Final answer:
The molar solubility of AgCl in 0.500 M NH₃ involves complex ion formation and is enhanced compared to its solubility in pure water due to the high Kf for Ag(NH₃)₂⁺⁴.
Step-by-step explanation:
The molar solubility of AgCl in 0.500 M NH₃ can be determined by considering the equilibrium involving the complex ion formation. The Ksp for AgCl and the Kf for Ag(NH₃)₂⁺⁴ are important for this calculation. Given that AgCl(s) dissolves in NH₃ to form Ag(NH₃)₂⁺⁴, we can set up the following equilibrium expressions:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) (Ksp = 1.80 × 10⁻¹⁰)
Ag⁺(aq) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺⁴(aq) (Kf = 1.7 × 10⁷)
Combining the equilibrium constants gives us an overall expression for the solubility of AgCl in the presence of NH₃. Since the concentration of NH₃ is high, the reaction is driven to the right, which enhances the solubility of AgCl beyond its solubility in pure water.