# What is the pH of a 0.0605 M solution of sodium hydroxide, NaOH?

Jun 8, 2016

$\text{pH} = 12.78$

#### Explanation:

You're dealing with a solution that contains a strong base, so right from the start you should know that sodium hydroxide dissociates completely in aqueous solution to produce sodium cations, ${\text{Na}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$. This means that for every mole of sodium hydroxide that you're dissolving in water, you will get one mole of hydroxide anions in aqueous solution.

Therefore, for a given volume of your solution, the concentration of the hydroxide anions will be equal to that of the strong base

["OH"^(-)] = ["NaOH"] = "0.0605 M"

Now that you know the concentration of hydroxide anions, you can calculate the pOH of the solution by using

color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))

Plug in your value to find

$\text{pOH} = - \log \left(0.0605\right) = 1.22$

You know that for aqueous solutions at room temperature you have

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{pOH " + " pH} = 14 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in the pOH of the solution to get

$\text{pH} = 14 - 1.22 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{12.78} \textcolor{w h i t e}{\frac{a}{a}} |}}}$