Balanced chemical equation :
[tex]2NH_3(aq)+AgCl(s) ->Ag(NH_3)_2^+(aq) + Cl^-(aq)[/tex], [tex]K_{sp}=2.9\times 10^{-3}[/tex]
So, solubility product will be equal to [tex]K_{sp}[/tex] :
[tex]K_{sp}=\dfrac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}\\\\2.9\times 10^{-3}=\dfrac{S.S}{(1.6-2S)^2}[/tex]
Solving above we get :
S = 0.041
Therefore, solubility of silver chloride is 0.041 .
Hence, this is the required solution.
