已知Ag(NH3)2+(aq)===Ag+(aq)+2NH3(aq)电力平衡常数Kd=8.0*10^-8且AgCl的Ks

AgCl(s) = Ag+(aq) + Cl-(aq) Ksp = 2.0×10^(-10)
Ag+(aq) + 2NH3(aq) = [Ag(NH3)2]+(aq) K = 1/Kd = 1.25×10^7
AgCl(s) + 2NH3(aq) = [Ag(NH3)2]+(aq) + Cl-(aq) K‘ = Ksp·K = 2.5×10^(-3)
设需要6mol/L氨水xL
AgCl(s) + 2NH3(aq) = [Ag(NH3)2]+(aq) + Cl-(aq)
6x 0 0
6x-0.04 0.02 0.02
K' = c{[Ag(NH3)2]+}×c(Cl-)÷c(NH3)^2 = (0.02/x)^2 ÷ [(6x-0.04)/x]^2 = 2.5×10^(-3)
解得：x = 0.073