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


1 Answer 1


There are two possible questions here, which have very different answers.

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.

  • 1
    $\begingroup$ I believe the teacher had in mind the total concentration, for strong and many weak acids equal at given dilution to concentration of the conjugate base. $\endgroup$
    – Poutnik
    Commented Nov 17, 2023 at 9:09

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