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








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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?




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.




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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.




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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.


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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 :)


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