# Conversion of Kc into Kw

In deriving the expression for $K_\mathrm w$, we often start with the equilibrium equation of water $$\ce{H2O <=> H+ + OH-}$$

The $\rm K_c$ expression is quoted as $$K_\mathrm c = \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]}$$ followed on by multiplying the $[\ce{H2O}]$ by $K_\mathrm c$.

The problem I face is I have been taught that the pure liquids aren't included in the expression of $K_\mathrm c$ since they have a concentration of 1. So why are we including in the above example?

• It's really a faulty derivation. This page gives a good explanation why. – Tyberius Apr 18 '17 at 18:19

"The problem I face is I have been taught that the pure liquids aren't included in the expression of $\mathrm{K}_{c}$ since they have a concentration of 1."

Liquids have a constant concentration (really activity), so in general, it is not helpful to include them in the equilibrium expression. However, that does not mean that the value is 1. Pure water has a concentration of roughly $55.5\ \mathrm{M}$. You can absorb this concentration into the equilibrium constant, since it is a constant. Therefore,

$$\mathrm{K}_{w} = \ce{[H+][OH-]} = K_{\mathrm{c}}\ce{[H2O]}$$

EDIT:

Just for the record, note that the autoionization of water is generally written:

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

since we don't have naked protons in water.

This changes the equilibrium constant slightly:

$$\mathrm{K} = \frac{\ce{[H3O+][OH-]}}{\ce{[H2O]}^{2}}$$

but $\mathrm{K}_{w}$ is still the same.