Question

Calculate the pH of the solution after the addition of the following amounts of 0.0568 M HNO3 to a 80.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

a) 0.00 mL of HNO3

b) 8.62 mL of HNO3

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

d) 102 mL of HNO3

e) Volume of HNO3 equal to the equivalence point

f) 89.2 mL of HNO3

thank you

Answer 1

Here, pKa = 8.04

Concentration of aziridine = 0.0750 M

We have, pKa = - log Ka

Therefore, Ka = 10^-pKa = 10^-8.04 = 9.12 x 10^-9

Also, Kb = Kw / Ka = 10^-14 / (9.12X10^-9) = 1.1 x 10^-6

Part a) 0.00 mL of HNO₃ is added.

Constructing the ICE

………………….C₂H₅N + H₂O  C₂H₆N⁺ + OH^-

Initial…………… 0.0750…………. 0              0

Final …………….0.0750-x ……….+ x………+x

Kb = [C₂H₆N⁺] [OH^-] / [C₂H₅N] = 1.1 x 10^-6 = (x)x(x) / (0.0750-x)

Here, x is small compared to 0.075. Therefore, x can be neglected

Therefore,

1.1 x 10^-6 x 0.0750 = x²

x = [OH-] = [C₂H₆N⁺] = 2.87 x 10^-4 M

pOH = - log [OH^-] = - log(2.87 x 10^-4 ) = 3.54

Now,

pH = 14 – pOH = 14 - 3.54 = 10.46

pH = 10.46

Part b) 8.62 mL of HNO₃ is added

Moles of C₂H₅N = 0.0750 x 0.08 L = 0.006 mole

Moles of C₂H₆N⁺ = (2.87 x 10^-4 M) x 0.08 L = 2.29 x 10^-5 moles

After the addition of 8.62 mL of 0.0568 M HNO₃

Moles of HNO₃ = 0.0568 M x 0.00862 L = 0.000489 moles

Moles of C₂H₅N in excess = 0.006 - 0.000489 = 0.00551 moles

Moles of C₂H₆N⁺ = 2.29 x 10^-5 moles + 0.000489 = 0.000511 moles

Total volume = 80 + 8.62 = 88.62 mL

[C₂H₅N] = 0.00551 moles / 0.08862 L= 0.0621 M

[C₂H₆N⁺ ] = 0.000511 mole / 0.08862 L = 0.00576 M

Applying the Hasselbalch equation

pOH = pKb + log [C₂H₆N⁺] / [C₂H₆N]

pKb = 5.96

pOH = 5.96 + log (0.00576/0.0621) = 5.96 + (- 0.879) = 4.927

pH = 14 - pOH = 14 - 4.927 = 9.07

pH = 9.07

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

At half the equivalence point,

pH = pKa

Therefore,

pH = 8.04

d) 102 mL of HNO₃ is added

Moles of C₂H₅N = 0.0750 x 0.08 L = 0.006 mole

Moles of C₂H₆N⁺ = (2.87 x 10^-4 M) x 0.08 L = 2.29 x 10^-5 moles

After the addition of 102 mL of 0.0568 M HNO₃

Moles of HNO₃ = 0.0568 M x 0.102 L = 0.00579 moles

Moles of C₂H₅N in excess = 0.006 - 0.00579 = 0.00021 moles

Moles of C₂H₆N⁺ = 2.29 x 10^-5 moles + 0.00579 = 0.00581 moles

Total volume = 80.0 + 102 = 182 mL

[C₂H₅N] = 0.00021 moles / 0.182 L = 0.00115 M

[C₂H₆N⁺ ] = 0.00581 mole / 0.182 L = 0.0319 M

Applying the Hasselbalch equation

pOH = pKb + log [C₂H₆N⁺] / [C₂H₆N]

pOH = 5.96 + log ( 0.0319 / 0.00115) = 7.403

pH = 14 – pOH = 14 - 7.403 = 6.596

pH = 6.596