# Question b8408

Mar 7, 2017

["H"_3"O"^(+)] = 1 * 10^(-2)"M"

["OH"^(-)] = 1 * 10^(-12)"M"

#### Explanation:

The thing to keep in mind here is that nitric acid is a strong acid, which means that it will ionize completely in aqueous solution to produce hydronium cations, ${\text{H"_3"O}}^{+}$, and nitrate anions, ${\text{NO}}_{3}^{-}$.

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

Notice that every mole of nitric acid that is dissolved in water produces $1$ mole of hydronium cations.

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

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

As you know, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}$

This means that you will have

["OH"^(-)] = 10^(-14)/(["H"_3"O"^(+)])

which will get you

["OH"^(-)] = 10^(-14)/0.01 = 1 * 10^(-12)color(white)(.)"M"#