Why don't we need to consider the $\ce{H+}$ ions of water when finding the $\ce{pH}$ of $0.01~\mathrm{M}$ of $\ce{HCl}$ solution?
The $\ce{pH}$ value of something is depending on the molarity of $\ce{H+}$ ions. But the teachers says the calculation is just $-\log(0.01)$, don't we need to consider the $\ce{H+}$ ions of water?
You could consider them, but what would be the result?
$0.01 M$ of $\ce{HCl}$ with $1\times10^{-7} M$ of $\ce{H^+}$ from water gives you $0.0100001 M$ of $H^+$. Find the pH of this concentration of $\ce{H^+}$ and compare with the pH you would obtain if you used $0.01 M$ directly.
You shouldn't see that the resulting pH differ significantly.
Do note, however, that the self-ionization of water technically must be taken into account. Though very often justifiable, it is an approximation to leave it out. Ask yourself, what is the pH of a $10^{-8}\ mol.L^{-1}$ solution of $\ce{HCl}$? It doesn't make sense to add a strong acid to pure water, however dilute, and obtain a basic pH! If you don't rely on the approximate formula, you will find that such a solution has a pH of about 6.98.
Here is a good source to get you going. The more proper way of solving for $\ce{[H^+]_{eq}}$ is based on the solution of a polynomial found from the combination of the charge balance equation and chemical entity balance equations along with equilibria constant definitions, and it is detailed here. Though used specifically to analyse the behaviour of a weak acid solution, the same equations can be used for the strong acid solution, with the added approximation that $\ce{[HA]_{eq}} \approx 0$ (i.e. $\ce{c_{HA}} \approx \ce{[A^{-}]}$).
