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?

Your 3.63E-4 = (OH^-) is correct but your ICE table needs some work.

..........M(OH)2 ==> M^2+ + 2OH^-
equil.......................3.63E-4
[note: If the pH is 10.58 and OH^- = 3.63E-4, then that's what OH IS. It isn't 2 times that. So
Ksp =- (M^2+)(OH^-)^2
(M^2+) = 1/2 x 3.63E-4 = 1.82E-4
(OH^-) = 3.63E-4
Substitute into Ksp expression and solve for Ksp.

Oh geez, thank you SO much! And so quick!

How did you find the concentration of [OH-]

14-pH = pOH

From the pOH, you can calculate [OH-] raising 10 to the opposite pOH power.

[OH-]=10^(-pOH)

why is OH/2 of M^2+

To find the Ksp of the compound, you need to use the equilibrium concentration of OH- and the stoichiometric relationship between M(OH)2 and OH- ions.

You correctly found the concentration of OH- ions, which is 3.63x10^-4 M.

Now, let's set up the ICE table correctly:

M(OH)2 <--> M^2+ + 2OH-

Initial concentration: excess 0 2(3.63x10^-4)
Change in concentration: +x +x +2x
Equilibrium concentration: excess + x x 2(3.63x10^-4) + x

Since the concentration of OH- ions is in excess, we can neglect the change in concentration of the M(OH)2 compound, so the equilibrium concentration of M(OH)2 can be approximated as "excess + x".

Now we can substitute these values into the Ksp expression:

Ksp = [M][OH-]^2
= (excess + x)(2(3.63x10^-4) + x)^2

Now, since the value of x is very small compared to the initial concentration, we can assume that the concentration of OH- ions is approximately equal to "2(3.63x10^-4) + x". This allows us to simplify the expression:

Ksp ≈ (excess + x)(2(3.63x10^-4) + x)^2
≈ (excess)(2(3.63x10^-4))^2

Since the question states that there is an "excess" of M(OH)2, we can assume that the concentration of M(OH)2 is much larger compared to the concentration of OH- ions. Therefore, we can omit the term "x" in the expression, resulting in:

Ksp ≈ (excess)(2(3.63x10^-4))^2
≈ (excess)(2(3.63x10^-4))^2
≈ (excess)*(2(3.63x10^-4))^2

Now you can calculate the Ksp by inserting the appropriate values for "excess" into the equation and performing the necessary calculations.

I hope this helps! Let me know if you have any further questions.