Why can the equilibrium reaction which exists in water be simply represented as

$$\ce{H2O <=> OH- + H+}$$

rather than

$$\ce{2 H2O <=> OH- + H3O+}$$

Doesn't this change both the equilibrium constant AND the enthalpy change?

Why would this still give the correct answer when you use the ionic product of water in solution?

  • 1
    $\begingroup$ There is no such thing as bare $\ce{H+}$ in water. When anybody talks about it, they actually mean $\ce{H3O+}$. That's why there is no difference. $\endgroup$ Aug 16, 2016 at 17:00

1 Answer 1


The former equation is just a simplified version of the latter.

The important factors are the concentrations (activities) of the positive and negative species.

However, since these values are small ($10^{-7}\,\mathrm{mol\cdot l^{-1}}$) compared to the concentration of water ($55,56\,\mathrm{mol\cdot l^{-1}}$) this distinction would not affect the value of the dissociation constant either.

  • $\begingroup$ It is said that water is incorporated in the equilibrium constant. [H2O] will DEFINITELY have a different value from [H2O]^2... $\endgroup$ Aug 17, 2016 at 4:05
  • $\begingroup$ I should probably delete my last sentence or stick to activities rather than concentrations. Since the activity of pure condensed phases is 1 per definition, and thus $[\ce{H2O}] = 1$, it would become clearer. I'll change that... en.wikipedia.org/wiki/Thermodynamic_activity $\endgroup$
    – aventurin
    Aug 17, 2016 at 9:20

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