# pH calculation - diluted strong acids and bases

In principle, when a strong acid or a strong base produces in solution a concentration of ions H3O+ or OH- greater than 10-6 M you can just neglect the contribution given to the concentration of these ions (and then to the pH) by the water autoionization (or self-ionization) . But what happens when the concentration of strong base or acid drops below this altitude?

Let's take an example. You want to calculate the pH of a 10-8 M solution of HNO3. Nitric acid is a strong acid (Arrhenius or Bronsted-Lowry theory) therefore completely dissociated in water.

If we calculate the pH as we have seen for strong acids and bases not particularly diluted we would have:

$[H_3O^+]&space;=&space;1\cdot10^{-8}M$

and then:

$pH&space;=&space;-log[H_3O^+]=-log[1\cdot&space;10^{-8}]&space;=&space;8$

Paradoxically, adding a strong acid to a neutral solution you would get an alkaline pH. But this is (obviously) not possible because we have a very dilute solution of strong acid!

In this case, we must take in account the$[H_3O^+]{\color{Purple}&space;}$ coming from water self-ionization:

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

Calling again with "x" the concentration of ions that comes from the self-ionization:

$x&space;=&space;[H_3O^+]_{self-ionization}&space;=&space;[OH^-]_{self-ionization}$

Let's say that:

$[H_3O^+]_{total}&space;=&space;[H_3O^+]_{HNO_3}&space;+&space;[H_3O^+]_{self-ionization}=&space;1\cdot10^{-8}&space;+&space;x$

Substituting in the expression of Kw  :

$K_w&space;=&space;[H_3O^+][OH^-]&space;=&space;(1\cdot10^{-8}&space;+&space;x)&space;\cdot&space;(x)&space;=10^{-14}$

shows that $x&space;=&space;9,5&space;\cdot&space;10^{-8}M$

We can now find the total hydrogen ion concentration:

$[H_3O^+]_{total}&space;=&space;1,0&space;\cdot&space;10^{-8}&space;+&space;9,5\cdot10^{-8}&space;=&space;1,05\cdot10^{-7}M\,$

and therefore the pH:

$pH&space;=&space;-log(1,05\cdot&space;10^{-7})&space;=&space;6,98$

As you can see, the very low concentration of our acid has very little impact on the pH, which is then only slightly below neutrality.

To calculate the pH of a very diluted strong base we can run the same type of calculation, this time finding $[OH^-]_{total}$ , which corresponds to $[OH^-]_{base}&space;+&space;[OH^-]_{self-ionization}$ exactly according to the same logic.