# How do you find pOH if [H+] is given?

Jun 17, 2016

You'll need two equations for this:

\mathbf("pH" = -log["H"^(+)])

$\setminus m a t h b f \left(\text{pH" + "pOH} = 14\right)$

So, all you need to do is take the negative base-10 logarithm of the concentration of ${\text{H}}^{+}$, and then subtract from $14$.

$- \log \left[{\text{H}}^{+}\right]$

$=$ $\text{pH}$

$\to \textcolor{b l u e}{\text{pOH" = 14 - "pH}}$

If you recall, the autoionization of water works like this:

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

Then, the equilibrium constant for water is:

$\setminus m a t h b f \left({K}_{w} = {10}^{- 14} = \left[{\text{H"^(+)]["OH}}^{-}\right]\right)$

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

So, when we take the logarithms and work with this equation, we'd get:

$- {\cancel{\log}}_{\cancel{10}} \left({\cancel{10}}^{- 14}\right)$

$= - {\log}_{10} \left(\left[{\text{H"^(+)]["OH"^(-)]) = -log(["H"^(+)]["OH}}^{-}\right]\right)$

Using the properties of logarithms, $\log x y = \log x + \log y$.

$\implies - \left(- 14\right) = 14$

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

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

But -log["H"^(+)] = "pH" and -log["OH"^(-)] = "pOH", so...

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