# HW Solutions #6

33. Assume that in the reaction of Cu^{2+} with ammonia, the only complex ion to form is the tetraammine species, [Cu(NH_{3})_{4}]^{2+}.

Given a solution where the initial [Cu^{2+}] is 0.10M, and the initial [NH_{3}] is 1.0M and that β4 (the final step) = 2.1 x 10^{13}, calculate the equilibrium concentration of the Cu^{2+} ion.

Assume that in the reaction of Cu^{2+} with ammonia, the only complex ion to form is the tetraammine species, [Cu(NH_{3})_{4}]^{2+}.

Given a solution where the initial [Cu^{2+}] is 0.10M, and the initial [NH_{3}] is 1.0M and that β4 = 2.1 x 10^{13}, calculate the equilibrium concentration of the Cu^{2+} 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:

Cu^{2+} + 4 NH_{3} <=> [Cu(NH_{3})_{4}]^{2+}

and the equilibrium constant can be expressed in terms of concentrations as:

[Cu(NH_{3})_{4}]^{2+}

β4= ------------------------ = 2.1 x 10^{13}

[Cu^{2+}] [NH_{3}]^{4}

Initially, the concentrations are given as:

Cu^{2+} + 4 NH_{3} <=> [Cu(NH_{3})_{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 Cu^{2+} 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(NH_{3})_{4}]^{2+}

b4 = -------------------- = 2.1 x 10^{13}

[Cu^{2+}] [NH_{3}]^{4}

0.1

= -------------------- = 2.1 x 10^{13}

(x) (0.6)^{4}

and by rearranging we get

0.1

[Cu^{2+}] = -----------------------

(0.6)^{4} (2.1 x 10^{13})

or [Cu^{2+}]= 3.7 x 10^{-14} M, a very small quantity indeed, which justifies our assumption that (0.1+x) is approximately 0.1.