Question

A 500.0 ml buffer solution is 0.10 m in benzoic acid and 0.10 m in sodium benzoate and has an initial ph of 4.19. part a what is the ph of the buffer upon addition of 0.025 mol of naoh?

Answer 1

Initial moles of C₆H₅COOH = 500/1000 × 0.10 = 0.05mol
Initial moles of C₆H₅COONa = 500/1000 × 0.10 = 0.05 mol
initial pH = Pka + log([C₆H₅COONa/ moles of C₆H₅COOH)
4.19 = pKa + log(0.05/0.05)
→pKa = 4.19
C₆H₅COOH + NaOH → C₆H₅COONa ₊ H₂o
moles of NaOH added = 0.010 mol
moles of C₆H₅COOH = 0.05 - 0.025 = 0.025 mol
Final pH = pKa + log([C₆H₅COONa)/[ C₆H₅COOH])
=pKa + log(moles of C₆H₅COONa/moles of C₆H₅COOH)
= 4.19 + log(0.025/0.075)
4.29