# The pH of a 2M solution of NaOH is...? (please include the complete solution of the problem)

May 30, 2017

$\text{pH} = 14.3$

#### Explanation:

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

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

This means that you express the $\text{pH}$ of the solution in terms of its $\text{pOH}$ by writing

$\text{pH" = 14 - "pOH}$

Now, the $\text{pOH}$ of the solution is defined as the negative log base $10$ of the concentration of hydroxide anions

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

This means that the $\text{pH}$ of the solution can be expressed as

"pH" = 14 - [- log(["OH"^(-)])]

"pH" = 14 + log(["OH"^(-)])" "color(darkorange)("(*)")

As you know, sodium hydroxide is a strong base, which implies that it ionizes completely in aqueous solution to produce sodium cations and hydroxide anions.

You can thus say that

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

Since every $1$ mole of sodium hydroxide that is dissolved in water produces $1$ mole of hydroxide anions, your solution will have

$\left[\text{OH"^(-)] = ["NaOH}\right]$

In your case, this is equal to

["OH"^(-)] = "2 M"

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ and calculate the $\text{pH}$ of the solution

$\text{pH} = 14 + \log \left(2\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{14.3}}}$

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the solution.

As a final note, don't get confused by the fact that you have

$\text{pH} > 14$

This is what you'll always get for basic solutions that have

$\left[{\text{OH}}^{-}\right] > 1$

and it means that $\text{pOH} < 0$.