Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0625 M HNO3 to a 80.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

1) 0.00ml of HNO3

2) 8.03ml of HNO3

3) volume of HNO3 equal to half the equivalence point volume

4) 92.7ml of HNO3

5) volume of HNO3 equal to the equivalence point

6)100.98ml of HNO3

Answer 1

pKa for azridinium = 8.04

So, Ka = 10^-pKa

= 10^-8.04

= 9.12 x 10^-9

The corresponding reaction is:

C2H5N + HBr ==> C2H5NH+... +... Br-

Since they react in a 1:1 mole ratio,

So, Using, M1V1 = M2V2

Where, M1, M2 = Molarities of azridine and HNO3

V1, V2 = Volumes of azridine and HNO3

Putting the respective values, we get

0.075 M x 80 mL = 0.0625 M x V2

96 mL = V2 = Volume of titrant ie HNO3 to reach equivalence point

(a) before addition of any HNO3 :

We have 0.0750 M azridine. For weak bases

[OH-] = (Kb x [HA]o)^0.5

[OH-] = ((9.12 x 10^-9)(0.0750))^0.5 = 2.61 x 10^-5 M

pOH = -log[OH-]

= -log (2.61 x 10^-5)

= 4.58

pH = 14 - 4.58

= 9.42

2) after addition of 8.03 mL of 0.0625 M HNO3 :

After adding 8.03 mL of the HNO3 , we are at the 8.36 % mark of completion ((8.03 / 96 )x 100 = 8.36 %) of the titration. So the ratio of [C2H5NH+] to [C2H5N ] is 8.36 / 91.64

Since we have a weak base(azridine ) and its conjugate acid ( C2H5NH+) in the same solution, which constitutes a buffer system. And at this point they are present in equal concentrations.So, using the Henderson-Hasselbalch equation for buffers,

pOH = pKb + log ([conj. Acid] / [weak base])

pOH = pKb+ log ([C2H5NH+] / [C2H5N ])

So, putting the value of pKa in above equation,

pOH = 8.04 + log 8.36 / 91.64 = 8.04 - 1.040 = 7.0

pH = 14 - pOH

= 14 - 7

= 7

3) after addition of half of equivalence point's volume ie 48 mL of HNO3

After adding 48 mL of the HNO3 we are at the 50% mark of completion of the titration. So the ratio of [C2H5NH+] to [C5H5N ] is 1 / 1.

pOH = 8.04 + log (50/50) = 8.04 + 0 = 8.04

pH = 14 - pOH

= 14- 8.04

= 5.96

4) after addition of 92.7 mL HNO3 :

After adding 92.7 mL of the HNO3 , we are at the 96.5 % mark of completion ((92.7/ 96 )x 100 = 96.5 %) of the titration. So the ratio of [C5H5NH+] to [C5H5N ] is 96.5 / 3.5

pOH = 8.04 + log (96.5 / 3.5 ) = 8.04 + 1.44 = 9.48

pH = 14 - pOH

= 14- 9.48

= 4.52