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

1 Answer
Dec 29, 2015

Answer:

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

Explanation:

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

The balanced chemical equation for this reaction looks like this

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

As you can see, this equation is unbalanced because it doesn't describe what actually happens in 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])#

The above equation should thus be written 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 equivalent to

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

Now it becomes clear that you're actually dealing with a neutralization reaction in which a weak acid neutralizes a strong base.

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

This means that you can write a complete ionic equation for this reaction

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

Eliminating spectator ions, which are those ions that are present on both sides of the equation, will get you the net ionic 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 look like this

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

So, if you want to use carbon dioxide gas in the equation, you can say that

#"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 give you

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

Therefore, you will have

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