Question

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?

Answer 1

The answer: `2.2 xx 10^-4 M`
The process: Write a balanced equation, establish mole ratios, relate the expression of `K_(sp)` to its value, and then solve.

Explanation:

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`