# Question #5fd98

##### 1 Answer

#### Explanation:

Your starting point here will be the *definition* of a solution's **negative log base**

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

This is equivalent to

#color(blue)(ul(color(black)(["OH"^(-)] = 10^(-"pOH"))))#

Now all you have to do is to plug in the value given to you for the

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

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