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^-8M solution of HNO_3. 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:

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

and then:

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

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 $[H_3O^+]{\color{Purple} }$ coming from water self-ionization:

$H_2O + H_2O \ rightleftharpoons H_3O ^ + ^ + OH -$

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

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

Let's say that:

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

Substituting in the expression of K_w :

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

shows that $x = 9.5 \ cdot 10 ^ {- 8} M$

We can now find the total hydrogen ion concentration:

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

and therefore the pH:

$pH = -log (1.05 \ cdot 10 ^ {- 7}) = 6.98$

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 $[OH^-]_{base} + [OH^-]_{self-ionization}$ exactly according to the same logic.