# Question 98abc

Dec 11, 2017

$2.6 \cdot {10}^{- 11}$ $\text{M}$

#### Explanation:

The trick here is to realize that at ${25}^{\circ} \text{C}$, 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 ${25}^{\circ} \text{C}$, 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, ${\text{H}}^{+}$.

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

$\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = 1.0 \cdot {10}^{- 14}$