Question

One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution:

Pb2+(aq) + 2I−(aq) → PbI2(s)

What volume of a 1.0 M KI solution must be added to 320.0 mL of a solution that is 0.22 M in Pb2+ ion to precipitate all the lead ion?

What mass of PbI2 should precipitate?

Answer 1

.320L * 0.22M = 0.070 moles of Pb
So, you will need 0.70 moles of I
x L * 1.0 M = 0.70 moles
Looks like 700 mL needed

Answer 2

Lead(II) iodide is PbI2. If you have 0.070 mols Pb wouldn't that need twice that or 0.140 mols I^-?
grams PbI2 pptd will be mols PbI2 x molar mass PbI2 = ?