12.2 Liquid-Liquid Extraction and Ternary Phase Equilibria

12.2 Liquid-Liquid Extraction and Ternary Phase Equilibria

When components in a liquid mixture have nearly identical boiling points, form close-boiling azeotropes, or degrade thermally at reboiler temperatures, distillation becomes economically unviable or physically impossible. Under these conditions, Liquid-Liquid Extraction (LLE)—also termed solvent extraction—is the separation method of choice. LLE exploits differences in the chemical structure and solubility of components between two partially miscible liquid phases.

Key industrial examples tested on the NCEES PE Chemical Exam include the extraction of aromatics (benzene, toluene, xylene) from aliphatic hydrocarbons using sulfolane, recovery of acetic acid from dilute aqueous streams using ethyl acetate or isopropyl ether, and extraction of antibiotics and active pharmaceutical ingredients (APIs) from fermentation broths.

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1. Terminology and Basic Phase Equilibrium Definitions

An extraction system involves at least three distinct chemical components:

1. Solute ($A$): The target species to be removed or recovered.
2. Carrier / Diluent ($B$): The liquid component initially containing the solute in the feed stream.
3. Solvent ($S$ or $C$): The added extracting liquid that is immiscible or partially miscible with the carrier $B$, and possesses high chemical affinity for solute $A$.

The feed stream ($F$) containing $A + B$ is contacted with solvent stream ($S$). The system separates into two liquid phases:

• Extract ($E$): The solvent-rich phase containing the extracted solute $A$ and a minor amount of carrier $B$.
• Raffinate ($R$): The carrier-rich phase containing the residual, unextracted solute $A$ and dissolved solvent $S$.

              Solvent Feed (S)
                    |
                    v
             +-------------+
  Feed (F) ->|  EXTRACTION |-> Extract (E) [Solvent-Rich + Solute A]
  [A + B]    |   COLUMN    |
             +-------------+
                    |
                    v
              Raffinate (R) [Carrier-Rich + Residual A]

1. Distribution Coefficient ($K_D$)

The distribution coefficient (also known as the partition coefficient) measures the equilibrium affinity of solute $A$ for the extract phase relative to the raffinate phase:

$K_D \equiv \frac{y_A}{x_A}$

Where:

• $y_A$ = mass or mole fraction of solute $A$ in the extract phase.
• $x_A$ = mass or mole fraction of solute $A$ in the conjugate raffinate phase.

For an effective extraction, $K_D > 1.0$ is highly desirable, as it minimizes the required solvent circulation rate. If $K_D < 1.0$, extraction is still possible but requires large solvent volumes ($S \gg F$).

2. Selectivity (Separation Factor, $\beta$)

Selectivity ($\beta$) measures the ability of solvent $S$ to separate solute $A$ from carrier $B$. It is directly analogous to relative volatility ($\alpha$) in distillation:

$\beta_{A/B} \equiv \frac{y_A / x_A}{y_B / x_B} = \frac{K_{D,A}}{K_{D,B}}$

Where:

• $y_B, x_B$ = mass or mole fractions of carrier $B$ in extract and raffinate phases, respectively.
• $K_{D,A}, K_{D,B}$ = distribution coefficients of solute $A$ and carrier $B$.

Selectivity Criteria:

• $\beta > 1.0$: The solvent preferentially dissolves solute $A$ over carrier $B$. Separation by extraction is feasible. Industrial systems typically target $\beta \ge 5$ to $10$.
• $\beta = 1.0$: The solvent dissolves $A$ and $B$ in exact proportion to their concentrations. Separation by extraction is thermodynamically impossible.
• $\beta < 1.0$: The solvent preferentially extracts carrier $B$ rather than solute $A$.

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2. Ternary Phase Diagrams: Equilateral and Right-Angled

Ternary liquid-liquid equilibria (LLE) are mapped using three-component phase diagrams governed by the Gibbs Phase Rule: $F = C - P + 2 = 3 - 2 + 2 = 3$ degrees of freedom. At fixed operating temperature and pressure, $F = 1$, meaning specifying one liquid-phase composition uniquely fixes all other conjugate equilibrium compositions.

             Equilateral Triangle               Right-Angled Triangle
                    Solute (A)                    Solute (A) ^
                       /\                                    |
                      /  \                                   |\ 
                     /    \                                  | \ 
                    / Dome \                                 |  \ Dome
                   /--------\                                |   \ 
                  /  Tie-    \                               |    \ 
                 /   Lines    \                              |-----\ 
                /              \                             |      \ 
  Carrier (B)  +----------------+ Solvent (S)     Carrier (B) +-------+---> Solvent (S)

Equilateral Triangular Diagrams

1. Each vertex represents $100%$ purity of a pure component ($A$, $B$, or $S$).
2. The side opposite a vertex represents $0%$ of that component.
3. Composition lines run parallel to the edges. At any interior point, $x_A + x_B + x_S = 1.00$ ($100%$).

Right-Angled Triangular Diagrams

Right-angled diagrams plot solute mass fraction ($x_A$) on the vertical y-axis and solvent mass fraction ($x_S$) on the horizontal x-axis. The origin $(0,0)$ represents $100%$ pure carrier $B$. The carrier fraction is determined by difference: $x_B = 1.0 - x_A - x_S$. This format is widely used on computer screens and engineering tests because standard Cartesian axes can be read directly.

Anatomical Features of the Ternary Diagram

• Binodal Curve (Solubility Envelope): The dome-shaped boundary separating the single-phase homogeneous liquid region (outside the dome) from the two-phase heterogeneous region (inside the dome). Any overall mixture falling beneath the binodal curve spontaneously splits into two conjugate liquid phases.
• Tie Lines: Straight lines connecting two equilibrium phases in thermodynamic contact inside the two-phase region. One end terminates at the Extract side of the binodal curve ($E$), while the other end terminates at the Raffinate side ($R$). Tie lines are generally non-horizontal and cannot cross each other.
• Plait Point ($P$): The critical point on the binodal curve where the tie lines shrink to zero length. At the plait point, the compositions and physical densities of the extract and raffinate phases become identical, and the interfacial tension vanishes.

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3. The Lever-Arm Rule in Extraction

When two streams are mixed, or when an unstable mixture splits into two equilibrium phases, the overall material balance obeys the Inverse Lever-Arm Rule.

Total and Component Mass Balances

Mixing Feed ($F$) with Solvent ($S$) produces an overall combined mixture ($M$):

$M = F + S = E + R$ $M \cdot x_{A,M} = F \cdot x_{A,F} + S \cdot x_{A,S} = E \cdot y_{A,E} + R \cdot x_{A,R}$ $M \cdot x_{S,M} = F \cdot x_{S,F} + S \cdot x_{S,S} = E \cdot y_{S,E} + R \cdot x_{S,R}$

Geometric Interpretation on the Phase Diagram

1. The mixture point $M$ must lie strictly on the straight line connecting $F$ and $S$.
2. By the inverse lever-arm rule on line $F-S$: $\frac{F}{S} = \frac{\overline{MS}}{\overline{FM}}$ $\frac{F}{M} = \frac{\overline{MS}}{\overline{FS}}, \quad \frac{S}{M} = \frac{\overline{FM}}{\overline{FS}}$
3. If point $M$ falls within the two-phase region, it splits along the unique equilibrium tie line passing through $M$, establishing conjugate phases $E$ and $R$.
4. By the inverse lever-arm rule on tie line $E-M-R$: $\frac{E}{R} = \frac{\overline{MR}}{\overline{EM}}$ $\frac{E}{M} = \frac{\overline{MR}}{\overline{ER}}, \quad \frac{R}{M} = \frac{\overline{EM}}{\overline{ER}}$

[!IMPORTANT] Lever-Arm Rule Direction:
The mass of a phase is proportional to the opposite line segment! For example, $E$ is proportional to the segment between $M$ and $R$ ($\overline{MR}$), NOT the segment adjacent to $E$ ($\overline{EM}$). Inverting this ratio is one of the most common calculation traps on the PE exam.

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4. Multistage Countercurrent Extraction: Hunter-Nash Method

To achieve high solute recovery while minimizing solvent consumption, industrial extractors use multistage countercurrent cascades.

          Extract E1 <--- [Stage 1] <--- [Stage 2] <--- ... <--- [Stage N] <--- Solvent S
                             |              |                        |
          Feed F   ---> [Stage 1] ---> [Stage 2] ---> ... ---> [Stage N] ---> Raffinate RN

Overall Cascade Balances and the Difference Point ($\Delta$)

Overall mass balances across the entire $N$-stage cascade:

$F + S = E_1 + R_N = M$

Rearranging shows that the difference in mass flow between adjacent countercurrent streams is constant across every stage in the column:

$\Delta = F - E_1 = R_N - S = R_j - E_{j+1}$

Where $\Delta$ is the operating difference point (or pole point). Geometrically:

1. The line through $F$ and $E_1$ must pass through $\Delta$.
2. The line through $R_N$ and $S$ must also pass through $\Delta$.
3. Therefore, point $\Delta$ lies at the intersection of line $F-E_1$ and line $R_N-S$.

The Hunter-Nash Stepping Procedure

1. Plot feed point $F$, solvent point $S$, desired raffinate $R_N$, and extract $E_1$.
2. Draw line $F-E_1$ and line $S-R_N$ and extend them until they intersect at the difference point $\Delta$.
3. Step 1 (Equilibrium Tie Line): Starting at $E_1$, follow the tie line across the two-phase region to determine the raffinate leaving stage 1, $R_1$.
4. Step 2 (Operating Line through $\Delta$): Connect $R_1$ to the pole point $\Delta$. The intersection of line $R_1-\Delta$ with the extract side of the binodal curve locates $E_2$.
5. Subsequent Steps: Follow the tie line from $E_2$ to find $R_2$, then draw line $R_2-\Delta$ to find $E_3$. Repeat until the calculated raffinate composition $R_j$ reaches or surpasses the design specification $R_N$.
6. The total number of tie lines traversed equals the number of theoretical stages ($N$).

Minimum Solvent Rate ($S_{min}$)

If the solvent rate is reduced, the mixture point $M$ shifts toward $F$, moving the operating point $\Delta$ closer to the binodal curve. The minimum solvent rate ($S_{min}$) occurs when an operating line passing through $\Delta$ coincides exactly with an equilibrium tie line that projects through the feed point $F$ (tie-line pinch). At this condition, an infinite number of stages ($N \to \infty$) is required.

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5. Immiscible Extraction and the Analytical Kremser Equation

When the carrier solvent ($B$) and extracting solvent ($S$) are virtually insoluble in one another (e.g., water and heavy hydrocarbons), the flow rates of carrier $B$ and solvent $S$ remain strictly constant through all stages.

Solute-Free Mass Ratios

Let:

• $B$ = mass flow rate of pure carrier liquid ($\text{kg/h}$ or $\text{lb/h}$).
• $S$ = mass flow rate of pure extracting solvent ($\text{kg/h}$ or $\text{lb/h}$).
• $X$ = solute-free mass ratio in raffinate: $X = \frac{x_A}{1 - x_A} = \frac{\text{mass solute } A}{\text{mass carrier } B}$.
• $Y$ = solute-free mass ratio in extract: $Y = \frac{y_A}{1 - y_A} = \frac{\text{mass solute } A}{\text{mass solvent } S}$.

The operating line is linear in solute-free coordinates:

$B (X_{in} - X_{out}) = S (Y_{out} - Y_{in})$ $Y = \frac{B}{S} (X - X_{out}) + Y_{in}$

If the equilibrium relationship is also linear ($Y = m \cdot X = K_D' \cdot X$):

The Extraction Factor ($E_{ext}$)

$E_{ext} \equiv \frac{m \cdot S}{B} = \frac{K_D' \cdot S}{B}$

[!TIP] Contrast with Absorption Factor:
Notice that in extraction, the equilibrium slope $m$ ($K_D$) is in the numerator ($E_{ext} = m S / B$), whereas in gas absorption, $m$ is in the denominator ($A = L / (m G)$). This is because in extraction, the extracting agent ($S$) receives solute from $B$, whereas in absorption, the liquid ($L$) scrubs solute from gas ($G$). For an effective countercurrent extraction cascade, $E_{ext} > 1.0$ (typically $1.3 \le E_{ext} \le 2.0$).

Analytical Theoretical Stage Sizing

For a countercurrent cascade with pure entering solvent ($Y_{in} = 0$):

$N = \frac{\ln\left[ \left( \frac{X_{in}}{X_{out}} \right) \left( 1 - \frac{1}{E_{ext}} \right) + \frac{1}{E_{ext}} \right]}{\ln(E_{ext})}$

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6. Industrial Extraction Equipment Selection

Selecting extraction equipment requires balancing stage efficiency, residence time, liquid-liquid settling rates, and mechanical shear.

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7. Comprehensive Worked Numerical Example: Countercurrent Extraction of Pyridine

Problem Statement

A wastewater stream of $F = 2,000.0\text{ kg/h}$ containing $25.0\text{ wt}%$ pyridine ($A$) and $75.0\text{ wt}%$ water ($B$) is to be treated in a countercurrent extraction cascade using pure benzene ($S$) at $25.0^\circ\text{C}$ to reduce the pyridine content in the exiting raffinate to $1.50\text{ wt}%$ (wet basis).

Thermodynamic & Process Data:

• Water ($B$) and benzene ($S$) are mutually completely immiscible.
• The equilibrium distribution of pyridine on a solute-free mass ratio basis is linear:
  $Y = m \cdot X = 0.80 \cdot X \quad (K_D' = 0.80)$
• Pure benzene solvent is fed to the extractor ($Y_{in} = 0$).

Calculate:

1. The mass flow rate of pure water carrier ($B$) and the initial solute ratio ($X_{in}$).
2. The required final solute ratio in the raffinate ($X_{out}$).
3. The minimum solvent flow rate $S_{min}$ in $\text{kg/h}$.
4. The actual solvent rate $S$ using a design margin of $1.40 \times S_{min}$.
5. The extraction factor $E_{ext}$.
6. The outlet extract solute ratio ($Y_{out}$) and total pyridine extracted in $\text{kg/h}$.
7. The required number of theoretical stages $N$ using the Kremser extraction equation.

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Step 1: Carrier Flow and Feed Solute Ratio

$B = F \times (1 - x_{A,F}) = 2,000.0\text{ kg/h} \times (1 - 0.250) = \mathbf{1,500.0\text{ kg/h of water}}$

Initial solute-free ratio in feed:

$X_{in} = \frac{x_{A,F}}{1 - x_{A,F}} = \frac{0.250}{0.750} = \mathbf{0.3333\text{ kg pyridine / kg water}}$

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Step 2: Target Raffinate Solute Ratio

The target raffinate concentration is $x_{A,R} = 0.0150$ ($1.50\text{ wt}%$):

$X_{out} = \frac{x_{A,R}}{1 - x_{A,R}} = \frac{0.0150}{1 - 0.0150} = \frac{0.0150}{0.9850} = \mathbf{0.01523\text{ kg pyridine / kg water}}$

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Step 3: Minimum Solvent Rate ($S_{min}$)

At the minimum solvent rate ($N \to \infty$), the extract leaving stage 1 ($Y_{out}^*$) reaches equilibrium with the incoming feed ($X_{in}$):

$Y_{out}^* = m \cdot X_{in} = 0.80 \times 0.3333 = 0.26667\text{ kg pyridine / kg benzene}$

Applying the overall solute balance at minimum solvent flow:

$B(X_{in} - X_{out}) = S_{min}(Y_{out}^* - Y_{in})$ $S_{min} = B \cdot \frac{X_{in} - X_{out}}{Y_{out}^* - 0} = 1,500.0 \times \frac{0.3333 - 0.01523}{0.26667} = 1,500.0 \times \frac{0.31810}{0.26667} = 1,500.0 \times 1.1929 = \mathbf{1,789.3\text{ kg/h}}$

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Step 4: Actual Solvent Rate ($S$)

Using $S = 1.40 \times S_{min}$:

$S = 1.40 \times 1,789.3\text{ kg/h} = \mathbf{2,505.0\text{ kg/h of benzene}}$

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Step 5: Extraction Factor ($E_{ext}$)

$E_{ext} = \frac{m \cdot S}{B} = \frac{0.80 \times 2,505.0\text{ kg/h}}{1,500.0\text{ kg/h}} = \frac{2,004.0}{1,500.0} = \mathbf{1.336}$

Because $E_{ext} = 1.336 > 1.0$, deep solute recovery is achieved without an internal equilibrium pinch.

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Step 6: Outlet Extract Composition and Solute Removal

From the operating line balance:

$Y_{out} = \frac{B}{S} (X_{in} - X_{out}) = \frac{1,500.0}{2,505.0} \times (0.3333 - 0.01523) = 0.59880 \times 0.31810 = \mathbf{0.19048\text{ kg/kg}}$

Total pyridine extracted:

$\dot{m}_{A,extracted} = B(X_{in} - X_{out}) = 1,500.0 \times (0.3333 - 0.01523) = 1,500.0 \times 0.31810 = \mathbf{477.15\text{ kg/h}}$ $\text{Recovery} = \frac{477.15\text{ kg/h}}{500.0\text{ kg/h}} \times 100\% = \mathbf{95.43\%}$

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Step 7: Number of Theoretical Stages ($N$)

Using the Kremser extraction equation for $Y_{in} = 0$:

$\frac{X_{in}}{X_{out}} = \frac{0.3333}{0.01523} = 21.884$ $1 - \frac{1}{E_{ext}} = 1 - \frac{1}{1.336} = 1 - 0.74850 = 0.25150$

Evaluating the numerator argument:

$\text{Arg} = \left( \frac{X_{in}}{X_{out}} \right) \left( 1 - \frac{1}{E_{ext}} \right) + \frac{1}{E_{ext}} = (21.884 \times 0.25150) + 0.74850 = 5.5038 + 0.74850 = 6.2523$ $\ln(\text{Arg}) = \ln(6.2523) = 1.83295$ $\ln(E_{ext}) = \ln(1.336) = 0.28968$

$N = \frac{1.83295}{0.28968} = \mathbf{6.33\text{ theoretical stages}}$

A standard design would specify an agitated column equivalent to $7$ theoretical stages.

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8. Critical PE Exam Traps & Pitfalls

Trap 1: Inverting the Lever-Arm Rule Ratio
In phase split calculations, remember that the length of the segment opposite a phase corresponds to that phase's mass. On tie line $E-M-R$, the mass of extract is $E = M \cdot (\overline{MR} / \overline{ER})$, and the mass of raffinate is $R = M \cdot (\overline{EM} / \overline{ER})$. Swapping the numerator segments inverts the phase split, yielding catastrophic errors.

Trap 2: Forgetting Downstream Solvent Recovery Costs
Extraction does not produce a pure final product; it merely transfers the solute from one liquid phase into another. The extract must subsequently be separated (typically via distillation) to recover the solvent for recycle and isolate pure solute. On PE exam design evaluations, the most volatile or easiest-to-distill solvent is chosen to minimize reboiler duty.

Trap 3: Misidentifying the Plait Point
The plait point is NOT necessarily at the apex (top) of the binodal curve! The apex represents the maximum solute concentration of the two-phase dome, whereas the plait point is where the tie line length shrinks to zero. Depending on the mutual solubility of carrier and solvent, the plait point can lie on the left or right slope of the curve.

Trap 4: Confusing Extraction Factor $E_{ext}$ with Absorption Factor $A$
In gas absorption, $A = L / (m G)$, where $m$ is in the denominator. In liquid extraction, $E_{ext} = m S / B$, where $m = K_D$ is in the numerator. Applying the absorption definition to extraction will cause you to compute $1/E_{ext}$, incorrectly concluding that an operable system has pinched.