Question #b8408

1 Answer
Mar 7, 2017

Answer:

#["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"#