# 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”##

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