# Question 5bde6

Mar 16, 2016

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, ${\text{Pb}}^{2 +}$, will react with hydrochloric acid, $\text{HCl}$, 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.

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

Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to release hydrogen cations, ${\text{H}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

${\text{HCl"_text((aq]) -> "H"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

The lead(II) cations will bond to the chloride anions and form the insoluble lead(II) chloride, ${\text{PbCl}}_{2}$.

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

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

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Pb"_text((aq])^(2+) + 2"Cl"_text((aq])^(-) -> "PbCl}}_{\textrm{2 \left(s\right]}} \downarrow} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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