Asked by Nguyen
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.)
Answers
Answered by
DrBob222
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.
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.
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