# What is the equilibrium constant of pure water at 25°C?

May 29, 2018

${K}_{w} = 1.0 \times {10}^{- 14}$ at 25ºC.

#### Explanation:

Sometimes, in pure water, one proton (${H}^{+}$ ion) of one water molecule will be attracted to one of the lone pairs in another water molecule.
This causes the ${H}^{+}$ ion to leave its current water molecule (leaving behind $O {H}^{-}$) to attach to the lone pair of another water molecule (forming ${H}_{3} {O}^{+}$, the hydronium ion).

This process is called the auto-ionization or self-ionization of water, since ${H}_{2} O$ is ionizing into $O {H}^{-}$ and ${H}^{+}$.

So, this reaction will sometimes occur:

${H}_{2} O + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + O {H}^{-}$

However, it's important to note that this reaction only happens very rarely. As such, the equilibrium constant, which is a measure of the concentration of products to the concentration of reactants, will be very low.

In fact, at 25ºC, this equilibrium constant will be $1.0 \times {10}^{- 14}$.