Ion product vs. solubility product

An expression such as [Ag⁺]²?[CrO₄^2–] in known generally as an ion product — this one being the ion product for silver chromate. An ion product can in principle have any positive value, depending on the concentrations of the ions involved. Only in the special case when its value is identical with K_s does it become the solubility product. A solution in which this is the case is said to be saturated. Thus when [Ag⁺]²?[CrO₄^2–] = 2.76E-12 at the temperature and pressure at which this value K_s of applies, we say that the "solution is saturated in silver chromate".

A solution must be saturated to be in equilibrium with the solid. This is a necessary condition for solubility equilibrium, but it is not by itself sufficient. True chemical equilibrium can only occur when all components are simultaneously present.

A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions.

Failure to appreciate this is a very common cause of errors in solving solubility problems.

Undersaturated and supersaturated solutions

If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be undersaturated.

A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.

How to know the saturation status of a solution

This is just a simple matter of comparing the ion product Q_s with the solubility product K_s. So for the system

Ag₂CrO₄(s)
2 Ag⁺ + CrO₄^2– (4b)

a solution in which Q_s?< K_s (i.e., K_s?/Q_s > 1) is undersaturated (blue shading) and the no solid will be present. The combinations of [Ag⁺] and [CrO₄^2–] that correspond to a saturated solution (and thus to equilibrium) are limited to those described by the curved line. The pink area to the right of this curve represents a supersaturated solution.

If Qsp<Ksp	Forward process is favored. No precipitation occurs, if solid is present, more solid can dissolve.
Qsp=Ksp	Solid and solution are in equilibrium; neither forward nor reverse process is favored.
Qsp>Ksp	Reverse process is favored. Precipitation occurs to form more solid.

Example 1A sample of groundwater that has percolated through a layer of gypsum (CaSO₄, K_s = 4.9E–5 = 10^–4.3) is found to have be 8.4E–5 M in Ca²⁺ and 7.2E–5 M in SO₄^2–. What is the equilibrium state of this solution with respect to gypsum?

Worked Problem Here

Subpages (1): Example 1