What is the "pH" of a 3.5 * 10^-3 "M" "HNO"_3 solution?

Nov 24, 2017

$\text{pH} = 2.46$

Explanation:

The important thing to keep in mind here is that nitric acid is a strong acid, which means that it ionizes completely in aqueous solution to produce hydronium cations and nitrate anions.

${\text{HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

This tells you that the concentration of the hydronium cations in the solution will be equal to that of the acid. In other words, you can expect all the moles of nitric acid present in your sample to ionize.

You can thus say that you have

["H"_3"O"^(+)] = 3.5 * 10^(-3)color(white)(.)"M"

Now, the $\text{pH}$ of the solution is simply a measure of the concentration of hydronium cations.

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

Plug in your value to find

$\text{pH} = - \log \left(3.5 \cdot {10}^{- 3}\right)$

$\text{pH} = - \log \left(3.5\right) - \log \left({10}^{- 3}\right)$

$\text{pH} = 3 \log \left(10\right) - \log \left(3.5\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 2.46}}}$

The answer is rounded to two decimal places because you have two sig figs for the concentration of the acid.