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QUESTION

# Calculate molar solubility of MgF2 in 0.10 M NaF at 25 C?

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"