How do you determine molar solubility if ksp is given? The question that I was asked was if the ksp of calcium fluoride CaF2 is 4.0x10^-11 what is the molar solubility in water?

First, let's start off with a balanced equation.
In this case, it would be:

#CaF_2(s) rightleftharpoons Ca^(2+)(aq) + 2F^(-)(aq)#

From this, we can establish a mole ratio: #1# mole of #CaF_2# corresponds to #1# mole of #Ca^(2+)# and #2# moles of #F^(-)#.

We also know that, for our equation, #K_(sp)# is associated with this expression:

#K_(sp) = [Ca^(2+)] xx [F^-]^2#

From the mole ratio that was established, we know that #1# mole of #Ca^(2+)# is #2# moles of #F^(-)#.
Therefore, #K_(sp)# can be re-written as:

#K_(sp) = [Ca^(2+)] xx (2xx[Ca^(2+)])^2#

The mole ratio also tells us that #1# mole of #Ca^(2+)# corresponds to #1# mole of #CaF_2#. So, #K_(sp)# can actually be rewritten again as:

#K_(sp) = [CaF_2] xx (2xx[CaF_2])^2#

#[CaF_2]# is the molar solubility of #CaF_2# in moles per liter. So, if we can find #[CaF_2]#, we know the molar solubility.

To do this, we'll just use the #K_(sp)# value that the question gives us!
The question tells us that #K_(sp)# is #4.0 xx 10^-11#. So, we can set that equal to the #K_(sp)# expression and solve for #[CaF_2]#:

#K_(sp) = [CaF_2] xx (2xx[CaF_2])^2 = 4.0 xx 10^-11#

#[CaF_2] xx 4xx[CaF_2]^2 = 4.0 xx 10^-11#

#4xx[CaF_2]^3 = 4.0 xx 10^-11#

#[CaF_2]^3 = (4.0 xx 10^-11)/4 = 1.0 xx 10^-11#

#[CaF_2] = root(3)(1.0 xx 10^-11) = 2.2 xx 10^-4#