Question

What is the pH of 40 mL of a 0.10M solution of HClO with a K_a of 3.0 x 10^-8 after 10mL of 0.20M NaOH has been added?

Answer 1

HOCl (aq) + OH- (aq) <---> OCl-(aq) + H2O (l)

Initial HClO moles = 0.1 x 40/1000 = 0.004

NaOH moles = OH- = M x v = 0.2 x 10/1000 = 0.002

HOCl moles after reacting with OH- moles =0.004-0.002 = 0.002

OCl- moles = OH- moles reacted = 0.002

total vol = 50 ml = 0.05 L , [OCl-] =0.002/0.05 = 0.04 , [HOCl] = 0.002/0.05 = 0.04

pH = pka + log [OCl-] /[HOCl]    ,   pka = -log Ka = -log ( 3 x 10^-8) = 7.523

pH = 7.523 + log ( 0.04/0.04)

           = 7.523