The precipitated AgCl dissolves only when a suitable complexing agent such as excess aqueous ammonia is added. In water alone, silver chloride remains mostly as a white precipitate because its solubility product is very small. In excess ammonia, dissolved silver ions are converted into the soluble complex ion [Ag(NH3)2]+, so more solid AgCl dissolves to replace the removed free Ag+.
Initial precipitation of silver chloride
Silver chloride forms when aqueous silver ions and chloride ions meet. The visible evidence is a white, curdy precipitate of AgCl. The net ionic precipitation equation is:
The reverse process is the slight dissolution of silver chloride. At equilibrium, a very small concentration of silver ions and chloride ions exists in solution:
The solubility product expression is:
Because \(K_{sp}\) for silver chloride is small, the equilibrium strongly favors solid AgCl in plain water. The precipitate therefore does not dissolve appreciably without a chemical process that removes one of the dissolved ions.
Effect of excess ammonia
Ammonia acts as a Lewis base. Its nitrogen atom has a lone pair that coordinates to Ag+. In excess aqueous ammonia, silver ions form the diamminesilver(I) complex:
The formation-constant expression is:
The formation of [Ag(NH3)2]+ lowers the concentration of free Ag+. According to Le Châtelier’s principle, the dissolution equilibrium of AgCl shifts toward the dissolved ions to restore some Ag+. The newly released Ag+ is again captured by ammonia, so the solid continues dissolving.
Overall equilibrium
The combined process can be written by adding the dissolution equilibrium and the complex-ion formation equilibrium:
After cancellation of Ag+, the net reaction is:
This net equation explains why the precipitated AgCl dissolves in excess ammonia even though it is only sparingly soluble in water.
Equilibrium interpretation
| Observation | Chemical explanation | Equilibrium effect |
|---|---|---|
| White precipitate forms after adding chloride to silver ion solution | Ag+ and Cl− form sparingly soluble AgCl(s) | The ionic product exceeds \(K_{sp}\), so precipitation occurs |
| Precipitate remains in water | AgCl has a small solubility product | The dissolution equilibrium favors the solid |
| Precipitate dissolves in excess ammonia | NH3 forms soluble [Ag(NH3)2]+ | Free Ag+ decreases, so more AgCl(s) dissolves |
| Clear solution after sufficient ammonia | Most silver is present as the complex ion, not as free Ag+ | Complex formation dominates the solubility equilibrium |
Role of Le Châtelier’s principle
The dissolution of AgCl is controlled by the concentration of dissolved silver ions. Ammonia removes free silver ions from solution by forming a complex ion. The system responds by dissolving more solid AgCl, because dissolving more solid replaces some of the silver ions removed by complex formation.
Important distinction
The precipitate does not dissolve because chloride disappears or because silver chloride becomes highly soluble by itself. The decisive factor is the formation of [Ag(NH3)2]+, a soluble complex ion. Complex-ion formation reduces the free silver ion concentration and changes the apparent solubility of AgCl.
Final answer: the precipitated AgCl dissolved in excess aqueous ammonia because Ag+ formed the soluble complex [Ag(NH3)2]+, shifting the AgCl solubility equilibrium toward dissolution.