How do you calculate Ksp from molar solubility?

Let's take a generic dissociation equilibrium to work with here

#"X"_ n "Y"_m rightleftharpoons color(blue)(n) * "X"^(m+) + color(purple)(m) * "Y"^(n-)#

Now, the molar solubility of this generic salt #"X"_n"Y"_m# tells you the number of moles of salt that can be dissolved in one liter of solution to form a saturated solution.

Let's assume that you are given a molar solubility equal to #s# #"mol L"^(-1)# for this salt in water at room temperature. This tells you that you can dissolve #s# moles of #"X"_n"Y"_m# per liter of solution at this temperature.

Now, notice that every mole of #"X"_n"Y"_m# that dissolves produces #color(blue)(n)# moles of #"X"^(m+)# cations and #color(purple)(m)# moles of #"Y"^(n-)# anions.

This means that the saturated solution will contain

#["X"^(m+)] = color(blue)(n) * s#

#["Y"^(n-)] = color(purple)(m) * s#

The solubility product constant, #K_(sp)#, for this dissociation equilibrium looks like this

#K_(sp) = ["X"^(m+)]^color(blue)(n) * ["Y"^(n-)]^color(purple)(m)#

Plug in the expressions you have for the concentrations of the two ions in terms of #s# to find

#K_(sp) = (color(blue)(n) * s)^color(blue)(n) * (color(purple)(m) * s)^color(purple)(m)#

#K_(sp) = color(blue)(n^n) * s^color(blue)(n) * color(purple)(m^m) * s^color(purple)(m)#

This is equivalent to

#color(green)(|bar(ul(color(white)(a/a)color(black)(K_(sp) = color(blue)(n^n) * color(purple)(m^m) * s^((color(blue)(n)+color(purple)(m)))color(white)(a/a)|))))#

Let's take a numerical example to test out this expression.

Magnesium hydroxide, #"Mg"("OH")_2#, has a molar solubility of #1.44 * 10^(-4)"M"# in pure water at room temperature. What is the #K_(sp)# of the salt?

The first thing to do is identify the values of #n# and #m# by writing the dissociation equilibrium for magnesium hydroxide

#"Mg"("OH")_ (2(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)#

As you can see, you have

#{(n=1), (m=2) :}#

This means that the #K_(ps)# of magnesium hydroxide is

#K_(sp) = 1^1 * 2^2 * (1.44 * 10^(-4)"M")^((1 + 2))#

#K_(sp) = 1.2 * 10^(-11)"M"^3#

The solubility product constant is usually given without added units, so you'd have

#K_(sp) = 1.2 * 10^(-11)#