Question

Question #b992c

Answer 1

`["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, `"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

`["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 `"M"` stands for 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 `>7` are 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"`