Question

What is the `"solubility equilibrium"`?

Answer 1

Well this could be anything. You most likely refer to `"solubility equilibria."`

Explanation:

Consider a sparingly soluble salt in aqueous solution...`MX_n`...

And according to the temperature it undergoes dissolution to some extent in water...

`MX_n(s) stackrel(H_2O)rightleftharpoonsM^(n+) +nX^-`

This is certainly a measurable equilibrium reaction, and we would write the solubility expression as...

`K_"sp"=[M^(n+)][X^-]^n`

And of course the solid does not appear in the expression, because as a solid it cannot express a concentration.

Now the expression depends on metal and anion from ALL sources. If we make `[X^-]` artificially high, by adding a soluble `X^-` source we may `"salt out"` the metal cation by adding the common ion. And if the metal cation were `Au^(3+)` or `Pt^(2+)` or `Rh^(+)`, we may well want to recover all the metal ions in solution... The salt precipitates on the basis of the acitibity of the `"common ion"`.

Under very high concentrations of `X^-`...we might get some formation of complex ions such as `[MX_(n')]^((n-n')^-)`...but this is another story.