Question

calculate the molar solubility of Fe(OH)2 in a buffer solution with PH= 9.50? For Fe(OH)2, Ksp = 4.9x10 -17

Answer 1

PH   = 9.5

POH   = 14-PH

          = 14-9.5

           = 4.5

POH   = 4.5

-log[OH-]   = 4.5

   [OH-]     = 10^-4.5   = 3.16*10^-5 M

                  Fe(OH)2 -------> Fe⁺²   + 2OH^-

                 Ksp      = [Fe⁺²][OH-]²

               4.9*10^-17   = [Fe⁺²](3.16*10^-5 )²

                [ Fe⁺² ]     = 4.9*10^-17/(3.16*10^-5)²    = 4.9*10^-8 M

   solubility of Fe(OH)2 is 4.9*10^-8 M