Three-Component Systems: Triangular Diagrams

Introduction

Three-component mixtures require two independent mole fractions because the third is fixed by their sum. A triangular diagram uses its plane to display all possible compositions at chosen temperature and pressure. Such diagrams are central to liquid–liquid extraction: they show when a solvent and feed form one liquid, when they split into two, and what compositions the two layers have. The geometry is only useful when each vertex and tie line is read carefully.

Core explanation

For components A, B and C, every mixture satisfies x A+x B+x C=1. Each vertex of an equilateral composition triangle represents a pure component. A side represents a binary mixture with zero of the opposite vertex's component. Interior points represent all three. To read x A, follow the family of lines parallel to the side opposite A; analogous parallel lines give x B and x C. The three values should sum to one within graphical precision. A point at the triangle centre has roughly one-third of each component, but visual horizontal distance alone is not a mole fraction unless the diagram's construction says so.

At fixed temperature and pressure, a ternary system may have a closed or lens-like liquid–liquid two-phase region bounded by a binodal. A point outside that region represents one homogeneous liquid. A point inside represents an overall composition that separates into two conjugate liquid phases. A tie line joins their equilibrium compositions at the binodal endpoints. The endpoints are not generally pure components or arbitrary points on the boundary. Different overall compositions on the same tie line share the same two equilibrium phase compositions and differ in phase amounts.

The material-balance equation is a vector relation, z=f α x^α+f β x^β, where each symbol contains the three component fractions and f α+f β=1. Because z lies on the straight segment between the two endpoints, graphical distances along that tie line provide the lever ratio. A mass-basis diagram requires mass-basis balances; a mole-fraction diagram requires mole amounts. One must not use distances across different tie lines or assume that the tie lines are all parallel unless measured data justify that approximation.

At a plait point the two conjugate liquid compositions approach one another and their tie-line length tends to zero. The two liquids become indistinguishable at that point. A plait point is analogous in some geometric respects to a critical solution point, but the ternary diagram is a composition projection at fixed T and pressure rather than a simple binary temperature axis. Near it, reading endpoints and phase amounts from a printed diagram becomes sensitive to small graphical errors.

In extraction, a feed containing solute and carrier is mixed with a solvent. The combined overall composition lies on the straight mixing line between feed and solvent points, with position set by their amounts. If that mixture falls in the two-phase region, it separates into solvent-rich extract and carrier-rich raffinate at the tie-line endpoints. The solute distribution between them determines extraction performance. Repeating contacts with fresh solvent can remove more solute, but each stage requires equilibrium and flow balances.

The ternary phase rule gives C=3 and P=2, hence F=3 in full T, pressure and composition space. At fixed T and pressure one remaining intensive freedom selects which tie line or endpoint pair is present; overall composition along a selected tie line changes amounts. This is why many tie lines can occupy one triangular diagram without contradicting phase-rule counting.

Step-by-step reasoning

Check the three vertex labels and whether fractions are molar or mass based. Read two fractions along their proper parallel grid lines and obtain the third by subtraction. Locate the overall mixture relative to the binodal. If inside, find the tie line through it, read the conjugate endpoints, and solve the vector material balance or measure lengths along that one line for phase amounts.

Visual explanation

Picture a triangle with A, B and C at the corners. A curved binodal encloses a two-liquid pocket. A short straight tie line within it touches the pocket boundary at a solvent-rich point and a carrier-rich point. The overall mixture point lies between them, and its distances to endpoints are opposite to the phase amounts.

Real-world analogy

A three-colour paint recipe can be marked by a point inside a triangle whose corners are pure colours. Mixing two existing recipes places the result on the straight segment between their points. The analogy captures composition geometry; actual liquid–liquid splitting requires a Gibbs-energy preference for two phases, which paint blending alone does not supply.

Real-world example

In solvent extraction of an organic solute from water, water, solute and extraction solvent form a ternary system. A suitable solvent may create two liquid layers with different solute fractions. A measured ternary binodal and tie lines let an engineer estimate extract and raffinate compositions after one equilibrium contact.

Why?

Why must an overall two-phase composition lie on its tie line? Each component's total amount is the sum of its amounts in the two layers. Dividing by total material makes the overall composition a weighted average of the two endpoint vectors. Weighted averages with nonnegative phase fractions occupy the straight segment joining those vectors.

Common misconception

A ternary point does not have three independent composition coordinates; the fractions sum to one. Another error is to use the nearest binodal points as equilibrium endpoints. Conjugate endpoints must lie on the measured equilibrium tie line through the overall composition, not merely on the same two-phase boundary.

Worked example

At fixed T and pressure, two equilibrium liquid endpoints have (A,B,C) fractions α=(0.70,0.20,0.10) and β=(0.10,0.50,0.40). If their mole amounts are equal, the overall point is the midpoint z=(0.40,0.35,0.25). Each coordinate is the average of the corresponding endpoint values, and 0.40+0.35+0.25=1. The material balance places z on the tie line connecting α and β.

Quick check

1. How many independent mole fractions describe an ordinary ternary composition? Answer: Two; the third is fixed by x A+x B+x C=1, so the triangular plot is two-dimensional.

Exam focus

State vertex labels, fraction basis and fixed temperature and pressure. Check that each composition triple sums to one. Separate a mixing line, which joins feed and solvent overall points, from an equilibrium tie line, which joins conjugate phase compositions after separation.

Advanced insight

The tie lines and binodal can be constructed from equality of each component's chemical potential in the two liquids. At a plait point the two solutions coalesce and the phase distinction vanishes. Activity-coefficient models or experimental measurements are needed to locate these curves; triangle geometry alone gives material balances but cannot predict equilibrium endpoints.

Summary

A triangular diagram maps ternary compositions with two independent fractions. A binodal encloses a possible two-liquid region, and each tie line links equilibrium layer compositions. Overall composition lies on the tie line by component balance, with phase fractions from the lever rule; extracting a solute uses both this geometry and measured phase equilibrium.

Practice questions

1. If x A=0.25 and x B=0.40, what is x C? Answer: x C=1−0.25−0.40=0.35 on the same mole-fraction basis. 2. Where is a mixture with x C=0 plotted in a triangle? Answer: On the side joining pure A and pure B, opposite vertex C. 3. What does a tie line join in a ternary liquid–liquid diagram? Answer: The two conjugate equilibrium liquid compositions at the stated temperature and pressure. 4. Can a point outside the binodal use a two-phase lever rule? Answer: No. Outside the two-phase region the equilibrium sample is one liquid, so there are no conjugate endpoints to balance.