Predict who is the acid and who is the base in an acid-base reaction using pKa

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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> 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.
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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.



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> [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
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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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