How to solve this problem on pH and buffer solutions....?

acid H+ & anion  , the NaOH removes the H+, & the reaction shift to the left increasing the concentration of the anion by "x",  & decreasing the concentration of the acid by "x"

the K for acetic acid is 1.8e-5, its pK = 4.74

the henderson Hasselbach equation says:
http://en.wikipedia.org/wiki/Henderson_Hasselbalch_equation

PH = Pka & log [anion] / [acid]
4.76  = 4.74  & log [anion] / [acid]
0.02  =  log [anion] / [acid]

10^x for both sides

1.047  =   [anion] / [acid]
1.047  =   [0.1+x] / [0.1-x]

1.047 [0.1-x] =   [0.1+x]
0.1047  - 1.047x =   0.1+x

0.1047  - 0.1  = 1.047x  +  x
0.0047 = 2.047x
x = 0.002296 molar change in concentration

your answer is: 0.002 Molar change