#K_(sp)# for iron (II) hydroxide, #Fe(OH)_2# has a value 1.8x#10^-15# at 25°C. What is the solubility of iron (II) hydroxide in mol/L and g/L at 25°C?

1 Answer
May 24, 2017

Answer:

#"Solubility"=7.66xx10^-6*mol*L^-1.#

Explanation:

We interrogate the equilibrium reaction..........

#Fe(OH)_2(s) rightleftharpoonsFe^(2+) + 2HO^-#

And we write the solubility expression,

#K_(sp)=[Fe^(2+)][HO^(-)]^2#.

Now cleary, #"S = solubility"=[Fe^(2+)]#

So #K_(sp)=Sxx(2S)^2=4S^3#.

And thus #S_"moles"=""^(3)sqrt((1.8xx10^-15)/4)=7.66xx10^-6*mol*L^-1.#

And thus #S_"mass"=89.86*g*mol^-1xx7.66xx10^-6*mol*L^-1#

#<1*mg*L^-1# under the given conditions...........