HW Solutions #6
- Page ID
- 2849
33. Assume that in the reaction of Cu2+ with ammonia, the only complex ion to form is the tetraammine species, [Cu(NH3)4]2+.
Given a solution where the initial [Cu2+] is 0.10M, and the initial [NH3] is 1.0M and that β4 (the final step) = 2.1 x 1013, calculate the equilibrium concentration of the Cu2+ ion.
Assume that in the reaction of Cu2+ with ammonia, the only complex ion to form is the tetraammine species, [Cu(NH3)4]2+.
Given a solution where the initial [Cu2+] is 0.10M, and the initial [NH3] is 1.0M and that β4 = 2.1 x 1013, calculate the equilibrium concentration of the Cu2+ ion.
The answers again were:
- 7.9 x 10-15
- 1.3 x 10-14
- 3.7 x 10-14
- 1.6 x 10-11
The reaction involved is:
Cu2+ + 4 NH3 <=> [Cu(NH3)4]2+
and the equilibrium constant can be expressed in terms of concentrations as:
[Cu(NH3)4]2+
β4= ------------------------ = 2.1 x 1013
[Cu2+] [NH3]4
Initially, the concentrations are given as:
Cu2+ + 4 NH3 <=> [Cu(NH3)4]2+
0.1 1.0 0
RHS 0 0.6 0.1
EQ. x 0.6+4x 0.1-x
RHS would correspond to the reaction going completely to the right-hand side. This is nearly true, given the large size of the equilibrium constant quoted.
EQ. corresponds to a slight shift back from the RHS values by an amount x which represents the equilibrium concentration of the free Cu2+ ions we are interested in finding.
Hence we can now solve for x to get the answer. To make matters much simpler we can assume that since x is very small, then (0.1-x) is approximately 0.1 and (0.6 + 4x) is roughly 0.6.
The equilibrium expression then turns out to be:
[Cu(NH3)4]2+
b4 = -------------------- = 2.1 x 1013
[Cu2+] [NH3]4
0.1
= -------------------- = 2.1 x 1013
(x) (0.6)4
and by rearranging we get
0.1
[Cu2+] = -----------------------
(0.6)4 (2.1 x 1013)
or [Cu2+]= 3.7 x 10-14 M, a very small quantity indeed, which justifies our assumption that (0.1+x) is approximately 0.1.