Q1(d). Phenol–water: two liquid layers and the phase rule

1. Components: phenol and water, no reaction: C=2. Phases: two liquid layers, conjugate solutions: L₁ (phenol dissolved in water) and L₂ (water dissolved in phenol): P=2. The vapour is not counted (liquids in an open vessel under the atmosphere).
2. Degrees of freedom: F=C-P+2=2-2+2, F=2 (temperature and pressure)
3. Condensed system: pressure has little effect on liquids and is held at 1 atm, so one variable is used up: F^′=C-P+1=1 The only variable left to choose is temperature; at 40°C the layers have fixed compositions L₁ and L₂ (ends of the tie line), whatever the overall proportions.
4. Reading the diagram: the mutual solubility of the two liquids rises with temperature; the two branches meet at the upper critical solution temperature (about 66°C, about 34 % phenol by mass). Above it, or outside the dome, there is one liquid.
5. Other regions at fixed pressure: one liquid (outside the dome): F^′=2-1+1=2 (temperature and composition). At the critical point the two layers become identical, an extra condition: F^′=0.
6. With the vapour included (closed vessel, three phases L₁ + L₂ + vapour): F=2-3+2=1; at each temperature the vapour pressure is fixed.

Lever rule: on a tie line the masses of the layers are in the inverse ratio of the distances of the overall composition from the two ends.

Application. the critical solution temperature is very sensitive to a third substance: a solute that dissolves in only one layer (for example NaCl in the water layer) raises the UCST of phenol–water. The sharp clouding temperature is therefore used as a test of purity.