Final answer:
The equilibrium constant for the solubility of Ag₂CO₃ in NH₃ is 8.1 x 10⁻⁵.
Step-by-step explanation:
The equilibrium constant for the solubility of Ag₂CO₃ in NH₃ can be determined using the solubility product constant (Ksp) and the formation constant (Kf) of the complex ion.
First, we need to write the balanced equation for the dissolution of Ag₂CO₃ in NH₃:
Ag₂CO₃(s) + 4NH₃(aq) <=> 2Ag(NH₃)₂⁺(aq) + CO₃²⁻(aq)
The equilibrium constant for this reaction can be calculated by multiplying the Ksp and Kf values:
K = Ksp * Kf = (8.1 x 10⁻¹²)(1.0 x 10⁷)
= 8.1 x 10⁻⁵
Therefore, the equilibrium constant for the solubility of Ag₂CO₃ in NH₃ is 8.1 x 10⁻⁵.