The number of moles of hydroxide is limiting/isn't in excess, so its moles is the minimum number of moles of salt that can be formed, as there isn't enough hydroxide left over to react with the excess acid.
n(NaOH) = 50/1000 x 0.185 = 9.25 x 10^-3 1 mole NaOH = 1 mole CH3CH2COONa So moles of salt is 9.25 x 10^-3 [Salt] = 9.25 x 10^-3/ 125/1000 = 0.074 mol dm-3 [Acid] = 7.25 x 10^-3/ 125/1000 = 0.058
Plugging this back into the equation gives pH buffer = 4.824 + log(0.074/0.058)
Ph = pka log[A-]/[HA] Ph = -log(1.5×10^-3) log(0.185/0.220) Put it in calculator Ph = 2.74
Your calculation assumes that you're adding butanoate ions to butanoic acid. In this case, NaOH is added, which partially neutralises the butanoic acid, generating butanoate ions.
If you re-read the thread, you'll find the values you should have used to get 4.93, which is the correct answer.