# How to you convert between pH, pOH, [H+] and [OH-]?

Jun 1, 2016

$\left[{H}^{+}\right] \to p H = - \log \left[{H}^{+}\right] = p H$
$\left[O {H}^{-}\right] \to p H = - \log \left[O {H}^{-}\right] = p O H$
$p H \to \left[{H}^{+}\right] = {10}^{- p H} = \left[{H}^{+}\right]$
$p O H \to \left[O {H}^{-}\right] = {10}^{- p O H} = \left[O {H}^{-}\right]$
$14 - p H = p O H$
$14 - p O H = p H$

#### Explanation:

$\left[{H}^{+}\right] \to p H = - \log \left[{H}^{+}\right] = p H$
$\left[O {H}^{-}\right] \to p H = - \log \left[O {H}^{-}\right] = p O H$
$p H \to \left[{H}^{+}\right] = {10}^{- p H} = \left[{H}^{+}\right]$
$p O H \to \left[O {H}^{-}\right] = {10}^{- p O H} = \left[O {H}^{-}\right]$
$14 - p H = p O H$
$14 - p O H = p H$

There are many steps involved.

Here is a video that breaks down the process.

There are a few key things to remember...

#### Explanation:

pH + pOH = 14

pH = -log[${H}^{+}$] AND pOH = -log[$O {H}^{-}$]

$\left[{H}^{+}\right] = {10}^{- p H}$

$\left[O {H}^{-}\right] = {10}^{- p O H}$

The video below explains how to use all this information so that you can complete these types of calculations.

Hope this helps!

Aug 11, 2017

Here's a concept map that shows how to get from any of these to the other.

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Really, the only things you need to remember are:

• "pX" = -log_10("X")
• ${\text{pK}}_{w} = 14$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$
• ${K}_{w} = \left[{\text{H"^(+)]["OH}}^{-}\right] = {10}^{- 14}$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$

You don't need to know that ${\text{pH" + "pOH" = "pK}}_{w}$... you can derive that by taking the $- {\log}_{10}$ of both sides of the ${K}_{w}$ expression!

...and the rest is actually putting forth the practice and reading your book examples... Change a few numbers around and test yourself...