Calculate the $\mathrm{pH}$ of pure water in which $\ce{HCl}$ was added. The final concentration of $\ce{HCl}$ is $\pu{10^{-6}M}$.

So what I did was to add the concentration of $\ce{H3O+}$ in pure water ($1.0\times10^{-7}$) to the concentration of $\ce{HCl}$ and calculating the $\mathrm{pH}$. But I got to $6.0$, which I don't feel is the correct answer. According to me, it should be lower right. I need help in this.


You are correct. The $\mathrm{pH}$ should be around six. The actual value can be found by calculating $-\log(1 \times 10^{-7} + 1\times 10^{-6})$, which gives me a final $\mathrm{pH}$ of 5.96. While you are correct that the $\mathrm{pH}$ of a strong acid solution tends to be significantly below 7, the concentration of $\ce{HCl}$ in your solution is so small that it does not have a very large effect on the overall $\mathrm{pH}$. By convention, you consider the pH of a solution to be determined by the largest source of hydrogen ions, meaning that the $\mathrm{pH}$ of this solution should be exactly six. Despite this, $\mathrm{pH}$ is actually a measure of the total concentration of $\ce{H+}$, so in this case, since the concentration provided by the acid is so small, both the hydrogen ions from the dissociation of water and those from the acid should be considered.

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    $\begingroup$ The question says the final concentration is $10^{-6}$. That will dominate the equilibrium in any case. Neglecting the deviation of the activity from the concentration, which is quite feasible with this low concentration, the pH becomes by definition 6. $\endgroup$ – Martin - マーチン May 18 '16 at 15:35

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