Let's suppose we have a sparingly soluble ionic salt, MX2. Its dissolution in water is governed by the following equilibrium expression:
MX2(s)⇌M2++2X−
We write the Ksolubility product expression in this way:
Ksp=[M2+][X−]2.
The larger the Ksp, clearly the GREATER the solubility of the salt MX2. Ksp values have been measured for many inorganic salts. Why? Well, if you are in the precious metal caper, and want to isolate gold or platinum or rhodium salts, then you would want to make sure that your experiments precipitate out all of the precious metal that is in solution. If you are working with Hg2+ or Cd2+, you would also want to make sure that you are NOT releasing these toxic ions into water waste.
Given the expression, Ksp=[M2+][X−]2, then if we represent the solubility of MX2 as S, then, clearly, Ksp=[S][2S]2=4S3. Now the equilibrium responds to ALL the X− and M2+ ions in solution. It does not differentiate the source of the ion. So suppose, you have artificially raised [X−] by adding a soluble source of this ion, which acts as a common ion. Because now added X− contributes to the ion product, you are at the same time REDUCING the solubility of MX2.
In your text you can read of salting out, and salting in which these solubility equilibria are manipulated.
Just to add, it may seem that using Ksp=[M2+][X−]2, you could reduce the solubility of MX2 TO ANY LEVEL, simply by loading the solution with a soluble form of X−. At very high anion concentrations, complex ion formation MAY occur:
M2++4X−⇌[MX4]2−
And this equilibrium would bring the metal ion into solution.
