Keeping in ming Arrhenius theory, write balanced chemical equations for: a) Aqueous hydrobromic acid and potassium hydroxide b) Aqueous sulphuric acid and barium hydroxide.?

The Arrhenius Theory of acids and bases categorizes substances that produce hydrogen cations in solution, #"H"^(+)#, as acids and substances that produce hydroxide anions, #"OH"^(-)#, in solution as bases.

For the first reaction, you have hydrobromic acid, #"HBr"#, and potassium hydroxide, #"KOH"#.

Hydrobromic acid is an Arrhenius acid because it ionizes to produce hydrogen cations in the solution.

#"HBr"_ ((aq)) -> "H"_ ((aq))^(+) + "Br"_ ((aq))^(-)#

Now, you can represent a hydrogen cation as a hydronium cation, #"H"_3"O"^(+)#, which means that the ionization of the acid can be represented as

#"HBr"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Br"_ ((aq))^(-)#

Potassium hydroxide is an Arrhenius base because it produces hydroxide anions in the solution.

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

When you mix these two solutions, a neutralization reaction occurs. You can have

# "H"_ ((aq))^(+) + "Br"_ ((aq))^(-) + "K"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l)) + "K"_ ((aq))^(+) + "Br"_ ((aq))^(-)#

Since potassium bromide, one of the products of the reaction alongside water, is soluble in water, you can eliminate the spectator ions

#"H"_ ((aq))^(+) + color(red)(cancel(color(black)("Br"_ ((aq))^(-)))) + color(red)(cancel(color(black)("K"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("K"_ ((aq))^(+)))) + color(red)(cancel(color(black)("Br"_ ((aq))^(-))))#

to get the net ionic equation.

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#

If you replace the hydrogen cations with the hydronium cations, you get

#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#

The same logic applies to the reaction between sulfuric acid and barium hydroxide. This reaction involves an Arrhenius acid that produces #2# hydrogen cations in the solution and an Arrhenius base that produces #2# hydroxide anions in the solution.

This time, you get

#"H"_ 2"SO"_ (4(aq)) + "Ba"("OH")_ (2(aq)) -> "BaSO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))#

because you have

#"H"_ 2"SO"_ (4(aq)) -> 2"H"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)#

#"Ba"("OH")_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)#

So when you mix these two solutions, the barium cations and the sulfate anions act as spectator ions

#2"H"_ ((aq))^(+) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + color(red)(cancel(color(black)("Ba"_ ((aq))^(2+)))) + 2"OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Ba"_ ((aq))^(2+)))) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + 2"H"_ 2"O"_ ((l))#

which means that you have

#2"H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#

which can be simplified to

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#