Question

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

Answer 1

`[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.

Here is a video that breaks down the process.

Answer 2

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!

Answer 3

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...