The toxic metal cadmium Cd2+ has a tendency to complex with as many as 4 Cl- ions. The complexation reactions can be written as:
Cd2+ + Cl- ---> CdCl+ ``````` log Kf,1 = 1.98
Cd2+ + 2Cl- ---> CdCl20 `````` log Kf,2 =2.60
Cd2+ + 3Cl- ---> CdCl3- `````` log Kf,3 = 2.40
Cd2+ + 4Cl- ----> CdCl42- ````` log Kf,4 = 2.50
What is the percentage of total Cd2+ in a solution that remains uncomplexed (free Cd2+ ) if the Cl- concentration is 0.005 M (ignore activity coefficient correction, assume activities equal concentrations).
To determine the percentage of total Cd2+ that remains uncomplexed, we need to calculate the concentration of Cd2+ that is complexed with each number of Cl- ions.
First, let's calculate the concentration of each complex:
For CdCl+:
[CdCl+] = [Cd2+] * [Cl-] (using stoichiometry, the concentration of [CdCl+] will be equal to the concentration of the Cd2+ and Cl- ions)
[CdCl+] = [Cd2+] * [Cl-] = [Cd2+] * (0.005 M) = (0.005 M) * [Cd2+]
[CdCl+] = (0.005 M) * [Cd2+]
Using the given log Kf,1 = 1.98, we can write the equation as:
log Kf,1 = log ([CdCl+]/[Cd2+][Cl-]) = 1.98
Now, rearrange the equation:
log ([CdCl+]/[Cd2+][Cl-]) = 1.98
([CdCl+]/[Cd2+][Cl-]) = 10^(1.98)
([CdCl+]/[Cd2+][Cl-]) = 100^0.99
Substitute [CdCl+] = (0.005 M) * [Cd2+] into the equation:
((0.005 M) * [Cd2+])/([Cd2+] * (0.005 M)) = 100^0.99
(0.005 M)/(0.005 M) = 100^0.99
1 = 100^0.99
1 = 57.89
This equation is not balanced, indicating an error in the calculation. Please revise the equation and try again.