What is the complete ionic equation for the reaction #2KOH_((aq)) + H_2SO_(4(aq)) -> 2H_2O_((l)) + K_2SO_(4(aq))#?

1 Answer
Mar 11, 2016


Here's what I got.


In order to find the complete ionic equation for the neutralization reaction, you must write the dissociation of the strong base and of the strong acid in aqueous solution.

As you know, strong bases and strong acids dissociate completely in aqueous solution. In this case, you are to assume that sulfuric acid acts as a strong acid in both steps of its ionization.

So, the two reactants will dissociate completely to form

#"KOH"_text((aq]) -> "K"_text((aq])^(+) + "OH"_text((aq])^(-)#

#"H"_2"SO"_text(4(aq]) -> 2"H"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

On the reactants' side, potassium sulfate, #"K"_2"SO"_4#, is a soluble ionic compound, which means that it exists as ions in aqueous solution.

#"K"_2"SO"_text(4(aq]) -> 2"K"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

Water, on the other hand, is a covalent compound that exists in molecular form.

Starting from the balanced molecular equation

#color(red)(2)"KOH"_text((aq]) + "H"_2"SO"_text(4(aq]) -> "K"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#

you can write

#color(red)(2)"K"_text((aq])^(+) + color(red)(2)"OH"_text((aq])^(-) + 2"H"_text((aq])^(+) + "SO"_text(4(aq])^(2-) -> 2"K"_text((aq])^(+) + "SO"_text(4(aq])^(2-) + 2"H"_2"O"_text((l])#

If you want,you can also write the net ionic equation by eliminating the spectator ions, i.e. the ions that are present on both sides of the equation

#color(red)(cancel(color(black)(color(red)(2)"K"_text((aq])^(+)))) + color(red)(2)"OH"_text((aq])^(-) + 2"H"_text((aq])^(+) + color(red)(cancel(color(black)("SO"_text(4(aq])^(2-)))) -> color(red)(cancel(color(black)(2"K"_text((aq])^(+)))) + color(red)(cancel(color(black)("SO"_text(4(aq])^(2-)))) + 2"H"_2"O"_text((l])#

This will get you

#color(red)(2)"OH"_text((aq])^(-) + 2"H"_text((aq])^(+) -> 2"H"_2"O"_text((l])#