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.?

a)
Denote Az as aziridine and thus AzH⁺ as aziridinium.
Ka for AzH⁺ = 10⁻⁸˙⁰⁴
Kb for Az = (1 × 10⁻¹⁴) / 10⁻⁸˙⁰⁴ = 1.10 × 10⁻⁶

When no HNO₃ is added, consider the dissociation of Az.
____________ Az(aq) + H₂O(l) ⇌ AzH⁺ ____ + ____ OH⁻ ____ Kb = 1.10 × 10⁻⁶
Initial: ____ 0.0750 M ________ 0 M ___________ 0 M
Change: _____ -x M __________ +x M __________ +x M
Eqm: ____ (0.0750 - x) M ______ x M ___________ x M

As Kb is very small, the dissociation of Az can be negligible.
Hence, [Az] at eqm = (0.0750 - x) M ≈ 0.0750 M

At eqm: Kb = [AzH⁺] [OH⁻] / [Az]
1.10 × 10⁻⁶ = x² / 0.0750
x = 2.87 × 10⁻⁴
pOH = -log[OH⁻] = -log(2.87 × 10⁻⁴) = 3.54
pH = 14.00 - 3.54 = 10.46


b)
No. of milli-moles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol

HNO₃ is a strong acid which completely ionizes to give H⁺.
No. of milli-moles of H⁺ added = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol < 6.00 mmol

When 5.27 ml of HNO₃ is added, a part of Az reacts with H⁺ to give AzH⁺. After reaction:
[Az] = (6.00 - 0.302) / (80.0 + 5.27) = 0.0349 M
[AzH⁺] = 0.302 / (80.0 + 5.27) = 0.00354 M

Consider the dissociation of AzH⁺:
AzH⁺(aq) ⇌ Az(aq) + H⁺(aq) …… pKa = 8.04

According to Henderson–Hasselbalch equation: pH = pKa + log₁₀([Az]/[AzH⁺])
pH = 8.04 + log₁₀(0.0349/0.00354) = 9.03


c)
At half equivalence point, [Az] ≈ [AzH⁺]

According to Henderson–Hasselbalch equation: pH = pKa + log₁₀([Az]/[AzH⁺])
pH = 8.04 + log₁₀(1) = 8.04


d)
No. of milli-moles of H⁺ added = (0.0574 mmol/ml) × (101 ml) = 5.80 mmol < 6.00 mmol

When 101 ml of HNO₃ is added, a part of Az reacts with H⁺ to give AzH⁺. After reaction:
[Az] = (6.00 - 5.80) / (80.0 + 101) = 0.00110 M
[AzH⁺] = 5.80 / (80.0 + 101) = 0.0320 M

According to Henderson–Hasselbalch equation: pH = pKa + log₁₀([Az]/[AzH⁺])
pH = 8.04 + log₁₀(0.00110/0.0320) = 6.58


e)
Volume of KNO₃ 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

Consider the dissociation of AzH⁺:
____________ AzH⁺(aq) ____ ⇌ ____ AzH⁺ ____ + ____ H⁺ ____ Ka = 10⁻⁸˙⁰⁴
Initial: _____ 0.0325 M ___________ 0 M ___________ 0 M
Change: _____ -y M _____________ +y M __________ +y M
Eqm: ____ (0.0325 - y) M _________ y M ___________ y M

As Ka is very small, the dissociation of AzH⁺ can be negligible.
Hence, [AzH⁺] at eqm = (0.0325 - y) M ≈ 0.0325 M

At eqm: Ka = [Az] [H⁺] / [AzH⁺]
10⁻⁸˙⁰⁴ = y² / 0.0325
y = 1.72 × 10⁻⁵
pH = -log[H⁺] = -log(1.72 × 10⁻⁵) = 4.76


f)
No. of milli-moles of H⁺ added = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol > 6.00 mmol

Az reacts with excess H⁺ to give AzH⁺.
After reaction, [H⁺] = (6.257 - 6.00) / (80.0 + 109) = 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