# pH of a solution of acetic acid and ammonium acetate

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 :

$\ce{CH3COOH}+\ce{NH4OH<=>CH3COONH4}+\ce{H2O}$

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?

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

• $\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{OH-}$? May 19 '14 at 12:53
• you mean + = -? May 19 '14 at 13:17
• Well, that is : $\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{H+}$ May 19 '14 at 13:18
• 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. May 19 '14 at 13:24
• @Awesome $\ce{NH3}+\ce{H2O<=>NH4+}+\ce{OH-}$ Again, $\ce{NH4OH}$ does not exist as a molecule. May 19 '14 at 13:25