Question

Question #5bde6

Answer 1

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.