Question

What is the molar solubility of magnesium fluoride in a solution that is `"0.1 M"` sodium fluoride?

Answer 1

In your case, the molar solubility of magnesium fluoride will be `6.4 * 10^(-7)"mol/L"`.

You need the value of the solubility product constant, `K_(sp)`, for magnesium fluoride; now, there are several values listed for `K_(sp)`, so I'll choose one `->` `6.4 * 10^(-9)`.

If this is not the value given to you, just replace it in the calculations with whatever value you have.

So, the equilibrium equation for the dissociation of magnesium fluoride is

`MgF_(2(s)) rightleftharpoons Mg_((aq))^(2+) + color(red)(2)F_((aq))^(-)` `color(blue)((1))`

The key to this problem is the fact that, even before adding the magnesium fluoride, your solution contains fluoride anions, `F^(-)`, from the dissociation of the sodium fluoride.

`NaF_((s)) -> Na_((aq))^(+) + F_((aq))^(-)`

Notice that 1 mole of sodium fluoride produces 1 mole of fluoride ions in solution; this means that the initial concentration of fluoride ions will be

`[F^(-)] = [NaF] = "0.1 M"`

Construct an ICE table for equation `color(blue)((1))`

`" "MgF_(2(s)) rightleftharpoons Mg_((aq))^(2+) + color(red)(2)F_((aq))^(-)`
I.......`-`....................0.............0.1
C.....`-`...................(+x)..........(+`color(red)(2)`x)
E......`-`.....................x...........0.1 + 2x

According to the definition of the solubility product constant, you'll get

`K_(sp) = [Mg^(2+)] * [F^(-)]^(color(red)(2))`

`6.4 * 10^(-9) = x * (0.1 + 2x)^(2)`

Since `K_(sp)` is so small, you can approximate `(0.1+2x)` with `0.1` to get

`6.4 * 10^(-9) = x * 0.1^2 => x = (6.4 * 10^(-9))/10^(-2) = color(green)(6.4 * 10^(-7)"mol/L")`

Answer 2

The molar solubility = `7.4xx10^(-8)"mol/l"`

`MgF_(2(s))rightleftharpoonsMg_((aq))^(2+)+2F_((aq))^(-)` `color(red)((1))`

You should be given a value for the solubility product `K_(sp)` which is given by :

`K_(sp)=[Mg_((aq))^(2+)][F_((aq))^-]^2=7.4xx10^(-10)mol^3.l^(-6)` `color(red)((2))`

From `color(red)((1))` you can see from LeChatelier's Principle that increasing `[F_((aq))^-]` will cause the equilibrium to shift to the left thus decreasing the solubility of the `MgF_2`.

This will also happen if we try to dissolve the salt in a solution which contains an ion which is common, in the case `F_((aq))^-`.

This is known as "The Common Ion Effect".

The molar solubility of the salt is also equal to `[Mg_((aq))^(2+)]` as `MgF_2` is 1 molar with respect to `Mg^(2+)`.

To make things simple we can assume that any `F_((aq))^-` from the `MgF_2` is tiny in comparison to the `F_((aq))^-` from the `NaF_((aq))` so we ignore it.

We'll give`[Mg_((aq))^(2+)]`the symbol `""s""`.

Now put the values into `color(red)((2))rArr`

`7.4xx10^(-10)=sxx(0.1)^2`

From which :

`s=7.4xx10^(-8)"mol/l"`