Which ions are the spectators ions in the reaction of #Mg(HCO_3)_2(aq)# and #HCl(aq)#?
1 Answer
Explanation:
You are dealing with a soluble ionic compound, magnesium bicarbonate,
Magnesium bicarbonate will dissociate to form magnesium cations and bicarbonate anions
#"Mg"("HCO"_3)_text(2(aq]) -> "Mg"_text((aq])^(2+) + 2"HCO"_text(3(aq])^(-)#
Hydrochloric acid will dissociate to form hydrogen cations and chloride anions
#"HCl"_text((aq]) -> "H"_text((aq])^(+) + "Cl"_text((aq])^(-)#
Now, when these two solutions are mixed, the ensuing reaction will produce magnesium chloride, a soluble salt, and carbonic acid,
This means that you can write
#"Mg"("HCO"_3)_text(2(aq]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + 2"H"_2"CO"_text(3(aq])#
Now, carbonic acid is unstable in aqueous solution and will decompose to give carbon dioxide and water. This will get you
#"Mg"("HCO"_3)_text(2(aq]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + 2"H"_2"O"_text((l]) + 2"CO"_text(2(g])uarr#
Now, the complete ionic equation will be
#"Mg"_text((aq])^(2+) + 2"HCO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"Cl"_text((aq])^(-) -> "Mg"_text((aq])^(2+) + 2"Cl"_text((aq])^(-) + 2"H"_2"O"_text((l]) + 2"CO"_text(2(g]) uarr#
The spectator ions are the ions present on both sides of the complete ionic equation. In this case, the magnesium cations and the chloride anions will be spectator ions.
The net ionic equation will look like this
#color(red)(cancel(color(black)("Mg"_text((aq])^(2+)))) + 2"HCO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2color(red)(cancel(color(black)("Cl"_text((aq])^(-)))) -> color(red)(cancel(color(black)("Mg"_text((aq])^(2+)))) + 2color(red)(cancel(color(black)("Cl"_text((aq])^(-)))) + 2"H"_2"O"_text((l]) + 2"CO"_text(2(g]) uarr#
This is equivalent to
#"HCO"_text(3(aq])^(-) + "H"_text((aq])^(+) -> "H"_2"O"_text((l]) + "CO"_text(2(g]) uarr#
