# How do you calculate the pOH of HCl?

May 11, 2016

$p O H = 14 - p H$

#### Explanation:

You can simply find the $p H$ from the concentration of ${H}^{+}$ which is equal to:

$\left[{H}^{+}\right] = {\left[H C l\right]}_{0}$

Then, $p H = - \log \left[{H}^{+}\right]$

From the water dissociation constant ${k}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = 1.0 \times {10}^{- 14}$ we can take the $- \log$ of both parties, we get:

$- \log {k}_{w} = - \log \left[{H}^{+}\right] - \log \left[O {H}^{-}\right] = 14$

knowing that $p H = - \log \left[{H}^{+}\right]$ and $p O H = - \log \left[O {H}^{-}\right]$, therefore,

$p H + p O H = 14 \implies p O H = 14 - p H$

Here is a video that further explains this topic:
Acids & Bases | Nature, Strength & pH Scale.