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Is ethanol considered to be an acid or a base? Or does it act as an amphoteric species? Is acidity and basicity a relative concept? Also, can someone please explain this on the basis of the Brønsted–Lowry acids/bases theory which is based on the protonic concept; will it donate or accept $\ce{H+}$ ions? I'm unsure which acid base theory should I follow in this case, any help will be appreciated.

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    $\begingroup$ Hello and welcome to Chemistry.SE. If you have any question about the site you can take the short tour or visit the help center. Also, as this is a homework type of question, you can read about the homework policy here. Basically you just need to show your thoughts and any work toward solving the problem yourself so that we are not just doing all of it for you. Best of luck with your problem! $\endgroup$
    – airhuff
    Apr 16, 2017 at 6:25
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    $\begingroup$ If you assume the question is about Brønsted–Lowry acids/bases, does ethanol contain a proton that can be donated? Does it contain a site that can accept a proton? $\endgroup$
    – airhuff
    Apr 16, 2017 at 6:27

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Because we are talking about Brønsted–Lowry acids and bases, we are only concerned with whether or not a molecule has (or the degree to which it has) the ability to donate a proton (acid) or the ability to accept a proton (base) or to do either (amphoteric).

The case of ethanol is rather simple to evaluate. Firstly, the hydroxyl proton could be donated to a strong base, although this only happens to a minimal degree in an aqueous solution as water is a much stronger acid than ethanol. Still, ethanol has the ability to act as an acid because of the ability to donate it's hydroxyl proton.

However, aqueous solutions of ethanol are slightly basic. This is is because the oxygen in ethanol has lone electron pairs capable of accepting protons, and thus ethanol can act as a weak base.

Summary - TL/DR:
As ethanol has the ability to both donate and accept protons, it should be considered an amphoteric compound with respect to Brønsted–Lowry acid-base theory.

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Acidity and basicity is a relative concept. A species can behave as an acid or a base but then that would depend on the reaction which we're talking about. Consider, for instance, acetic acid $(\ce{CH3COOH})$. It can be amusing that it ofttimes doesn't behave as an acid, contrary to its name! As an example you might want to look the reaction of its conjugate base, acetate ion, with $\ce{HCl}$. $$\ce{HCl + CH3COONa -> NaCl + CH3COOH}$$ In this case, it can be said that acetate ion was forced to accept $\ce{H+}$ from $\ce{HCl}$, thus it was behaving as a base.

But things like can be downright confusing and thus we usually refer to species being an acid or a base in comparison to water. Alcohols (in general) are considered to be basic in nature, with the notable exception of phenol. The decreasing acidic strength order would then be summed up as:

$$\ce{Phenol \gt H2O \gt ROH}$$

Thus ethanol can be termed as a base.

Such comparisons can be easily made if you're ready to look into the $\ce{pK_a}$ values (you might want to look it up if you're unsure what it means). Water has a $\ce{pK_a}$ of $\ce{15.7}$. Alcohols generally have a $\ce{pK_a}$ in the range of $\ce{16-19}$. Thus they are weaker acids than water. A controversial case, however, is of methanol $\ce{(CH3OH)}$. It has a $\ce{pK_a=15.54}$ which makes it a little acidic than water.

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    $\begingroup$ Are you sure about that pKa of water? Shouldn't it be 14? Also, you shouldn't use the \ce command on pKa. For one it isn't a chemical and it also seems to have turned your decimal points into multiplication. $\endgroup$
    – Tyberius
    Apr 16, 2017 at 17:39
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    $\begingroup$ chem.libretexts.org/Core/Organic_Chemistry/Fundamentals/… That value of pKa doesn't work when we take into account that equilibrium constants are really defined in terms of activities. I think the maker of \mchem had you in mind when they made it, because you can use \pu instead for what you were doing and it should format nicer. $\endgroup$
    – Tyberius
    Apr 16, 2017 at 19:03
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    $\begingroup$ @berryholmes You're contradicting yourself in your first comment... $\endgroup$
    – Mithoron
    Apr 16, 2017 at 19:57
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    $\begingroup$ @berryholmes You say that both $K_a=K_{eq}\cdot \ce{H2O}$ and that $K_w=K_{eq}\cdot \ce{H2O}$. This would of course imply (correctly, but not necessarily derived correctly) that $K_a=K_w=1\cdot 10^{-14}$. $\endgroup$
    – Tyberius
    Apr 16, 2017 at 21:35
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    $\begingroup$ @Tyberius My apologies, I made a blunder in my comment, $K_a = K_{eq}$ so when we take $\ce{[H+] = [OH-] = 10^{-7}}$, and evaluate $\frac{10^{-7} \times 10^{-7}}{55.56}$, it comes out to be $ \approx 1.8 \times 10^{-16}$ which is the value we were looking for. So $ K_a = K_{eq} = \frac{K_w}{[H_2O]}$. I want to correct my mistake but i'm unable to edit my previous comment :( $\endgroup$ Apr 17, 2017 at 5:09

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