Question

Question #b8408

Answer 1

`["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, `"H"_3"O"^(+)`, and nitrate anions, `"NO"_3^(-)`.

`"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-)`

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^@"C"` has

`["H"_3"O"^(+)] * ["OH"^(-)] = 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"`