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
The reaction forms lead(II) sulfide.
Explanation:
Lead(II) nitrate,
"Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)Pb(NO3)2(aq]→Pb2+(aq]+2NO−3(aq]
In order to precipitate the lead(II) cations, hydrogen sulfide,
This will result in the formation of lead(II) sulfide,
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])^(-)Pb2+(aq]+2NO−3(aq]+H2S(aq]→PbS(s]⏐⏐↓+2H+(aq]+2NO−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.
