# Solubility of a solid equilibria

1. Apr 9, 2010

I'm having some trouble understanding heterogenous equilibria. I read that solids, pure liquids and solvents are omitted from the equilibrium expression. Lets say I have some AgCl in a litre of water. The tiny amount of AgCl that dissociates and dissolved is given by its Ksp. I understand that once a sufficient amount of solid AgCl is present it will give a constant amount of dissociated ions regardless of how much more AgCl is added. What I don't understand is how the solvent (water in this case) is omitted. If I double the volume of water wouldn't I be doubling the amount of dissolved Ag+ and Cl- ions because I would be doubling the amount of solvent for them to dissolve in?

2. Apr 9, 2010

### Staff: Mentor

In AgCl solution solvent is not taking place in the equilibrium - whatever happens is described by Ksp. However, system is described not only by Ksp, but also by mass balances. Amount of AgCl put into the system is constant (that's just mass conservation). That means if you start with 1L of saturated solution in equilibrium with some small amount of solid AgCl, and you add more water, it may happen that all solid dissolves.

In some other reactions - like hydrolysis (A- + H2O <-> HA + OH-) water is one of the reactants - that means it is part of the equilibrium. In general equilibrium constant is given by

$$K = \frac {[HA][OH^-]} {[A^-][H_2O]}$$

however, concentration of water is (almost) constant - it is not changing, as there is a huge excess of water. That means we can multiply both sides by water concentration (something like 55.56 M) and write

$$K' = \frac {[HA][OH^-]} {[A^-]}$$

where K' = 55.56K

We can also assume water activity equals 1 - as it doesn't change, it doesn't matter what the value will be. It works for not too concentrated solutions, sometimes it may lead to troubles. But these are rare situations.

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3. Apr 9, 2010