In the previous sections we calculated the pH in solutions of strong acids and bases. In these cases, the concentration of protons or hydroxide ions coincides with the concentration of the acid or the base, and we only have to take the logarithm of said concentration to obtain the pH or pOH. But we are surprised that a $10^{-8}M$ HCl solution has a basic pH of 8. Logic makes us think that said solution should have a pH slightly below 7, although very close to this value .
The explanation for this "surprise" lies in neglecting the autoinoculation of water, since it contributes more protons to the medium than the acid itself. Calculation of pH in this situation should be done as follows.
$HCl+H_2O\rightarrow \underbrace{H_3O^+}_{10^{-8}}+ \underbrace{Cl^-}_{10^{-8}}$
$H_2O+H_2O\rightleftharpoons \underbrace{H_3O^+}_{x+10^{-8}} +\underbrace{OH^-}_{x}$
Since the concentration of protons and hydroxides must satisfy the $K_w$
\begin{equation}10^{-14}=[H_3O^+][OH^-]=(x+10^{-8})(x)\end{equation}
Solving the quadratic expression we obtain: $x=9.5x10^{-8}\;M$
$[H_3O^+]=10^{-8}+x=1.05x10^{-7}$M
pH=6.98
The pH obtained is in accordance with the weakly acid solution that we have treated.
As can be seen in the data obtained, the autoininization of water contributes to the concentration of protons almost 10 times more than acid. In these cases it makes no sense to neglect this equilibrium. This is not the case with more concentrated solutions of acid or base in which the contribution of water is negligible.