What is the molar solubility of Mg(OH)₂ in a basic solution with a pH of 12.50? K_spfor Mg(OH)₂ is 5.6 × 10^-12. A)1.8 × 10^-10 mol L^-1 B)5.6 × 10^-9 mol L^-1 C)2.4 × 10^-6 mol L^-1 D)1.1 × 10^-4 mol L^-1

Question

Quizplus · Accepted Answer

The first step in solving this problem is to write the balanced equation for the dissolution of Mg(OH)2 in water:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The equilibrium constant expression for this reaction is:
Ksp = [Mg2+][OH-]2

Since Mg(OH)2 is a base, it will partially dissolve in a basic solution (pH > 7). Specifically, the OH- ion concentration will be higher than in pure water, which will lead to an increase in the solubility of Mg(OH)2.

To calculate the molar solubility, we need to find the concentration of Mg2+ and OH- ions in solution. Since the pH is given, we can use the fact that:
pH + pOH = 14

Therefore, pOH = 14 - pH = 1.5.

From this, we can calculate the OH- ion concentration as:
[OH-] = 10^-pOH = 3.2 × 10^-2 M

Using the Ksp expression, we can solve for [Mg2+]:
Ksp = [Mg2+][OH-]2
5.6 × 10^-12 = [Mg2+](3.2 × 10^-2)^2
[Mg2+] = 5.6 × 10^-9 M

Finally, we can calculate the molar solubility as the concentration of Mg(OH)2 in equilibrium with the solution:
Molar solubility = [Mg(OH)2] = [Mg2+] = 5.6 × 10^-9 M

The only answer choice that is closest to this value is B, which is therefore the correct answer.

What Is the Molar Solubility of Mg(OH)2 in a Basic

What Is the Molar Solubility of Mg(OH)₂ in a Basic