a) 10.457.
b) 9.32.
c) 8.04.
d) 6.58.
e) 4.76.
f) 2.87.
Explanation:
Aziridine is an organic compounds containing the aziridine functional group, a three-membered heterocycle with one amine group (-NH-) and two methylene bridges (-CH2-). The parent compound is aziridine (or ethylene imine), with molecular formula C2H5N.
Aziridine has a basic character.
It has pKa = 8.04.
So, pKb = 14 – 8.04 = 5.96
Kb = 1.1 x 10⁻⁶.
If we denote Aziridine the symbol (Az), It is dissociated in water as:
Az + H₂O → AzH⁺ + OH⁻
a) 0.00 ml of HNO₃:
There is only Az,
[OH⁻] = √(Kb.C)
Kb = 1.1 x 10⁻⁶. & C = 0.0750 M.
[OH⁻] = √(1.1 x 10⁻⁶)(0.075) = 2.867 x 10⁻⁴.
∵ pOH = - log[OH-] = - log (2.867 x 10⁻⁴) = 3.542.
∴ pH = 14 – pOH = 14 – 3.542 = 10.457.
b) 5.27 ml of HNO₃
To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol.
The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (0.302).
This will form a basic buffer in the presence of weak base (Az).
pOH = pKb + log[salt]/[base]
[salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (0.302) / (85.27) = 3.54 x 10⁻³ M.
[base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 0.302 mmol) / (85.27 ml) = 0.0668 M.
pOH = pKb + log[salt]/[base] = 5.96 + log[3.54 x 10⁻³]/[ 0.0668 M] = 4.68.
pH = 14 – pOH = 14 – 4.68 = 9.32.
c) Volume of HNO₃ equal to half the equivalence point volume
:
At half equivalence point, the concentration of the salt formed is equal to the concentration of the remaining base (aziridine), [salt] = [base].
pOH = pKb + log[salt]/[base] = 5.96 + log[1.0] = 5.96.
pH = 14 – pOH = 14 – 5.96 = 8.04.
d) 101 ml of HNO₃:
To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (101.0 ml) = 5.7974 mmol.
The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (5.7974).
This will form a basic buffer in the presence of weak base (Az).
pOH = pKb + log[salt]/[base]
[salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (5.7974) / (181.0) = 0.032 M.
[base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 5.7974 mmol) / (181.0 ml) = 0.00112 M.
pOH = pKb + log[salt]/[base] = 5.96 + log[0.032]/[ 0.00112] = 7.416.
pH = 14 – pOH = 14 – 7.416 = 6.58.
e) Volume of HNO₃ equal to the equivalence point
:
At the equivalence point the no. of millimoles of the base is equal to that of the acid.
Volume of HNO₃ needed for the equivalence point = (6.00 mmol) / (0.0574 mmol/ml) = 104.5 ml
At the equivalence point:
[Az] = 0.[AzH⁺] = (6.00 mmol) / (80.0 + 104.5 ml) = 0.0325 M.As Ka is very small, the dissociation of AzH⁺ can be negligible.
Hence, [AzH⁺] at eqm ≈ 0.0325 M.
[H+] = √(Ka.C) = √(10⁻⁸˙⁰⁴ x 0.0325) = 1.72 x 10⁻⁵.
pH = - log[H+] = - log(1.72 x 10⁻⁵) = 4.76.
f) 109 ml of HNO₃:
No. of milli-moles of H⁺ added from HNO₃ = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol.
Which is higher than the no. of millimoles of the base (Az) = 6.0 mmol.
After the addition, [H⁺] = (6.257 - 6.00) / (80.0 + 109 mL) = 0.00136 M.
As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible. pH = -log[H⁺] = -log(0.00136) = 2.87.
