Question

a.Calculate the solubility (in grams per 1.00×102mL of solution) of magnesium hydroxide in a solution buffered at pH = 10.

b.How does this compare to the solubility of Mg(OH)2 in pure water?

Answer 1

a)

Mg(OH)2(s)<-> Mg+2 + 2Oh-

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

Mg(OH)2 5.61×10–12

also

pOH = 14-p´H = 14-10 = 4

[OH-] = 10^-pOH = 10^-4

then

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

5.61*10^-12 = [Mg+2] * (10^-4)^2

[Mg+2] = (5.61*10^-12) / ((10^-4)^2) = 0.000561 M

1 mol of MG+2 = 1 mol of Mg(OH)2

mas = mol*MW = (0.000561)(58.3197 ) = 0.032717 g of Mg(OH)2

this is per liter

per 100 mL -->  0.032717 g of Mg(OH)2 / L * 1000 mL / 1L =  0.032717 g of Mg(OH)2 per 1000 mL

per 100 mL --> 100/1000*0.032717 =0.0032717 g/100 mL

b)

this is much lower than common solubility of Mg(OH)2

in water, there is lower [OH-] (10^-7) than [OH-] at pH = 10

then expect muchm ore Mg+2 in solution