# How do I find pH of an extremely low molarity strong acid? [duplicate]

Given $$[HNO_{3}] = 4.53(10)^{-10},$$ find the pH of the acid. $$K_{a}$$ is not given, and it is at 25 degrees celsius.

Because the molarity is very low, as in less than 10^-6, the teacher tells us that the hydronium ions from water affect the pH, and so the pH is not exactly $$-log(4.53(10)^{10})$$. How do I calculate the pH then?

I remember the teacher creating a reaction based off of nitric acid and water, and that because nitric acid is a strong acid, that all the HNO3 would dissociate leading to [H3O+] = 4.53e-10, and then I tried putting this answer into the standard autoionization of water reaction (as 25 celsius), so $$10^{-14} = (4.53e-10 + x)x$$ solve for x, then pH would be -log(4.53e-10+x), but this creates an imaginary answer

Question 1 is what you actually posed here: What is the pH of a solution of nitric acid where $$\ce{[HNO3]} = 4.53 \times \pu{10^{-10} M}$$? The problem with question 1 is that it really cannot be solved without a pKa for nitric acid. To answer it, you have to know how much nitric acid to add to a solution so that $$4.53 \times \pu{10^{-10} M}$$ of the acid remains undissociated, and the answer is very sensitive to the pKa.
Question 2 is: what happens if $$4.53 \times \pu{10^{-10} mol}$$ of $$\ce{HNO3}$$ is dissolved in water? This question can be solved (approximately) even if all you know is that nitric acid is strong. The rough argument goes like this: Pure water has $$\ce{[H+]} = \pu{10^{-7} M}$$. Even if the acid fully dissociates, it will add only $$\pu{4.53 \times 10^{-10} M}$$ more protons, which is negligible (only 0.45% of the concentration already present) so the pH is still roughly 7. Feel free to do the calculation more precisely than that.
The problem is, in the resulting solution, $$[\ce{HNO3}]$$ is not $$\pu{4.53 \times 10^{-10}}$$, but actually much smaller, because nearly all of the acid has dissociated. So question 2 is actually very different from the question you've stated. That said, it might be the question your teacher intended to ask.