What is the concentration of Cu2+ in a solution that is 0.10 M Cu(NH3)4^2+? (Kf = 1.2 x 10^12)
The answer is 8.0 x 10^-4
How do I solve the problem to get the answer
Write the equation and set up an ICE chart.
.........Cu^2+ + 4NH3 ==> Cu(NH3)4^2+
I........0........0.........0.1
C........x........4x........-x
E........x.......4x.........0.1-x
Kf = 1.2E12 = [Cu(NH3)4^2+]/(x)(4x)^4
Solve for x.
8E-4 M is correct.
To solve this problem, we need to use the equilibrium constant of the formation reaction for Cu(NH3)4^2+. The equilibrium constant, Kf, is given as 1.2 x 10^12.
The formation reaction for Cu(NH3)4^2+ is: Cu^2+ + 4NH3 ⇌ Cu(NH3)4^2+
Let's assume the concentration of Cu^2+ in the solution is x M. As per the reaction stoichiometry, the concentration of Cu(NH3)4^2+ will be 0.10 M (given). The concentration of NH3 will be 4 * 0.10 M since the stoichiometric coefficient of NH3 is 4.
Using the equilibrium constant expression:
Kf = [Cu(NH3)4^2+] / ([Cu^2+] * [NH3]^4)
Substituting the given values:
1.2 x 10^12 = (0.10) / (x * (4 * 0.10)^4)
Simplifying the equation:
1.2 x 10^12 = 1 / (x * 0.10^4)
Cross-multiplying and simplifying:
1.2 x 10^12 * x * 0.10^4 = 1
Dividing both sides of the equation:
x = 1 / (1.2 x 10^12 * 0.10^4)
Evaluating the expression:
x ≈ 8.3 x 10^-13
Therefore, the concentration of Cu^2+ in the solution is approximately 8.3 x 10^-13 M.