**Molar solubility** represents the number of moles of solute that can be dissolved per liter of solution before the solution becomes saturated.

The *solubility product constant*, or #K_(sp)#, expresses the product of the molar concentrations of ions raised to the power of their respective stoichiometric coefficients from the equilibrium reaction.

Usually, you'll get either the value of #K_(sp)# and have to determine the molar solubility of the dissociated ions, or vice versa. I'll demonstrate both cases. A general equation of a salt dissolved in aqueous solution is

#A_aB_(b(s)) rightleftharpoons aA_((aq)) + bB_((aq))#

The expression for this reaction's #K_(sp)# is

#K_(sp) = [A]^(a) * [B]^(b)#

Let's say you know #K_(sp)#. You can determine the molar solubility of both ions by using the fact that *both concentrations increase proportional to their respective stoichiometric coefficients*.

For A, the increase in the concentration of the ion will be #a*x# (there are #"a"# moles of #"A"# dissociated in solution); for #"B"#, this increase will be #b * x#. Therefore,

#K_(sp) = (a*x)^(a) * (b * x)^(b) = a^(a) b^(b) * x^((a+b))#

Since #"a"# and #"b"# are derived from the equilibrium equation, solving for #x# will produce the molar solubilities of both ions.

LIkewise, if you know the concentrations of the dissociated ions, #K_(sp)# can then be determined by plugging these values in the equation

#K_(sp) = [A]^(a) * [B]^(b)#

Here are some links to actual examples:

http://socratic.org/questions/find-the-max-concentration-of-mg-2-ions-permissible-in-1l-0-01m-naoh-solution-if

http://socratic.org/questions/how-can-you-use-the-solubility-product-constant-to-calculate-the-solubility-of-a