Mg(OH)₂ ( K _sp = 8.9×10^{- 12}) is precipitated by the addition of strong base to a 0.10 M solution of MgCl₂. At what [OH^-] will Mg(OH)₂ begin to precipitate?

Question

Quizplus · Accepted Answer

The solubility product constant expression for Mg(OH)2 is:

Ksp = [Mg2+][OH-]^2

Adding a strong base (in this case, OH-) to a solution containing Mg2+ ions will increase the concentration of OH- ions until they reach a concentration that is equal to or greater than the solubility product constant. At this point, Mg(OH)2 will begin to precipitate.

Starting with a 0.10 M solution of MgCl2, the initial concentration of Mg2+ ions will also be 0.10 M. We can use this value to calculate the minimum concentration of OH- ions needed to reach the solubility product constant and initiate precipitation:

Ksp = [Mg2+][OH-]^2
8.9×10^-12 = (0.10 M)([OH-]^2)
[OH-]^2 = (8.9×10^-12) / (0.10 M)
[OH-] = √(8.9×10^-12 / 0.10) = 9.4×10^-7 M

Therefore, the minimum concentration of OH- ions needed to initiate precipitation is 9.4×10^-7 M, which corresponds to option C.

Mg(OH)2 ( K sp = 8

Mg(OH)₂ ( K _sp = 8