If the concentration of mg2+ in the solution were 0.021 m, what minimum [oh−] triggers precipitation of the mg2+ ion? (ksp=2.06×10−13.)

According to the balanced equation for this reaction:

Mg(OH)2(s) ⇄ Mg+2(aq)  +   2OH-(aq)

when Ksp = [Mg+2][OH]^2
when we have Ksp = 2.06x10^-13  & [Mg+2] = 0.021 M
so by substitution:

2.06x10^-13 = 0.021 * [OH]^2
[OH]^2 = (2.06x10^-13) / 0.021 = 9.8x10^-12 
∴ [OH] = 3x10^-6 M 
So 3x10^-6 M is the minimum concentration of [OH -] required to precipitate Mg+2

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