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)`