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.
