Solubility of a Substance in solution

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SUMMARY

The solubility of Mg(OH)2 in a 0.55 mol/L KOH solution can be determined using its solubility product constant (Ksp) of 9.0 x 10-12. An ICE table reveals that the concentration of Mg2+ is represented as x, while the concentration of OH- is approximated as 0.55 mol/L due to the dominance of KOH. The Ksp equation simplifies to Ksp = [Mg2+][OH-]2, leading to the conclusion that the hydroxide concentration can be accurately approximated as (0.55)2.

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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?
 
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Since the Ksp of MgOH2 is only 9X10-12, you see that the concentration of OH- is ABSOLUTELY DOMINATED by the concentration of OH- due to KOH, which is completely soluble at that concentration. Thus, the hydroxide term for your solubility expression is quite accurately approximated as (0.55)2.
 

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