Question

So I am given this.
1. MgF2 <-> Mg^(+2) + 2F^(-) [Mg^(+2)] = 1.21 x 10-5 M
a) Write the Ksp Equation
- Found this to be [Mg][F]^2
b) Calculate the Ksp Value
found this to be 7.09x10^-15
c) If you have 3.0 L of Solution, find grams of MgF2 that will dissolve
Found this to be 2.25x 10^-3 g, or 0.00225 g MgF2
d) If you had 0.10 M KF originally in the solution, find M of MgF2 that will dissolve
e) If you add 100mL(0.1 L) of 0.10 M KF and 100mL(0.1 L) of 0.05M MgCl2, find Q. Will it precipitate?

The problem is, I have no idea how to start with d and F, can anyone help??

Answer 1

And why? I am thinking that I am suppose to have the KF to completely dissociate and form a certain amount of K+ and F-, which is then added to the given 2X which is the [] of F- from MgF2. Then I find the molarity of F- in total, but then what??

Answer 2

For d, let the amount of Mg2+ be n, then the amount of F- will be 0.1 + 2n ≈ 0.1 Then use Ksp to find n.

Answer 3

By the way, your answers for a, b and c are correct :)

Answer 4

SO the total M of F- I found to be 0.1000242. But do I just use the Ksp for the MgF2 ? I mean, doesn't adding the KF change anything about the validity of that equation since there's another compound present?

Answer 5

Also, thank you for letting me know about a b and c :P

Answer 6

Q: doesn't adding the KF change anything about the validity of that equation since there's another compound present?
A: the validity is universal, as long as the solid is present


Q: I found to be 0.1000242
A: definitely not! F- is 0.1 before you add MgF2. It will remain almost 0.1 since an extremely small amount of MgF2 will dissolve.

Answer 7

Oh, right, so I just ignore that small amount, haha. Ok, I will keep that in mind that adding another substance doesn't change the dissociation equation, thanks.
so the [] of Mg I got from the Ksp is 7.09x10^-13
which is also the amount that MgF2 dissolved

Answer 8

so for e) I found the equation of it to be 2KF + MgCl2 <-> 2KCl + MgF2

Answer 9

so the Q is just [KCl]^2[MgF2]/[KF]^2[MgCl2] right? Am I missing anything? or should MgF2 be ignored because it's a solid?

Answer 10

"so the [] of Mg I got from the Ksp is 7.09x10^-13 which is also the amount that MgF2 dissolved"

Correct!

Answer 11

Thanks for following me throughout this whole question, can I get anything on e?

Answer 12

I think you must use the net ionic equation:
Mg2+ = 2 F- -> MgF2

Then Q = [Mg2+].[F-]²

Answer 13

OH... So I am supposew to add it to the ORIGINAL equation? So do I just get the 0.10M F- from KF and the 0.05M Mg from MgCl2... then just
Q = [0.05][0.10] = 0.005 Since it's bigger than the Ksp... It will have to shift to the reactant, and so it will form a percipitate??

Answer 14

Your reasoning is correct, but your values are not.

[Mg2+] = 0.05 x (100/100+100) = 0.05/2 = 0.025 M
[F-] = 0.1 x (100/100+100) = 0.1/2 = 0.05 M

Then Q = [Mg2+].[F-]²

Answer 15

Alright... But if KF already is in molarity, why do I still have have x(100/100+100)? And where is that 100+100 coming from? Wait... OH. I see, I just have to add the Liters together since the total volume is changed, right? The same moles are there just the concentration relative is changed.

Thanks a lot! Just one more question. Ksp, being a constant, is the reason why it will be the same for this dissociation, right? Even though non-MgF2 substances are present, the relation just "cares" that there are so much of the dissociated ions there right?

Answer 16

The relation is valid whether other substances are present or not.

The same holds for pH = -lg[H3O+] which holds whatever the compounds in the solution.

Answer 17

Ok, that relation of dissociation would be valid, would it be true for Ka and Koh? One more thing I noticed, adding them said nothing about the MgF2. So would that mean that if the Q would be higher than Ksp, it would mean there are already enough products of MgCl2, so that MgCl2 would not dissolve, and MgCl2 would be formed in solid, aka percipitation

But similarly, if the Q was lower than Ksp, then that would mean there aren't enough dissolved to follow the Ksp to be at equilibrium, so that more would dissolve and that no percipitation would be formed, right?

If I am right then I'm done, thanks for your support. :D

Answer 18

MgCl2 will always dissolve. Its solubility is about half a kilogram per litre.

If Q < Ksp, you will end up with only ions in the solution.

Answer 19

Ok, that's good. So... thanks!