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+)