The Ksp of magnesium fluoride is 6.4E-9. Calculate the solubility (in grams per liter) of magnesium fluoride in water.
MgF2 <---> Mg^2+(aq) + 2F^-(aq)
[Mg^2+] = x & [2F^-] = 2x
(x)(2x^2) = 6.4E-9
4x^3 = 6.4E-9
x^3 = 1.6E-9
x = 1.2E-3 [Mg^2+]
2x = [F^-] = 2.4E-3
?g MgF2/L = (1.2E-3 mol MgF2/L)(64.3018 g MF2/1 mol MgF2) = 8.4E-2g/L dissolved MgF2.
Again I have a feeling I messed somthing toward the final calculations. I used 1.2E-3 mol, which is really Mg^2+. I was following another example in book.
The Ksp of magnesium fluoride is 6.4E-9. Calculate the solubility (in grams per liter) of magnesium fluoride in water.
MgF2 <---> Mg^2+(aq) + 2F^-(aq)
[Mg^2+] = x & [2F^-] = 2x
No, you have let (MgF2) = x so (Mg^+2)=x and (F^-)= 2x NOT (2F^-)=2x. THE F^- is x. Twice the (F^-) would be 4x.
(x)(2x^2) = 6.4E-9
4x^3 = 6.4E-9
x^3 = 1.6E-9
x = 1.2E-3 [Mg^2+] This is also (MgF2).
2x = [F^-] = 2.4E-3
You see that 2x is (F^-) so you have done this step properly. Had you been consistent, then you would have said 2F^- = 4x. What you did was to make two errors but they compensated for each other and you have the correct answer for x, 2x, Mg and F.
?g MgF2/L = (1.2E-3 mol MgF2/L)(64.3018 g MF2/1 mol MgF2) = 8.4E-2g/L dissolved MgF2.
Again I have a feeling I messed somthing toward the final calculations. I used 1.2E-3 mol, which is really Mg^2+. Yes, you are correct but it also is the (MgF2). Although you didn't put that (I added it at the beginning). You do have an error in the molar mass of MgF2. I didn't do it exactly but it is closer to 62 something since Mg is about 24.3 and F is 19 each for 38 and 24.3 + 38 is about 62.3/ I was following another example in book. I hope this clears up the slight misunderstanding of the Ksp process.<
I had another thought which might help. Many students make the same mistake of calling (2F^-) = 2x and they will write
Ksp = (Mg^+2)(2F^-)^2 which is not correct.
It IS correct that since
MgF2 ==> Mg^+2 + 2F^-, then
(F^-)=(2Mg^+2) and since you called (MgF2) = x and (Mg^+2) = x; it should be obvious that (F^-) = (2Mg^2+)= 2x.
That is correct! The correct expression for the solubility product constant, Ksp, of magnesium fluoride would be:
Ksp = (Mg^2+)(F^-)^2
Since the concentration of fluoride ions, [F^-], is 2x, you can substitute it into the equation:
Ksp = (Mg^2+)(2x)^2
Simplifying further:
Ksp = 4x^3
And since the given value for Ksp is 6.4E-9, you can set up the equation:
6.4E-9 = 4x^3
Solving for x:
x^3 = (6.4E-9)/(4)
x^3 = 1.6E-9
x = (1.6E-9)^(1/3)
x = 1.2E-3 M
So, the concentration of magnesium ions, [Mg^2+], and therefore the solubility of magnesium fluoride, is 1.2E-3 M.
To calculate the solubility in grams per liter, you need to convert the concentration into moles per liter by multiplying by the molar mass of MgF2:
Mass of MgF2 (g/L) = (1.2E-3 mol/L)(62.3 g/mol)
Mass of MgF2 (g/L) = 7.5E-2 g/L
Therefore, the solubility of magnesium fluoride in water is 7.5E-2 grams per liter.