Why do we take active mass of water 1 but while calculating pKa of water as 55.345? [duplicate]

Question

First, I should mention that the question was already asked here:Why is active mass of a pure solid or liquid always taken as unity? But while calculating $\mathrm{p}K_\mathrm{a}$ of pure water we take active mass of pure at $\pu{25 ^{\circ}C}$ as 55.345 and therefore get the $\mathrm{p}K_\mathrm{a} = 15.74$.

$$K_\mathrm{a}=\frac{[\ce{OH−}][\ce{H+}]}{[\ce{H2O}]}=\frac{\pu{e−14}}{55.345}=\pu{1.807E−16}=10^{−15.74}$$

Shouldn't we take active mass of water as one the do the calculation as

$$\mathrm{p}K_\mathrm{a}=−\log{([\ce{H+}][\ce{OH−}])}$$

For $\pu{25 ^{\circ}C}$, $[\ce{H+}][\ce{OH−}]=\pu{e−14}=𝐾_\mathrm{w}$

Thus follows $\mathrm{p}K_\mathrm{a}=14$

Why do we use different active masses in the two situation?Any help would be appreciated.

A similar answer is given but why we take active mass of water as 55.345 is not discussed - What is the real pKa of water?

Answer 1

The activity of the solvent is not 1. but is just reasonably constant and the concentration is included in such equilibrium constants such as Kw, Ksp, and Ka and Kb. This works because these constants are useful in dilute solutions and the concentration of solvent does not change much. The changes are also accommodated by the incorporation of activity coefficients of ions as a function of concentration and ionic strength. The only case that I can think of where pure liquid and solid have the same activity each defined as 1. is the freezing-melting point at standard pressure or possibly the three phases at the triple point.

One of the reasons for arguments about the acidity of water, methanol, ethanol etc. is inability to consider the different solvent concentrations.