Respuesta :

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
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