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

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)#