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_{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?