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

3 Answers
Jun 1, 2016

Answer:

#[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.

Answer:

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!

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")#
  • #"pK"_w = 14# at #25^@ "C"# and #"1 atm"#
  • #K_w = ["H"^(+)]["OH"^(-)] = 10^(-14)# at #25^@ "C"# and #"1 atm"#

You don't need to know that #"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...