Calculate [Cu2+] in a 0.10 M CuSO4 (aq) solution that is also 6.4 M in free NH3 .

Cu2+(aq) + 4NH3(aq) --->[Cu(NH3)4]2+ (aq)
kf = 1.1 x 10^13

To calculate the concentration of Cu2+ in the solution, we need to use the equilibrium constant (Kf) and the initial concentrations of Cu2+ and NH3.

The equilibrium constant (Kf) expression for the reaction is:
Kf = [Cu(NH3)4]2+ / ([Cu2+] * [NH3]^4)

Given that Kf = 1.1 x 10^13, [Cu2+] = 0.10 M, and [NH3] = 6.4 M, we can rearrange the equation to solve for [Cu(NH3)4]2+.

1.1 x 10^13 = [Cu(NH3)4]2+ / (0.10 * (6.4)^4)

Next, we can multiply both sides of the equation by the denominator to move it to the other side:

1.1 x 10^13 * 0.10 * (6.4)^4 = [Cu(NH3)4]2+

Now, calculate the right side of the equation:

1.1 x 10^13 * 0.10 * (6.4)^4 ≈ 4.941 x 10^11

Therefore, the concentration of Cu(NH3)4]2+ is approximately 4.941 x 10^11 M.

To calculate the concentration of Cu2+ in the given solution, we can use the equilibrium constant expression for the formation of the complex ion [Cu(NH3)4]2+:

[Cu(NH3)4]2+ = [Cu2+][NH3]^4

First, let's determine the concentration of free NH3 in the solution. We are given that the solution is 6.4 M in NH3.

Now, let's assume that the initial concentration of Cu2+ is x M.

Using the equilibrium constant expression, we have:

1.1 x 10^13 = x * (6.4)^4

Simplifying the equation, we get:

1.1 x 10^13 = 107.3741824x

Dividing both sides by 107.3741824, we find:

x = (1.1 x 10^13) / 107.3741824

Now, calculate the value of x:

x ≈ 1.024 x 10^11 M

So, the concentration of Cu2+ in the 0.10 M CuSO4 solution, which is also 6.4 M in free NH3, is approximately 1.024 x 10^11 M.

I think the easiest way to solve this type problem is to assume that the reaction goes to completion and forms the complex, as follows:

............Cu + 4NH3 ==> Cu(NH3)4^2+
I...........0.1...6.4.......0
C..........-0.1..-0.4......+0.1
E............0....6.0......0.1
--------------------------------
and considering the large Kf for Cu(NH3)4^2+ it isn't unreasonable that Cu(NH3)4 will be essentially 0.1M. THEN we reverse the whole thing and calculate the small component of the Cu that ionizes. We make the E line the new I line.
-------------------------------------
I........0.......6.0........0.1
C.......+x......+4x........0.1-x
E.......+x......6.0-4x....0.1-x

Then Kf = 1.1E13 = [Cu(NH3)2]/(Cu^2+)((NH3)^4
Substitute the E line and solve for x = (Cu^2+). You will have some complicated math if you go with (0.1-x)/(x)(6+x)^4.
I would make the simplifying assumptions that 0.1-x = 0.1 and 6.0+x = 6.0 which makes the final equation look like this.
1.1E13 = 0.1/(x)(6)^4 and solve for x.