Question

Acid-base titrations. You have 60.0 mL of a weak base (0.0305 M) and titrate it with a strong acid (0.0500M). pKa of the weak conjugate acid is 4.36.

a) What is the pH before the acid is added?

b) What volume of the acid is needed at the equivalence point (in mL)?

c) What is the pH at the equivalence point?

d) What is the pH after 5.00 mL of the acid is added beyond the equivalence point?

Answer 1

weak base vs strong acid(S.A)

a) before addition of S.A

pkb of Weak base = 14-4.36 = 9.64

C = concentration of weak base = 0.0305 M

pOH = 1/2(pkb-logC)

      = 1/2(9.64-log0.0305)

      = 5.58

pH = 14-5.58 = 8.42

b) equivalence point

no of mol of weak base = 60*0.0305 = 1.83 mmol

no of mol of strong acid = 1.83 mmol

   volume of strong acid needed = n/M = 1.83/0.05 = 36.6 ml

c)

concentration of salt formed at euivalence point = n/V

                                    = 1.83/(60+36.6)

                    = 0.019 M

                                  
pkb of Weak base = 14-4.36 = 9.64

C = concentration of salt = 0.019 M

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

    = 7-1/2(9.64+log0.019)

    = 3.04

d) after addition of 5.0 ml of S.A beyond equivalence point

concentration of excess s.a = (5*0.05)/(60+41.6)

                              = 0.0025 M

pH = -log(H3O+)

     = -log0.0025

     = 2.6