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.
(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
---------------------------------
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+).
how do i do the ^4 when there is an x and it is a bionomial. i am having trouble
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.
dec
Roll Tide
To calculate the concentration of Cu2+ in the solution, you can use the formation constant equation for the formation of the complex ion and the known concentrations of the reactants.
The formation constant (Kf) equation is as follows:
Kf = [Cu(NH3)4^2+] / [Cu^2+][NH3]^4
Let's represent the concentration of Cu2+ as [Cu^2+], and the concentration of NH3 as [NH3].
Given:
Amount of CuSO4 added = 4.8 * 10^-3 mol
Volume of NH3 solution = 0.500 L
Molarity of NH3 solution = 0.32 M
To find the concentration of Cu2+ ([Cu^2+]), we need to determine the value of [Cu(NH3)4^2+], which can be found by rearranging the formation constant equation:
[Cu(NH3)4^2+] = Kf * [Cu^2+] * [NH3]^4
Now let's plug in the given values:
[Kf] = 5.6 * 10^11
[NH3] = 0.32 M
To find the concentration of Cu2+ ([Cu^2+]), we first need to determine the moles of Cu(NO3)2 added:
moles of Cu(NO3)2 = (4.8 * 10^-3 mol)
Since the volume of the NH3 solution is 0.500 L, the moles of NH3 can be calculated using the relationship:
moles of NH3 = volume (in L) * molarity
moles of NH3 = (0.500 L) * (0.32 M)
Now we can plug these values into the formation constant equation to find the concentration of Cu2+ ([Cu^2+]):
[Cu(NH3)4^2+] = (5.6 * 10^11) * [Cu^2+] * (moles of NH3 ^ 4)
[Cu^2+] = [Cu(NH3)4^2+] / [(moles of NH3 ^ 4)] [Divide both sides by (moles of NH3 ^ 4)]
[Cu^2+] = (5.6 * 10^11) * [Cu^2+] * [(moles of NH3) ^ 4] / [(moles of NH3 ^ 4)]
Simplifying the equation, we can cancel out (moles of NH3 ^ 4):
[Cu^2+] = (5.6 * 10^11) * [Cu^2+]
Now we can solve for [Cu^2+]:
[Cu^2+] = [Cu^2+] / (5.6 * 10^11)
Simplifying further, we obtain:
1 = 1 / (5.6 * 10^11)
Therefore, [Cu^2+] = 1 / (5.6 * 10^11)
Substituting the value and evaluating, we get:
[Cu^2+] = 1.79 * 10^-12 M
So the concentration of Cu2+ in the solution is 1.79 * 10^-12 M.