Asap calculate the ph of the solution after the addition of the following amounts of 0.0574 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) 5.27 ml of hno3
c) volume of hno3 equal to half the equivalence point volume
d) 101 ml of hno3
e) volume of hno3 equal to the equivalence point
f) 109 ml of hno3

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.

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