Question

This question was posted: At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.56. What is the Ksp of the compound at 22 °C?
I have found my [OH-] concentration (3.63x10^-4)

but I am looking at it and cannot find how they got the concentration of [OH-]. Everything that needs to be done after that I understand. I just do not get how they got the concentration

Answer 1

Your post isn't entirely clear as to what you do and don't understand. Here is how you do it.
........M(OH)2 ==> M^2+ + 2OH^-
The problem tells you pH = 10.56 which makes pOH = 3.44 and from that
pOH = -log(OH^-) = 3.44
Therefore, (OH^-) = 3.63E-4

Ksp = (M^2+)(OH^-)^2 = ?
(M^2+) = 1/2 x 3.63E-4
(OH^-) = 3.63E-4
Ksp = about 2E-11

Answer 2

I was just having trouble with getting the [OH-] concentration. I had the answer just didn't know how to get to it. Thank you, it helped.