Question

Show how to get the following pH’s (on left) from the titration problem

A 1.8 M solution of HCI(aq) is used to titrate 49.1ml of a 0.88M solution of a weak base. If the Kb for the weak base is 7.4x10, calculate the pH of the initial solution and after 6.0 ml additions of HCI up to 30 ml of total HCl added. - 7.4ml 6 4.35 287 8 8.39 24 483 30 0.86 1 8 02 (o ? 1015202.59 35

Answer 1

a) initial pH

pkb = - logkb

      = -log(7.4*10^-6)

      = 5.13
pOH = 1/2(pkb-logC)

    = 1/2(5.13-log0.88)

    = 2.593

pH = 14-poH

     = 14-2.593

     = 11.407

b)after addition of HCl

no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol

no of mol of HCl = 6*1.8 = 10.8 mmol

pH = 14 - (pkb + log(salt/base))

    = 14 - (5.13+log(10.8/(43.21-10.8)))

    = 9.35

c) after addition of HCl

no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol

no of mol of HCl = 12*1.8 = 21.6 mmol

pH = 14 - (pkb + log(salt/base))

    = 14 - (5.13+log(21.6/(43.21-21.6)))

    = 8.87

d) after addition of HCl

no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol

no of mol of HCl = 18*1.8 = 32.4 mmol

pH = 14 - (pkb + log(salt/base))

    = 14 - (5.13+log(32.4/(43.21-32.4)))

    = 8.39

e) after addition of HCl

no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol

no of mol of HCl = 24*1.8 = 43.2 mmol

reaches to equivalencepoint

concentration of salt = 43.2/(49.1+24) = 0.591 M

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

    = 7-1/2(5.13+log0.591)

    = 4.55

f) after addition of HCl

no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol

no of mol of HCl = 30*1.8 = 54 mmol

concentration of excess HCl = (54-43.21)/(54+49.1) = 0.105 M

pH = -log(H3O+)

     = -log(0.105)

     = 0.98