# Why isn't Kw = ([H+][OH-]) / [H2O], when H2O is the reactant?

For the hypothetical reaction:

$$\ce{A <=> B + C}$$

the equilibrium constant expression is written as

$$\ K= \frac{[\ce B][\ce C]}{[\ce A]}$$

So, why is $$K_\mathrm w$$ written as:

$$K_\mathrm w = [\ce{H^+}][\ce{OH^-}]$$

I would expect:

$$K_\mathrm w = \frac{[\ce{H^+}][\ce{OH^-}]}{[\ce{H_2O}]}$$

• The assumption is that the equilibrium equation is for a dilute solution so that $\ce{[H2O]}$ is a constant. – MaxW Oct 14 at 19:25
• Because the amount of $H_2O$ is large relative to the products, making $[H_2O]$ close to 1? – BSkagen Oct 14 at 19:29
• Related: What is the real pKa of water? – Loong Oct 14 at 19:45
• – Jan Oct 15 at 4:55