Skip to main content
Chemistry LibreTexts

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:

    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.