What is the conjugate acid of each of the following? What is the conjugate base of each?

\({\rm{a) }}{{\rm{H}}_2}S\)

\({\rm{b) }}{{\rm{H}}_2}{\rm{PO}}_4^ - \)

\({\rm{c) P}}{{\rm{H}}_3}\)

\({\rm{d) }}H{S^ - }\)

\({\rm{(e)HSO}}\;_3^ - \)

\({\rm{\;(f)\;}}{{\rm{H}}_3}{{\rm{O}}_2}^ + \)

\({\rm{\;(g)\;}}{{\rm{H}}_4}{{\rm{N}}_2}\)

\({\rm{\;(h)\;C}}{{\rm{H}}_3}{\rm{OH}}\)

To determine what is the conjugated base or acid, we will revise definitions of Bronsted-Lowry's acid and base: acid is proton\(\left( {{H^ + }} \right)\)donor and base is proton\(\left( {{H^ + }} \right)\)acceptor. Conjugate means that it is a product of acid-base reaction.Let's look at the example:

\({\rm{HA}}(aq) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {A^ - }(aq) + {{\rm{H}}_3}{{\rm{O}}^ + }(aq)\)

HA is acid because it donates the proton to the water. Hence, water is base because it will take that proton. \({A^ - }\)is a product of acid-base reaction of HA, so it means we shall name it conjugate. Now we will determine whether it is a base or an acid. \({A^ - }\), to go back to the HA, needs to take a proton from \({{\rm{H}}_3}{{\rm{O}}^ + }\), hence it is a base, conjugate base. The explanation is same for the \({{\rm{H}}_3}{{\rm{O}}^ + }\)- it is conjugate acid.