Question

The dibasic compound B (pKb1 = 4.03, pKb2 = 8.06) was titrated with 1.18 M HCl. The initial solution of B was 0.118 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: Va = 0.00, 2.00, 5.00, 9.00, 10.00, 11.00, 15.00, 19.00, 20.00, and 22.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

Answer 1

dibasic base = pKb1 = 4.03

pKb2 = 8.06

millimoles of base = 100 x 0.118 = 11.8

(a) before addition of any HCl

pOH = 1/2 [pKb1- logC]

pOH = 1/2 [pKb1 - log 0.118]

          = 1/2 (4.03 - log 0.118)

          = 2.48

pH + pOH = 14

pH = 14 - pOH

      = 11.52

pH = 11.52

(b) after addition of 2.00 mL of HCl

millimoles of HCl = 2 x 1.18 = 2.36

millimoles of base = 9.44

millimoles of salt = 2.36

pOH = pKb1 + log [salt / base]

        = 4.03 + log (2.36 / 9.44)

       = 3.43

pH = 10.57

(c) after addition of 5.00 mL of HCl

millimoles of HCl = 5 x 1.18 = 5.9

it is half equivalence point. so

pOH = pKb1

pOH = 4.03

pH = 9.97

(d) after addition of 9.00 mL of HCl

millimoles of HCl = 9 x 1.18 = 10.62

millimoles of base = 1.18

millimoles of salt = 10.62

pOH = pKb1 + log [salt / base]

        = 4.03 + log (10.62 / 1.18)

       = 4.98

pH = 9.02

(d) after addition of 10.0 mL of HCl

millimoles of HCl = 10 x 1.18 = 11.8

it is equivalence point . at equivalence point

pOH = (pKb1 + pKb2 )/ 2

         = 4.03 + 8.06 / 2

         = 6.04

pH = 7.96