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Solubility of a Substance in solution

  1. Jun 18, 2012 #1
    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?
     
  2. jcsd
  3. Jun 18, 2012 #2

    chemisttree

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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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