# Question #230c9

Jun 17, 2016

${\text{PbCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "PbCl"_ (2(s)) darr + "CO"_ (2(g)) uarr + "H"_ 2"O}}_{\left(l\right)}$

#### Explanation:

Lead(II) carbonate, ${\text{PbCO}}_{3}$, will react with hydrochloric acid, $\text{HCl}$, to form lead(II) chloride, ${\text{PbCl}}_{2}$, carbon dioxide, ${\text{CO}}_{2}$, and water.

It's worth noting that the reaction produces carbonic acid, ${\text{H"_2"CO}}_{3}$, which then decomposes to produce carbon dioxide and water.

The balanced chemical equation that describes this reaction looks like this

${\text{PbCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "PbCl"_ (2(s)) darr + "CO"_ (2(g)) uarr + "H"_ 2"O}}_{\left(l\right)}$

The idea here is that lead(II) carbonate is insoluble in aqueous solution

${\text{PbCO"_ (3(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + "CO}}_{3 \left(a q\right)}^{2 -}$

As you know, hydrochloric acid is a strong acid, which means that it ionizes completely in aqueous solution to release hydrogen ions, ${\text{H}}^{+}$, which you'll also see listed as hydronium ions, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

These hydrogen ions will react with the carbonate anions, ${\text{CO}}_{3}^{2 -}$, to form carbonic acid, which then decomposes to form carbon dioxide and water.

The lead(II) cations will combine with the chloride anions to form the insoluble lead(II) chloride, ${\text{PbCl}}_{2}$, which will precipitate out of solution.

This will cause lead(II) carbonate's equilibrium reaction to shift to the right, which in turn will cause more of the solid to dissolve in solution.