If a solution of lead iodide has [I-] = 1.1x10-4 M, what is the lead(II)ion concentration at equilibrium?

Ksp = (Pb^2+)(I^-)^2

Substitute Ksp and I (I^- = 1.1E-4M) and solve for (Pb^2+)

To find the lead(II) ion concentration at equilibrium, we first need to know the balanced chemical equation for the dissolution of lead iodide (PbI2) in water. The balanced equation is:

PbI2(s) ⇌ Pb2+(aq) + 2I−(aq)

From the equation, we can see that one molecule of PbI2 produces one Pb2+ ion and two I− ions.

Given that the concentration of I− ions is 1.1x10-4 M, we can conclude that the concentration of Pb2+ ions is half of the concentration of I− ions because each PbI2 molecule dissociates into one Pb2+ ion. Therefore, the lead(II) ion concentration at equilibrium is:

[Pb2+] = 1.1x10-4 M / 2 = 5.5x10-5 M.

So, the lead(II) ion concentration at equilibrium is 5.5x10-5 M.

5.86*10^-9