Question

A 1.481.48 L buffer solution consists of 0.1220.122 M butanoic acid and 0.3160.316 M sodium butanoate. Calculate the pH of the solution following the addition of 0.0780.078 moles of NaOHNaOH. Assume that any contribution of the NaOHNaOH to the volume of the solution is negligible. The ?aKa of butanoic acid is 1.52×10−51.52×10−5.

Answer 1

to Henderson canation CH₃ COOH + OH CH, CO 8 +420 Initial 0.122 Conc 0.053 Final 0.122 Conc 0.053M 0.0539 0.053 0.069M OM 0.053M 0.053M. After addition d nach buffer solution Contains [cz Hy CoolH] = 0.069M. and [C3 H7 Coo = [Cathy Coona] 0.3160+ 0.053 - 0.369 M. pH pla+log Csalt] Caud] p4 pica of C3 Hy Coolt + log [C3 H Coona] [C3 Hy COOH

Given ka for 3 H₂ Coolt 1.52 x105 plaz – log log ka TO -log 1152 x 105 - logrosa – log 105 LO 10 - 0.18 + 5 log to 5-0.18 = 4.82 p4 4.82 + log 0.369 10 0.069 ру = 4.82 + log 5.35 = 4.82 +0.73 5:55 pHd buffer solution 0.078 moles Naoth canal to 5.55 after addition of