Question

The Ksp of PbI2 is 8.73 × 10-9. Find the solubility of PbI2 in M

Please guide me , the I C E table still confuses me, please help me!!

Answer 1

The ice table is the easiest way to do this. It's done this way.
...........PbI2 ==> Pb^2+ + 2I^-
I...........solid.........0..............0
C.........solid..........x...............2x
E..........solid..........x..............2x
Note: I is initially. Initially you have solid and before anything happens you have zero Pb ion and zero I ion. because initially nothing has dissolved.
Change. For every mole of PbI2 that dissociates, you get 1 mol Pb ion an 2 mols I ion.
Equilibrium line you add I line + C line to get E line.

Ksp = (Pb^2+)(I^-)^2
Plug the E line into Ksp expression and solve for x = (Pb^2+) = PbI2