Question

A 110.0-mL aliquot of 0.120 M weak base B (pKb = 4.55) was titrated with 1.20 M HClO4. Find the pH at the following volumes of acid added: Va = 0.00, 1.30, 5.50, 10.00, 10.90, 11.00, 11.10, and 14.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

Answer 1

millimoles of weak base = 110 x 0.120 = 13.2

pKb = 4.55

a) 0.00 ml the addition of any HClO4

pOH = 1/2 [pKb -logC] = 1/2 [4.55 -log0.120] = 2.74

pH + pOH = 14

pH = 11.26

b) after the addition of 1.30 mL HClO4

millimoles of HClO4 = 1.30 x 1.20 = 1.56

millimoles of base = 13.2 - 1.56 = 11.64

millimoles of salt = 1.56

pOH = pKb + log [salt /acid ]

        = 4.55 + log [1.56 / 11.64]

        = 3.68

pH = 10.32

c) after the addition of 5.50 mL HClO4

millimoles of HClO4 = 5.50 x 1.20 = 6.6

it is half equivalence point . so

pOH = pKb

pOH = 4.55

pH = 9.45

d) after the addition of 10 mL HClO4 :

millimoles of HClO4 = 10.00 x 1.20 = 12

millimoles of base = 1.2

millimoles of salt = 12

pOH = 4.55 + log [12 / 1.2]

        = 5.55

pH = 8.45

e) after the addition of 10.90 mL HClO4 :

millimoles of HClO4 = 10.90 x 1.20 = 13.08

millimoles of base = 0.12

millimoles of salt = 13.08

pOH = 4.55 + log [13.08 / 0.12]

        = 6.59

pH = 7.41

f) after the addition of 11.00 mL HClO4 :

millimoles of HClO4 = 11.00 x 1.20 = 13.2

it is equivalence point only salt is formed

salt concentration = millimoles / total volume = 13.2 / (110 + 11) = 0.109 M

salt is from strong acid weak base so pH <7

pH = 7 -1/2 [pKb + logC]

pH = 7 - 1/2 [4.55 + log 0.109]

pH = 5.21