Lead (II) Iodide Concentration Calculation

27 Jul 2021

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

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