Question

An analytical chemist is titrating 223.1 mL of a 0.5700 M solution of benzoic acid (HC_6 H_5 CO_2) with a 0.9000 M solution of KOH. The PK_a of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 142.2 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 =

Answer 1

milli moles of weak acid = 223.1 x 0.57 = 127.2

millimoles of base KOH = 0.98 x142.2 = 139.4

C6H5COOH + KOH ----------------------> C6H5COOK + H2O

127.2 139.4 0           0 --------------------------> initial

0             12.23   127.2     127.2----------------------> after reaction

here strong base remained

remaining base millimoles = 12.23

base concentration = millimoles / total volume

                                 = 12.23 / (223.1+142.2)

                                =0.0335 M

[OH-] = 0.0335 M

pOH = -log [OH-]

pOH = -log (0.0335)

pOH = 1.475

pH + pOH = 14

pH = 14-pOH

pH = 12.53