Final answer:
Calculating the solubility of silver chloride in a 0.130 M NH₃ solution requires knowledge of the chemical equilibrium and the complexation of silver with ammonia. The solubility cannot be gleaned from the provided options as detailed calculations are needed to consider the solubility product constant and the formation constant.
Step-by-step explanation:
To calculate the solubility of silver chloride in a solution that is 0.130 M in NH₃, we need to consider the chemical equilibrium and solubility product constant (Ksp) of silver chloride, as well as the complexation equilibrium of silver with ammonia. Silver chloride's Ksp is low, meaning it is generally considered insoluble in water, but the presence of ammonia can increase its solubility because ammonia forms a complex ion with silver ions.
The solubility is not directly given by the concentration of NH₃ because we need to consider the reactions involved. Hence, the solubility could not be simply 0.130 M as the ammonia could form complex ions with silver, thus increasing its solubility. Without detailed calculations involving both the Ksp value for AgCl and the formation constant for the Ag(NH₃)⁺ ion, we cannot arrive at an exact value. So the correct answer cannot be determined from the options provided (A, B, D) without further information.