This answer presents a derivation of the value of ionic product of water at $25^{\circ}\text{C}$.

The relation $K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G^{\circ}}{RT}}$ is used for the derivation.

But why is the equilibrium constant $\mathrm{K_w=[H^+][OH^-]}$ and not $\mathrm{K_d=\frac{[H^+][OH^-]}{[H_2O]}}$ or $\mathrm{K_i=\frac{[H_3O^+][OH^-]}{[H_2O]^2}}$?

After all, the reaction $\ce{H2O <=> H+ + OH-}$ is used for calculating $\Delta_\text{r}G^{\circ}$, so intuitively it appears that $\mathrm{K_d}$ should be used.
How can using any of the equations be equivalent when we have specified which reaction we are using while calculating $\Delta_\text{r}G^{\circ}$? What am I missing?

  • 3
    $\begingroup$ There is this habit to implicitly involve [H2O] in the constant, assuming diluted solutions. It obviously fails for concentrated solutions, but then concentrations fail as well, not following activities any more. $\endgroup$
    – Poutnik
    Commented Jan 28 at 12:14
  • $\begingroup$ Yes, but how is this assumption accounted for in $\mathrm{\Delta_r G ^{\circ}}$? I am not aware a similar analogue of this assumption for free energy. $\endgroup$
    – Dodo
    Commented Jan 28 at 12:25
  • 2
    $\begingroup$ These constants are not true thermodynamic constants. But ones using explicit concentrations are not either. $\endgroup$
    – Poutnik
    Commented Jan 28 at 12:47
  • $\begingroup$ @Poutnik Could you please elaborate by answering this question? I don't understand. $\endgroup$
    – Dodo
    Commented Jan 29 at 21:19
  • 1
    $\begingroup$ From a strictly mathematical sense, all those equations are equivalent for pure water. Now, which one you decide to label as the autodissociation constant (as in, preferentially over the others) is another matter (and one which is largely less interesting). Don't get too worked up over labels, nature doesn't care what we call things. $\endgroup$ Commented Jan 30 at 12:17

1 Answer 1


The thermodynamic constant of water dissociation can be written as$$K_\mathrm{D}=\dfrac{a(\ce{H3O+(aq)}) \cdot a(\ce{OH-(aq)})}{\ce{(a(H2O(l)}))^2} \approx \\ \approx \dfrac{a(\ce{H3O+(aq)}) \cdot a(\ce{OH-(aq)})}{1^2} \approx \dfrac{[\ce{H3O+(aq)}][\ce{OH-(aq)}]}{1^2}=[\ce{H+}][\ce{OH-}]= K_\text{w} $$

$a$ as the thermodynamic activity is defined as

$$\mu = \mu^{\circ} + RT \ln a$$

where $\mu$ resp. $\mu^{\circ}$ is chemical potential defined as

$$\mu = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_j, j \ne i}$$

The standard chemical potential is defined for pure water, therefore activity of the pure water $a=1$.
OTOH the standard chemical potential of ions is defined as extrapolation from very diluted solutions to concentration $\pu{1 mol L-1}$, putting $$\lim_{c \to 0}{(a)} = c$$

  • 1
    $\begingroup$ minor correction - denominator should not be squared since you used H+ in numerator rather than H3O+, $\endgroup$
    – Andrew
    Commented Jan 30 at 14:10
  • $\begingroup$ @Andrew Thank you. Well, $\ce{H+(aq)}$ implies $\ce{H3O+(aq)}$. Just because we are formally used to H+ or H+(aq), it does not mean we do not need 2 H2O for it. $\endgroup$
    – Poutnik
    Commented Jan 30 at 14:21

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