Given buffer contains:
Acetic acid: CH3COOH.
Sodium Acetate : CH3COONa.
Mass of CH3COONa taken = 11.1 g
Molar mass of CH3COONa = 82.03 g/mol
# moles of CH3COONa = 11.1g/82.03 g.mol-1 = 0.1353 moles
Volume of water = 200 ml = 0.2 L
Therefore, the initial concentration of CH3COONa = 0.1353/0.2 = 0.677 M
Buffer pH = 6.5
pKa (CH3COOH) = 4.76
Based on Henderson-Hasselbalch equation:
pH = pKa + log [CH3COONa]/[CH3COOH]
Now, initially there is only 0.677 M CH3COONa . If 'x' is the concentration of the protonated species (i.e CH3COOH) formed, then the amount of CH3cooNa remaining = 0.677-x
6.5 = 4.76 + log [0.677-x]/[x]
x = 0.011 M
i.e. [CH3COOH] = 0.011 m
[CH3COONa] = 0.677-0.011 = 0.666 M