Liquid-liquid extraction: triangular diagrams and stage calculations

Why it matters

Liquid–liquid extraction separates components by their different solubilities in two immiscible or partly miscible liquids rather than by volatility. It is chosen when distillation is impossible or expensive: close boilers, azeotropes, heat-sensitive products (antibiotics, vitamins), dilute solutes in water (acetic acid, phenol), aromatics from paraffins, and metal recovery in hydrometallurgy. Triangular diagrams and stage balances are how extraction columns and mixer-settlers are designed.

Key ideas

Terminology. The feed contains solute A in a carrier (diluent) B. The solvent C is added. After contact and settling, the solvent-rich phase is the extract (E) and the carrier-rich phase is the raffinate (R). Solvent is later recovered from both (often by distillation).

Triangular (ternary) diagrams. Each corner is a pure component; any point inside is a mixture whose three mass fractions add to 1 (equilateral or right-angled triangles are both used).

• The binodal (solubility) curve encloses the two-phase region. Mixtures inside it split into an extract and a raffinate.
• Tie lines join the two phases in equilibrium. They usually slope, because the solute is not equally distributed.
• The plait point is where the two conjugate phases become identical and the tie lines shrink to a point; it is found by extrapolating a conjugate (tie-line correlation) curve, not by an intersection of tie lines.
• Type I systems have one partially miscible pair (B–C) and a plait point (acetone–water–MIBK); Type II have two partially miscible pairs and no plait point.

Equilibrium measures.

• Distribution coefficient K = y_A/x_A (solute fraction in extract / in raffinate). K > 1 means less solvent is needed.
• Selectivity (separation factor) β = (y_A/y_B)/(x_A/x_B), which must exceed 1 – it plays the role of relative volatility.

Lever rule and mixing point. Mixing feed F with solvent S gives a mixture M on the straight line FS, with F/S = (MS length)/(FM length). If M lies in the two-phase region, it splits along the tie line through M into E and R, again in amounts given by the lever rule.

Contacting schemes.

• Single stage: one mixer-settler.
• Cross-current (multistage, fresh solvent each stage): simple but uses much solvent.
• Counter-current multistage: feed and solvent enter at opposite ends; the most efficient use of solvent. On the triangle, all stage balances pass through a difference point Δ = F − E₁ = R_N − S, which lies outside the triangle. Stages are stepped off alternately with tie lines and lines through Δ.
• Minimum solvent: the solvent rate at which a line through Δ coincides with a tie line (infinite stages).

Immiscible carrier and solvent. If B and C are practically insoluble in each other, work on a solute-free basis: X = kg A/kg B, Y = kg A/kg C, with equilibrium Y = K′·X. Then the problem is identical to dilute absorption, and the extraction factor ε = K′·S/B plays the role of the absorption factor.

Equipment. Mixer-settlers (high stage efficiency, large holdup), spray and packed columns, sieve-tray columns, agitated columns (rotating-disc contactor, Scheibel, Kühni), pulsed columns, and centrifugal extractors for short contact or small density differences.

Formulas

Overall balance and mixing point: F + S = M = E + R; x_A,M = (F·x_A,F + S·y_A,S)/(F + S) Lever rule: E/R = (RM length)/(ME length) along the tie line; E/M = (x_A,M − x_A,R)/(y_A,E − x_A,R) Distribution coefficient: K = y_A/x_A; selectivity: β = (y_A/y_B)/(x_A/x_B)

Immiscible liquids, solute-free basis (Y = K′·X):

• Single stage, pure solvent: X₁ = X_F / (1 + ε), ε = K′·S/B
• N cross-current stages, equal fresh solvent S/N each: X_N = X_F / (1 + K′·S/(N·B))^N
• Counter-current (Kremser form), pure solvent: X_N/X_F = (ε − 1)/(ε^(N+1) − 1) (ε ≠ 1); N = ln[((X_F − Y_S/K′)/(X_N − Y_S/K′))(1 − 1/ε) + 1/ε]/ln ε
• F, S, E, R, M – masses (kg) or flows (kg/h); x, y – mass fractions; X, Y – mass ratios, kg A/kg B and kg A/kg C; K′ – dimensionless (on the ratio basis).

Worked examples

Example 1 – single, cross-current and counter-current (standard). Given: feed = 100 kg B carrying 20 kg A (X_F = 0.20); 150 kg pure solvent C in total; B and C immiscible; equilibrium Y = 1.5X.

1. Single stage, 150 kg: ε = 1.5 × 150/100 = 2.25. X₁ = 0.20/3.25 = 0.0615 → 69.2 % of A extracted.
2. Two cross-current stages, 75 kg each: ε = 1.125 per stage. X₂ = 0.20/2.125² = 0.0443 → 77.9 % extracted.
3. Two counter-current stages, 150 kg: ε = 2.25. X₂/X_F = (2.25 − 1)/(2.25³ − 1) = 1.25/10.39 = 0.120 → X₂ = 0.0241 → 88.0 % extracted.

Recovery: 69.2 % (single), 77.9 % (cross-current), 88.0 % (counter-current) for the same total solvent.

Example 2 – lever rule on the triangle (GATE level). Given: 100 kg of feed with 30 wt % A and 70 wt % B is mixed with 80 kg of pure solvent C in one stage. The tie line through the mixture joins a raffinate R (A 0.10, C 0.02, B 0.88) and an extract E (A 0.20, C 0.657, B 0.143). Find E, R, K and β.

1. M = 180 kg; x_A,M = 30/180 = 0.1667; x_C,M = 80/180 = 0.4444.
2. Lever rule on A: E/M = (0.1667 − 0.10)/(0.20 − 0.10) = 0.667 → E = 120 kg; R = 60 kg.
3. Checks: A: 120 × 0.20 + 60 × 0.10 = 30 kg ✓; C: 120 × 0.657 + 60 × 0.02 = 80 kg ✓.
4. K = 0.20/0.10 = 2.0; β = (0.20/0.143)/(0.10/0.88) = 12.3.
5. Fraction of A extracted = 24/30 = 0.80.

E = 120 kg, R = 60 kg, K = 2.0, β ≈ 12.3; 80 % of A extracted in one stage.

Common mistakes

• Forgetting that extract and raffinate both contain some of all three components in partly miscible systems.
• Applying the lever rule with lengths measured on the wrong side of the mixing point.
• Using K (fraction basis) in the immiscible-liquid formulas, which need K′ on a solute-free ratio basis.
• Placing the difference point inside the triangle – it normally lies outside, on the extension of the lines through F, E₁ and S, R_N.
• Assuming a large K is enough: selectivity β > 1 is also needed, or the solvent drags carrier along.
• Comparing schemes at different total solvent amounts.

For GATE CH

Expect lever-rule and mixing-point problems on the triangle, single-stage extraction with immiscible liquids, cross-current versus counter-current recovery, minimum solvent concepts, and definitions of distribution coefficient, selectivity and plait point. Practise solute-free-basis balances – they make most numericals one-line problems.

Quick check

1. What is the plait point?
2. Define selectivity in extraction.
3. For immiscible liquids with ε = 1 in a single stage with pure solvent, what fraction of solute is extracted?
4. For the same total solvent, which scheme gives the highest recovery?

Answers: 1. the point on the binodal curve where the two conjugate phases become identical; 2. β = (y_A/y_B)/(x_A/x_B); 3. one half; 4. counter-current.