calculate Cu2+ in a 0.15M CuSO2 (aq) solution that is also 6.5M in free NH3.

Cu^2+(aq)+ 4NH3(aq)<=> [Cu(NH3)4]^2+ kf=1.1*10^13

Do you mean CuSO4?

Kf = 1.1E13 = [Cu(NH3)4]^2+/(Cu^2+)(NH3)^4.

Do this ICE chart in two steps. Here is the first one. With Kf so large the reaction goes to completion (almost) like this.
...Cu^2+ + NH3 ==>Cu(NH3)2+4
I..0.15.....6.5.......0
C..-0.15...-0.6.....+0.15
E....0......5.9......0.15

Now redo the ICE chart BUT showing the small amount of reverse reaction like this.
....Cu^2+ + 4NH3 ==> Cucomplex^2+
I...0.......5.9.......0.15
C...+x......+4x........-x
E...x......5.9+4x.....0.15-x

Now substitute this last E line into the kf expression and solve for x = (Cu^2+)
Post your work if you get stuck.

To calculate the concentration of Cu2+ in the given solution, we can use the equilibrium constant expression:

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

Let's assume the concentration of Cu2+ is x M. Since the solution is 6.5M in free NH3, the concentration of NH3 is 6.5 M.

Using the equilibrium constant expression and given equilibrium constant (kf = 1.1 x 10^13), we can write:

1.1 x 10^13 = (x) * (6.5)^4

Let's solve this equation to find the value of x, which represents the concentration of Cu2+:

1.1 x 10^13 = 840.5 * x^4

By dividing both sides of the equation by 840.5, we get:

x^4 = (1.1 x 10^13) / 840.5

Now, we can take the fourth root of both sides to isolate x:

x = ((1.1 x 10^13) / 840.5)^(1/4)

Using a calculator, we find:

x ≈ 0.003191 M

Therefore, the concentration of Cu2+ in the 0.15 M CuSO2 (aq) solution that is also 6.5 M in free NH3 is approximately 0.003191 M.

To calculate the concentration of Cu2+ in the given solution, we can use the concept of equilibrium and the equilibrium constant (Kf) of the reaction. Here's how you can solve it step by step:

Step 1: Write down the balanced equation for the reaction between Cu2+ and NH3:
Cu^2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]^2+ (aq)

Step 2: Define the variables:
Let's assume the initial concentration of Cu2+ is 'x' M. Thus, its concentration at equilibrium will also be 'x' M.
Since NH3 is a weak base, we need to consider it in excess. Thus, the concentration of NH3 remaining will be 6.5 M.

Step 3: Apply the equilibrium constant expression:
The equilibrium constant (Kf) expression for this reaction can be written as:
Kf = [Cu(NH3)4]^2+ / (Cu^2+ * (NH3)^4)

Step 4: Substitute the known values into the expression:
Kf = (x) / (0.15 * (6.5)^4)

Step 5: Solve for x:
Rearrange and solve for x:
x = Kf * 0.15 * (6.5)^4

Step 6: Calculate the value of x:
Plug in the given value of Kf and evaluate the expression to find the concentration of Cu2+:
x = (1.1 * 10^13) * 0.15 * (6.5)^4

After performing the calculations, the value of x will be the concentration of Cu2+ in mol/L in the given CuSO4 solution.

Please note that the concentration of Cu2+ will depend on the value of Kf, which was provided in the question.