Question #d99ec

1 Answer
May 1, 2017

#"Pb"_ ((aq))^(2+) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s))#


Lead(II) nitrate and potassium iodide are soluble in water, which implies that they dissociate completely in aqueous solution.

The lead(II) cations and the iodide anions will combine to form lead(II) iodide, an insoluble ionic compound that will precipitate out of solution.

The other product of the reaction will be aqueous potassium nitrate, a soluble ionic compound that will exist as ions in the resulting solution.

You will thus have

#"Pb"("NO"_ 3)_ (2(aq)) + 2"KI"_ ((aq)) -> "PbI"_ (2(s)) darr + 2"KNO"_ (3(aq))#

The complete ionic equation looks like this

#"Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"K"_ ((aq))^(+) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) + 2"K"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-)#

The net ionic equation, which you get by eliminating the spectator ions, i.e. the ions that are present on both sides of the chemical equation

#"Pb"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#

looks like this

#"Pb"_ ((aq))^(2+) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s))#