calculate the molar concentration of uncomplexed zn2+(aq) in a solution that contains .22M Zn(NH3)4^2+ and .4978M NH3 at equilibrium. Kf for Zn(NH3)4^2+ is 2.9 x 10^9

To calculate the molar concentration of uncomplexed Zn2+(aq) in the solution, we can use the equilibrium expression for the formation of Zn(NH3)4^2+ and the given Kf value.

The chemical equation for the formation of Zn(NH3)4^2+ is:

Zn2+(aq) + 4NH3(aq) ⇌ Zn(NH3)4^2+(aq)

The equilibrium expression is:

Kf = [Zn(NH3)4^2+]/([Zn2+][NH3]^4)

Given that Kf = 2.9 x 10^9, [Zn(NH3)4^2+] = 0.22 M, and [NH3] = 0.4978 M, we need to find [Zn2+] which is the concentration we want to calculate.

Rearranging the equilibrium expression, we have:

[Zn2+] = [Zn(NH3)4^2+] / ([NH3]^4/Kf)

Substituting the given values, we get:

[Zn2+] = 0.22 M / (0.4978^4 / 2.9 x 10^9)

Calculating the value inside the parentheses first:

0.4978^4 = 0.097976

Dividing by Kf:

0.097976 / (2.9 x 10^9) ≈ 3.378 x 10^-11

Now, substituting this value into the equation:

[Zn2+] ≈ 0.22 M / 3.378 x 10^-11

Calculating:

[Zn2+] ≈ 6.508 x 10^11 M

Therefore, the molar concentration of uncomplexed Zn2+(aq) in the solution is approximately 6.508 x 10^11 M.

To calculate the molar concentration of uncomplexed Zn2+(aq), we first need to understand the equilibrium reaction involving Zn(NH3)4^2+.

The equilibrium reaction is: Zn(NH3)4^2+ ⇌ Zn2+(aq) + 4NH3(aq)

The equilibrium constant (Kf) for Zn(NH3)4^2+ is given as 2.9 x 10^9.

Let's assume the molar concentration of uncomplexed Zn2+(aq) as x M. Since we have 4 moles of NH3 for every 1 mole of Zn(NH3)4^2+, the concentration of NH3 will be (4x) M in the equilibrium equation.

Using the equilibrium constant expression, we can write:

Kf = [Zn2+(aq)][NH3(aq)]^4 / [Zn(NH3)4^2+]

Plugging in the known values:

2.9 x 10^9 = x * (4x)^4 / 0.22

Simplifying the equation:

2.9 x 10^9 = 256x^5 / 0.22

Multiply both sides by 0.22:

0.22 * 2.9 x 10^9 = 256x^5

Divide both sides by 256:

(0.22 * 2.9 x 10^9) / 256 = x^5

Take the fifth root of both sides to solve for x:

x = (0.22 * 2.9 x 10^9)^(1/5)

Using a calculator, we can evaluate this expression to find the value of x, which represents the molar concentration of uncomplexed Zn2+(aq) in the solution.

.....................Zn^2+ + 4NH3 ==> Zn(NH3)4^2+

Equil...............x..........-.497.............0.22

Kf = 2.9E9 = [Zn(NH3)4]^2+/(Zn^2+)(NH3)^4
Substitute the E line into Kf expression and solve for (Zn^2+)