Calculate the solubility of silver chloride in a solution that is .170M in NH3

To calculate the solubility of silver chloride (AgCl) in a solution of ammonia (NH3), we need to consider the formation of a complex ion called the silver-ammonia complex or silver diamine ion (Ag(NH3)2+). This complex is formed when silver chloride dissolves in ammonia.

The solubility product equilibrium expression for the dissolution of AgCl in NH3 is given by:

Ksp = [Ag(NH3)2+] * [Cl-]

Where [Ag(NH3)2+] represents the concentration of silver diamine complex, and [Cl-] represents the concentration of chloride ion.

Given that the concentration of ammonia is 0.170M, we need to convert it to the concentration of the silver diamine complex. The stoichiometry of the reaction tells us that for every AgCl molecule that dissolves, one Ag(NH3)2+ molecule is formed. Therefore, the concentration of Ag(NH3)2+ is equal to the concentration of AgCl that dissolves.

Hence, the concentration of Ag(NH3)2+ = 0.170M.

Since AgCl dissociates completely into Ag+ and Cl- ions, the concentration of Cl- is equal to the concentration of Ag+ formed.

Hence, the concentration of Cl- = 0.170M.

Finally, substituting these values into the solubility product expression, we have:

Ksp = (0.170M) * (0.170M)

To get the solubility (S) of AgCl, we need to solve for S in the equation:

Ksp = S * S

Solving for S, we find:

S = sqrt(Ksp)

Therefore, the solubility of AgCl in a solution that is 0.170M in NH3 is given by the square root of the solubility product constant (Ksp).