How do you find pOH if [H+] is given?
1 Answer
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
-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,
=> -(-14) = 14
= -(log["H"^(+)] + log["OH"^(-)])
= -log["H"^(+)] + (-log["OH"^(-)])
But
=> color(blue)("pH" + "pOH" = 14)