What is the solubility of Mg(OH)2 in a 0.55mol/L solution of KOH if the Ksp of Mg(OH)2 is 9.0x10-12? I figured out that the initial concentration of OH- in the solution is 0.55mol/L. I did an ICE table for Mg(OH)2 and (assume x represents an unknown concentration) found that the concentration of Mg2+ is x, and OH- is 0.55mol/L+2x. So if I continue with it, I get to this point: Ksp=[Mg2+][OH-]^2 9.0x10-12=x(2x+0.55)^2 When i expand it and move everything to the right, i get a cubic equation that takes a lot of time to solve and so I get the feeling I'm not approaching this correctly. Any help?