Answer to Question #102207 in Physical Chemistry for Michael

Question

If 2.40g of CuSO4 is dissolved in 9.95*10^2 ml of 0.350 M NH3, calculate the concentrations of the following species at equilibrium. (a) Cu2+ (b)NH3 (c)Cu(NH3)4^2+

Answer 1

C₀(NH₃)=0.35M

n(CuSO₄)=2.40/159.6=0.015 mol

C₀(Cu²⁺)=0.015/0.995=0.0151 M

Cu²⁺+4NH₃=Cu(NH₃)₄²⁺

k₁₂₃₄ = [Cu²⁺]*[NH₃]⁴/[Cu(NH₃)₄²⁺] = 9.33*10^-13 (instability constant of this complex)

(0.0151-x)*(0.35-4x)⁴/x=9.33*10^-13

x=0.0151 = [Cu(NH₃)₄²⁺] (ammost all Cu²⁺ react with NH₃)

[NH₃] = 0.35-0.0151*4=0.2896M

(0.0151-x)*0.29⁴/0.0151=9.33*10^-13

(0.0151-x)=1.99*10^-12 = [Cu²⁺]

Answer: [Cu²⁺]=1.99*10^-12M; [NH₃] =0.2896M; [Cu(NH₃)₄²⁺]=0.0151M