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 precipitateMg+2