# What is the pH of a "0.150-M" "NaOH" solution?

## The answer is 13.18 but I'm not sure how to get that answer? From what I know the formula is pH = -log[H+] so I just put -log(.150) which = .82 but that's not the correct answer. Can someone please correct me and show steps?

Dec 6, 2017

$13.176$

#### Explanation:

The $\text{pH}$ of the solution is given by

color(blue)(ul(color(black)("pH" = - log (["H"_3"O"^(+)]))))

so you can't use

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\text{pH} = - \log \left(0.150\right)}}}$

because that's the concentration of the hydroxide anions, ${\text{OH}}^{-}$, not of the hydronium cations, ${\text{H"_3"O}}^{+}$. In essence, you calculated the $\text{pOH}$ of the solution, not its $\text{pH}$.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a $1 : 1$ mole ratio.

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

["OH"^(-)] = ["NaOH"] = "0.150 M"

Now, the $\text{pOH}$ of the solution can be calculated by using

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

$\text{pOH} = - \log \left(0.150\right) = 0.824$

Now, an aqueous solution at ${25}^{\circ} \text{C}$ has

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

This means that you have

$\text{pH} = 14 - 0.824 = 13.176$

You should leave the answer rounded to three decimal places because you have three sig figs for the concentration of the solution.