What are the equilibrium concentrations of the dissolved ions in a saturated solution of #Fe(OH)_2# at 25°C?

Question

Naomi Aronson · Accepted Answer

We need #K_"sp"# for #Fe(OH)_2#.........

And we evaluate the balance. #Fe(OH)_2(s)rightleftharpoonsFe^(2+) + 2HO^-# For which this site quotes #K_"sp"=8xx10^-16#. So we set the solubility of #Fe(OH)_2-=S#. And thus #K_"sp"=[Fe^(2+)][HO^-]^2=Sxx(2S)^2=4S^3# And thus......#S=""^(3)sqrt((8.0xx10^-16)/4)# #=5.85xx10^-6*mol*L^-1# And so #Fe^(2+)=5.85xx10^-6*mol*L^-1# And #[HO^-]=1.17xx10^-5*mol*L^-1# Can you work out the solubilities in #g*L^-1#?

Mateo Aguilar · Answer

undefined In a saturated solution of Fe(OH)2 at 25°C, the equilibrium concentrations of the dissolved ions can be determined using the solubility product constant (Ksp) for Fe(OH)2. At equilibrium, Fe(OH)2 dissociates into Fe2+ ions and OH- ions. The equilibrium expression for the dissociation of Fe(OH)2 is given by:

Fe(OH)2 (s) ⇌ Fe2+ (aq) + 2OH- (aq)

The Ksp expression for Fe(OH)2 is:

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

Since Fe(OH)2 is a sparingly soluble salt, we can assume that the concentration of Fe2+ ions produced by the dissociation of Fe(OH)2 is negligible compared to the concentration of OH- ions. Therefore, we can approximate that:

[OH-] ≈ 2 * [Fe(OH)2]

Using this approximation, we can express the equilibrium concentration of OH- ions in terms of the solubility of Fe(OH)2 (let's call it "S"):

[OH-] ≈ 2S

Substituting this expression into the Ksp expression, we get:

Ksp = [Fe2+] * (2S)^2
Ksp = 4S^3

Given the Ksp value for Fe(OH)2 at 25°C, we can solve for the solubility "S" and then calculate the equilibrium concentrations of Fe2+ and OH- ions using the relationship [OH-] ≈ 2S.