If the concentration of Mg in sea water is 1400mg/l what OH concentration is required to precipitate Mg(OH)2?

ksp= 1.5E-11
what is the relationship between Q the reaction quotient and the ksp for Mg(OH)2 to precipitate

Answer:
The condition for the equilibrium
Mg(OH)2 <> Mg2+ + 2OH-
is that
Ksp = [Mg2+] [ OH-]^2

[OH-] = square root Ksp / [Mg2+]

1400 mg = 1.4 g
1.4 / 24.312 = 0.0576 moles Mg2+ in 1 L of solution

[OH-] = square root 1.5 x 10^-11 / 0.0576 = 0.0000161 M

If [OH-] > 0.0000161 M precipitation should occur

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