What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M solution of aqueous ammonia? Assume that there is no volume change upon the addition of the solid, and that the reaction goes to completion and forms Cu(NH3)42+.

Explanation:

Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺  + 2 NO₃⁻

187.5 gm      4M           1 M

187.5 gm reacts with 4 M ammonia

18.8 g     reacts with  .4 M ammonia

ammonia remaining left after reaction

= .8 M - .4 M = .4 M .

187.5 gm reacts with 4 M ammonia   to form 1 M Cu(NH₃)₄²⁺

18.8 g reacts with .4 M ammonia  to form 0.1 M Cu(NH₃)₄²⁺  

At equilibrium , the concentration of Cu²⁺ will be zero .

concentration of ammonia will be .4 M

concentration of  Cu(NH₃)₄²⁺ formed will be 0.1 M

iron and sleel

make siroeel

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explanation: