# The [OH-] of an aqueous solution is 4.62 * 10^-4 M. What is the pH of this solution?

Dec 19, 2015

$\text{pH} = 10.66$

#### Explanation:

For pure water at ${25}^{\circ} \text{C}$, the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$, is equal to the concentration of hydroxide ions, ${\text{OH}}^{-}$.

More specifically, water undergoes a self-ionization reaction that results in the formation of equal concentrations of hydronium and hydroxide anions.

$2 {\text{H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

At room temperature, the self-ionization constant of water is equal to

${K}_{W} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}$

This means that neutral water at this temperature will have

["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"

As you know, pH and pOH are defined as

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

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

and have the following relationship

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

In your case, the concentration of hydroxide ions is bigger than ${10}^{- 7} \text{M}$, which tells you that you're dealing with a basic solution and that you can expect the pH of the water to be higher than $7$.

A pH equal to $7$ is characteristic of a neutral aqueous solution at room temperature.

So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first

$\text{pOH} = - \log \left(4.62 \cdot {10}^{- 4}\right) = 3.34$

This means that the solution's pH will be

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

$\text{pH} = 14 - 3.34 = \textcolor{g r e e n}{10.66}$

Indeed, the pH is higher than $7$, which confirms that you're dealing with a basic solution.