# What is the pH of a 0.01 M solution of the strong acid HNO_3 in water?

Jan 4, 2016

$\text{pH} = 2$

#### Explanation:

The important thing to keep in mind when dealing with strong acids is that they dissociate completely in aqueous solution to form hydronium ions, ${\text{H"_3"O}}^{+}$, and their conjugate base.

Complete dissociation means that every molecule of acid will ionize, i.e. donate its acidic proton to a water molecule.

That tells you that a strong monoprotic acid will produce hydronium ions in a $1 : 1$ mole ratio, that is, every mole of acid will produce one mole of hydronium ions. In this case, nitric acid will dissociate to form

${\text{HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

This means that the concentration of hydronium ions will be equal to that of the nitric acid

["H"_3"O"^(+)] = ["HNO"_3] = "0.01 M"

As you know, a solution's pH is simply a measure of its concentration of hydronium ions

color(blue)("pH" = - log( ["H"_3"O"^(+)]))

In this case, the pH of the solution will be

$\text{pH} = - \log \left(0.01\right) = \textcolor{g r e e n}{2}$