By defintion, $\mathrm pK_\mathrm a$ value is defined as $$\mathrm pK_\mathrm a = -\lg ([\ce{H2O}] K_\mathrm{eq})$$

Consider the reaction: $$\ce{HCl + H2O <=> H3O+ + Cl-}$$

The Eq constant of $\ce{HCl}$ is defined as

$$K_\mathrm{eq} = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{H2O}][\ce{HCl}]}$$

Why is molar concentration of water fixed at 55.5? Wouldn't it depend on the extent of ionization?

  • 1
    $\begingroup$ What does HCl have to do with the pKa of water? $\endgroup$ – jerepierre Mar 23 '16 at 17:08
  • $\begingroup$ @jerepierre edited to be more specific. $\endgroup$ – doodle1234 Mar 23 '16 at 18:49

For most practical purposes, we assume that the aqueous solution is dilute and that the concentration of water is constant, that is, $[\ce{H2O}] \approx 55~\mathrm{M}$. This is the standard that is used to define the ionization constant of water, $K_\mathrm{w}$.

If the equation $1\times 10^{-14}~\mathrm{M^2} = [\ce{H3O+}][\ce{OH-}]$ didn't hold (at $25~\mathrm{^\circ C}$) then we would have to come up with other equations for acid-base equilibrium. Even for strong acid/base dissociations we approximate the equilibrium equation with full dissociation (i.e. one forward arrow).

I recommend reading pp 12-15 of the following if you want a more theoretical treatment of strong acid/base dissociation where activities are used: http://www.chem1.com/acad/pdf/c1xacid2.pdf

  • $\begingroup$ Please use \ce{} for chemical formulae. (See edit) $\endgroup$ – orthocresol Mar 23 '16 at 18:41

The truth is convoluted. In reality the equilibrium "constant" isn't really a constant. The value is really a function that depends on temperature, pressure, and ionic strength of the solution. The gist is that in concentrated solutions, where the concentration of "free" water molecules isn't 55 molar, the equilibrium constant needs such a significant correction due to ionic strength that making the correction to water alone doesn't improve the accuracy of the formula.


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