Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = 3.79x10⁻⁴M
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = 5.35x10⁻⁴M
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
Q = 1.08x10⁻¹⁰
As Q > Ksp, the equilibrium will shift to the left producing Co(OH)₂(s) the precipitate
