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.