What is the conjugate base for #"H"_2"S"#?

1 Answer
Dec 5, 2015



As you know, Bronsted - Lowry acids are defined as proton donors and Bronsted - Lowry bases are defined as proton acceptors.

When a Bronsted - Lowry acid reacts with a Bronsted - Lowry base, a proton, #"H"^(+)#, is being transferred from the acid to the base.

This reaction will leave the acid with one less proton than what it had before the reaction, and the base with one extra proton than what it had before the reaction.

The compound that's left behind after an acid donates a proton is called a conjugate base because it can reform the original acid by accepting a proton.

In this case, when hydrogen sulfide, #"H"_2"S"#, donates a proton, the compound left behind will be the bisulfide anion, #"HS"^(-)#, its conjugate base.

For example, when dissolved in aqueous solution, hydrogen sulfide, which in this context is known as hydrosulfuric acid, donates a proton to water to form bisulfide anions and hydronium cations

#overbrace("H"_2"S"_text((aq]))^(color(blue)("acid")) + "H"_2"O"_text((l]) rightleftharpoons overbrace("HS"_text((aq])^(-))^(color(red)("conjugate base")) + "H"_3"O"_text((aq])^(+)#