Write a balanced molecular, complete ionic, and net ionic equation for the reaction between "HBr" and #"NaOH"# ?
1 Answer
Here's what I got.
Explanation:
Hydrobromic acid,
This should tell you that the net ionic equation will involve the hydrogen cations--or hydronium cations,
When an acid and a base react, the neutralization reaction produces water and an aqueous salt. In this case, the salt is sodium bromide,
The balanced molecular equation looks like this
#"HBr"_ ((aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaBr"_ ((aq))#
Now, because hydrobromic acid is a strong acid, it will ionize completely in aqueous solution, which means that you can write
#"HBr"_ ((aq)) -> "H"_ ((aq))^(+) + "Br"_ ((aq))^(-)#
Similarly, the fact that sodium hydroxide is a strong base means that you can write
#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#
This means that the complete ionic equation looks like this
#"H"_ ((aq))^(+) + "Br"_ ((aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l)) + "Na"_ ((aq))^(+) + "Br"_ ((aq))^(-)#
Sodium bromide is soluble in water, which is why it exists as dissociated ions in aqueous solution.
To get the net ionic equation, you must eliminate the spectator ions, i.e. the ions that are present on both sides of the equation.
#"H"_ ((aq))^(+) + color(red)(cancel(color(black)("Br"_ ((aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("Br"_ ((aq))^(-))))#
This will get you
#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#
This is the net ionic equation that characterizes a neutralization reaction that takes place between a strong acid and a strong base.
