After studying ionic equilibrium, I was just making some wild questions and I was unable to calculate the $\ce{pH}$ of this solution. Assume, 1 mole of each was added to $1L$ solution of water.

My attempt :


This reaction will happen till all $\ce{CH3COOH}$ is consumed by $\ce{NH4OH}$. More of $\ce{NH4OH} $ will be formed by Le Chatelier's principle. So, the answer must be same as that of a weak acid and weak base salt : $pH=\dfrac{1}{2}(pK_w+pK_a-pK_b)$. But how will we calculate if there was $2$ mole of acetic acid?


1 Answer 1


You have some mistaken assumptions.

$\ce{NH4OH}$ and $\ce{CH3COONH4}$ are not molecules that exist.

The assuption "all $\ce{CH3COOH}$ is consumed" is incorrect.

Why do you say more $\ce{NH4OH}$ will be formed? How is this possibe? What can it be formed from?

Review that Henderson–Hasselbalch equation

  • $\begingroup$ $\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{OH-}$? $\endgroup$
    – evil999man
    May 19, 2014 at 12:53
  • $\begingroup$ you mean + = -? $\endgroup$
    – DavePhD
    May 19, 2014 at 13:17
  • $\begingroup$ Well, that is : $\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{H+}$ $\endgroup$
    – evil999man
    May 19, 2014 at 13:18
  • $\begingroup$ Again, ammonium hydroxide isn't something that is stable. The two components undergo a large extent reaction to make ammonia. As a matter of fact, this is what Wikipedia has to say: "Ammonia solution, also known as ammonium hydroxide." So what Dave is suggesting is that you write the net ionic equation, or in other words - the only right equation you can write for this system. Leave out the spectator ions to help you better see the chemistry. $\endgroup$
    – Dissenter
    May 19, 2014 at 13:24
  • 1
    $\begingroup$ @Awesome $\ce{NH3}+\ce{H2O<=>NH4+}+\ce{OH-}$ Again, $\ce{NH4OH}$ does not exist as a molecule. $\endgroup$
    – DavePhD
    May 19, 2014 at 13:25

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