# Question b992c

Nov 7, 2016

["H"_3"O"^(+)] = 5.8 * 10^(-9)"M"

#### Explanation:

This one is pretty straightforward, meaning that all you have to do here is use the definition of the pH.

As you know, the pH of a solution is simply a measure of how many hydronium cations, ${\text{H"_3"O}}^{+}$, it contains. In other words, the pH of a solution is a measure of the concentration of hydronium cations,

By definition, the pH is equal to

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

Now, the problem provides you with the pH and asks for the corresponding concentration of hydronium cations.

To figure that out, rewrite the equation as

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

Use both sides of the equation as powers of $10$ to get rid of the common log term

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

This is equivalent to

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

Plug in your value to find

["H"_3"O"^(+)] = 10^(-8.24) = 5.8 * 10^(-9)"M"

TYhe answer is rounded to two sig figs because you have two decimal places for the pH value.

The $\text{M}$ stands for moles per liter. Now, does this value make sense?

As you know, a neutral solution kept at room temperature has

$\text{pH} = 7$

That value corresponds to

["H"_ 3"O"^(+)]_"neutral sol" = 1.0 * 10^(-7)"M"#

Solutions that have pH values $> 7$ are basic, meaning that the concentration of hydronium cations is lower than what is present in a neutral solution.

$\text{pH" = 8.24 > "pH} = 7$
$5.8 \cdot {10}^{- 8} \text{M" < 1.0 * 10^(-7)"M}$