# Predicting the vapor pressure of water ammonia solution

I am working on a project that involves the compressors for a vapor compression refrigeration cycle of R717 (ammonia) and came across something that has been boggling my mind a little bit for no good reason. I suspect there is water entering the system and making the compressor work harder than it needs to and thus causing higher energy costs. I was reading an article that stated

For example, pure ammonia at -4°F (-20°C) has a saturation pressure of 12.9 psig (0.90 barg). An ammonia-water solution of 10% water and 90% ammonia by weight, at the same -4°F (-20°C) has a saturation pressure of 10.2 psig (0.72 barg). Therefore to maintain an evaporator temperature of -4°F (-20°C) in a system with 10% water contamination, the suction pressure would have to be run at 12.9 – 10.2 = 2.7 psi (0.18 bar) lower than if the water were not present.

Now I remember from thermodynamics that I can use Antoine's equation to determine the vapor pressure of a substance at a given temperature. I have gone to the NIST and obtained coefficients for water and ammonia. Problem is, at -4F, ammonia is a liquid, but water would be a solid. My issue is, how do I calculate the vapor pressure when, at the given temperature, one of the species is a solid? Or is that not a concern, because according to the phase diagram, , the solution is a fluid assuming at most 20% water in the solution. If that is the case, the how would vapor pressure be calculated as the antoine's coefficients at only valid for liquid water about 0C? I think I am having a brain fart honestly, but maybe you all could help me out.