# Acidic Strength comparison

I wanted to compare the acidic strength of aniline and water.

Water on losing hydrogen gives $$OH^-$$ and aniline on releasing H gives $$\ce{Ph-NH-}$$ and this -ve charge goes into resonance so the -ve charge is distributed, then which one is has more acidic strength??

If it is aniline then it should react with $$\ce{NaOH}$$, since water, which is comparatively weaker acid is formed as a product, then why does it not do so

Please tell me if i am wrong somewhere in my logic and which one is more acidic.

[NOTE: I am a class 12th student, so please try to keep the answer within the scope of a 12th grade student's knowledge]

• You need two phenyl rings to get amine with acidity comparable to water (diphenylamine). Commented Sep 2, 2023 at 17:09
• Usually what we call the Ka of aniline is the value of the anilinium $\ce{C6H5NH3^+}$ cation Commented Sep 2, 2023 at 20:15
• The pKa of aniline is ~30 that of water is 14. pKa is logarhythmic scale so the aniline proton is approx 10^16 less acidic than water Commented Sep 2, 2023 at 21:05

Actually whether aniline will react with $$\ce{NaOH}$$ or not will depend on what medium we choose for the reaction.
If it is $$\ce{dil. NaOH}$$ then the $$\ce{OH-}$$ dissociated will be solvated so that it will no longer be a base however will be a neucleophile but as $$\ce{-NH2}$$ is attached at a $$\ce{sp2}$$ centre so can't do substitution reactions.
And if it is $$\ce{conc. NaOH}$$ or the medium is alcohol or polar aprotic then it can do acid-base reaction.