The solubility of mg(oh)2 (ksp = 8.9 x 10–12) in 1.0 l of a solution buffered (with large capacity) at ph 9.47 is:

When Ksp is the solubility product constant for a solid substance when it dissolved in the solution. and measure how a solute dissolves in the solution.

and when the solubility is the maximum quantity of solute which can dissolve in a certain solution.

since the reaction equation is:

Mg(OH)2 → Mg2+  +  2OH-

∴ Ksp expression = [Mg2+][OH]^2

when PH = 9.47

and PH+ POH = 14 

∴ POH =  4.53 

when POH = -㏒[OH-]

∴[OH-] = 2.95 x 10^-5

by substitution on Ksp expression:

8.9 x 10^-12 = [Mg2+] (2.95 x 10^-5)^2

∴ [Mg2+] = 0.01 M

∴ the molar solubility = 0.01 M

The answer should be c.