# What is the conjugate acid of NaOH using the Brønsted-Lowry definition of acids?

To the best of my knowledge, a conjugate acid of a base is the base after it has accepted a proton, or a $\ce{H+}$ ion. In this case:

$$\ce{NaOH + H+ -> Na+ + H2O}$$

Is the conjugate acid of $\ce{NaOH}$ the sodium ion, or the water? Common sense tells me it can't be the $\ce{Na+}$ ion, because it has no protons to donate, so how could it ever be an acid? So I am thinking that the conjugate acid is $\ce{H2O}$. However, wouldn't that mean that the conjugate acid of any base of the form

$$\ce{(something)OH + H+ -> (something)+ + H2O}$$ would be water, and that seems unsettling to me. I also believe that since $\ce{NaOH}$ undergoes the following reaction:

$$\ce{NaOH -> Na+ + OH-}$$

the $\ce{Na+}$ is something of a 'spectator ion' (not sure if that's the correct term), this seems to imply that $\ce{H2O}$ should be the conjugate acid. Basically, I'm really confused, and could use a little help sorting all this out.

• In Bronsted theory OH- is a base not NaOH like in Arrhenius theory. Aug 30 '15 at 11:19

Your first equation is more properly written as

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

in aqueous media. However, we can do better if we explicitly show the dissociation of $\ce{NaOH}$ as

$$\ce{NaOH -> Na+ + OH-}$$

and substitute that into the first expression (note that I write $\ce{2H2O}$ as $\ce{H2O + H2O}$) to get

$$\ce{Na+ + \underbrace{OH^{-}}_{base} + \underbrace{H3O^{+}}_{acid} -> Na+ + \underbrace{H2O}_{conjugate\;acid} + \underbrace{H2O}_{conjugate\;base}}$$

where we see that $\ce{H2O}$ is the conjugate acid of $\ce{OH-}$ as well as the conjugate base of $\ce{H3O+}$. The last bit - where water plays 2 roles - is due to water being amphoteric, or able to act as an acid or a base.