Calculate the solubility (in grams per 100 mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.

I can get the answer in grams per mole, but i am having difficulty changing it to g/100ml.

1.2×10−8

To calculate the solubility of magnesium hydroxide (Mg(OH)2) in grams per 100 mL of solution, we'll first need to convert the given pH value to hydroxide ion concentration ([OH-]).

A solution buffered at pH 12 would have a hydroxide ion concentration of 10^(-pOH), where pOH is equal to 14 - pH. So for pH = 12, pOH = 14 - 12 = 2, and [OH-] = 10^(-2) = 0.01 M (molar).

Now, let's use the balanced equation for the dissociation of magnesium hydroxide:
Mg(OH)2 ⇌ Mg2+ + 2OH-

The solubility of Mg(OH)2 can be determined from its solubility product constant (Ksp), which is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. The Ksp expression for Mg(OH)2 is:
Ksp = [Mg2+][OH-]^2

Since Mg(OH)2 is a strong electrolyte, it fully dissociates into its ions when it dissolves. Therefore, the concentration of Mg2+ will be equal to the concentration of OH-. So, we can write:
Ksp = [Mg2+][OH-]^2 = (0.01 M)(0.01 M)^2 = (0.01 M)(0.0001 M) = 1 × 10^(-4) M^3

Now, to convert this from molar to grams per 100 mL, we need to consider the molar mass of Mg(OH)2, which is 58.33 g/mol.

The solubility in grams per 100 mL can be calculated using the following steps:
1. Calculate the molar solubility of Mg(OH)2 using the Ksp value:
Molar solubility (M) = sqrt(Ksp) = sqrt(1 × 10^(-4) M^3)
2. Convert the molar solubility to grams per 100 mL:
Grams per 100 mL = (molar solubility in g/L) × (100 mL/1000 mL)

Let's perform the calculations:
1. Molar solubility (M) = sqrt(1 × 10^(-4) M^3) ≈ 0.01 M
2. Grams per 100 mL = (0.01 g/L) × (100 mL/1000 mL) = 0.001 g/100 mL

Therefore, the solubility of magnesium hydroxide in a solution buffered at pH 12 is approximately 0.001 grams per 100 mL of solution.

1.2x10^-6