# How do you change pOH to pH?

Jun 1, 2016

Remember this:

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

I'll tell you how to derive this below.

$\text{pH}$ indirectly tells you the concentration of hydrogen ions (${\text{H}}^{+}$) in solution:

color(green)("pH" = -log["H"^(+)])

where $\left[\text{X}\right]$ is the concentration of $\text{X}$ in $\text{M}$.

$\text{pOH}$ tells you something similar, but for hydroxide ions instead:

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

These can be shown to relate if you recall that water slightly ionizes according to the following equilibrium reaction:

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

For this, we have the equilibrium constant for the autoionization of water

${K}_{w} = \left[{\text{H"^(+)]["OH}}^{-}\right] = {10}^{- 14} ,$

which heavily favors the production of water, since ${10}^{- 14}$ is very, very small.

Now, let's try taking the negative (base 10) logarithm of both sides.

$- \log \left({K}_{w}\right) = - \log \left(\left[{\text{H"^(+)]["OH}}^{-}\right]\right) = 14$

But $- \log \left({K}_{w}\right) = \text{p} {K}_{w} = 14$, so what we have, using the additive argument properties of logarithms and our definitions of $\text{pH}$ and $\text{pOH}$ above, is

$\setminus m a t h b f \left(\textcolor{b l u e}{14}\right) = - \log \left(\left[{\text{H"^(+)]["OH}}^{-}\right]\right)$

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

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

That tells us that:

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

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