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

3 Answers
BRIAN M.
Jun 1, 2016

[H+]→pH=−log[H+]=pH
[OH−]→pH=−log[OH−]=pOH
pH→[H+]=10−pH=[H+]
pOH→[OH−]=10−pOH=[OH−]
14−pH=pOH
14−pOH=pH

Explanation:

[H+]→pH=−log[H+]=pH
[OH−]→pH=−log[OH−]=pOH
pH→[H+]=10−pH=[H+]
pOH→[OH−]=10−pOH=[OH−]
14−pH=pOH
14−pOH=pH

There are many steps involved.

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Here is a video that breaks down the process.

mrpauller.weebly.com
Aug 11, 2017

There are a few key things to remember...

Explanation:

pH + pOH = 14

pH = -log[H+] AND pOH = -log[OH−]

[H+]=10−pH

[OH−]=10−pOH

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

Hope this helps!

Truong-Son N.
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=−log10(X)
  • pKw=14 at 25∘C and 1 atm
  • Kw=[H+][OH−]=10−14 at 25∘C and 1 atm

You don't need to know that pH+pOH=pKw... you can derive that by taking the −log10 of both sides of the Kw expression!

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

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