Lead (II) Iodide Concentration Calculation
What is the concentration of iodide in solution if the lead concentration is measured at 0.0003 M?
The concentration of iodide in solution is 1.5 x 10-5 M.
Given the Ksp of lead (II) iodide is 7.1x10-9, we can calculate the concentration of iodide in solution when the lead concentration is 0.0003 M.
The equilibrium expression for the dissolution of PbI2 is:
PbI2 ā Pb2+ + 2Iā
Using the Ksp expression for lead (II) iodide, Ksp = [Pb2+] [I-]2, we can set up an equation where:
[Pb2+] = 0.0003 M
Let x be the concentration of iodide in solution.
By applying the changes at equilibrium:
Initial concentrations: 0 0
Change: -x + x + 2x
At equilibrium: (0-x) (0+ x) (2x)
Putting the values into the Ksp expression, we get:
Ksp = [Pb2+] [I-]2
0.0003 (2x)2 = 7.1x10-9
x = 1.5 x 10-5 M
Therefore, the concentration of iodide in solution is 1.5 x 10-5 M.
An alternate method to solve the problem is using the quadratic equation:
0.0003 (2x)2 = 7.1x10-9
2x2 = 7.1x10-9 / 0.0003
x = 1.5 x 10-5 M