Use the ksp values in the table to calculate the molar solubility of fe(oh)2 in pure water

The solubility equilibrium for Fe(OH)2 is:

Fe(OH)2 (aq) ↔ Fe²⁺ (aq) + 2OH⁻ (aq)

Ksp(Fe(OH)2) = [Fe²⁺][OH⁻]²

If 's' is the molar solubility of Fe(OH)2 then:

Ksp = (s)(2s)² = 4s³

the solubility constant Ksp for Fe(OH)2 can be obtained from any online database= 4.87*10⁻¹⁷

4.87*10⁻¹⁷ = 4s³

s³ = 1.218*10⁻¹⁷

s = ∛1.218*10⁻¹⁷ = 2.33*10⁻⁶ M

Molar solubility of Fe(OH)2 = 2.33*10⁻⁶ M

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890