Calculate the molar concentration of uncomplexed Zn2+ (aq) in a solution that contains 0.22 mol of Zn(NH3)4 2+ per liter and 0.3109 M NH3 at equilibrium. Kf for Zn(NH3)4 2+ is
2.9 X 10^9
I started this way...
(0.3109 + 4)2/0.22-x)= 2.9 X 10^9
I'm lost and cant figure out how to do this problem
To solve this problem, we will use the concept of the formation constant (Kf) and the equilibrium constant expression.
First, let's write the balanced equation for the formation of the complex ion Zn(NH3)4 2+:
Zn2+ + 4NH3 ⇌ Zn(NH3)4 2+
The equilibrium constant expression for this reaction is:
K = [Zn(NH3)4 2+] / ([Zn2+] * [NH3]^4)
Given that Kf = 2.9 X 10^9 and the concentration of NH3 is 0.3109 M, we need to determine the molar concentration of uncomplexed Zn2+ (aq).
Let's assume that the concentration of uncomplexed Zn2+ (aq) is 'x' M.
Considering the stoichiometry of the reaction, the concentration of the complex formed will be '4x' M.
Now, substitute the values into the equilibrium constant expression:
2.9 X 10^9 = (4x) / (x * (0.3109)^4)
Simplifying the above equation, we get:
2.9 X 10^9 = 4 / (0.3109^4)
Now, solve the equation to find the value of 'x':
2.9 X 10^9 * 0.3109^4 = 4x
x = (2.9 X 10^9 * 0.3109^4) / 4
x = 4.4627 X 10^-3 M
Therefore, the molar concentration of uncomplexed Zn2+ (aq) in the solution is approximately 4.4627 X 10^-3 M.
To solve this problem, we need to use the concept of equilibrium and the equilibrium constant (Kf) to determine the molar concentration of uncomplexed Zn2+ in the solution.
The given equilibrium reaction equation is:
Zn(NH3)4^2+ (aq) ⇌ Zn^2+ (aq) + 4 NH3 (aq)
The equilibrium constant expression (Kf) can be written as:
Kf = [Zn^2+] / [Zn(NH3)4^2+][NH3]^4
Here's how we can proceed:
1. Since we have a solution with 0.22 mol of Zn(NH3)4^2+ per liter, the initial concentration of Zn(NH3)4^2+ is 0.22 M.
2. The concentration of NH3 at equilibrium is given as 0.3109 M.
3. Let's assume the concentration of uncomplexed Zn^2+ as x M. Therefore, the new concentration of Zn^2+ after dissociation will be (0.22 - x) M.
4. Since the stoichiometric ratio between Zn(NH3)4^2+ and Zn^2+ is 1:1, the concentration of Zn^2+ produced will also be x M.
5. The concentration of NH3 after dissociation will be 4x M because of the stoichiometric ratio.
6. Now, substitute the values into the equilibrium constant expression:
Kf = (x) / (0.22 - x)(0.3109)^4
7. Rearrange and simplify the equation:
Kf(0.22 - x)(0.3109)^4 = x
8. Now, you can solve this equation for x by substituting the value of Kf, rearranging, and solving for x.
Note: The value of Kf (2.9 × 10^9) is very large, indicating that the forward reaction is favored, and most of the Zn^2+ forms the complex Zn(NH3)4^2+. Therefore, the value of x will be relatively small compared to 0.22 M.
Once you find the value of x, you can calculate the molar concentration of uncomplexed Zn^2+ by (0.22 - x).