition chromatography (Martin and Synge Nobel Prize work).
§8.7 Solid-Phase Extraction (SPE): Sorbents, Cartridge Conditioning, Elution Mechanics & Preconcentration
Solid-Phase Extraction (SPE) is a modern chromatographic sample preparation technique that replaces messy, solvent-intensive liquid-liquid extraction with rapid, automated, high-recovery partitioning onto a solid sorbent packed into a disposable polypropylene cartridge or 96-well plate.
``` The Four Sequential Steps of Solid-Phase Extraction
- Condition 2. Load Sample 3. Wash Matrix 4. Elute Analyte
[Solvent] [Sample] [Rinse] [Eluent] | | | | v v v v +---------+ +---------+ +---------+ +---------+ | Sorbent | Active | ####### | Analyte | ####### | Matrix | | Analyte | Bed | Chains | | Retained! | | Washed Out! | | Desorbed! +---------+ +---------+ +---------+ +---------+ | | | | v Waste v Breakthrough? v Impurities v Pure Analyte ```
Chemistry of SPE Sorbents
1. Reversed-Phase Sorbents: Octadecylsilane ($\text{C}_{18}$), octylsilane ($\text{C}_8$), or phenyl groups chemically bonded to porous spherical silica ($40\text{--}60\,\mu\text{m}$). Nonpolar organic analytes are retained from polar aqueous matrices via hydrophobic van der Waals interactions.
2. Normal-Phase Sorbents: Unmodified silica, alumina, florisil, or cyano/amino-bonded phases. Polar analytes are retained from nonpolar organic solvents via dipole-dipole, hydrogen bonding, and $\pi\text{--}\pi$ interactions.
3. Ion-Exchange Sorbents: Strong cation exchange (SCX, sulfonic acid $-\text{SO}_3^-$), weak cation exchange (WCX, carboxylic acid $-\text{COO}^-$), strong anion exchange (SAX, quaternary amine $-\text{N}^+(\text{CH}_3)_3$), and weak anion exchange (WAX, primary/secondary amines).
4. Polymeric Sorbents: Macroporous poly(divinylbenzene-co-N-vinylpyrrolidone) (e.g., Oasis HLB), containing balanced hydrophobic and hydrophilic moieties, capable of retaining both polar and nonpolar compounds over a wide $\text{pH}$ range ($1\text{--}14$).
The Four-Step SPE Workflow
1. Conditioning and Equilibration:
- The cartridge is first wetted with a water-miscible organic solvent (methanol or acetonitrile) to solvate and "open up" the tangled alkyl chains of the bonded phase.
- It is then rinsed with an aqueous buffer matching the sample matrix to equilibrate the bed without allowing it to dry out.
2. Sample Loading:
- The aqueous sample is percolated through the bed at a controlled flow rate ($1\text{--}5\text{ mL}\cdot\text{min}^{-1}$). The analyte partitions strongly onto the sorbent ($k' \gg 1000$), while non-retained matrix species pass directly through to waste.
3. Washing (Rinsing):
- A wash solution of intermediate solvent strength (e.g., $5\text{--}10\%$ methanol in water) is passed through to displace weakly bound matrix interferents (salts, proteins, sugars) without desorbing the analyte.
4. Elution:
- A small volume ($0.5\text{--}2.0\text{ mL}$) of a strong organic eluting solvent (e.g., pure methanol, ethyl acetate, or acidified acetonitrile) is applied to desorb the analyte quantitatively.
Preconcentration and Enrichment Factor ($EF$)
Because a large volume of dilute sample ($V_{\text{sample}}$, e.g., $1000\text{ mL}$ of environmental water) can be loaded onto an SPE cartridge and eluted into a tiny final volume ($V_{\text{eluent}}$, e.g., $2.0\text{ mL}$), SPE provides massive physical preconcentration:
For example, loading $1000\text{ mL}$ and eluting into $2.0\text{ mL}$ with $95\%$ recovery yields an enrichment factor of $EF = (1000 / 2) \times 0.95 = 475$, dramatically lowering analytical detection limits.
§8.8 Supercritical Fluid Extraction (SFE) & Microwave-Assisted Extraction (MAE): Green Analytical Separation
Driven by the 12 principles of Green Analytical Chemistry (GAC), classical liquid-liquid extraction and Soxhlet extraction—which consume liters of toxic, flammable volatile organic solvents ($\text{CHCl}_3, \text{CCl}_4, \text{CH}_2\text{Cl}_2$, hexane)—are increasingly replaced by advanced, solvent-minimized green extraction technologies: Supercritical Fluid Extraction (SFE) and Microwave-Assisted Extraction (MAE).
``` Supercritical Fluid Phase Diagram & SFE State Pressure (P) ^ Liquid Phase | / | / | Solid Phase / | / P_c ---------•---------------+---------------> SUPERCRITICAL FLUID REGION | / / Density of Liquid (~0.7-0.9 g/mL) | / Triple Pt / Critical Pt Viscosity of Gas (~10⁻⁴ P) | /-------•-------/ (31.1°C, 73.8 bar) Diffusivity of Gas (~10⁻³ cm²/s) | / / Gas Phase +-----+-------+---------------+------------> Temperature (T) 0 T_c (31.1 °C) ```
Physical Principles of Supercritical Fluid Extraction (SFE)
A substance is in a supercritical state when both its temperature and pressure exceed its critical point ($T > T_c, P > P_c$). Carbon dioxide ($\text{CO}_2$) is the near-universal choice for SFE:
- Mild critical parameters: $T_c = 31.1^\circ\text{C}$, $P_c = 73.8\text{ bar}$ ($7.38\text{ MPa}$).
- Nontoxic, nonflammable, chemically inert, environmentally benign, and inexpensive.
- High volatility allows instantaneous solvent removal: depressurizing to atmospheric pressure converts $\text{CO}_2$ into a gas, leaving pure, concentrated, solvent-free analyte extracts.
Unique Transport Properties of Supercritical $\text{CO}_2$
| Physical Property | Gas | Supercritical $\text{CO}_2$ | Liquid | | :--- | :--- | :--- | :--- | | Density ($\text{g}\cdot\text{cm}^{-3}$) | $10^{-3}$ | $0.2\text{--}0.9$ | $1.0$ | | Diffusion Coefficient ($\text{cm}^2\cdot\text{s}^{-1}$) | $10^{-1}$ | $10^{-4}\text{--}10^{-3}$ | $10^{-5}$ | | Viscosity ($\text{g}\cdot\text{cm}^{-1}\cdot\text{s}^{-1}$) | $10^{-4}$ | $10^{-4}$ | $10^{-2}$ |
Because supercritical $\text{CO}_2$ possesses a density approaching that of a liquid, it exhibits high solvating power (dissolving nonpolar organic lipids, pesticides, and hydrocarbons). Concurrently, its gas-like viscosity and elevated molecular diffusivity allow it to penetrate deep inside solid micropores, accelerating extraction kinetics from hours (Soxhlet) to under $15\text{--}30\text{ minutes}$.
- Chemical Modifiers (Entrainers): Because pure $\text{CO}_2$ is nonpolar (dielectric constant $\varepsilon_r \approx 1.2\text{--}1.5$), extracting moderately polar compounds (pharmaceuticals, phenols, alkaloids) requires adding $1\text{--}10\text{ mol}\%$ of an organic modifier (typically methanol, ethanol, or water).
Microwave-Assisted Extraction (MAE)
In MAE, the solid sample and a polar solvent (or solvent mixture) are irradiated with microwave radiation in closed vessels:
- Direct Volumetric Dielectric Heating: Polar solvent molecules (or intracellular water) absorb microwave energy via dipole rotation, rapidly heating the internal cell matrix.
- Cellular Rupture and Mass Transfer: Rapid localized heating creates internal steam pressure that ruptures plant cell walls and cellular membranes, flushing analytes out into the extraction solvent with $> 95\%$ recovery in a fraction of conventional extraction times.
## Advanced University Honors Research Monograph: Hildebrand Solubility Parameters & Cavitation Free Energy The thermodynamic partition coefficient $K_D$ of a neutral solute between an aqueous phase and an organic solvent can be modeled from first principles using Regular Solution Theory and the concept of Hildebrand solubility parameters ($\delta$):
where $c_{\text{coh}}$ is the cohesive energy density, $\Delta H_{\text{vap}}$ is the enthalpy of vaporization, and $V_m$ is the molar volume of the pure liquid. The standard Gibbs free energy of transfer $\Delta G_{\text{transfer}}^\circ$ is partitioned into three physical components:
1. Cavitation Free Energy ($\Delta G_{\text{cavitation}}^\circ$): The reversible work required to create a cavity within the solvent network large enough to accommodate the solute molecule. Because water possesses an immense cohesive energy density ($\delta_{\text{water}} = 47.9\text{ MPa}^{1/2}$) held by strong hydrogen bonding networks, creating a cavity in water carries a massive free energy penalty ($\Delta G_{\text{cav,aq}} \gg 0$).
2. Hydrophobic Squeezing Driving Force: Nonpolar organic solvents (e.g., hexane $\delta = 14.9\text{ MPa}^{1/2}$, chloroform $\delta = 19.0\text{ MPa}^{1/2}$) have much lower cohesive energy densities. The system minimizes total Gibbs free energy by expelling the hydrophobic solute from water into the organic solvent, allowing water molecules to re-form hydrogen bonds.
This fundamental cavitation thermodynamic driving force explains why distribution ratios $K_D$ increase exponentially with solute molecular surface area and hydrophobic alkyl chain length.
Rigorous Tiered Solved Examination Problems
Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Intermediate, Advanced, and Honors tiers.
A $100.0\text{ mL}$ aqueous solution contains $0.500\text{ g}$ of an organic analyte $A$. The distribution ratio between diethyl ether and water is $D = 8.50$.
- Calculate the mass of analyte remaining in the aqueous phase and the percent extracted ($\%E$) after a single batch extraction using $60.0\text{ mL}$ of ether.
- Calculate the mass of analyte remaining in the aqueous phase and the percent extracted after three successive extractions using $20.0\text{ mL}$ of ether each (same total volume of $60.0\text{ mL}$).
- Calculate the minimum number of $10.0\text{ mL}$ ether extractions required to achieve at least $99.90\%$ total extraction recovery.
Part 1: Single Extraction with 60.0 mL Ether
Given $w_0 = 0.500\text{ g}, V_{aq} = 100.0\text{ mL}, V_{org} = 60.0\text{ mL}, D = 8.50$: The unextracted fraction is:
The remaining mass is:
The percent extracted is:
Part 2: Three Successive Extractions with 20.0 mL Ether Each
Here $n = 3, V_{org} = 20.0\text{ mL}$:
The fraction remaining after 3 extractions is:
The remaining mass is:
The percent extracted is:
Using three $20.0\text{ mL}$ portions increases recovery from $83.61\%$ to $94.92\%$ while consuming identical total solvent!
Part 3: Minimum Extractions for 99.90% Recovery with 10.0 mL Portions
For $V_{org} = 10.0\text{ mL}$:
We require $\%E \ge 99.90\% \implies q_n \le 0.0010$:
Rounding up to the nearest integer, $12$ successive extractions of $10.0\text{ mL}$ each are required.
A novel pharmaceutical drug $HA$ is a weak monoprotic acid with acid dissociation constant $K_a = 2.00 \times 10^{-5}$ ($pK_a = 4.70$). Its distribution constant between chloroform and water is $K_D = \frac{[HA]_{org}}{[HA]_{aq}} = 140.0$. Assume that the ionized conjugate base $A^-$ does not extract into chloroform.
- Derive the expression for the distribution ratio $D$ as a function of pH.
- Calculate the numerical value of $D$ at $\text{pH } 2.00, 4.70, 6.00,$ and $8.00$.
- A $50.0\text{ mL}$ aqueous buffer containing $25.0\text{ mg}$ of drug is extracted once with $25.0\text{ mL}$ of chloroform. Calculate the percent extraction ($\%E$) at $\text{pH } 2.00$ and at $\text{pH } 6.00$.
Part 1: Derivation of D(pH)
Part 2: Calculation of D at Specific pH Values
- At $\text{pH } 2.00$:
- At $\text{pH } 4.70$ ($\text{pH} = pK_a$):
- At $\text{pH } 6.00$:
- At $\text{pH } 8.00$:
Part 3: Extraction Recovery at pH 2.00 and pH 6.00
Given $V_{aq} = 50.0\text{ mL}, V_{org} = 25.0\text{ mL} \implies V_{org}/V_{aq} = 0.500$:
- At $\text{pH } 2.00$ ($D = 139.72$):
- At $\text{pH } 6.00$ ($D = 6.68$):
Zinc ($\text{Zn}^{2+}$) and lead ($\text{Pb}^{2+}$) form extractable chelates with dithizone ($\text{H}_2\text{Dz}$) in chloroform according to:
The extraction equilibrium constants are:
- For lead: $K_{\text{ex},\text{Pb}} = 1.00 \times 10^1$
- For zinc: $K_{\text{ex},\text{Zn}} = 1.00 \times 10^{-2}$
The dithizone concentration in chloroform is fixed at $[\text{H}_2\text{Dz}]_{org} = 1.00 \times 10^{-3}\text{ M}$. Equal phase volumes are used ($V_{org} = V_{aq}$).
- Calculate the $\text{pH}_{1/2}$ for $\text{Pb}^{2+}$ and for $\text{Zn}^{2+}$.
- Calculate the distribution ratios $D_{\text{Pb}}$ and $D_{\text{Zn}}$ at $\text{pH } 3.50$.
- Calculate the separation factor $\beta = D_{\text{Pb}} / D_{\text{Zn}}$ at $\text{pH } 3.50$ and determine the percentage of lead and zinc extracted in a single contact.
Part 1: Calculation of $\text{pH}_{1/2}$
For divalent metal cations ($n = 2$):
Given $[\text{H}_2\text{Dz}]_{org} = 1.00 \times 10^{-3}\text{ M} \implies \log [\text{H}_2\text{Dz}]_{org} = -3.00$:
- For Lead:
- For Zinc:
Part 2: Distribution Ratios at pH 3.50
The distribution ratio formula is:
At $\text{pH } 3.50$:
- For Lead:
- For Zinc:
Part 3: Separation Factor and Extraction Percentages
The separation factor is:
With equal phase volumes ($V_{org} / V_{aq} = 1$):
At $\text{pH } 3.50$, lead is quantitatively extracted ($99\%$) while $91\%$ of zinc remains in the aqueous phase, demonstrating clean separation.
A $50.0\text{ mL}$ aqueous solution containing $1.20\text{ g}$ of iron(III) and $50.0\text{ mg}$ of nickel(II) in $6.5\text{ M HCl}$ is extracted with $25.0\text{ mL}$ of methyl isobutyl ketone (MIBK). Under these conditions, the distribution ratios are $D_{\text{Fe}} = 850.0$ and $D_{\text{Ni}} = 0.00040$.
- Calculate the mass of iron(III) and the mass of nickel(II) extracted into the MIBK layer in a single batch extraction.
- Calculate the purity of the iron in the organic phase (as percent of total metal extracted).
- If the organic layer is scrubbed by shaking with $10.0\text{ mL}$ of fresh $6.5\text{ M HCl}$, calculate the residual nickel mass remaining in the organic layer.
Part 1: Single Batch Extraction Masses
Given $V_{aq} = 50.0\text{ mL}, V_{org} = 25.0\text{ mL} \implies V_{org}/V_{aq} = 0.500$:
- Iron(III) ($w_{0,\text{Fe}} = 1.20\text{ g}, D_{\text{Fe}} = 850.0$):
Mass unextracted in aqueous: $w_{\text{aq},\text{Fe}} = 1.20 \times 2.347 \times 10^{-3} = 2.82 \times 10^{-3}\text{ g} = 2.82\text{ mg}$. Mass extracted in MIBK: $w_{\text{org},\text{Fe}} = 1.20\text{ g} - 0.00282\text{ g} = 1.1972\text{ g}$.
- Nickel(II) ($w_{0,\text{Ni}} = 0.0500\text{ g}, D_{\text{Ni}} = 0.00040$):
Mass unextracted in aqueous: $w_{\text{aq},\text{Ni}} = 50.0\text{ mg} \times 0.99980 = 49.99\text{ mg}$. Mass extracted in MIBK: $w_{\text{org},\text{Ni}} = 50.0\text{ mg} \times 0.00020 = 0.0100\text{ mg} = 10.0\,\mu\text{g}$.
Part 2: Purity of Iron in Organic Extract
Part 3: Scrubbing with 10.0 mL of Fresh 6.5 M HCl
During scrubbing, the organic layer ($V_{org} = 25.0\text{ mL}$) is equilibrated with fresh aqueous scrub solution ($V_{\text{scrub}} = 10.0\text{ mL}$): The fraction of nickel remaining in the organic layer after scrubbing is:
The residual nickel in the scrubbed MIBK layer is:
Scrubbing eliminates over $99.9\%$ of the co-extracted trace nickel while iron losses to the scrub solution are under $0.05\%$, achieving spectroscopic-grade matrix elimination.
A $0.2500\text{ g}$ sample of an aluminum-zinc alloy is dissolved in acid, treated with ammonium citrate and EDTA to mask zinc and iron, buffered to $\text{pH } 9.0$, and treated with sodium diethyldithiocarbamate ($\text{Na-DDTC}$). The resulting $\text{Cu(DDTC)}_2$ complex is extracted quantitatively into $50.00\text{ mL}$ of carbon tetrachloride ($\text{CCl}_4$). The absorbance of the organic layer in a $1.000\text{ cm}$ cell at $436\text{ nm}$ is $A_{436} = 0.520$. A calibration curve prepared with pure copper standards treated under identical conditions yielded the linear regression equation:
- Calculate the mass of copper (in $\mu\text{g}$) present in the alloy sample.
- Calculate the mass percentage ($\% \text{ w/w}$) of copper in the alloy.
- Calculate the molar absorptivity ($\varepsilon$) of $\text{Cu(DDTC)}_2$ at $436\text{ nm}$ ($M(\text{Cu}) = 63.546\text{ g}\cdot\text{mol}^{-1}$).
Part 1: Mass of Copper in Alloy
From the calibration line:
Part 2: Weight Percentage of Copper in Alloy
Part 3: Molar Absorptivity of Cu(DDTC)2
The concentration of copper in the $50.00\text{ mL}$ organic extract is:
The net absorbance (blank subtracted) is $A_{\text{net}} = 0.520 - 0.0080 = 0.512$.
Two organic compounds $X$ and $Y$ are separated using a Craig countercurrent distribution apparatus with equal phase volumes ($V_{org} = V_{aq} = 10.0\text{ mL}$). The distribution ratios are $D_X = 3.00$ and $D_Y = 0.333$. The apparatus undergoes $n = 100$ transfers.
- Calculate the partition probabilities $p$ and $q$ for solute $X$ and solute $Y$.
- Calculate the tube number of maximum concentration ($r_{\text{max}}$) and the peak standard deviation ($\sigma$) for both solutes.
- Calculate the chromatographic resolution ($R_s$) between solute peaks after 100 transfers and determine if baseline separation is achieved.
Part 1: Partition Probabilities
Since $V_{org} / V_{aq} = 1$:
- For Solute X ($D_X = 3.00$):
- For Solute Y ($D_Y = 0.333 = 1/3$):
Part 2: Peak Center ($r_{\text{max}}$) and Width ($\sigma$) after $n = 100$ Transfers
- For Solute X:
- For Solute Y:
Part 3: Resolution Between Peaks
The separation between peak maxima is:
The resolution is:
Because $R_s = 2.89 > 1.50$, the two components are separated with complete baseline resolution ($> 99.9\%$ purity in their respective fractions).
A $1000\text{ mL}$ surface water sample containing a pesticide pollutant at an unknown trace concentration is processed via solid-phase extraction using a $500\text{ mg}$ $\text{C}_{18}$ cartridge. After loading and washing with $5.0\text{ mL}$ of water/methanol ($95:5$), the pesticide is eluted quantitatively with $2.00\text{ mL}$ of HPLC-grade acetonitrile. A $20.0\,\mu\text{L}$ injection of the SPE eluate into an RP-HPLC-UV system yielded a chromatographic peak area of $48,200\text{ counts}$. A calibration standard of the pesticide at $1.25\,\mu\text{g}\cdot\text{mL}^{-1}$ injected at the same volume yielded a peak area of $60,250\text{ counts}$. An independent spike recovery study on the same matrix showed that the method extraction recovery is $92.5\%$.
- Calculate the pesticide concentration in the $2.00\text{ mL}$ SPE eluate.
- Calculate the original pesticide concentration in the $1000\text{ mL}$ water sample in $\text{ng}\cdot\text{L}^{-1}$ (parts-per-trillion, $\text{ppt}$).
- Calculate the nominal and effective enrichment factors achieved by the SPE procedure.
Part 1: Concentration in the SPE Eluate
Using linear HPLC detector response:
Part 2: Original Water Concentration
The total mass of pesticide measured in the $2.00\text{ mL}$ eluate is:
Correcting for the $92.5\%$ extraction recovery:
The concentration in the original $1000\text{ mL}$ ($1.000\text{ L}$) water sample is:
Part 3: Nominal and Effective Enrichment Factors
- Nominal Enrichment Factor:
- Effective Enrichment Factor (incorporating recovery):
The SPE protocol concentrated the analyte by a factor of $462.5$, transforming a trace ppt-level pollutant into an easily quantifiable ppm-level HPLC peak.
Uranyl ion ($\text{UO}_2^{2+}$) is extracted from an aqueous nitric acid phase into benzene using thenoyltrifluoroacetone ($\text{HTTA}$) and tributyl phosphate ($\text{TBP}$). The individual and mixed extraction equilibrium constants are:
- Extraction with $\text{HTTA}$ alone:
- Adduct formation in the organic phase:
The extraction is performed at $\text{pH } 2.50$ with $[\text{HTTA}]_o = 0.0500\text{ M}$.
- Calculate the distribution ratio $D_0$ in the absence of $\text{TBP}$ ($[\text{TBP}]_o = 0$).
- Calculate the synergistic distribution ratio $D_{\text{syn}}$ in the presence of $[\text{TBP}]_o = 0.0200\text{ M}$.
- Calculate the Synergistic Enhancement Factor ($SEF = D_{\text{syn}} / D_0$) and determine the percent extraction ($\%E$) for equal phase volumes.
Part 1: Distribution Ratio without TBP ($D_0$)
At $\text{pH } 2.50 \implies [\text{H}^+] = 10^{-2.50} = 3.162 \times 10^{-3}\text{ M}$:
The distribution ratio with HTTA alone is:
Part 2: Synergistic Distribution Ratio with TBP ($D_{\text{syn}}$)
In the presence of TBP, both unadducted and adducted complexes coexist in the organic phase:
Given $\beta_{\text{adduct}} = 4.00 \times 10^4$ and $[\text{TBP}]_o = 0.0200\text{ M}$:
Part 3: Synergistic Enhancement Factor and Extraction Yield
- Synergistic Enhancement Factor:
- Percent Extraction:
Without TBP:
With TBP:
The addition of a tiny amount of neutral donor ligand raises uranyl extraction from an unusable $20\%$ to a quantitative $99.5\%$!
Neodymium ($\text{Nd}^{3+}$) and praseodymium ($\text{Pr}^{3+}$) are separated industrially by countercurrent liquid-liquid extraction from an aqueous nitric acid feed into kerosene containing di-(2-ethylhexyl)phosphoric acid ($\text{D2EHPA}$, $[\text{HA}]_2$). Under operational conditions:
- Distribution ratio of Neodymium: $D_{\text{Nd}} = 1.80$
- Distribution ratio of Praseodymium: $D_{\text{Pr}} = 1.20$
The phase volume ratio is adjusted to $V_{org} / V_{aq} = 0.680$.
- Calculate the extraction factors $E_{\text{Nd}} = D_{\text{Nd}} \left(\frac{V_{org}}{V_{aq}}\right)$ and $E_{\text{Pr}} = D_{\text{Pr}} \left(\frac{V_{org}}{V_{aq}}\right)$.
- Calculate the fraction $p$ and $q$ in the organic and aqueous phases per stage for both lanthanides.
- In a multi-stage countercurrent cascade with $N = 12$ theoretical extraction stages, calculate the recovery of neodymium in the organic extract and the residual contamination of praseodymium using Kremser's equation:
Part 1: Extraction Factors
With $V_{org} / V_{aq} = 0.680$:
Notice the brilliant engineering choice of flow ratio: $E_{\text{Nd}} > 1.0$ (drives Nd into organic phase), while $E_{\text{Pr}} < 1.0$ (keeps Pr in aqueous phase)!
Part 2: Single-Stage Phase Fractions
- Neodymium:
- Praseodymium:
Part 3: Kremser Cascade Performance after N = 12 Stages
- For Neodymium ($E_{\text{Nd}} = 1.224, N = 12$):
Fraction extracted into organic product:
- For Praseodymium ($E_{\text{Pr}} = 0.816, N = 12$):
Fraction remaining in aqueous raffinette: $19.82\% \implies$ Fraction entering organic: $80.18\%$. To reach $> 99.9\%$ purity, a center-feed scrub section is coupled to the cascade, washing out the co-extracted praseodymium with dilute acid.