Question

Part A 2+ which has a In an excess of NH3 (aq), Cu24 ion forms a deep blue complex ion, Cu(NH3)4 formation constant Kf = 5.6 x 101" Calculate the concentration of Cu2+ in a solution prepared by adding 4.6 x 10-3mol of CuSO4 to 0.470 L of 0.39 M NH3 Express your answer using two significant figures. ν ΑΣΦ ? [Cu2]= M Request Answer Submit

Answer 1

Initial concentration of Cu²⁺ = ( 4.6 ×10^-3mol/0.470L) × 1L = 9.79×10^-3M

Initial concentration of NH3 = 0.39M

consider the complete formation of Cu(NH3)4²⁺

Cu²⁺(aq) + 4NH3(aq) ----------> Cu(NH3)4²⁺

0.00979 M of Cu²⁺ react with 0.03916M of NH3 to give 0.00979M of Cu(NH3)4²⁺

remaining concentration of NH3 = 0.39M - 0.03916M = 0.3508M

Now , consider the dissociation equillibrium of Cu(NH3)4²⁺

Cu(NH3)4²⁺ (aq) <-------> Cu²⁺(aq) + 4NH3(aq)

K_d = [Cu²⁺] [NH3]⁴/[Cu(NH3)4²⁺]

K_d = 1/K_f = 1/5.6 ×10¹¹= 1.79×10^-12

Initial concentration

[Cu(NH3)4²⁺] = 0.00979

[Cu²⁺] = 0

[NH3] = 0.3508

change in comcentration

[ Cu(NH3)4²⁺] = -x

[Cu²⁺]= +x

[NH3] = + 4x

equillibrium concentration

[Cu(NH3)4²⁺] = 0.00979 - x

[Cu²⁺] = x

[NH3] = 0.3508 + 4x

so,

x(0.3508 + 4x )⁴ / (0.0079 - x) = 1.79 ×10^-12

we can assume 0.3508 + 4x = 0.3508 and 0.0079 - x = 0.0079

x(0.3508)⁴/(0.0079) = 1.79 ×10^-12

x 1.917 = 1.79×10^-12

x = 9.34 × 10^-13

Therefore, at equillibrium

concentration of Cu²⁺ = 9.34×10^-13M