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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]

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  • $\begingroup$ You need two phenyl rings to get amine with acidity comparable to water (diphenylamine). $\endgroup$
    – Mithoron
    Commented Sep 2, 2023 at 17:09
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    $\begingroup$ Usually what we call the Ka of aniline is the value of the anilinium $\ce{C6H5NH3^+}$ cation $\endgroup$
    – Maurice
    Commented Sep 2, 2023 at 20:15
  • $\begingroup$ 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 $\endgroup$
    – Waylander
    Commented Sep 2, 2023 at 21:05

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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.

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    $\begingroup$ That's a good point. Even if OP's issue is more of "lets just get a hold on water pKa", it may be good to find its limitations early on. $\endgroup$
    – Mithoron
    Commented Sep 2, 2023 at 18:21

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