The solubility equilibrium of Co(OH)2 is:-
Co(OH)2 (s) ↔ Co2+ (aq) + 2OH- (aq)
The solubility product is given as:
Ksp = [Co2+][OH-]²
If 's' is the molar solubility of cobalt(II) hydroxide then we have:
[Co2+] = s
[OH-] = 2s
Ksp = (s)(2s)² = 4s³
3.50*10⁻¹⁶ = 4s³
s = [tex]\sqrt[3]{3.50*}10^{-16} /4[/tex] = 4.49*10⁻⁶ M
Ans: Solubility of Cobalt(II) hydroxide is 4.49*10⁻⁶ M
