# What is the [OH-] concentration of NaOH solution with a pH of 9.4?

## (this was an example we did in class but i ddint take it down so idk how to go about doing this type of problem HELP PLEASE!!)

##### 1 Answer
Dec 16, 2016

["OH"^(-)] = 2.5 * 10^(-5)"M"

#### Explanation:

For starters, you know that an aqueous solution kept at room temperature has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH " + " pOH} = 14}}}$

so use this equation to find the pOH of the solution.

$\text{pOH} = 14 - 9.4 = 4.6$

Now, the pOH of the solution is given by the concentration of hydroxide anions, ${\text{OH}}^{-}$

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

To find the concentration of hydroxide anions starting from this equation, rewrite it as

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

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

this will get you

$\left[\text{OH"^(-)] = 10^(-"pOH}\right) \to$ remember this equation!

In your case, you have

$\left[{\text{OH}}^{-}\right] = {10}^{- 4.6}$

color(darkgreen)(ul(color(black)(["OH"^(-)] = 2.5 * 10^(-5)"M")))

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