Question

Calculate the pH of the solution after the addition of the following amounts of 0.0656 M HNO3 to a 70.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

a) 0.00 mL of HNO3; Ph=

b) 9.06 mL of HNO3; Ph=

c) Volume of HNO3 equal to half the equivalence point volume; Ph=

d) 77.0 mL of HNO3; Ph=

e) Volume of HNO3 equal to the equivalence point; Ph=

f) 83.8 mL of HNO3; Ph=

Answer 1

Concentration of aziridine = 0.0750 M

pKa = 8.04

First we need to calculate Kb from the pKa

We know,

pKa = - log Ka

Ka = 10^-pKa

      = 10^-8.04

        = 9.12X10^-9

Kb = Kw / Ka

= 10^-14 / 9.12X10^-9

= 1.1X10^-6

a) pH when 0.00 mL of HNO₃ is added.

It is the weak base, so we need to put ICE chart

C₂H₅N + H₂O C₂H₆N⁺ + OH^-

Initially 0.0750 0              0

Finally 0.0750-x + x            +x

Kb for C₂H₅N = 1.1X10^-6

Kb = [C₂H₆N⁺] [OH^-] / [C₂H₅N]

1.1X10^-6 = (x) * (x) / (0.0750-x)

The x in the 0.0750-x can be neglected due to Kb value is too small

1.1X10^-6 *0.0750 = x²

x² = 8.22X10^-8

   x = 2.87X10^-4 M

x = [OH^-] = 2.87X10^-4 M

x = [C₂H₆N⁺] = 2.87*10^-4 M

pOH = - log [OH^-]

         = - log (2.87x10^-4 )

= 3.54

pH = 14 - pOH

      = 14 - 3.54

      = 10.46

b) pH after 9.06 mL of HNO₃ is added.

We need to calculate moles of C₂H₅N = 0.0750 *0.070 L = 0.00525 mole

moles of C₂H₆N⁺ = 2.87X10^-4 M * 0.070 L = 2.009X10^-5 moles

When we added 9.06 mL of 0.0656 M HNO₃

So, there in conjugate base added moles and decrease the number of moles of base.

Moles of HNO₃ = 0.0656 M *0.00906 L = 0.000594 moles

moles of C₂H₅N in excess = 0.00525 - 0.000594 = 0.00465 moles

Moles of C₂H₆N⁺ = 2.009X10^-5 moles + 0.000594 = 0.000614 moles

New molarity-

Total volume = 70 + 9.06 = 79.06 mL

[C₂H₅N] = 0.00465 moles / 0.07906 L

              = 0.0588 M

[C₂H₆N⁺ ] = 0.000614 mole / 0.07906 L = 0.00776 M

Now, using Henderson Hasselbalch equation

pKb = 5.96

pOH = pKb + log [C₂H₆N⁺] / [C₂H₆N]

         = 5.96 + log (0.00776/0.0588

         = 5.96 + (- 0.879)

         = 5.081

pH = 14 - pOH

       = 14 - 5.081

        = 8.919

c) Volume of HNO3 equal to half the equivalence point volume

We know at half the equivalence point volume

pH = pKa

so, pH = 8.04

d) pH after 77.0 mL of HNO₃ is added.

moles of C₂H₅N = 0.00525 moles

moles of C₂H₆N⁺ = 2.009X10^-5 moles

When we added 77.0 mL of 0.0656 M HNO₃

So, there in conjugate base added moles and decrease the number of moles of base

Moles of HNO₃ = 0.0656 M *0.077 L = 0.00505 moles

moles of C₂H₅N in excess = 0.00525 - 0.00505 = 0.00020 moles

Moles of C₂H₆N⁺ = 2.009X10^-5 moles + 0.00505 = 0.00507 moles

New molarity-

Total volume = 70.0 + 77.0 = 147 mL

[C₂H₅N] = 0.00020 moles / 0.147 L

              = 0.00136 M

[C₂H₆N⁺ ] = 0.00507 mole / 0.147 L = 0.0344 M

Now, using Henderson Hasselbalch equation

pKb = 5.96

pOH = pKb + log [C₂H₆N⁺] / [C₂H₆N]

         = 5.96 + log ( 0.0344 / 0.00136)

         = 5.96 + 1.403

         = 7.363

pH = 14 - pOH

       = 14 7.363

        = 6.637