Examples of Solving Complex Ion Equilibrium
A solution has initial [Cu+2] = 0.010 M and initial [NH3] = 1.000 M.
The complex ion Cu(NH3)4+2 forms in solution with Kf = 5.01 X 1013.
Calculate the [Cu+2] when equilibrium is achieved.
Cu(NH3)4+2
Cu+2 + 4 NH3
0.010 M
1.000 M
formation
0.0 M
Since the formation equilibrium constant is so large,
solving the usual "ICE" table will give an equilibrium molarity of Cu+2 equal to 0.
Kf =
Cu(NH3)4+2
4
NH3
Cu+2
To find the actual molarity (a very small number), one must solve this kind of problem
for the reverse reaction, a dissociation. This is set up by first converting all metal ion into
complex, the product of formation, and then using the reverse reaction in a "SPICE" table.
Start with Product Initially
Cu(NH3)4+2
SPI
C
E
Cu+2 + 4 NH3
0.010 M
0.0 M
-x
+x
x
0.010 M
Now plug into Kd
and solve for x
dissociation
0.960 M = 1.000 - 4(0.010)
+4x
0.960 M
Kd =
Cu+2
x is a very small number
4
NH3
Cu(NH3)4+2
A solution has initial [Fe+3] = 0.0050 M and initial [C2O4-2] = 0.075 M.
The complex ion Fe(C2O4)3-3 forms in solution with Kf = 2.00 X 1020.
Calculate the [Fe+3] when equilibrium is achieved.
Fe+3
+
3 C2O4-2
Fe(C2O4)3-3
formation