Silver chloride dissolves readily in \(2 \mathrm{M} \mathrm{NH}_{3}\), but is quite insoluble in \(2 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3}\). Explain.

First, we will write the balanced equilibrium equation for silver chloride dissolving in water: \[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

Next, we'll look at the interaction between silver chloride and ammonia (NH₃). Ammonia reacts with the silver ion (Ag⁺) to form a soluble complex ion, [Ag(NH₃)₂]⁺. The balanced equation for this reaction is: \[ \text{Ag}^+ (aq) + 2 \, \text{NH}_3 (aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+ (aq) \] Due to the formation of this soluble complex ion, the presence of ammonia (NH₃) helps to remove the Ag⁺ ions from the solution. According to Le Chatelier's principle, this will shift the equilibrium of the dissolution of AgCl to the right, causing more AgCl to dissolve, thus increasing the solubility of AgCl in NH₃.

Now, let's explore the effect of ammonium nitrate (NH₄NO₃) on the solubility of AgCl. When NH₄NO₃ dissolves in water, it dissociates into NH₄⁺ and NO₃⁻ ions: \[ \text{NH}_4 \text{NO}_3 (aq) \rightarrow \text{NH}_4^+ (aq) + \text{NO}_3^- (aq) \] The presence of excess NH₄⁺ ions in the solution suppresses the dissociation of NH₃ from the equilibrium: \[ \text{NH}_4^+ (aq) \rightleftharpoons \text{NH}_3 (aq) + \text{H}^+ (aq) \] Therefore, there is not enough free NH₃ to react with the Ag⁺ ions and form the soluble complex ion [Ag(NH₃)₂]⁺. As a result, very little AgCl dissolves in NH₄NO₃. Moreover, the presence of high concentrations of the common ion (Cl⁻) in the solution from NH₄Cl formed by NH₄⁺ and Cl⁻ would further suppress the dissolution of AgCl according to Le Chatelier's principle. This will make AgCl less soluble in 2 M NH₄NO₃. To summarize, silver chloride is more soluble in 2 M NH₃ due to the formation of a soluble complex ion, [Ag(NH₃)₂]⁺, which removes Ag⁺ ions from the equilibrium and shifts the dissolution equilibrium to the right, while it is less soluble in 2 M NH₄NO₃ due to the lack of free NH₃ and the presence of a common ion, Cl⁻.