# HW Solutions #6

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.

1. 7.9 x 10-15
2. 1.3 x 10-14
3. 3.7 x 10-14
4. 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.