How do you write a balanced chemical equation for the removal of lead (II) nitrate using hydrogen sulfide? What is the precipitate formed?

1 Answer
Mar 18, 2016

Answer:

The reaction forms lead(II) sulfide.

Explanation:

Lead(II) nitrate, #"Pb"("NO"_3)_2#, is a soluble ionic compound that dissociates completely in aqueous solution to form lead(II) cations, #"Pb"^(2+)#, and nitrate anions, #"NO"_3^(-)#.

#"Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#

In order to precipitate the lead(II) cations, hydrogen sulfide, #"H"_2"S"#, is bubbled through the solution.

This will result in the formation of lead(II) sulfide, #"PbS"#, an insoluble ionic compound that will precipitate out of solution.

The complete ionic equation will look like this

#"Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + "H"_2"S"_text((aq]) -> "PbS"_text((s]) darr + 2"H"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)#

To get the net ionic equation, remove the spectator ions, i.e. the ions that exist on both sides of the equation

#"Pb"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) + "H"_2"S"_text((aq]) -> "PbS"_text((s]) darr + 2"H"_text((aq])^(+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-))))#

This will get you

#"Pb"_text((aq])^(2+) + "H"_2"S"_text((aq]) -> "PbS"_text((s]) darr + 2"H"_text((aq])^(+)#

The color of the solution will change from colorless to black once the lead(II) sulfide precipitates out of solution.

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