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 6.4107mol/L.

You need the value of the solubility product constant, Ksp, for magnesium fluoride; now, there are several values listed for Ksp, so I'll choose one 6.4109.

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

MgF2(s)Mg2+(aq)+2F(aq) (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 (1)

MgF2(s)Mg2+(aq)+2F(aq)
I...........................0.............0.1
C........................(+x)..........(+2x)
E...........................x...........0.1 + 2x

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

Ksp=[Mg2+][F]2

6.4109=x(0.1+2x)2

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

6.4109=x0.12x=6.4109102=6.4107mol/L

Apr 13, 2015

The molar solubility = 7.4×108mol/l

MgF2(s)Mg2+(aq)+2F(aq) (1)

You should be given a value for the solubility product Ksp which is given by :

Ksp=[Mg2+(aq)][F(aq)]2=7.4×1010mol3.l6(2)

From (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 MgF2.

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 [Mg2+(aq)] as MgF2 is 1 molar with respect to Mg2+.

To make things simple we can assume that any F(aq) from the MgF2 is tiny in comparison to the F(aq) from the NaF(aq) so we ignore it.

We'll give[Mg2+(aq)]the symbol s.

Now put the values into (2)

7.4×1010=s×(0.1)2

From which :

s=7.4×108mol/l