The solubility of Mg(OH)₂ (K_sp = 8.9 $\times$ 10^-¹²)in 1.0 L of a solution buffered (with large capacity)at pH 9.73 is:

Question

Quizplus · Accepted Answer

The solubility of Mg(OH)2 can be calculated using the following equation:
Ksp = [Mg2+][OH-]^2
pOH = 14 - pH = 4.27
[OH-] = 10^-pOH = 5.01 x 10^-5 M
Substituting into the Ksp expression:
8.9 x 10^-12 = [Mg2+](5.01 x 10^-5)^2
[Mg2+] = 0.177 M
Therefore, the solubility of Mg(OH)2 in 1.0 L of the buffered solution is:
0.177 mol/L x 1.0 L = 0.177 mol
which is equivalent to 1.1 moles when rounded to one decimal place. 
The answer is B.

The Solubility of Mg(OH)2 (Ksp = 8 ×\times× 10-12)in 1

The Solubility of Mg(OH)₂ (K_sp = 8 $\times$ 10^-¹²)in 1