AgCl + 2NH3 =[Ag(NH3)2](+) + Cl(-)
x0 0
x-0.1*2 0.1 0.1
K = [ [Ag(NH3)2](+) ]*[Cl(-)]*[Ag(+)]/{[NH3]^2 * [Ag(+)]
= Kf * KspAgCl = 1.1*10^7 * 1.77*10(-10) = 0.00195
= 0.1^2/(x - 0.2)^2
x ~ = 2.47 mol/L