Question
A saturated solution of MgF2 at 23 degrees C was prepared by dissolving solid MgF2 in 1 L water. The Ksp of MgF2 is 1.5e-5.
a.)Calculate the mass of MgF2 dissolved.
b.) When .1 mols of solid KF was dissolved in the MgF2 solution, precipitation was observed. Find the mass of the additional precipitate.
I think I get a, I just found the molarity of Mg2+ using the Ksp and then converted to grams; but, I have no idea how to do b. Thanks in advance.
a.)Calculate the mass of MgF2 dissolved.
b.) When .1 mols of solid KF was dissolved in the MgF2 solution, precipitation was observed. Find the mass of the additional precipitate.
I think I get a, I just found the molarity of Mg2+ using the Ksp and then converted to grams; but, I have no idea how to do b. Thanks in advance.
Answers
The b part is a common ion problem; i.e., F^- is the ion common to both KF and MgF2.
For part b, Ksp = (Mg^2+)(F^-)^2
The common ion shifts the solubility to the left. MgF2 ==> Mg^2+ + 2F^-
because of Le Chatlier's Principle. The new solubility will = new (Mg^2+) and you can convert that to g MgF2 in the 1L. Subtraction from the initial amount will give you the amount pptd.
For part b, Ksp = (Mg^2+)(F^-)^2
The common ion shifts the solubility to the left. MgF2 ==> Mg^2+ + 2F^-
because of Le Chatlier's Principle. The new solubility will = new (Mg^2+) and you can convert that to g MgF2 in the 1L. Subtraction from the initial amount will give you the amount pptd.
I'm sorry but I am still so confused on pretty much everything. How do you calculate the new solubility? And how do you convert that to g MgF2 in 1L? And what is the initial amount that you subtract from?