Question

Question #98abc

Answer 1

`2.6 * 10^(-11)` `"M"`

Explanation:

The trick here is to realize that at `25^@"C"`, an aqueous solution has

`color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1.0 * 10^(-14)color(white)(.)"M"^2)))`

This means that for an aqueous solution at `25^@"C"`, you will have

`["OH"^(-)] = (1.0 * 10^(-14)color(white)(.)"M"^2)/(["H"_3"O"^(+)])`

So all you have to do here is to plug in your value and find the concentration of the hydroxide anions.

`["OH"^(-)] = (1.0 * 10^(-14)color(white)(.)"M"^color(red)(cancel(color(black)(2))))/(3.9 * 10^(-4)color(red)(cancel(color(black)("M")))) = 2.6 * 10^(-11)color(white)(.)"M"`

The answer is rounded to two sig figs, the number of sig figs you have for the concentration of hydronium cations, which, in your case, are given as hydrogen ions, `"H"^(+)`.

Keep in mind that you will oftentimes see this equation written without added units.

`["H"_3"O"^(+)] * ["OH"^(-)] = 1.0 * 10^(-14)`