Question #57f4c

For a compound that acts as a Bronsted - Lowry acid, the conjugate base is the species formed after the initial compound donates a proton, #"H"^(+)#.

You can look at this from another perspective - if by adding a proton to a chemical species you end up with the initial Bronsted - Lowry acid, then that species is the acid's conjugate base.

Simply put, you get the conjugate base of a Bronsted - Lowry acid by removing one proton, which is the same as a hydrogen ion, #"H"^(+)#, from its chemical formula.

Take a look at the first compound, hydrochloric acid. In aqueous solution, hydrochloric acid donates its proton to water, which acts as a base, to form hydronium ions, #"H"_3"O"^(+)#, and chloride anions, #"Cl"^(-)#

#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#

So, which chemical species is the conjugate base of hydrochloric acid?

As you can see, the hydrochloric acid molecule donated its proton to a water molecule. So what remained behind? The chloride anion, #"Cl"^(-)#.

This means that you have

#overbrace("HCl"_text((aq]))^(color(red)("acid")) + overbrace("H"_2"O"_text((l]))^(color(blue)("base")) -> overbrace("H"_3"O"_text((aq])^(+))^(color(blue)("conjugate acid")) + overbrace("Cl"_text((aq])^(-))^(color(red)("conjugate base"))#

How about the conjugate base for hydrogen sulfide, #"H"_2"S"#?

The exact same principle applies. Take away of its protons to get

#"H"_2"S"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "HS"_text((aq])^(-)#

The conjugate base for hydrogen sulfide is the bisulfite anion, #"HS"^(-)#. So what happens when you add a proton to the conjugate base, #"HS"^(-)#?

You reform the acid #-># #"H"_2"S"#.

I'll leave the third one to you as practice.

Keep in mind, for neutral compounds, taking away one proton will always result in a negative charge on the resulting anion, so don't forget about the charge when writing the chemical formula of the conjugate base.