# Question 44d2f

Dec 27, 2017

For all intents and purposes, $\text{pH} = 7.0$.

#### 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.

${\text{H"_ 2"O"_ ((l)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

At ${25}^{\circ} \text{C}$, these concentrations are equal to

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

This is why pure water at ${25}^{\circ} \text{C}$ is said to be neutral with a $\text{pH}$ equal to $7$.

Consequently, you can say that pure water at ${25}^{\circ} \text{C}$ 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 $x$ $\text{M}$ 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

${x}^{2} + 1 \cdot {10}^{- 11} x - 1 \cdot {10}^{- 14} = 0$

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

$x = 9.9995 \cdot {10}^{- 8}$

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"^(+)])#

$\text{pH} = - \log \left(1.00005 \cdot {10}^{- 7}\right)$

$\text{pH} = 6.999978$

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 $\text{pH}$ of the solution will be

$\text{pH} = 6.999978 \approx 7.0$

Keep in mind that the $\text{pH}$ of an acidic solution, no matter how dilute, can never be $> 7$ at ${25}^{\circ} \text{C}$.