Answer to Calculation One

The Correct answer was 3

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:

  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+
  β4  = --------------------  = 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.
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