(Review complex formation before trying this.)

Concentrated NH₄OH was added to 1.00 L of 0.200 M CuSO₄ until [NH₃] = 0.600.  All of the copper was in solution, mostly in the form of Cu(NH₃)_n²⁺ complexes.  What were the concentrations of the various copper species and how much NH₄OH was added?  (Table  11.5 might prove useful.) (As short a question as this is, there are several steps involved.)

Answers:(Revived...the species had inadvertently been reversed vs the answers.)

Cu²⁺ + 4 NH₃ --> Cu(NH3)₄²⁺ K_f=1.1 X 10 ¹²

(0.2-x)/(0.6+x)⁴(x) = 1.1 X 10 ¹²

x = [Cu²⁺]/1M ~ 0.2/(0.6)⁴(1.1 X 10 ¹²) = 1.4 X 10 ^-12

[Cu(NH₃)₄²⁺] = 0.196 M

[Cu(NH₃)₃²⁺] = 3.7 X 10^-3 M

[Cu(NH₃)₂²⁺] = 1.2 X 10^-5 M

[Cu(NH₃)²⁺] = 1.0 X 10^-8 M

[Cu²⁺] = 1.4 X 10^-12 M

Total NH₃ = 0.600 free and 4(.200) bound to copper in various combinations = 1.40 moles