Question

An analytical chemist is titrating 148.0 mL of a 1.100 M solution of butanoic acid (HC,H,CO2 with a 0.7400 M solution of KOH. The pK, of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 259.6 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. pH = 0 x 6 ?

Answer 1

butamic and is a vesh aut, it is strong base Kort rs Weak acid titration. The amount of base needed to fully consumed acid 148. mt x 1.100 (M) VM, = 52 M2 8.7400 (0) is used. 2220 m. The chemist asted more them that i.l.; 259.6m2 woch do after egnivalence point only low is present in the solution = (259.6-220)2 39.6mb. So the concentration of nou 39. bmx 8.74 (m) 28.0719 (M) Total volume 248 +259.6)m. (148+259.6m So, the. [on-J2 0.0719() ...poiten h3 [-32 1:14 pH + por 214; PH214-POH = 14-1.14=12.86