Question #44d2f

The idea here is that pure water already contains equal concentrations of hydronium cations and hydroxide anions produced by the auto-ionization of water.

#"H"_ 2"O"_ ((l)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

At #25^@"C"#, these concentrations are equal to

#["H"_3"O"^(+)] = ["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M"#

This is why pure water at #25^@"C"# is said to be neutral with a #"pH"# equal to #7#.

Consequently, you can say that pure water at #25^@"C"# has

#["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)color(white)(.)"M"^2#

Now, hydrochloric acid is a strong acid that ionizes completely in aqueous solution to produce hydronium cations in a #1:1# mole ratio.

This means that you have

#["H"_3"O"^(+)] = ["HCl"] = 1 * 10^(-11)color(white)(.)"M"#

Now, keep in mind that the concentration of hydronium actions you add to the solution will impact the auto-ionization of water. Assuming that the auto-ionization reaction produces #x# #"M"# of hydronium cations and of hydroxide anions in this solution, you will have

#overbrace((x + 1 * 10^(-11)color(white)(.)color(red)(cancel(color(black)("M")))))^(color(blue)("total concentration of H"_3"O"^(+)]) * overbrace(xcolor(white)(.)color(red)(cancel(color(black)("M"))))^(color(blue)("concentration of OH"^(-)]) = 1 * 10^(-14)color(white)(.)color(red)(cancel(color(black)("M"^2)))#

Rearrange to quadratic equation form to get

#x^2 + 1 * 10^(-11)x - 1 * 10^(-14) = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since #x# represents concentration, you can discard the negative one to get

#x = 9.9995 * 10^(-8)#

This means that your solution has--remember, you need to add the concentration of the hydronium cations that you get from the auto-ionization of water and the concentration of hydronium cations that you add to the solution from the ionization of hydrochloric acid!

#["H"_3"O"^(+)] = 9.9995 * 10^(-8)color(white)(.)"M" + 1 * 10^(-11)color(white)(.)"M"#

#["H"_3"O"^(+)] = 1.00005 * 10^(-7)color(white)(.)"M"#

You can thus say that your solution has

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

#"pH" = - log(1.00005 * 10^(-7))#

#"pH" =6.999978#

For all intents and purposes, an especially given the number of sig figs that you have for the concentration of the hydrochloric acid--one significant figure means one decimal place for the answer--you can say that the #"pH"# of the solution will be

#"pH" = 6.999978 ~~ 7.0#

Keep in mind that the #"pH"# of an acidic solution, no matter how dilute, can never be #>7# at #25^@"C"#.