Pb^2+ + 2I^- ==> PbI2.
mols Pb^2+ = 1.50g/atomic mass Pb
mols I^- = twice that.
g NaI = mols I^- x molar mass NaI
A solution contains 1.50 g of dissolved Pb
2+ ions. How many grams of NaI must be added to the solution to completely precipitate all of the dissolved Pb
2+ as PbI2?
1 answer