# Question 07916

Mar 30, 2016

Here's why that is the case.

#### Explanation:

That is simply how pH is defined.

You see, pH is actually a measure of concentration. More specifically, a solution's pH will tell you that solution's concentration of hydronium ions, ${\text{H"_3"O}}^{+}$.

The pH is calculated by taking the negative log base 10 of the concentration of hydronium ions

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

Since you're dealing with the negative log, solutions that have a relatively high concentration of hydronium ions, i.e. acidic solutions, will have a low pH.

Likewise, solutions that contain a relatively low concentration of hydronium ions, i.e. basic solutions, will have a high pH.

These concentrations are relative to the concentration of hydroxide anions, ${\text{OH}}^{-}$. A solution that contains equal concentrations of hydronium and hydroxide ions will be neutral.

In pure water at room temperature, based on water's self-ionization reaction, you have

color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14))color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M")color(white)(a/a)|))

This gives a pH for pure water equal to

$\text{pH"_"pure water} = - \log \left({10}^{- 7}\right) = 7$

Now, your solution has a pH of $9.73$. Since the pH is higher than that of pure water, you can say that the concentration of hydronium ions will be lower than what you get in pure water at room temperature.

Now, set up the equation for pH

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

Rearrange to get

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

To get rid of the log, you can write

10^(log(["H"_3"O"^(+)])) = 10^(-"pH")

This will be equivalent to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \left[\text{H"_3"O"^(+)] = 10^(-"pH}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now all you have to do to get the answer is plug in the value you have for the pH of the solution

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 9.73} = 1.862 \cdot {10}^{- 10}$

Rounded to two sig figs, the answer will indeed be

["H"_3"O"^(+)] = 1.9 * 10^(-10)"M"#

Notice that this solution contains a lower concentration of hydronium ions that pure water, which is why its pH is higher than $7$.