In an excess of NH3(aq), Cu2+ ion forms a deep blue complex ion, Cu(NH3)42+, which has a formation constant Kf = 5.6 1011. Calculate the Cu2+ concentration in a solution prepared by adding 4.8 10-3 mol of CuSO4 to 0.500 L of 0.32 M NH3.

Answer 1

(Cu^2+) = 4.8E-3/0.5 = 0.0096M Cu^2+.
(NH3+) = 0.32M
.........Cu^2+ + 4NH3 ==> Cu(NH3)4^2+
I......0.0096....0.32......0
C.....-0.00960..-4*0.0096..0.00960
E......0........0.282......0.00960
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What I have done above is form an ICE chart and with Kf so huge for Cu(NH3+4^2+, I have shown it going to completion. That won't be exact but good enough for a start. THEN we redo the ICE chart but this time in reverse, starting with the E line and calculate the Cu^2+ at equilibrium.
........Cu^2+ + 4NH3 ==> Cu(NH3)4^2+
I.......0.......0.282.....0.00960
C.......x......+4x.........-x
E.......x......0.282+4x....0.00960-x

Kf = [Cu(NH3)4^2+]/(Cu^2+)(NH3)^4
Substitute Kf and the E line of the last ICE chart and solve for x = (Cu^2+).

Answer 2

how do i do the ^4 when there is an x and it is a bionomial. i am having trouble

Answer 3

x is small; therefore,you can use the simplifications used in solving Ka and Kb problems; i.e., 0.1 or so -x = 0.1 or that x is so small in comparison to the number that you can ignore .
5.6E11 = (0.00960-x)/(x)(0.282+4x)^4
5.6E11 = (0.00960)/(x)(0.282)^4
Now you have a simple equation that can be solved easily. What you have done is assumed 0.00960-x = 0.00960 and 0.282+4x = 0.282. If x is truly small, and it is, 0.282+x is still just 0.282 and 0.00960-x is just 0.00960. When you finish with the value of x look at the original values to confirm that x is small enough to neglect in these two instances.

Answer 4

dec

Answer 5

Roll Tide