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

1 Answer
Jun 17, 2016

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")


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)