The precipate is Lead(II)iodide, hopefully, and you have .545grams lead iodide.
The easy way is to figure the number of moles that you have.
moles= massPbI2/molmassPbI2
Next, because the moles of Pb+2 ion is the same as the moles of PbI2 which was formed, then
Molarityoriginal= molesPb+2/.032
If 32.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.545 g precipitate, what is the molarity of lead(II) ion in the original solution?
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