Question

Calculate the pH during the titration of 30.00 mL of 0.200 M pyridine (C5H5N; Kb=1.8x10- 9 ) with 0.200 M HCl after a) 30.00 mL and b) 60.00 mL have been added.

Answer 1

a)

pkb = -logkb

      = -log(1.8*10^-9)

      = 8.74

no of pyridine taken = 30*0.2 = 6 mmol

no of mol of HCl added = 30*0.2 = 6 mmol

now the reaction is at equivalence point

   C5H5N + HCl ----> C5H5NH^+Cl^-(salt)

pH of salt formed = 7-1/2(pkb+logC)

concentration of salt = n/v = 6/(30+30) = 0.1 M

                   = 7-1/2(8.74+log0.1)

                 pH = 3.13

b) after 60.00 mL of 0.2 M HCl added

concentration of excess HCl = (60-30)* 0.2/(60+30) = 0.067 M

pH = -log(H3O+)

    = -log0.067

= 1.174