# If the poH of a solution is 10, what is the pH of this solution? Is this solution acidic or basic?

May 14, 2016

Consider the autoionization reaction of water with itself. This is an equilibrium that is heavily favored towards water, but nevertheless, it occurs.

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

Or, this is the same thing:

$\setminus m a t h b f \left({\text{H"_2"O"(l) rightleftharpoons "H"^(+)(aq) + "OH}}^{-} \left(a q\right)\right)$

From this, we have the equilibrium constant known as the autoionization constant, ${\text{K}}_{w}$, equal to ${10}^{- 14}$. Thus, we have the following equation (remember to not use a liquid in the expression):

color(green)("K"_w = ["H"^(+)]["OH"^(-)] = 10^(-14))

where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion and $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide polyatomic ion in $\text{M}$.

Next, let's take the base-10 negative logarithm of this. Recall that -log("K"_w) = "pK"_w. We then get:

"pK"_w = 14 = -log(["H"^(+)]["OH"^(-)])

= -log(["H"^(+)]) + (-log(["OH"^(-)]))

Similar to what happened with -log("K"_w) = "pK"_w, -log(["H"^(+)]) = "pH" and -log(["OH"^(-)]) = "pOH". Thus, we have:

$\textcolor{b l u e}{\text{pK"_w = "pH" + "pOH} = 14}$

So, once you know this equation, what you get is:

color(blue)("pH") = 14 - "pOH"

$= 14 - 10$

$= \textcolor{b l u e}{4}$

Acidity is basically (pun?) when the $\text{pH}$ is less than $7$. We know that $4 < 7$, thus the solution is acidic. The overall spectrum is:

$\text{pH} < 7$ $\to$ acidic
$\text{pH} = 7$ $\to$ neutral
$\text{pH} > 7$ $\to$ basic