Question

Calculate [Cu2+] in a 0.15 M CuSO4(aq) solution that is also 6.3 M in free NH3. Cu2+(aq)+4NH3(aq)⇌[Cu(NH3)4]2+(aq)Kf=1.1×1013

Answer 1

Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)        Kf=1.1×10¹³

As Kf is very high all the copper ion will complex with ammonia to give [Cu(NH₃)₄]²⁺(aq)

So we can say that the concentration of [Cu(NH₃)₄]²⁺(aq) = concentration of copper ion = 0.15 M

The concentration of ammonia used for reaction = four times of conc of copper ion = 4 X 0.15 = 0.6 M

So ammonia left = 6.3 - 0.6 = 5.7M

Putting values

Kf=1.1×10¹³ = (0.15) / ([Cu⁺²] [5.7]⁴

[Cu⁺²] = 1.29x 10^-17 M