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The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.400...

The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.

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Expert Solution

Cu^2+(aq) + 4NH3(aq) <---> Cu(NH3)4^2+(aq)

Kf, of Cu(NH3)42 is 1.70 × 10^13

Cu2+             + 4 NH3 -------> [Cu(NH3)42+]

I                       0.0400             0.400                      0.0
c                  - 0.0400            - (4*0.400)                         0.0 400

E                      0                                  0.24               0.0400

Thus at equilibrium concentration of [Cu2+]= 0

                                    Cu2+ +        4 NH3 <--------> [Cu(NH3)4 2+]
I                                   0                      0. 24                       0.04
c                                  +x                      +4x                       -x
E                                  x                 0.24 +4x                   0.04-x

Kf= [Cu(NH3)4 2+]   / [Cu2+][ NH3]^4


1.70 x 10^13 = 0.04-x / x ( 0.24+4x)^4

1.70 x 10^13 = 0.04/ x ( 0.24)^4 = 12.06/x

x =7.09* x 10^-13M

[Cu2+]= 7.09 x 10^-13 M

[NH3]= 0.24 +4x = 0.24

Cu(NH3)4 2+]   = 0.04-x = 0.03999 or 0.04

queen_honey_blossom answered 2 years ago
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