# Question 29a9b

May 27, 2017

["H"_3"O"^(+)] = 2.50 * 10^(-2)"M"

#### Explanation:

As you know, the $\text{pH}$ of a solution is given by the following equation

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

In order to find the concentration of hydronium cations that corresponds to a given $\text{pH}$, start by rewriting the equation as

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

This will be equivalent to

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

which will get you

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)}}}$

Plug in your values to find

["H"_3"O"^(+)] = 10^(-1.602) = color(darkgreen)(ul(color(black)(2.50 * 10^(-2)color(white)(.)"M")))#

The answer is rounded to three sig figs, the number of decimal places you have for the $\text{pH}$ of the solution.