When 18.0 mL of a 8.43×10-4 M cobalt(II) fluoride solution is combined with 22.0 mL of a 9.72×10-4 M sodium hydroxide solution does a precipitate form? no (yes or no) For these conditions the Reaction Quotient, Q, is equal to .

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

its first and third is your answer i did the qustion the fourth isnt right

elements make up everything, and they combine to make compounds.

the answer is c