Why is water not part of the equilibrium constant?

Question

In the acid base reaction $$\ce{NH4+ + H2O <-> NH3 + H3O+}$$

the acidity constant, which is a relation between concentrations is given by $$K_{\mathrm{a}}=\frac {\ce{[NH3]} \cdot \ce{[H3O+]}}{\ce{[NH4+]} \cdot \ce{[H2O]}}$$

Since $\ce{[H2O]}$ is constant, it can be included inside the constant to obtain a new $K_\mathrm{a}$ which is $6.3 \times 10^{-10}$.

(Being a constant allows us to relate the molar concentration to the density of water.)

And so, even though some quantity of water reacts with $\ce{NH4+}$, but the volume of water varies in a way that $\ce{[H2O]}$ would still be constant.

What I don't understand is why the molar concentration of water is constant. Why isn't it handled the same way as the other reactant? I understand that if $\ce{H2O}$ was the only liquid and the rest of the reactants and products were of different state (gaseous or solid), then water would exist alone in the beaker and its concentration would therefore remain constant, but this is not the case.

That was about heterogeneousness. Trying another guess; in the reaction of esterification $$\ce{{alcohol} + {carboxylic acid} <-> {ester} + H2O}$$ $\ce{H2O}$ is considered as a product which concentration is not constant, so it is counted explicitly in the equilibrium constant.

Could the answer be that $\ce{NH4+}$, $\ce{NH3}$ and $\ce{H3O+}$ are dissolved in water, while the alcohol, the carboxylic acid, and the ester are not?

I'm looking for an answer that clearly shows which volume is considered (water or the solution), because I'm confused as to why something would or wouldn't be included in the equilibrium constant.

Answer 1

There is no IUPAC definition of $K_a$, however, many reputable texts like Levine's Physical Chemistry define $K_a$ as:

$$\frac {a(\ce{A-})a(\ce{H3O+})}{ a(\ce{HA})a(\ce{H2O})}$$

where "a(X)" is the activity of species "X".

Then various approximations can be made, such as approximating $a(\ce{H2O}) = 1$ and approximating the activity of the solutes as the concentration of the solutes.

These approximations are only reasonable in dilute solutions.

Answer 2

The concentration of water in water (sounds weird but this is just liquid water) is found as follows:$$[\ce{H2O}]= \mathrm{\frac{1.0\frac{g}{cm^3}}{18.0148\frac{g}{mol}}}=55.5\mathrm{\frac{mol}{L}}$$

Compare this with the usual concentrations of reactants and products which is much smaller than that. Thus, the change in concentration of water throughout reactions which take place in water is considered to be constant.

The above is a very good approximation. While it is true that water's density varies some with temperature, at $98~\mathrm{^\circ C}$ the density of water is still $0.96\mathrm{\frac{g}{cm^3}}$, so the concentration of water will still be much much more than that of any of the reactants.

So, as you speculated, whether or not water is considered in the equilibrium constant entirely depends on whether or not water is a product and if water is the solvent.

If water is the solvent, it would be very normal to ignore the water which is produced or the water which might act as a reactant.

Similarly, any reaction which takes place by solvolysis will exclude the solvent from the equilibrium constant.

Also, you say that if water were alone in a beaker, then obviously all you have is water so the concentration is constant. Consider that water auto-ionizes by:$$\ce{2H2O <=> H3O+ + OH-}$$In this case, even though we know that we don't have just water, we know that the concentration of the other things is tiny compared to the concentration of water, so we just ignore the other stuff and drink the water.