Question

what are the concentrations of Cu2+, NH3, and Cu(NH3)4^2+ at equilibrium when 18.8g of cu(NO3)2 iS added to 1.0L of a .800M solution of aqueous ammonia? Assume that the reaction goes to completion and forms Cu(NH3)4^2+

Kf=2.0e12

Answer 1

Initial concentration of aqueous ammonia = 0.800 M

Volume of aqueous ammonia = 1.0 L

Mass of Cu(NO₃)₂ = 18.8 g

Molar mass of Cu(NO₃)₂ = 187.56 g/mol

Moles of Cu(NO₃)₂ = 18.8 /187.56

= 0.10

[Cu(NO₃)₂] = 0.10 / 1.0

= 0.10 M

Cu(NO₃)₂ + 4NH₃        Cu(NH₃)₄²⁺   
Initial 0.10 0.80 0
Change -x -4x x
Final 0.10 - x 0.80 - 4x x

Kf = [Cu(NH₃)₄²⁺] / [Cu(NO₃)₂][NH₃]⁴  

2.0*10¹² = x / (0.10 - x)(0.80 - 4x)⁴

x = 0.10

So, at equilibrium:

[Cu²⁺] = 0.10 - 0.10 = 0 M

[NH₃] = 0.80 - 4*0.10 = 0.40 M

[Cu(NH3)₄²⁺] = 0.10 M