What is the pH of a 1.4 * 10^-2 M NaOH solution?

Dec 28, 2015

$\text{pH} = 12.15$

Explanation:

Even before doing any calculations, you can say that since you're dealing with a strong base, the pH of the solution must be higher than $7$.

The higher the concentration of the base, the higher the pH will be.

In your case, you're dealing with a solution of sodium hydroxide, $\text{NaOH}$, a strong base that dissociates completely in aqueous solution to form sodium cations, ${\text{Na}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$

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

Notice that the salt dissociates in a $1 : 1$ mole ratio with the hydroxide anions, you you can say that

["OH"^(-)] = ["NaOH"] = 1.4 * 10^(-2)"M"

Now, the pH of the solution is determined by the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$. For aqueous solutions, the concentration of hydronium ions is related to the concentration of hydroxide ions by the ion product constant of water, ${K}_{W}$

${K}_{W} = \left[{\text{OH"^(-)] * ["H"_3"O}}^{+}\right]$

At room temperature, you have

${K}_{W} = {10}^{- 14}$

This means that the concentration of hydronium ions can be determined by using

["H"_3"O"^(+)] = K_W/(["OH"^(-)])

Plug in your values to get

["H"_3"O"^(+)] = 10^(-14)/(1.4 * 10^(-2)) = 7.14 * 10^(-13)"M"

The pH of the solution is equal to

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

$\text{pH} = - \log \left(7.14 \cdot {10}^{- 13}\right) = \textcolor{g r e e n}{12.15}$

As predicted, the pH is not only higher than $7$, but it is significantly higher than $7$.

Alternatively, you can use the pOH of the solution to find its pH. As you know,

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

$\text{pOH} = - \log \left(1.4 \cdot {10}^{- 2}\right) = 1.85$
$\textcolor{b l u e}{\text{pH" + "pOH} = 14}$
$\text{pH} = 14 - 1.85 = \textcolor{g r e e n}{12.15}$