Solubility Help Please!?

Question

What is the solubility of MX2
in 0.11 M M(NO3)2?

MX2 --> M + 2X (ksp=4.79e-13)

i...enough....11.........0
d..-x..............+x..........+2X
f...some........(.11+x)....2X

I am solving

Ksp=[M]*[X]^2
4.79e-13=(.11+X)(2x)^2
I keep getting the answer 1e-6

I don't know what I am doing wrong. Please explain if you can help. Any help is appreciated :)

Answer 1

As 1st appxn, neglect X compared to 0.11. Then you have 0.44 X^2 = 4.79 x 10-13
Then, X^2 = 11 x10 -13 (appx) or about 1x10-6 mole/liter. So what's the problem? You have a "common ion effect", where the presence of a large amount of [M] in solution prevents much of [X] from dissolving.