I'll discuss how to determine pH given #"pKa"# for a monoprotic acid, which is an acid that only donates one proton per molecule when placed in aqueous solution.
The general equation for a monoprotic acid in aqueous solution is
#HA_((aq)) rightleftharpoons H_(aq)^(+) + A_(aq)^(-)#
If you're dealing with a buffer, then you are dealing with a weak acid. In this case, the Henderson-Hasselbalch equation can take you directly from #"pKa"# to the solution's pH (assuming you know the concentrations of the weak acid and its conjugate base)
#pH = pKa + log(([A^(-)])/([HA]))#
If you're not dealing with a buffer, then you must use the acid dissociation constant, #"K"_a#, to help you determine the pH of the solution. In this case, you need to determine #[H^(+)]# in order to determine pH, since
#pH = -log([H^(+)])#
The value of the acid dissociation constant can be derived from #"pKa"#
#K_a = 10^("-pKa")#
For a strong acid, #"pKa" <1# and #"K"_a>1# ; strong acids dissociate completely in aqueous solution, so #[H^(+)]# = #[HA]#, which means
#pH = -log([HA])#
If you're dealing with a weak acid, you have to use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart). The initial concentration of the acid is #C#, so
......#HA rightleftharpoons H^(+) + A^(-)#
I:......C.........0.........0
C:...(-x).........(+x)......(+x)
E:..(C-x).......(x).......(x)
Remember that #"K"_a# is defined as
#K_a = ([H^(+)] * [A^(-)])/([HA])#, which means that you'll get
#K_a = (x * x)/(C-x) = x^(2)/(C-x)#
At this point, the only unknown you'll usually have is #x#; solve for #x# and pick the positive solution (remember that #x# represents the concentration of #H^(+)# and #A^(-)#, so it must be a positive number).
As a result, #[H^(+)] = x => pH = -log([H^(+)])#
