Determine the concentration of magnesium ions, [Mg 2+], when solid MgF 2 (K sp = 4.780x10^-9) is added to a 8.

Question

Determine the concentration of magnesium ions, [Mg 2+], when solid MgF 2 (K sp = 4.780x10^-9) is added to a 8.34x10^-2 M NaF( aq) solution.

Can anyone help me with this? I posted this question yesterday and no one has been able to help yet.. I am really lost.

Answer 1

Some of the MgF2 will dissolve to produce the two ionic species, but you must add in the F- concentration (common ion effect). For whatever amount of MgF2 dissolves, you'll get twice as much F- as you will Mg2+.

Ksp = [Mg2+][F-]^2
4.78 * 10^-9 = [x][0.0834 + 2x]^2
4.78 * 10^-9 = [x](0.00696 + 0.1668x + 4x^2)
0 = 4x^3 + 0.1668x^2 + 0.00696x - 4.78*10^-9

Using the cubic equation calculator (link below), the only real answer you get for x is 6.87 * 10^-7. To verify, you can plug it back in for x in the original equation to verify (6.87* 10^-7 <<< 0.0834, so you can ignore it)

6.87 * 10^-7 * 0.0834^2 = 4.78 * 10^-9

In case where you know x will be negligible in comparison to the ion already present because that value is much higher than the Ksp value, you can ignore it in the solution and make it easier (similar to Ka calculations). If we do that, we get:

4.78^10-9 = x * 0.0834^2
x = 6.87 * 10^-7

Unfortunately, that doesn't always work.

maurizia forgot to square the expression for F-, which is why that answer is wrong.

Answer 2

Let x = moles/l MgF2 that dissolve to estabilish equilibrium.
This will produce x moles/l mg2+ and 2x moles/L F- via the reaction

MgF2 >> Mg2+ + 2 F-

But there is arleady some F- in the solution corresponding to 8.34 x 10^-2 M NaF.

Therefore

[ Mg2+ ] = x

[ F- ] = 2x + 8.34 x 10^-2

The equilibrium

MgF2(s) <> Mg2+ + 2 F-

requires thet

Ksp = [ Mg2+ ][ F- ]^2

Substituing the above concentrations we get

4.780 x 10^-9 = x ( 2x + 8.34 x 10^-2)

Since 2x is small compared to 8.34 x 10^-2 we can assume

2x + 8.34 x 10 ^-2 = 8.34 x 10^-2

x = [ Mg2+ ] = 5.7 x 10^-8

Answer 3

MgF2 = Mg(2+) + 2F-
so Ksp=[Mg2+]*[F-]^2 (or x*(2x)^2 based on stoichiometry)
[Mg]=Ksp/[F-]^2
[F-]=8.34x10^-2 (sodium fluoride will dissociate completely) + the amount from the dissociation of MgF2:
Ksp=x*(2x)^2=4x^3, so x=(4.78x10^-9/4)^(1/3)=1.06x10^-3. so total fluoride concentration= [F-]=2x1.06x10^-3 + 8.34^-2=2.12x10^-3=8.55x10^-2.

thus [Mg]=4.78x10^-9/(8.55x10^-2)^2
=6.54x10^-7 M. (you don't need to multiply [F-] by 4 like above

Answer 4

Mr.Weasley Your carelessness will taken 5 points from Gryffindor. Be more careful next time.