The pH of the #"NaOH/CH"_3"COOH"# buffer will be equal to #"4.75"#.
The first thing you have to do is determine how many moles of #NaOH# and of #CH_3COOH# you'll have in solution
#n_(NaOH) = 50 * 10^(-3)"L" * "0.1 M" = "0.005 moles"#, and
#n_(CH_3COOH) = 100 * 10^(-3)"L" * "0.1 M" = "0.01 moles"#
Since #NaOH# is a strong base, it will react with the acetic acid completely. Use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart) to determine how many moles of each species will be left after the reaction
#NaOH_((aq)) + CH_3COOH_((aq)) -> CH_3COONa_((aq)) + H_2O_((l))#
I......0.005.................0.01.................................0
C......(-0.005)............(-0.005)........................(+0.005)
E........0.......................0.005..........................0.005
The #NaOH# will be completely consumed in the reaction; the concentrations of #CH_3COOH# and #CH_3COONa# wil be
#C_(CH_3COOH) = n/V_("total") = ("0.005 moles")/((100 + 50) * 10^(-3)"L") = "0.033 M"#, and
#C_(CH_3COONa) = ("0.005 moles")/(150 * 10^(-3)"L") = "0.033 M"#
Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution
#pH_("solution") = pKa + log(([CH_3COONa])/([CH_3COOH]))#
#pH_("solution") = 4.75 + log(("0.033 M")/("0.033 M")) = 4.75 + log(1) #
#pH_("solution") = 4.75 + 0 = 4.75#
