the answer: is yes, It is a buffer solution.
first, we need to get moles of sodium hydroxide and propanoic acid:
moles NaOH = molarity * volume
= 0.5M * 0.1 L = 0.05 moles
moles propanoic acid = molarity * volume
= 0.75 M * 0.1 L = 0.075 moles
[NaOH] at equilibrium = 0.05 m
[propanoic acid ] at equilibrium = 0.075 - 0.05 = 0.025 m
when Pka for propanoic acid (given) = 4.89
so by substitution:
∴PH = Pka + ㏒[NaOH]/[propanoic acid ]
∴ PH = 4.89 + ㏒ 0.05 / 0.025
= 5.19