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 ⇌ Pb2+ + 2I–

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