Given the following information:

Question

Ag^+(aq) + Cl^-(aq)<-->AgCl(s) Keq=1.8x10^-1

Ag^+(aq) + 2NH3(aq)<->Ag(NH3)2^+(aq) Keq=8.2x10^3

Explain why the AgCl(s) dissolved when NH3 was added to the test tube.

Answer 1

eqn 1. Ag^+(aq) + Cl^-(aq)<-->AgCl(s) Keq=1.8x10^-1

eqn 2. Ag^+(aq) + 2NH3(aq)<->Ag(NH3)2^+(aq) Keq=8.2x10^3

I think you made a typo in Keq (= Ksp) = 1.8 x 10^-10 and not -1.
Note the equation 2 has a large Keq which means the equilibrium is far to the right (more products and fewer reactants). Equation 1 is a small K so the reaction is far to the left (which is why AgCl is not very soluble and forms a ppt in water solvent. From Le Chatelier's Principle, equation 2 is far to the right which means (Ag^+) is decreased significantly. That makes Ag^+ small in equation 1 and the reaction shifts to the right (meaning the AgCl solid dissolves to increase the Ag^+ removed by equation 2. That continues until all of the AgCl solid is dissolved. Usually a relatively large excess of NH3 is used which shifts that equilibrium even farther to the right and that ends up dissolving all of the AgCl solid from equation 1.