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