Solubility Equilibria

Solubility is very important

                           When a typical ionic solid dissolves in water, it dissociates completely into cations and anions.
CaF2 (s) -----> Ca 2+ (aq) + 2F- (aq)
             In the above reaction the calcium fluoride dissolves in water and dissociates into calcium ions and fluoride ions.
As the reaction occurs the concentration of the calcium ions and fluoride ions increase, making it more likely that these ions will collide again and re-form the solid.
Thus two competing actions are being performed.
Ca 2+ (aq) + 2F- (aq) ----> CaF2 (s)
Eventually equilibrium is reached and no more solid dissolves or forms.

An equilibrium expression for this process is Ksp = [Ca2+] [F-]2
[Ca2+] and [F-]2 are expressed in mol/L.
Ksp is called the solubility product constant.
Since CaF2 is a pure solid, it is not added in this expression.
Pure liquids and pure solids are never included in an equilibrium expression.

Calculating Solubility Products

Copper(I) Bromide, CuBr, has a measured solubility of 2.0 x 10 -4 mol/L at 25 degrees celsius.  That is when excess CuBr(s) is placed in 1.0 L of water, we can determine that 2.0 x 10-4 mol of a solid dissolves to produce a saturated solution.                 Calculate the solid's Ksp value.
                                                       Both Cu+ and Br- have a concentration of 2.0 x 10-4 when they become ions.
                                                                                  [2.0 x 10-4] [2.0 x 10-4] = 4.0 x 10-8
                                                                                                       Ksp = 4.0 x 10-8