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

2 Answers
Apr 13, 2015

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")

Apr 13, 2015

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"