Explain the Liquid Water aspect of using phase rule?

For a single phase of pure water, P equals 1. If we treat the water as the single species H₂O, s is 1 and r is 0. The phase rule then predicts two degrees of freedom:

F = 2 + s - r - P

= 2 + 1 - 0 - 1 = 2

Since F is the number of intensive variables that can be varied independently, we could for instance vary T and p independently, or T and ρ, or any other pair of independent intensive variables.

Next let us take into account the proton transfer equilibrium

2H₂O (l) ↔H₃OC (aq)

and consider the system to contain the three species H₂O, H₃O⁺, and OH^-. Then for the species approach to the phase rule, we have s = 3. We can write two independent relations:

1. For reaction equilibrium, -2μ_H2O + μ_H3O + μOH^- = 0;

2. for electroneutrality, m_H3O+ = mOH^-.

Thus, we have two relations involving intensive variables only. Now s is 3, r is 2, P is 1, and the number of degrees of freedom is given by

F = 2 + s - r - P = 2  

This is the same value of F as before

If we consider water to contain additional cation species (e.g., H₅O₂⁺), each such species would add 1 to s and 1 to r, but F would remain equal to 2. Thus, no matter how complicated are the equilibria that actually exist in liquid water, the number of degrees of freedom remains 2.

Applying the components approach to water is simple. All species that may exist in pure water are formed, in whatever proportions actually exist, from the single substance H2O. Thus, there is only one component: C = 1. The component version of the phase rule, F = 2 + C - P, gives the same result as the species version: F = 2.