Question

How do you find pOH if [H+] is given?

Answer 1

You'll need two equations for this:

`\mathbf("pH" = -log["H"^(+)])`

`\mathbf("pH" + "pOH" = 14)`

So, all you need to do is take the negative base-10 logarithm of the concentration of `"H"^(+)`, and then subtract from `14`.

`-log["H"^(+)]`

`=` `"pH"`

`-> color(blue)("pOH" = 14 - "pH")`

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If you recall, the autoionization of water works like this:

`\mathbf("H"_2"O" (l) rightleftharpoons "H"^(+)(aq) + "OH"^(-)(aq))`

Then, the equilibrium constant for water is:

`\mathbf(K_w = 10^(-14) = ["H"^(+)]["OH"^(-)])`

where `["H"^(+)]` is the concentration of `"H"^(+)` in `"M"` and `["OH"^(-)]` is the concentration of `"OH"^(-)` in `"M"`.

So, when we take the logarithms and work with this equation, we'd get:

`-cancel(log)_cancel(10)(cancel(10)^(-14))`

`= -log_10(["H"^(+)]["OH"^(-)]) = -log(["H"^(+)]["OH"^(-)])`

Using the properties of logarithms, `log xy = log x + log y`.

`=> -(-14) = 14`

`= -(log["H"^(+)] + log["OH"^(-)])`

`= -log["H"^(+)] + (-log["OH"^(-)])`

But `-log["H"^(+)] = "pH"` and `-log["OH"^(-)] = "pOH"`, so...

`=> color(blue)("pH" + "pOH" = 14)`