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In my book (Bruice) it is specified how it is possible to compare the pKa of two substances to understand who acts as an acid and who as a base. Here's the example in the book: $\ce{NH3 + H2O}$.

$\mathrm{p}K_\mathrm{a}$(NH3) = 36

$\mathrm{p}K_\mathrm{a}$(H2O) = 15.7

Now it is explained that, being the $\mathrm{p}K_\mathrm{a}$ of water lower than that of ammonia, water will be the substance that will behave as an acid, so $\ce{NH4+}$ and $\ce{OH-}$ will be the products.

But this reasoning, in my opinion, does not make much sense, as the $\mathrm{p}K_\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\ce{NH2-}$ and $\ce{H3O+}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\ce{H3O+}$ and $\ce{OH-}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\mathrm{p}K_\mathrm{a}$ of two different equilibria to predict who the acid is.

Furthermore, also $\ce{NH3}$, and not necessarily $\ce{H2O}$, can act as an acid. Indeed, ammonia can give both acid hydrolysis and basic hydrolysis to which it is possible to associate a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$.

Reference (1) Paula Yurkanis Bruice, Organic Chemistry, 8th ed.; Pearson Education, Inc., 2011, pp. 58.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Karsten
    Jan 14, 2023 at 14:51
  • $\begingroup$ If you already know all that, then why ask? $\endgroup$
    – Mithoron
    Jan 14, 2023 at 15:15
  • $\begingroup$ Because i was in doubt with what my book said. $\endgroup$
    – Luckenberg
    Jan 14, 2023 at 15:29

1 Answer 1

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But this reasoning, in my opinion, does not make much sense, as the $\mathrm{p}K_\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\ce{NH2-}$ and $\ce{H3O+}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\ce{H3O+}$ and $\ce{OH-}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\mathrm{p}K_\mathrm{a}$ of two different equilibria to predict who the acid is.

You are absolutely right, the argument does not work like that. It is possible to make a different, correct argument based on $\mathrm{p}K_\mathrm{a}$ values, though.

First, you write down the two possible reactions:

$$\ce{H2O(l) + NH3(aq) <=> OH-(aq) + NH4+(aq)}\tag{1}$$

$$\ce{NH3(aq) + H2O(l) <=> NH2-(aq) + H3O+(aq)}\tag{2}$$

I chose the order so that the reactant acting as acid is written first (on the far left) and the conjugate acid produced from the reactant acting as base is written last (on the far right).

Then, you compare the strength of the acids on the far left and the far right.

(1) $\ce{compare H2O with NH4+}$

(2) $\ce{compare NH3 with H3O+}$

For reference, here are the $\mathrm{p}K_\mathrm{a}$ values:

$\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$

$\mathrm{p}K_\mathrm{a}(\ce{NH4+}) = 9.8$

$\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 36$

$\mathrm{p}K_\mathrm{a}(\ce{H3O+}) = -1.7$

For (1), $\ce{NH4+}$ is a stronger acid than $\ce{H2O}$ (difference of about 6 in $\mathrm{p}K_\mathrm{a}$), so the equilibrium constant will be smaller than 1. For (2), $\ce{H3O+}$ is a much stronger acid than $\ce{NH3}$ (difference of about 38 in $\mathrm{p}K_\mathrm{a}$), so the equilibrium constant will be much smaller than 1. Neither reaction will be on the side of the products if both reactants are initially present at reasonably high concentrations.

Experience confirms this. The species $\ce{NH2-}$ does not exist in aqueous solution, it would get protonated immediately. Ammonia in water acts as a base, but most of the ammonia stays deprotonated (i.e. ammonia is a weak base). Of course, you can add acid to lower the hydroxide concentration and increase the ratio of ammonium to ammonia to the point where ammonium is the major species.

[OP in comments] when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions

No, we are looking at the net reaction of one species donating and one accepting a proton, without showing the proton explicitly. For reaction (1), for example:

$$\ce{H2O(l) <=> H+ + OH-(aq)}$$

plus

$$\ce{NH3(aq) + H+ <=> NH4+(aq)}$$

gives (after cancelling the protons):

$$\ce{H2O(l) + NH3(aq) <=> OH-(aq) + NH4+(aq)}$$

If you know the equilibrium constants of the "half reactions" (or the $\mathrm{p}K_\mathrm{a}$ values), you can get the equilibrium constant of the acid/base reaction.

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  • $\begingroup$ Thanks for your answer! Therefore, since equilibrium (2) is shifted to the left more than equilibrium (1), the pKb is lower than the pKa, so that ammonia prefers to act as a base rather than an acid, but this does not exclude that acid hydrolysis can still take place. Does this apply to all acid-base reactions? Even for example by reacting ammonia and a carboxylic acid? Furthermore, when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions. So Is It possible? $\endgroup$
    – Luckenberg
    Jan 14, 2023 at 14:44
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    $\begingroup$ @Luckenberg You can't just compare the acid and base strength of a molecule, you have to know what it reacts with. Substances behave differently depending on whether they are in water, in glacial acetic acid or in liquid ammonia. However, with our usual bias towards aqueous solutions, you can say ammonia acts as a base rather than an acid. I would have to see the quantitative argument before I can say whether it is sound or not. $\endgroup$
    – Karsten
    Jan 14, 2023 at 14:49
  • $\begingroup$ @Luckenberg "when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions." No, I think we are just separating the acid/base reaction into half reactions, where one species gains a proton and the other loses it. I added a bit to the answer to illustrate. $\endgroup$
    – Karsten
    Jan 14, 2023 at 14:49
  • $\begingroup$ brilliant! just one last quick example to evaluate if I understood correctly. Consider the reaction between ammonia and acetic acid: if ammonia acts as a base, then the reaction is shifted to the right, since NH4+ is a weaker acid than CH3COOH [pKa(NH4+)= 9.8; pKa(CH3COOH)=4]. Conversely, if acetic acid were to act as a base, the equilibrium would be enormously shifted to the left, this because NH3 is an extremely weaker acid than CH3COOH2+. So this explains why ammonia actually acts as a base, since the other reaction is practically negligible. And this is true for every acid-base reaction. $\endgroup$
    – Luckenberg
    Jan 14, 2023 at 16:12
  • $\begingroup$ @Luckenberg The discussion of ammonia and acetic acid is correct. They react almost to completion. CH3COOH2+ is such a good acid that it does not exist in water. In vacuum, doing mass spec on acetic acid, it might fly, though. $\endgroup$
    – Karsten
    Jan 14, 2023 at 16:17

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