A 25.0 mL sample of drinking water suspected to contain lead(II) ions is treated with excess 0.800 M sodium chloride. A precipitate forms, which is filtered, dried, and weighed. 1.36 g of precipitate are collected. What was the concentration of lead in the drinking water sample? (Assume that the lead(II) ions were in the form of lead(II) nitrate in the water.)

Question

A 25.0 mL sample of drinking water suspected to contain lead(II) ions is treated with excess 0.800 M sodium chloride.  A precipitate forms, which is filtered, dried, and weighed.  1.36 g of precipitate are collected.  What was the concentration of lead in the drinking water sample?  (Assume that the lead(II) ions were in the form of lead(II) nitrate in the water.)

DrBob222 · Answer

Convert 1.36g PbCl2 to Pb.
1.36 x (atomic mass Pb/molar mass PbCl2) = ? g Pb in 25 mL. Convert that to what ever concentration units you want. The problem doesn't specify. Note: I don't think it makes any difference what form the Pb is as long as it's soluble.