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Take the following simple equilibrium: $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

Ethanoic acid will act as a typical Brønsted-Lowry acid and donate a proton to water, forming a conjugate acid-base pair. In this case, H3O+ is the conjugate acid, and CH3COO- is the conjugate base.

Logic tells us that if we have a solution of ethanoic acid and water, increasing the amount of acid will decrease the solution's pH. This makes sense, since there'd literally be a higher acid concentration.

However, the interpretation I get due to the equilibrium is the opposite. I see it as increasing the concentration of ethanoic acid will mean there's a larger concentration of conjugate base, hence the solution will become more basic.

Is this flawed logic or simply a misunderstanding of Le Chatelier's principle?

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  • $\begingroup$ Flawed logic. Ignoring autoionization of water, $\ce{[H^+] = [CH3COO^-]}$. $\ce{H^+}$ is a strong acid, $\ce{CH3COO^-}$ is a weak conjugate base, so the solution is acidic. $\endgroup$ – MaxW Jul 25 '18 at 22:05
  • $\begingroup$ @MaxW That doesn't explain much. $\ce{[H+] = [CH3COO-]}$ is meant to suggest what exactly? How does that statement suggest that the solution is acidic? $\endgroup$ – holbo2 Jul 25 '18 at 22:09
  • $\begingroup$ Sorry that you don't understand. Assuming that $\ce{[H+] =[CH3COO−]}$ implies that the concentration of $\ce{CH3COOH}$ should be above about $1\times 10^{-5}$ molar. In reality $\ce{[H+] =[CH3COO−] + [OH^-]}$. $\endgroup$ – MaxW Jul 25 '18 at 22:16
  • $\begingroup$ @MaxW I've been reading about how conjugate acids/bases affect equilibria. I read that weak acids produce strong conjugate bases; in this case, ethanoate is a strong conjugate base. Strong conjugate bases shift equilibria to the left since the tendency of them to accept protons is much greater than the tendency of weak acids to donate them. So is this why the solution will become more acidic? $\endgroup$ – holbo2 Jul 25 '18 at 22:35
  • $\begingroup$ $\ce{CH3COO−}$ is a weak base compared to $\ce{OH−}$. $\endgroup$ – MaxW Jul 25 '18 at 22:53
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You are neglecting the hydronium ions produced during dissociation of acetic acid.

We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.

  1. We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.
  2. Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.

Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 also involves the production of a $\ce{H+}$. Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*


*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.

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  • $\begingroup$ Could you explain in a bit more detail as to why step 2 occurs at a lesser degree than step 1? You only said that it 'has to'. My interpretation of your answer is that since step 2 is happening less than step 1, there will be an increasingly greater $\ce{[H+]}$ in the solution. However, the ethanoate produced will form more ethanoic acid, which in turn produces even more $\ce{[H+]}$... Basically, $\ce{[H+]} > \ce{[OH-]}$ or something? $\endgroup$ – holbo2 Jul 26 '18 at 10:28
  • $\begingroup$ @holbo2, I clearly stated that it has to "because step 2 consumes the acetate ions produced in step 1." Your interpretation is incorrect. Here, I have consolidated the entire equilibrium process into a two-step sequence. If instead you consider a sequence of steps that goes 1 2 1 2 1 2 1 2 ..., then by a similar argument as mine you can say that each successive step partially counteracts the act of the step before. Certainly there will not be "an increasingly greater $\ce{[H+]}$ in the solution." $\endgroup$ – a-cyclohexane-molecule Jul 26 '18 at 11:01
  • $\begingroup$ Fair enough, but even the fact that you 'clearly stated' it doesn't explain why it occurs lesser; to someone who clearly doesn't understand it well, that's a statement and not an explanation. I see it simply because the equilibrium will be shifted towards the right in this case, since this is the side of acid dissociation. I get that the steps counteract each other, but would one step happen at a faster rate than the other? $\endgroup$ – holbo2 Jul 26 '18 at 17:45
  • $\begingroup$ (> Would one step happen at a faster rate than the other?) @holbo2, yes, that's the basis of my answer. To elaborate on my explanation, step 1 must occur at a faster rate than does step 2 to produce the excess acetate ions required in step 2. Otherwise, there wouldn't be excess acetate ions for step 2 to consume. $\endgroup$ – a-cyclohexane-molecule Jul 26 '18 at 18:25
  • $\begingroup$ @a-cyclohexane-molecule Makes sense. Am I correct in saying - the forward reaction happens at a faster rate than the reverse reaction, hence the equilibrium is shifted towards the right which means that the rate of $\ce{[H+]}$ production is happening faster than $\ce{[CH3COOH]}$ production? Is this why stronger acids are even more acidic in solution... Something like that, anyway? That kinda correlates with how buffer systems stay in equilibrium with small additions of acid/base, right? $\endgroup$ – holbo2 Jul 26 '18 at 18:31
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Ok, let me come at this from a different angle using math.

As you say, take the following simple equilibrium: $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}\tag{1}$$

Now the equilibrium equation is given by:

$$K_\mathrm{a} = \frac{\ce{[H^+][CH3COO^-]}}{\ce{[CH3COOH]}}\tag{2}$$

Let's assume some initial concentration of acetic acid $x$. Let $\ce{[H^+]_x}$ and $\ce{[CH3COO^-]_x}$ be the corresponding concentration of $\ce{H^+}$ and $\ce{CH3COO^-}$.

$$K_\mathrm{a} = \frac{\ce{[H^+]_x[CH3COO^-]_x}}{x}\tag{3}$$

Now let's double the initial concentration of acetic acid to $2x$, using $\ce{[H^+]_{2x}}$ and $\ce{[CH3COO^-]_{2x}}$ so:

$$K_\mathrm{a} = \frac{\ce{[H^+]_{2x}[CH3COO^-]_{2x}}}{2x}\tag{4}$$

thus

$$\frac{\ce{[H^+]_x[CH3COO^-]_x}}{x} = \frac{\ce{[H^+]_{2x}[CH3COO^-]_{2x}}}{2x}\tag{5}$$

and

$$\ce{[H^+]_{2x}[CH3COO^-]_{2x}} = 2\times (\ce{[H^+]_x[CH3COO^-]_x})\tag{6}$$

now if we assume that $\ce{[H^+] = [CH3COO^-]}$ then

$$\ce{[H^+]_{2x} = \sqrt{2}\times [H^+]_x}\tag{7}$$

Since acetic acid is a weak acid, the assumption $\ce{[H^+] = [CH3COO^-]}$ is valid roughly so long as $[\ce{CH3COOH}] > 1\times10^{-5}$. In reality the charge in the solution must be neutral so:

$$\ce{[H^+] = [CH3COO^-] + [OH^-]}\tag{8}$$

So another way to look at the assumption is that

$$\ce{[H^+]} \gg \ce{[OH^-]}\tag{9}$$

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  • $\begingroup$ That's a nice way of looking at it - the mathematics from steps 1 to 7 make sense. I have one question though: you said that in reality, the assumption will be $\ce{[H+] = [CH3COO-] + [OH-]}$. However, how does this mean that $\ce{[H^+]} \gg \ce{[OH^-]}$? Sorry if this is a silly question. $\endgroup$ – holbo2 Jul 26 '18 at 17:55
  • $\begingroup$ The solution is acidic so $\ce{[H^+] \gg [OH^-]}$ must be true. $\ce{K_w = 1\times10^{-14} = [H^+] [OH^-]}$ $\endgroup$ – MaxW Jul 26 '18 at 18:03
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While you are right that by increasing the concentration of acetic acid in a solution we also increase the concentration of acetate, this does not factor into the pH calculation. Since:$$\mathrm{pH=-log[\ce{H3O+}]}$$

It only matters that by adding more acetic acid, there is more $\ce{H3O+}$ in the solution. This is why the pH decreases.

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I will do my best here, If more acid is added, the pH should remain the same due to the continuous reactions with water creating base so that equilibrium is once again reached. This should remain the scenario until such time as more acid is added than the amount of water. When this stage is reached the additional acid will no longer be able to react with water and so will remain is its acidic state thus decreasing the pH.

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  • $\begingroup$ That makes sense, however I have one problem. You said that the pH should remain the same as the equilibrium is always restored. However, adding water to an acid will dilute it, hence its pH will increase, so this doesn't agree with your statement and instead kind of agrees with what I said about a more basic solution being formed? $\endgroup$ – holbo2 Jul 25 '18 at 22:27
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    $\begingroup$ Sorry your best isn't good enough. This is wrong. The more acetic acid you add, the more acidic the solution gets. $\endgroup$ – MaxW Jul 25 '18 at 22:51

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