Consider a solution of $\text{HCl}$ with a concentration of $10^{-10}$ $\text{M}$. Now, if I find it's pH: $$\begin{aligned}\text{pH} &= -\log([\text{H}^+])\\ \text{pH} &= -\log(10^{-10}) = 10\end{aligned}$$ At room temperature, $\text{pH} + \text{pOH} = 14 \implies \text{pOH}$ of the given acid is $4$. This means that the $[\text{OH}^-]$ ions is $10^{-4}$ $\text{M}$. But from where do these $\text{OH}^-$ ions come?

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    $\begingroup$ At this dilution rates you can no longer get away with using concentration instead of activity, and you have to calculate $\mathrm{pH}$ as $\mathrm{pH} = -\log a(\ce{H+}).$ Note that $\mathrm{pH}$ also heavily relies on solvent/medium. $\endgroup$
    – andselisk
    May 31, 2019 at 7:54
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    $\begingroup$ I object. At dilutions this great, you can more than ever rely on using concentration instead of activity. Now to the point. What is the pH of pure water? $\endgroup$ May 31, 2019 at 8:24
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    $\begingroup$ @andselisk right answer, but to another question ;-) $\endgroup$
    – Karl
    May 31, 2019 at 8:24
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    $\begingroup$ In one word: Autoprotolyis $\endgroup$
    – Karl
    May 31, 2019 at 8:26
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    $\begingroup$ In acid base reactions in water there is always the reaction $\ce{H2O <=> H^+ + OH^-}$ occurring and this cannot be ignored. The equilibrium constant $K_w=10^{14}$ so the $\ce{[H^+]}=10^{-7}$ from this reaction far exceeds that from $10^{-10} $ molar acid. see answer chemistry.stackexchange.com/questions/100346/… $\endgroup$
    – porphyrin
    May 31, 2019 at 10:32