Question

Consider the titration of 20.0 mL of 0.0800 M C5H5N (a weak base; Kb = 1.70e-09) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =

(b) 4.0 mL pH =

(c) 8.0 mL pH =

(d) 12.0 mL pH =

(e) 16.0 mL pH =

(f) 22.4 mL pH =

Answer 1

millimoles of pyridine = 20 x 0.08 = 1.6

kb= 1.7x10^-9

pKb = -logKb = -log (1.7x10^-9) = 8.77

a) before the addition of any HCl

pOH = 1/2 [pKb -logC] = 1/2 [8.77 -log0.100] = 4.88

pH + pOH = 14

pH = 9.12

b) after the addition of 4.0 mL HCl

millimoles of HBr = 4 x 0.1 = 0.4

C6H5N + HBr ----------------------> C6H5NH+Br-

1.6          0.4                                0

1.2           0                                   0.4

pOH = pKb + log [0.4 / 1.2]

pOH = 8.29

pH = 5.71

c) after the addition of 8 mL HCl

millimoles of acid = 8 x 0.1 = 0.8

it is half equivalence point . so

pOH = pKb

pOH = 8.77

pH +pOH =14

pH = 5.23

d) after the addition of 12 mL HBr

millimoles of HBr = 12 x0.1 = 1.2

C6H5N + HBr ----------------------> C6H5NH+Br-

1.6          1.2                                0

0.4           0                                   1.2

pOH = pKb + log [1.2 / 0.4]

pOH = 9.25

pH = 4.75

e) after the addition of 16 mL HCl

millimoles of HCl = 16 x 0.1 = 1.6

it is equivalence point only salt is formed

salt millimoles = 1.6

salt concentration = millimoles / total volume = 1.6 / (20 + 16) = 0.044 M

salt is from strong acid weak base so pH <7

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

pH = 7 - 1/2 [8.77 + log 0.044]

pH = 3.29

e) after the addition of 22.4 mL HBr

millimoles of HBr = 22.4 x 0.1 = 2.24

C6H5N + HBr ----------------------> C6H5NH+Br-

1.6           2.24                               0

0               0.64                                 1.6

strong acid remained in the solution

[H+] = 0.64 / (20 + 22.4) = 0.0151 M

pH = -log(0.0151)

pH = 1.82