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UWI-Mona, Home PageAssume 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 beta4 = 2.1 · 1013, calculate the
equilibrium concentration of the Cu2+ ion.
The answers again were: 1. 7.9 · 10-15 2. 1.3 · 10-14 3. 3.7 · 10-14 4. 1.6 · 10-11
The reaction involved is: Cu2+ + 4 NH3 <=> [Cu(NH3)4]2+
and the equilibrium constant can be expressed as:
[Cu(NH3)4]2+
beta4 = ------------------------ = 2.1 · 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+ beta4 = -------------------- = 2.1 · 1013 [Cu2+] [NH3]4 0.1 = -------------------- = 2.1 · 1013 (x) (0.6)4 or 0.1 x = ----------------------- (0.6)4 (2.1 · 1013)
from which we get x= 3.7 · 10-14 M, a very small quantity, which justifies our assumption that (0.1+x) is approximately 0.1.
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