# Water self ionization and PH

We know that the neutralization reaction:

$H_3O^+&space;+&space;OH^-&space;\rightarrow&space;H_2O$

It is practically quantitative. In fact, as all the chemical reactions are actually equilibrium reactions, and it will remain a very small amount of H3O+ and OH- unreacted.

$H_3O^+&space;+&space;OH^-&space;\rightleftharpoons&space;H_2O$

The same reaction, roles reversed, takes place in any aqueous solution:

$H_2O&space;+&space;H_2O&space;\rightleftharpoons&space;H_3O^+&space;+&space;OH^-$

This equilibrium is called water self ionization or water autoionization. Water can either donate or accept a proton. In the reaction of self ionization, a water molecule accepts a proton, while the other donates a proton. Considering that, as we said, the neutralization reaction H3O+/OH- is virtually quantitative, the self-ionization equilibrium is shifted to the left.

Indeed it is so, but there's always a minimum concentration of H3O+ and OH- in solution. This is experimentally demonstrated with electrical conductivity experiments.

The constant auto-ionization of water: Kw

What is really important is that we are also able to calculate the equilibrium constant of this reaction.

$K&space;=&space;\frac{[H_3O^+][OH^-]}{[H_2O]^2}$

Since [H 2 O] can be considered constant and equal to about 55.6 M, you can incorporate in the equilibrium constant.
$K\cdot[H_2O]^2&space;=&space;K_w&space;=&space;[H_3O^+][OH^-]&space;=&space;1&space;\cdot10^{-14}$

The value of the equilibrium constant depends only on the temperature, and this is the value for the temperature of 25 ° C.

This expression is very important. It tells us that in any aqueous solution are H 3 O + and OH -. But since H 3 O + and OH - are respectively the acid and the strongest base that can exist in aqueous solutions, we can also deduce that:

When $[H_3O^+]&space;>&space;[OH^-]$ → the solution is acidic

When $[H_3O^+]&space;<&space;[OH^-]$ → the solution is basic

When $[H_3O^+]&space;=&space;[OH^-]$ → the solution is neutral

Moreover, if the solution is neutral you can calculate that semplcemente $[H_3O^+]&space;=&space;[OH^-]&space;=&space;1\cdot10^{-7}$ , For any volume of solution.

For example:

I a 0.1 M aqueous solution of hydrochloric acid.

$HCl&space;+&space;H_2O&space;\rightleftharpoons&space;H_3O^+&space;+&space;Cl^-$

H 3 O +, which is derived from the balance of waterself dissociation will be irrelevant compared to what comes from HCl (being completely dissociated H 3 O + will 0.1M).

Since Kw is a constant and must remain so, the concentration of OH - is calculated as:

$K_w&space;=&space;[H_3O^+][OH^-]$ then $[OH^-]&space;=\frac{&space;K_w}{[H_3O^+]}$ replacing or $[OH^-]&space;=\frac{&space;1\cdot&space;10^{-14}}{10^{-1}}&space;=&space;10^{-13}$

It then observes, that when in solution increases the concentration of H 3 O +, that of OH - must necessarily decrease.