The solubility of Mg(OH)2 in a particular buffer solution is 0.66g/L .

What must be the pH of the buffer solution?

Mg(OH)2 ==> Mg2+ + 2OH-

Ksp = (Mg2+)(OH-)^2
Look up Ksp, (Mg2+) = M = moles/L and moles = 0.66 g/molar mass.
Sove for OH- and convert to pH.

DrBob is right for once

Thanks a ton!!:D

To determine the pH of the buffer solution, we need to consider the equilibrium reaction between the magnesium hydroxide compound and water:

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH^-(aq)

First, let's find the concentration of Mg2+ and OH^- ions in the solution based on the given solubility of Mg(OH)2.

Given: Solubility of Mg(OH)2 = 0.66 g/L

The molar mass of Mg(OH)2 is:
24.31 g/mol (for Mg) + 2(16.00 g/mol + 1.01 g/mol) (for 2 O + 2 H) = 58.33 g/mol

Hence, the molarity of Mg(OH)2 in the solution is calculated as follows:
0.66 g/L ÷ 58.33 g/mol = 0.0113 mol/L

According to the stoichiometry of the balanced equation, the concentration of Mg2+ ions is equal to the concentration of Mg(OH)2, which is 0.0113 mol/L.

Similarly, as two hydroxide ions are produced for every one Mg(OH)2 molecule that dissolves, the concentration of OH^- ions can be calculated:
2 × 0.0113 mol/L = 0.0226 mol/L

Now, to find the pH of the buffer solution, we need to determine the pOH value using the concentration of OH^- ions. Remember, pH + pOH = 14.

Let's calculate the pOH:
pOH = -log(OH^- concentration)

pOH = -log(0.0226) ≈ 1.65

Finally, we can find the pH:
pH = 14 - pOH
pH = 14 - 1.65 ≈ 12.35

Therefore, the pH of the buffer solution must be approximately 12.35.