# Question #b992c

##### 1 Answer

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

This is equivalent to

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

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 *moles per liter*. Now, does this value make sense?

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

#"pH" = 7#

That value corresponds to

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

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

In your case,

#"pH" = 8.24 > "pH" = 7#

and

#5.8 * 10^(-8)"M" < 1.0 * 10^(-7)"M"#