Question

If 0.248 mg of MgF2 is the maximum quantity that can be dissolved in 0.5 L of 0.1 Μ KF, what will be the Ksp of MgF2 under the same conditions?

Answer 1

We are provided with a solubility, which means we need an ICE table to solve this. Let's write out the net ionic equation, but with the solid as the reactant.

`MgF_2 -> Mg^(2+) + 2 F^-`

Initially, the concentration of Mg is 0M and the concentration of F is already 0.1M

Then at equilibrium, we subtract variable `s` from `MgF_2` and add `s` to both other values.

This gives us

`MgF_2 = -S`
`Mg^(2+) = +S`
`2 F^- = 0.1 + S`

Since we know that S is in mol / L, we must take the given mass and turn it into moles, then divide by L.

So...

`(0.000248g) / (62.3018(g/(mol))) = 0.00000398 mol MgF_2`

`(0.00000398 mol) / (0.5 L) = 0.00000796 M`

And...

`K_(sp) = [Products] / 1 = [S * (0.1 + S)] = 0.1S + S^2`

S is already 0.00000796 M, so we just plug this in

`0.1 * 0.00000796 M + 0.00000796 M^2`

This means that the Ksp is 0.000000796

Or, in scientific notation: `7.96*10^7`