# Can Acids have pH of more than 7? [duplicate]

Consider a solution of $$\text{HCl}$$ with a concentration of $$10^{-10}$$ $$\text{M}$$. Now, if I find it's pH: $$\text{pH} = -\log([\text{H}^+])$$ $$\text{pH} = -\log(10^{-10}) = 10$$ 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?

• 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. – andselisk May 31 at 7:54
• 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? – Ivan Neretin May 31 at 8:24
• @andselisk right answer, but to another question ;-) – Karl May 31 at 8:24
• In one word: Autoprotolyis – Karl May 31 at 8:26
• ^ That said, I think this is a useful duplicate for search, concisely worded. – M.A.R. May 31 at 8:42