When 15.0 mL of a 6.42×10-4 M sodium sulfide solution is combined with 25.0 mL of a 2.39×10-4 M manganese(II) acetate solution does a precipitate form? (yes or no) For these conditions the Reaction Quotient, Q, is equal to .

Q = 3.59x10⁻⁸

Yes, precipitate is formed.

Explanation:

The reaction of Na₂S with Mn(CH₃COO)₂ is:

Na₂S(aq) + Mn(CH₃COO)₂(aq) ⇄ MnS(s) + 2 Na(CH₃COO)(aq).

The solubility product of the precipitate produced, MnS, is:

MnS(s) ⇄ Mn²⁺(aq) + S²⁻(aq)

And Ksp is:

Ksp = 1x10⁻¹¹= [Mn²⁺] [S²⁻]

Molar concentration of both ions is:

[Mn²⁺] = 0.015Lₓ (6.42x10⁻⁴mol / L) / (0.015 + 0.025)L = 2.41x10⁻⁴M

[S²⁻] = 0.025Lₓ (2.39x10⁻⁴mol / L) / (0.015 + 0.025)L = 1.49x10⁻⁴M

Reaction quotient under these concentrations is:

Q = [2.41x10⁻⁴M] [1.49x10⁻⁴M]

Q = 3.59x10⁻⁸

As Q > Ksp, the equilibrium will shift to the left producing MnS(s) the precipitate