Question #29a9b

1 Answer
May 27, 2017

#["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.