My question was: 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.
Your answer was: 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.
New question: 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?
You said you understood part a. That's the initial amount of MgF2 dissolved in the 1L of water. You should have done this for part a.
............MgF2 ==> Mg^2+ + 2F^-
I...........solid.....0.......0
C...........solid.....x.......2x
E...........solid.....x.......2x
Ksp = (Mg^2+)(F^-)^2
1.5E-5 = (x)(2x)^2 = 4x^3
x = solubility MgF2 = about 0.016 but you should do it a little more accurately. That works out to approx 0.97grams MgF2/L
For part b you have two things to consider. The F from MgF2 and the F from KF.
(Mg^2+) from MgF2 = x
(F^-) = 0.032 from MgF2 and 0.1 from KF You may want to assume that 0.1 + 0.032 = 0.1 (the easy way but not too accurate) but at least do this first to see what the answer is.
(x)(0.1)^2 = 1.5E-5
x = (Mg^2+) = (MgF2) = about 1.5E-3 M which converts to 0.0934g MgF2. You had about 0.97 dissolved form part a so the difference is what ppts in part b.
I think the assumption is not valid and I believe you should go through the quadratic formula to find (Mg^2+)=(Mg^2+) as above. That quadratic on second thought it isn't a quadratic but a cubic) you get from (Mg^2+)(F^-)^2 = Ksp
1.5E-5 = (x)(2x+0.1)^2
Solve for x, convert to grams/L, subtract from your number from part a and that will be the amount pptd.
Th whole point of these problems is to show you that by adding a common ion you can decrease the solubility of a sparingly soluble salt dramatically.
You can find calculators on the web that will solve a cubic or you can use that assumption (modified of course) so that 0.1 + 0.032 = 0.132 and use that instead of the 0.1 I used above. Hope this helps.
No worries! I'll break it down step by step to help you understand.
To calculate the new solubility, you have to consider the common ion effect. In this case, the common ion is F^- from both KF and MgF2. The addition of KF introduces extra F^- ions which will shift the equilibrium and reduce the solubility of MgF2.
The balanced equation for the dissociation of MgF2 is:
MgF2(s) ⇌ Mg^2+(aq) + 2F^-(aq)
Initially, the solution is in equilibrium with Mg^2+ and F^- ions at the saturation point (maximum solubility). Adding KF will add more F^- ions to the solution, which will shift the equilibrium to the left, resulting in precipitation of additional MgF2.
To calculate the new solubility, you need to first determine how many moles of F^- are added to the solution by dissolving 0.1 moles of KF. Since each mole of KF yields two moles of F^-, the number of moles of F^- added is 2 * 0.1 = 0.2 moles.
Now, you need to consider the new equilibrium concentration of Mg^2+. Since F^- is in excess, the concentration of F^- ions will stay approximately the same as before. The new concentration of Mg^2+ will be equal to the initial concentration plus the additional concentration from the KF:
New (Mg^2+) = Initial (Mg^2+) + Additional (Mg^2+) from KF
To convert the new concentration of Mg^2+ to grams of MgF2 in 1L, you can use molar mass and stoichiometry. The molar mass of MgF2 is 62.3 g/mol. So, multiply the molarity of Mg^2+ (new concentration) by the molar mass of MgF2 to get the mass in grams.
The initial amount refers to the initial moles of MgF2 dissolved in 1L of water at the saturation point. This can be calculated using the given Ksp value of 1.5e-5 and the stoichiometry of the equation.
To summarize the steps for part b:
1. Calculate the moles of F^- ions added by dissolving KF.
2. Use the moles of F^- ions to calculate the new concentration of Mg^2+.
3. Convert the new concentration of Mg^2+ to grams of MgF2 using molar mass and stoichiometry.
4. Subtract the grams of MgF2 initially dissolved from the calculated mass of MgF2 to determine the mass of the additional precipitate.
I hope this clarifies the process for you! Let me know if you have any further questions.