Question #c729d

As you know, a Bronsted-Lowry acid is characterized by the fact that it can act as a proton donor. Likewise, a Bronsted-Lowry base is characterized by the fact that it can accept a proton.

A proton, #"H"^(+)#, is simply the nucleus of a hydrogen atom, #"H"#.

In a given chemical reaction, you can identify a Bronsted-Lowry acid by looking for the chemical species that loses a proton. Consequently, the chemical species that accepts the proton lost by a Bronsted-Lowry acid will act as a Bronsted-Lowry base.

Now, the chemical species that is left behind after an acid donates a proton is called a conjugate base because it can accept a proton to reform the original acid.

Likewise, the chemical species that is formed when a base accepts a proton is called a conjugate acid because it can donate this proton to reform the original base.

In your case, you have

#color(red)("H")"NO"_ (3(aq)) + "SO"_ (4(aq))^(2-) -> color(red)("H")"SO"_ (4(aq))^(-) + "NO"_(3(aq))^(-)#

Here #"HNO"_3# donates its proton to #"SO"_4^(2-)#, which accepts it to form #"HSO"_4^(-)#.

As a result, you can say that #"HNO"_3# acts as an acid and #"SO"_4^(2-)# acts as a base.

#overbrace(color(red)("H")"NO"_ (3(aq)))^(color(purple)("acid")) + overbrace("SO"_ (4(aq))^(2-))^(color(green)("base")) -> color(red)("H")"SO"_ (4(aq))^(-) + "NO"_(3(aq))^(-)#

Now, #"HNO"_3#, which is called nitric acid, donates its proton to leave behind the nitrate anion, #"NO"_3^(-)#, which means that the nitrate anion will be the acid's conjugate base.

On the other hand, #"SO"_4^(2-)#, which is the sulfate anion, accepts a proton to form #"HSO"_4^(-)#, the hydrogen sulfate anion, which means that the hydrogen sulfate anion is the base's conjugate acid.

#overbrace(color(red)("H")"NO"_ (3(aq)))^(color(purple)("acid")) + overbrace("SO"_ (4(aq))^(2-))^(color(green)("base")) -> overbrace(color(red)("H")"SO"_ (4(aq))^(-))^(color(green)("conjugate acid")) + overbrace("NO"_(3(aq))^(-))^(color(purple)("conjugate base"))#