What are the products of #KOH(s) + CO_2(g)#?

Question

Mateo Aguilar · Accepted Answer

#"KOH"_text((aq]) + "CO"_text(2(g]) -> "KHCO"_text(3(aq])#

Potassium hydroxide, #"KOH"#, will react with carbon dioxide, #"CO"_2#, to form potassium bicarbonate, #"KHCO"_3#, and water.

This is the balanced chemical equation for the reaction.

#"KOH"_text((aq]) + "CO"_text(2(g]) -> "KHCO"_text(3(aq]) + "H"_2"O"_text((l])#

As you can see, this equation is out of balance because it fails to explain the true nature of this reaction.

The idea here is that carbon dioxide gas is actually acidic when dissolved in aqueous solution. More specifically, aqueous carbon dioxide will exist in equilibrium with carbonic acid, #"H"_2"CO"_3#, a weak acid

#"CO"_text(2(g]) rightleftharpoons "CO"_text(2(aq])#

#"CO"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_2"CO"_text(3(aq])#

Therefore, the equation above should be expressed as

#"KOH"_text((aq]) + overbrace("CO"_text(2(aq]) + "H"_2"O"_text((l]))^(color(blue)("H"_2"CO"_3)) -> "KHCO"_text(3(aq]) + "H"_2"O"_text((l])#

This is the same as

#"KOH"_text((aq]) + "H"_2"CO"_text(3(aq]) -> "KHCO"_text(3(aq]) + "H"_2"O"_text((l])#

It is now evident that what you have here is essentially a neutralization reaction, where a strong base is neutralized by a weak acid.

Because it's a strong base, potassium hydroxide will dissociate completely in aqueous solution to form potassium cation,#"K"^(+)#, and hydroxide anions, #"OH"^(-)#.

This implies that an ionic equation for this reaction can be written in its entirety.

#"K"_text((aq])^(+) + "OH"_text((aq])^(-) + "H"_2"CO"_text(3(aq]) -> "K"_text((aq])^(+) + "HCO"_text(3(aq])^(-) + "H"_2"O"_text((l])#

The net ionic equation can be obtained by removing spectator ions, or ions that are present on both sides of the equation.

#color(red)(cancel(color(black)("K"_text((aq])^(+)))) + "OH"_text((aq])^(-) + "H"_2"CO"_text(3(aq]) -> color(red)(cancel(color(black)("K"_text((aq])^(+)))) + "HCO"_text(3(aq])^(-) + "H"_2"O"_text((l])#

which will have this appearance

#"OH"_text((aq])^(-) + "H"_2"CO"_text(3(aq]) -> "HCO"_text(3(aq])^(-) + "H"_2"O"_text((l])#

Therefore, you can state that if you want to include carbon dioxide gas in the calculation

#"OH"_text((aq])^(-) + "CO"_text(2(g]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) -> "HCO"_text(3(aq])^(-) + color(red)(cancel(color(black)("H"_2"O"_text((l]))))#

This will provide you with

#"OH"_text((aq[)^(-) + "CO"_text(2(g]) -> "HCO"_text(3(aq])^(-)#

Consequently, you will possess

#"KOH"_text((aq]) + "CO"_text(2(g]) -> "KHCO"_text(3(aq])#

Peyton Adams · Answer

undefined The products of the reaction between solid KOH and gaseous CO₂ are potassium carbonate (K₂CO₃) and water (H₂O).