Answer:
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 = <em>2.41x10⁻⁴M</em>
[S²⁻] = 0.025Lₓ (2.39x10⁻⁴mol / L) / (0.015 + 0.025)L = <em>1.49</em>x10⁻⁴M
Reaction quotient under these concentrations is:
Q = [2.41x10⁻⁴M] [1.49x10⁻⁴M]
<em>Q = 3.59x10⁻⁸</em>
As Q > Ksp, <em>the equilibrium will shift to the left producing MnS(s) </em>the precipitate
