Question

Question #44d2f

Answer 1

For all intents and purposes, `"pH" = 7.0`.

Explanation:

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"`.