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

Mar 18, 2016

#### Explanation:

Lead(II) nitrate, "Pb"("NO"_3)_2, is a soluble ionic compound that dissociates completely in aqueous solution to form lead(II) cations, ${\text{Pb}}^{2 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$.

${\text{Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

In order to precipitate the lead(II) cations, hydrogen sulfide, $\text{H"_2"S}$, is bubbled through the solution.

This will result in the formation of lead(II) sulfide, $\text{PbS}$, an insoluble ionic compound that will precipitate out of solution.

The complete ionic equation will look like this

${\text{Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + "H"_2"S"_text((aq]) -> "PbS"_text((s]) darr + 2"H"_text((aq])^(+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

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

${\text{Pb"_text((aq])^(2+) + "H"_2"S"_text((aq]) -> "PbS"_text((s]) darr + 2"H}}_{\textrm{\left(a q\right]}}^{+}$

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