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The solubility product of Mg(OH)2 is Ksp = 1.0 x 10^-11 at 298 K.…
Question
The solubility product of Mg(OH)2 is Ksp = 1.0 x 10^-11 at 298 K. Calculate the molar solubility (in mol L^-1) of Mg(OH)2 in a solution buffered at pH = 9.0. Give the answer in scientific form.
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Step-by-step solution
At pH = 9.0, pOH = 5.0, so [OH-] = 1.0e-5 M (fixed by the buffer).
For Mg(OH)2, Ksp = [Mg2+][OH-]^2, so [Mg2+] = Ksp/[OH-]^2 = 1.0e-11/(1.0e-5)^2 = 1.0e-11/1.0e-10 = 0.10 M.
Since each formula unit gives one Mg2+, molar solubility = 0.10 mol L^-1.
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