# Question 46084

Apr 13, 2016

["H"_3"O"^(+)] = 5.6 * 10^(-8)"M"

#### Explanation:

As you know, a solution's pH is a measure of the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, found in that solution. More specifically, a solution's pH is equal to

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

In order to find the concentration of the hydronium cations when given the pH of the solution, use the fact that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{10}^{\log} \left(a\right) = a} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange the equation that defines the pH to get

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

This will be equivalent to

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

Finally, rearrange to find

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

Plug in the value given to you for the pH of the solution to get

["H"_3"O"^(+)] = 10^(-7.25) = color(green)(|bar(ul(color(white)(a/a)5.6 * 10^(-8)"M"color(white)(a/a)|)))

It's worth noting that a neutral aqueous solution at room temperature has equal concentrations of hydronium and hydroxide ions

["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"#

As you can see, decreasing the concentration of hydronium ions results in an increase in pH. As a result, the solution goes from being neutral to being slightly basic.