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"#