Question #29a9b

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Answer 1 · 2017-05-27T00:03:02

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

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.