As observed in the lab, when concentrated $\ce{HCl}$ is added to a solution of deep blue $\ce{[Cu(NH3)4]^2+}$ a green color appears indicating the partial formation of the yellow colored $\ce{[CuCl4]^2-}$. However, when diluted $\ce{HCl}$ is added, a light blue color appears because the tetrachlorocuprate complex doesn't form. Instead, $\ce{[Cu(NH3)4]^2+}$ reacts with $\ce{H+}$ to form free $\ce{Cu^2+}$ ions in water.
I wanted to prove this by calculations, so I wrote the reactions happening.
$$ \ce{[Cu(NH3)4(H2O)2]^{2+} + 4H2O <=> [Cu(H2O)6]^2+ + 4NH3} \tag{1}\\ K_1 = 10^{-13}\\\\$$ $$4×(\ce{H3O+ + NH3 -> NH4+ + H2O}) \tag{2}\\ K_2 = 10^{9.25}\\\\$$ $$\ce{[Cu(H2O)6]^{2+} + 4Cl- -> [CuCl4(H2O)2]^2- + 4H2O}b\tag{3}\\ K_3 = 4×10^5$$
Overall: $$\ce{[Cu(NH3)4(H2O)2]^{2+} + 4H3O+ + 4Cl- -> 4NH4+ +4H2O + [CuCl4(H2O)2]^2-}\\ K= (10^{-13}) \cdot (10^{9.25})^4 \cdot (4×10^5) = 4×10^{29}$$
But as I obtained, $\ce{[CuCl4]^2-}$ must form regardless whether the $\ce{HCl}$ is concentrated or diluted, because the constant of the overall reaction is very large. I know this is wrong but I don't know where have I mistaken.
I speculated that the stability constant of $\ce{[CuCl4]^2-}$ is indeed not $4×10^{5}$ but a number much, much smaller. In this case, the constant of the overall reaction will also be very small, and $\ce{HCl}$ has to be concentrated in order for the equilibrium to be shifted forward.
But actually I searched for long on the internet and that was the only stabilty constant that I found on a site that I don't know whether is credible or not.