Question

Calculate the solubility (in grams per 1.00×102mL of solution) of magnesium hydroxide in a solution buffered at pH = 12. Express your answer using two significant figures. I got S1 = 1.2×10^-8 which is correct but for this part i did not get it "How does this compare to the solubility of Mg(OH)2 in pure water? "Express your answer using two significant figures. please solve it i could not get the answer.

Answer 1

Ksp of Mg(OH)2 = 2.06 x 10^-13

At pH = 12

pOH = 2

[OH-] = 1.0 x 10^-2

Mg(OH)2 (s)   -----------> Mg2+ (aq) + 2 OH- (aq)

                                           S                  1.0 x 10^-2

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

2.06 x 10^-13 = S x (1.0x 10^-2)^2

S = 2.06 x 10^-9 M

solubility = 2.06 x 10^-9 x 58.32

              = 1.20 x 10^-7 g/ 1000 mL

solubility S1 = 1.2 x 10^-8 g/ 100 mL

in pure water :

Mg(OH)2 (s)   -----------> Mg2+ (aq) + 2 OH- (aq)

                                           S                    2 S

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

2.06 x 10^-13 = S x (2 S)^2

S = 3.72 x 10^-5 M

solubility = 2.17 x 10^-3 g/1000 mL

solubility = 2.2 x 10^-4 g/100 mL

S = 2.2 x 10^-4

S1 / S = 1.2 x 10^-8 / 2.2 x 10^-4

           = 5.4 x 10^-5

solubility at pH 12 is less than pure water