Question

A 1.35 L buffer solution consists of 0.115 M butanoic acid and 0.331 M sodium butanoate. Calculate the pH of the solution following the addition of 0.079 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52 x 10-5. pH =

Answer 1

Given that - Butanoic and Molarity M = 0.11519 Volume v=1.352 Molas = MXV=0.115 MX1352 015525 ha of Butawic acide Sodium Butanoate ME 0:33) VE 1.354 MXV=0.331 X 135. = 0.44685 Nooh ka=1,52x10? pka = -log ka Moles = 0.0 79 = -10g (1.szx10) 249818 = 4.82 Moles CH₃CH₂) Coottt Nooth > CH3 (CH2) Coona + H20. Intial o. 15525 0.079 0:44685 Chawle -0.079 -0.079 +0.079 Equilibrium 0.07625 o 0.52585. pot = pka + log estetin (salt) CH3 Cut _)_coonas Net moins = 0,35 Total Volume Taude] = CH3 C.CH₂)2 CooH = Net moles 0.07625 1.35 Total Volume .=0.015648 895

p*= pha trog Calls = 4:82 t log Co tlog C03895) To 05648). = 4.82 + 0.8386. = 5,65 Auswas 1 ptr 50651