Extraction, distillation and sample preparation

Total distribution can change when the chemical form changes

A partition ratio for one definite form and a distribution ratio for all forms need not be equal. Consider symbolic acid HA⇌H⁺+A⁻ in water, with only neutral HA allowed in an extracting phase. Define K=[HA]_e/[HA]_a and D=[HA]_e/([HA]_a+[A⁻]_a). A change in aqueous ionization changes D even if K remains fixed.

In an ideal concentration model with standard concentration c°=1 mol L⁻¹ and activity coefficients assigned one, pH=−log₁₀([H⁺]/c°) and pK_a=−log₁₀ K_a. The aqueous ratio [A⁻]/[HA]=10^(pH−pK_a) gives D=K/(1+10^(pH−pK_a)). Fixed pH, rapid exchange, unchanged volumes and exclusion of ion pairing, complexes and aggregation are assumptions.

Use invented pK_a=5.00, K=12.0, initial amount 2.00 mmol, V_a=50.0 mL and V_e=10.0 mL at assigned 298.15 K. At pH 4, 5 and 6, D is 10.9091, 6.0000 and 1.09091. Extracted amounts are respectively 1.37143, 1.09091 and 0.358209 mmol. These are given hypothetical properties, not an acid’s measured behavior or a pH adjustment procedure.

Derive the fraction remaining from a mass balance

At equilibrium, n_0=c_aV_a+c_eV_e and c_e=Dc_a. Therefore c_a=n_0/(V_a+D V_e), and the fraction remaining in the aqueous phase is q_a=V_a/(V_a+D V_e). The extracted fraction is 1−q_a. This finite-inventory equation prevents confusing a large concentration ratio with complete extraction.

The simple balance excludes reaction, adsorption, volatilization, saturation, phase-volume changes and mechanical transfer losses. A real preparation may violate several assumptions simultaneously. Measured recovery should cover the procedure and analyte actually being reported rather than borrowing an ideal equilibrium fraction.

Successive fresh contacts multiply remaining fractions

If each fresh portion begins without solute and reaches the same defined equilibrium, its remaining fraction acts on what survived the previous contact. For n equal contacts sharing a fixed total extracting volume V_total, q_series=[V_a/(V_a+D V_total/n)]ⁿ. Multiplication here follows successive inventories; it does not presume statistically independent measurements.

With invented V_a=60.0 mL, V_total=60.0 mL, D=3.00 and initial 1.00 mmol, one contact leaves 0.25000 mmol. Three fresh 20.0 mL contacts each leave half of their incoming amount, producing 0.12500 mmol at the end. Six fresh 10.0 mL contacts give (2/3)⁶=0.0877915 of the initial amount.

Thus recoveries in this restricted model are 75.0%, 87.5% and 91.2209%. Additional contacts also take time and can introduce loss, emulsions or environmental constraints. The arithmetic is a conceptual allocation comparison, not an instruction to choose or handle a solvent.

Pool amounts and volumes, not the last contact concentration

A second invented model assigns initial 2.00 mmol, aqueous volume 40.0 mL, D=2.00 and three fresh 10.0 mL extracting portions. Each contact leaves two thirds of its incoming amount. Extracted amounts are 0.666667, 0.444444 and 0.296296 mmol; the aqueous remainder is 0.592593 mmol.

Pooling the separated extracting portions gives amount 1.407407 mmol in 30.0 mL, so concentration is 0.0469136 mol L⁻¹. The three individual concentrations are 0.0666667, 0.0444444 and 0.0296296 mol L⁻¹. The final pooled concentration is not the concentration of the last portion.

The pooled extract is not placed back into contact with the aqueous remainder in this model. If it were, another equilibrium problem would follow. Keep phase separation, transfer and any later dilution as distinct inventory steps.

A vapor can have a different composition from its liquid

Distillation uses vapor–liquid behavior rather than the two-liquid ratio above. For an ideal liquid and ideal vapor at one fixed temperature, Raoult’s law gives partial pressures p_i=x_iP_i*, where x_i is liquid mole fraction and P_i* is the pure-component vapor pressure at that temperature. Total pressure P is their sum, and vapor mole fractions are y_i=p_i/P.

Assign liquid x_A=0.400 and x_B=0.600 with hypothetical pure pressures P_A*=120 kPa and P_B*=60 kPa. Partial pressures are 48.0 and 36.0 kPa; P=84.0 kPa. Vapor y_A=4/7≈0.571429 and y_B=3/7≈0.428571. A is more concentrated on a mole-fraction basis in the equilibrium vapor than in the liquid.

For a one-mole feed divided into 0.800 mol equilibrium liquid and 0.200 mol equilibrium vapor of these compositions, feed fraction is z_A=0.8x_A+0.2y_A=0.434286. It is not the assigned liquid fraction 0.400. This is one equilibrium stage and component balance, with no calculated heat duty, operating sequence or claim of complete separation.

Real mixtures can be nonideal. At an azeotropic point, vapor and liquid have the same composition under the specified conditions, even though other properties can differ. Ordinary distillation at those conditions cannot enrich that equilibrium vapor relative to its liquid composition. A positive deviation can produce a minimum-boiling azeotrope; it does not imply that every positive-deviation mixture has one. No azeotrope is represented by the ideal example.

Preparation can change both inventory and analyte identity

Extraction, cleanup, drying, concentration and derivatization can prepare material for a measurement, but each needs an analyte and denominator. If a derivatization changes the chemical entity, response calibration and conversion accounting must follow that product back to the intended quantity. Adsorption or degradation is not described by an equilibrium fraction alone.

A clean-looking extract does not establish representative sampling or quantitative recovery. Appropriate blanks, matrix checks, recovery evidence and documented transfer factors belong to the full procedure. Choosing a preparation means matching chemical form and mechanism to the question, then checking that the relevant inventory survives.

Worked examples

Try it yourself

1. In a new one-contact model, V_a=40.0 mL, V_e=20.0 mL and D=2.00. Starting from 1.00 mmol, find the extracted amount.

   Show the answer and reasoning

   0.500 mmol.

   The remaining fraction is 40/(40+2×20)=1/2; the other half enters the extracting phase.

2. Use the three-contact pooling model above but stop after two fresh portions. Find pooled amount and concentration.

   Show the answer and reasoning

   1.111111 mmol in 20.0 mL; 0.0555556 mol L⁻¹.

   Remaining amount is 2.00×(2/3)²=0.888889 mmol; subtract from initial and divide the pooled amount by 0.0200 L.

3. Why cannot three contacts with already solute-loaded portions be represented by the same q_series formula without further information?

   Show the answer and reasoning

   Each contact includes an incoming solute amount in the extracting phase.

   The original balance assumed every fresh portion initially had zero solute. A loaded portion changes the contact inventory and may change the net transfer direction.

4. At a specified azeotropic point, what composition comparison limits enrichment by that equilibrium vapor?

   Show the answer and reasoning

   The liquid and vapor mole fractions are equal.

   The enrichment step requires a difference between x and y. Equality at the azeotropic conditions removes that composition difference; it does not make all phase properties identical.

What to carry forward

• A form-specific partition ratio and a total distribution ratio can differ.
• Close each contact inventory before computing recovery.
• Repeated extraction requires the stated fresh-phase and equilibrium assumptions.
• Pooled concentration uses pooled amount and volume.
• Liquid, vapor and feed composition need separate symbols and balances.