chem 1046

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Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 9. Enter your answer to 2 significant figures.) Ksp = 2.06e-13

  • chem 1046 -

    pH = 9 means pOH = 5 and (OH^-) = 1E-5

    ........Mg(OH)2 ==> Mg^2+ + 2OH^-
    Ksp = (Mg^+)(OH^-) = 2.06E-13
    Substitute 1E-5 for OH and solve for Mg. That gives solubility in mols/L. Convert to mols/100 mL, the convert mols to grams.

  • chem 1046 -

    does the 2 in front of the OH^- get corporate into the Ksp making it (OH)^2??

  • chem 1046 -

    Yes, the OH^- is squared.
    2.06E-13 = (Mg^2+)(1E-5)^2
    Solve for Mg^2+. I see I made a typo and omitted the square of OH on my first response. It is
    Ksp = (Mg^2+)(OH^-)^2 etc. Sorry about that.

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