Question

Consider the titration of 100.0 mL of 0.400 M HONH₂ by 0.200 M HCl.

(K_b for HONH₂ = 1.1×10^-8)

Part 1
Calculate the pH after 0.0 mL of HCl added.

pH =

Part 2
Calculate the pH after 40.0 mL of HCl added.

pH =

Part 3
Calculate the pH after 75.0 mL of HCl added.

pH =

Part 4
Calculate the pH at the equivalence point.

pH =

Part 5
Calculate the pH after 300.0 mL of HCl added.

pH =

Part 6
At what volume of HCl added, does the pH = 6.04?

mL

Answer 1

millimoles of NH2OH = 100 x 0.400 = 40

kb= 8.7×10^–9

pKb = -logKb = -log (8.7x10^-9) = 8.06

part 1

pOH = 1/2 [pKb -logC]

          = 1/2 [8.06 -log 0.400] = 4.23

pH + pOH = 14

pH = 9.77

Part 2

millimoles of HCl = 40 x 0.2 = 8

mmoles of base = 40

NH2OH +   HCl   ------------> NH2OH2+

40              8                             0

32               0                              8

pOH = pKb + log [salt / base]

         = 8.06 + log [8 / 32]

         = 7.46

pH = 6.54

Part 3

mmoles of HCl = 75 x 0.2 = 15

NH2OH +   HCl   ------------> NH2OH2+

40            15                             0

25               0                             15

pOH = 8.06 + log (15 / 25)

       = 7.84

pH = 6.16

Part 4

volume of HCl = 200

concentration of salt = 40 / 100 + 200 = 0.133 M

pH = 7 - 1/2 (pKb + log C)

      = 7 - 1/2 (8.06 + log 0.133)

pH = 3.41

Part 5

mmoles of HCl = 60

pH = 1.30