### K_{w}

In module 9 was spoken of water as an ampholyte, so a very weak base ánd a very weak acid.

If no other substances are dissolved in water, than the pH of that water, at 25 degrees C is exactly 7 as also the pOH = 7.

With the (water)equilibrium (H_{2}O + H_{2}O H_{3}O^{+} + OH^{-}) comes of course also an equilibrium condition K:

The waterconcentration is constant, does not change.

The water concentration is about 55.5 mol/l and the dissociation practically does not change anything.

From this you can draw a relation:

K.[H_{2}O]^{2} = [H_{3}O^{+}].[OH^{-}]
K·55,5 = K_{w} = 10^{-7}·10^{-7} = 10^{-14}

Kw is called the water constant and is, at 25°C : 10^{-14}

As in all equilibria, also the water constant depends only on the temperature.

The above formulas can also be written with p-values:

pK_{w} = pH + pOH

or in numbers:

pK_{w} = 14 = 7 + 7

If the pH and the pOH equal each other, so the [H_{3}O^{+}] and the [OH^{-}] are equal,
than the solution is neutral.

Because K_{w} is a constant (so pK_{w} too) pH and pOH toghether must be 14.

example: pH = 5 and pOH = 9. [att: only at room temperature!]

If the pH becomes bigger, the pOH becomes smaller and v.v.

- adding acid makes the [H
_{3}O^{+}] larger, so the pH <.
- adding base makes the [OH
^{-}] larger, zo the pOH <.