Question

Question #29a9b

Answer 1

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

Explanation:

As you know, the `"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 `"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

`color(blue)(ul(color(black)(["H"_3"O"^(+)] = 10^(-"pH"))))`

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 `"pH"` of the solution.