I was thinking about what exactly happens at the triple point of water, and what could be the effect of adding solutes/solvents. I feel we can represent the triple point using the following equilibrium reaction: $$\ce{H2O (s) <=> H2O (l) <=> H2O (g)}$$

Is this correct? How would this change if I add solutes?

Note: I know that adding solutes would increase the concentration of the solution, but how exactly will it shift the equilibrium?

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    $\begingroup$ Quibbles - (1) If the liquid isn't pure, then the system can't be at the triple point. (2) There is also an equilibrium between the gas and the solid. That's why it is a triple point. Three equilibriums at once. // The salt would only exist in the solution. So ice would melt (have you ever made ice cream by hand?), the water gets colder, and the vapor pressure of the gas phase goes down. $\endgroup$
    – MaxW
    Jan 20, 2020 at 7:42
  • $\begingroup$ @MaxW Are you saying that even if I prepared a ~5M solution of NaCl +water and the solution is homogenous, it would not have a triple point? $\endgroup$ Jan 20, 2020 at 7:51
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    $\begingroup$ The triple point will no longer be a point. Ditto for the melting and boiling points, BTW. $\endgroup$ Jan 20, 2020 at 7:55
  • $\begingroup$ The phase diagram in chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/… shows a simple transition of the phase lines when a solute is added. Is this wrong? $\endgroup$ Jan 20, 2020 at 7:59
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    $\begingroup$ It is an oversimplification. $\endgroup$ Jan 20, 2020 at 8:03

2 Answers 2


At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria are shown below.

enter image description here

Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a frozen over pond. There must be some liquid water in contact with the gas phase too.

A subtlety here... You can't make a solid phase of ice that is homogeneously say 1 molar NaCl. If you take a 1 molar NaCl and try to freeze it you get relatively pure water first, then saltier and saltier water freezing as the solid phase gets colder and colder. Since the impure ice has doesn't have a homogeneous composition, it doesn't have a specific melting point.

So for the system you proposed there is pure water ice, a 1 molar NaCl solution, and a gas phase of water vapor. Unfortunately humans can't make a perfectly adiabatic container, so as the ice melts, the 1 molar NaCl solution becomes more dilute. Thus there is no specific triple point.

In other words... if you just throw some ice into the 1 molar NaCl solution, the final temperature of the theoretical system will depend on the relative amounts of the three phases - How much 1 molar NaCl solution, how much ice, and the volume of the gas phase.


The triple point of water is an example of Thermodynamic Equilibrium, which is explained by Wikipedia as:

simultaneously in mutual thermal, mechanical, chemical, and radiative equilibria

Your equation only covers the chemical facet of this equilibrium and hence is not the correct way of representing total thermodynamic equilibrium. I am also not sure of the correct way of representing these kinds of equilibria which deal with both energy and physical states.

If this equilibrium is disturbed chemically i.e. by adding more water, according to Le Chatelier's principle, the equilibrium will shift towards the solid and gaseous states. As the enthalpy of vapourization is more than enthalpy of fusion, more ice will be formed as compared to the amount of steam evolved in order to balance the energies involved. Hence, it would tend to shift to the left.

  • $\begingroup$ No voodoo here. Ice from pure water, a salt solution and water vapor would be a well well studies system. The whole system depends on temperature. There has to be enough ice to melt which will cool the water and the gas phase. $\endgroup$
    – MaxW
    Jan 20, 2020 at 7:44

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