The pH of the NaOH/CH3COOH buffer will be equal to 4.75.
The first thing you have to do is determine how many moles of NaOH and of CH3COOH you'll have in solution
nNaOH=50⋅10−3L⋅0.1 M=0.005 moles, and
nCH3COOH=100⋅10−3L⋅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)+CH3COOH(aq)→CH3COONa(aq)+H2O(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 CH3COOH and CH3COONa wil be
CCH3COOH=nVtotal=0.005 moles(100+50)⋅10−3L=0.033 M, and
CCH3COONa=0.005 moles150⋅10−3L=0.033 M
Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution
pHsolution=pKa+log([CH3COONa][CH3COOH])
pHsolution=4.75+log(0.033 M0.033 M)=4.75+log(1)
pHsolution=4.75+0=4.75
