In a qualitative analysis, the test for the presence of Cu^2+ ion is the formation of the bright blue complex ion Cu(NH3)42+. What is the equilibrium concentration of Cu2+ when 1.0 mL of 0.200 M Cu^2+ is combined with 1.0 mL of 15.0 M NH3? (Kf Cu(NH3)4^2+ = 5.0 x 10^12).

(NH3) = 15 x 1/2 = 7.5M

(Cu^2+)= 0.200 x 1/2 = 0.100

.....Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+
I....0.1.....7.5........0
C...-0.1....-4*0.1.....+0.1
E....0.......7.1........0.1
Because the Kf for the complex is so high ESSENTIALLY all of the Cu^2+ will be transformed into the complex. Now let's turn this backwards.

.....Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+
.....0........7.1......0.1....I
.....x.......+4x.......-x.....C
.....x.......7.1+4x...0.1-x...E

Substitute the E line into the Kf for the complex and solve for x.

To find the equilibrium concentration of Cu2+ when 1.0 mL of 0.200 M Cu2+ is combined with 1.0 mL of 15.0 M NH3, we can use the concept of equilibrium constants and the given formation constant (Kf) for Cu(NH3)42+.

Let's follow these steps to solve the problem:

Step 1: Write the balanced equation for the formation of Cu(NH3)42+.
Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq)

Step 2: Define the initial and final concentrations.
The initial concentration of Cu2+ is given as 0.200 M in a volume of 1.0 mL, which can be converted to L by dividing by 1000.
Initial concentration of Cu2+ (Ci) = 0.200 M (since 1 mL = 0.001 L)

The concentration of NH3 is given as 15.0 M in a volume of 1.0 mL.
Initial concentration of NH3 (Co) = 15.0 M (since 1 mL = 0.001 L)

The final volume after combining both solutions will be 1.0 mL + 1.0 mL = 2.0 mL = 0.002 L.

Step 3: Calculate the moles of Cu2+ and NH3.
Moles of Cu2+ (nCu2+) = initial concentration (Ci) * volume (V)
= 0.200 M * 0.001 L
= 0.0002 moles

Moles of NH3 (nNH3) = initial concentration (Co) * volume (V)
= 15.0 M * 0.001 L
= 0.015 moles

Step 4: Define the change in moles.
Since the stoichiometric ratio between Cu2+ and Cu(NH3)42+ is 1:1, the change in moles of Cu2+ is the same as the change in moles of Cu(NH3)42+ formed.

Change in moles of Cu2+ (ΔnCu2+) = -nCu2+
= -0.0002 moles

Change in moles of NH3 (ΔnNH3) = -4 * nCu2+ (according to the stoichiometric ratio)
= -4 * (-0.0002) moles
= 0.0008 moles

Step 5: Calculate the equilibrium concentration of Cu2+.
Using the formation constant (Kf) and the relationship between the concentrations of Cu2+ and Cu(NH3)42+ at equilibrium, we have:
Kf = [Cu(NH3)42+]/([Cu2+][NH3]^4)

Let's assume x is the equilibrium concentration of Cu2+ (Cu(NH3)42+).
At equilibrium, [Cu2+] = Ci - ΔnCu2+ / V
= 0.200 M - (-0.0002 moles) / 0.002 L
= 0.200 M + 0.1 moles / 0.002 L
= 0.200 M + 50 M
= 50.200 M

Substituting these values into the formation constant equation:
Kf = x / (50.200 M * (15.0 M)^4)

Solving for x:
x = Kf * 50.200 M * (15.0 M)^4
x = (5.0 x 10^12) * 50.200 M * (15.0 M)^4
x = 37.65 M

Therefore, the equilibrium concentration of Cu2+ is 37.65 M.