Question

Calculate the solubility (g/L) of Iron(II) hydroxide in (a) water (b) 0.010M FeBr2 (c) at pH 11.00. Ksp of (OH)2=4.9x10^-17

Please show all work for each part

Answer 1

a) water :

Fe(OH)2 molar mass = 89.86 g / mole

Fe(OH)2 ---------------> Fe2+ + 2OH-

                                    S           2S

Ksp = [Fe2+][OH-]^2

Ksp = (S)(2S)^2 = 4S^3

4.9x10^-17 = 4S^3

S = 2.035 x 10^-6 mole / L

    = ( 2.035 x 10^-6 mole / L ) x (89.86 g / mole)

   = 2.07 x 10^-4 g / L

solubility = 2.07 x 10^-4 g / L

b)

Fe(OH)2 molar mass = 89.86 g / mole

Fe(OH)2 ---------------> Fe2+ + 2OH-

                                   0.01         2S

Ksp = [Fe2+][OH-]^2

Ksp = (0.01)(2S)^2

4.9x10^-17 = (0.01)(2S)^2

S = 3.5 x 10^-8 mole / L

    = ( 3.5 x 10^-8 mole / L ) x (89.86 g / mole)

   = 3.14 x 10^-6 g / L

solubility = 3.14 x 10^-6 g / L

c)

pH = 11.00

pH + pOH = 14

pOH = 14-11 = 3

[OH-] = 10^-3 M

Fe(OH)2 molar mass = 89.86 g / mole

Fe(OH)2 ---------------> Fe2+ + 2OH-

                                   S         10^-3

Ksp = [Fe2+][OH-]^2

Ksp = (S)(10^-3)^2

4.9x10^-17 = (S)(10^-3)^2

S = 4.9 x 10^-11 mole / L

    = ( 4.9 x 10^-11 mole / L ) x (89.86 g / mole)

   = 3.14 x 10^-6 g / L

solubility = 4.40 x 10^-9 g / L