# Clarification in a particular application of Le Chatelier principle [closed]

There is something I don't understand about a particular application of the Le Chatelier principle. The following system is in equilibrium: $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$ If I add some $\ce{H3O+}$ the reaction will shift left. Now if I keep adding $\ce{H3O+}$, it will keep shifting left until the $\ce{CH3COO-}$ would deplete? And would the $\ce{H3O+}$ will remain at its initial concentration? So I can't understand how a buffer work, because if you have a reaction like above and you are adding acid, the added $\ce{H3O+}$ and the $\ce{CH3COO-}$ will bind together to form $\ce{CH3COOH}$ and water thus keeping the $\ce{H3O+}$ exactly the same. Where am I wrong?

• chemistry.stackexchange.com/questions/81192/… Mar 15 '18 at 21:54
• This question does not add any value for future visitors unless you clarify what was the need to bring in a buffer here, when the corresponding sodium ethanoate salt is not even present in the solution. Till then, I'm voting to close the question as unclear. You may always edit the question later to reflect changes. Mar 16 '18 at 2:29
• @GaurangTandon There is no requirement of "sodium" for something to be a buffer. Other spectator ions would do. Why does OP need to include spectator ions to be understood? Mar 16 '18 at 11:08
• @DavePhD Oh, I am sorry. I was about to write "metal ethanoate salt" or simply "ethanoate salt", but then substituted it with "sodium" for some reason. My fault. The reason why I said the question is unclear, instead, is that the OP says "The following system is in equilibrium:" but didn't ever say whether it was only acetic acid or acetic acid and its salt (only latter works). Moreover, the part after "So I can't understand how a buffer work..." simply wasn't clear. That said, I tried my best to answer the OP based on what I understood, and advised the OP to edit their question for clarity. Mar 16 '18 at 11:19
• @GaurangTandon as long as there is some acetic acid and some acetate (some acid and some conjugate base) it is a buffer. It don't think you should be distracted by the spectator ions, whether they be metal ions or quaternary amine ions or whatever. Mar 16 '18 at 11:29

For starters, I wouldn't classify your equilibrium as a buffer. A buffer requires excess weak acid, and it also requires either a strong base or the salt of the weak acid (which you don't have). The equilibrium you stated is the dissociation of CH3COOH into its conjugate base and a conjugate acid.

Say you had the following buffer:

$$CH_{3}COOH + NaOH ⇌ CH_{3}COONa + H_{2}O$$

The equilibrium of this would be:

$$CH_{3}COOH ⇌ CH_{3}COO^{-} + H^{+}$$

And the ionic equation would be:

$$H^{+} + OH^{-} → H_{2}O$$

As per definition, a buffer is a system that minimises pH change on small additions of acid or alkali. How? All down to equilibrium...

If you were to increase $[H^{+}]$:

• In terms of the equilibrium, the added $H^{+}$ would react with the base, $CH_{3}COO^{-}$, to form $CH_{3}COOH$. This is Le Chatelier's principle in the fact that increasing $[H^{+}]$ shifts the equilibrium to the left.

If you were to increase $[OH^{-}]$:

• In terms of the ionic equation, the added $OH^{-}$ would react with the $H^{+}$ to form $H_{2}O$. This is Le Chatelier's principle in the fact that increasing $[OH^{-}]$ shifts the equilibrium to the right.

In both of these, it's clear that the addition of $H^{+}$ or $OH^{-}$ would just be an acid-base reaction. The products formed minimise the $pH$ of the system due to their relative strengths compared with the added acids/alkalis.

The $pH$ does change though, but just not as much as it would if it wasn't in the buffer system. A buffer just keeps the equilibrium central.

Your question is good-intentioned but there are quite a few things that would need clarifying.

First, the above setup is not a buffer. You need additional metal ethanoate for a buffer to form. The above dissociation is only of a simple weak acid dissociating in water.

Secondly, yes, adding more $\ce{H3O+}$ to the above solution will shift the reaction to the left ("suppress the dissociation of acetic acid") but it would still increase the overall concentration of $\ce{H3O+}$, contrary to "the $\ce{H3O+}$ will remain at its initial concentration" (note that the backward shift is almost always negligible compared to the moles of hydronium we add (corrected thanks to @TanYongBoon!)).

If this doesn't make sense, consider waiting a bit more for your course work to proceed towards completion. Because sometimes, certain problems automatically clear themselves as you proceed later in the coursework :)