# Question 5fd98

Jan 10, 2017

["OH"^(-)] = 2.24 * 10^(-5)"M"

#### Explanation:

Your starting point here will be the definition of a solution's $\text{pOH}$, which as you know is defined as the negative log base $10$ of the concentration of hydroxide anions, ${\text{OH}}^{-}$

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))

Now, in order to solve for the concentration of hydroxide anions, you must rearrange this equation as

log(["OH"^(-)]) = - "pOH"

and rewrite it using exponents of base $10$

10^log(["OH"^(-)]) = 10^(-"pOH")

This is equivalent to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[\text{OH"^(-)] = 10^(-"pOH}\right)}}}$

Now all you have to do is to plug in the value given to you for the $\text{pOH}$ of the solution

["OH"^(-)] = 10^(-4.65) = color(darkgreen)(ul(color(black)(2.24 * 10^(-5)"M")))#

I'll leave the answer rounded to three sig figs.