Question #5bde6

1 Answer
Mar 16, 2016

Answer:

Here's what I got.

Explanation:

You're essentially dealing with a double replacement reaction in which a soluble ionic compound that contains the lead(II) cations, #"Pb"^(2+)#, will react with hydrochloric acid, #"HCl"#, to form lead(II) chloride, an insoluble ionic compound that precipitates out of solution.

An example of soluble ionic compound that can deliver the lead(II) cations to the solution is lead(II) nitrate, #"Pb"("NO"_3)_2#.

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

Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to release hydrogen cations, #"H"^(+)#, and chloride anions, #"Cl"^(-)#.

#"HCl"_text((aq]) -> "H"_text((aq])^(+) + "Cl"_text((aq])^(-)#

The lead(II) cations will bond to the chloride anions and form the insoluble lead(II) chloride, #"PbCl"_2#.

The complete ionic equation would be - I will use lead(II) nitrate as an example here

#"Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"Cl"_text((aq])^(-) -> "PbCl"_text(2(s]) darr + 2"H"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)#

To get the net ionic equation, you need to eliminate spectator ions, which are ions that are present on both sides of the solution.

In this case, you would have

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

which will give you

#color(green)(|bar(ul(color(white)(a/a)color(black)("Pb"_text((aq])^(2+) + 2"Cl"_text((aq])^(-) -> "PbCl"_text(2(s]) darr)color(white)(a/a)|)))#

Lead(II) chloride is a white insoluble solid that will precipitate out of solution.

http://fphoto.photoshelter.com/image/I0000.zNLnN3BmBE