Question

Caloulote the 4 of the solution after the addition of each of the given amounts of 0.0559 M HNO, to a 60.0 mL solution of next question 0.0750 M aziridine. The pK, of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL HNO,? pH = What is the pH of the solution after the addition of 5.91 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = What is the pH of the solution after the addition of 75.9 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? pH = What is the pH of the solution after the addition of 84.43 mL HNO,? pH =

Answer 1

Answer:

Ó-H NH

Aziridine (Az) is a heterocyclic organic compound containing a three-membered ring in which one amine group (-NH-) and two methylene bridges (-CH₂-). The structure of aziridine (C₂H₅N) is shown above:

Aziridine has a basic character; given-

pKa = 8.04

So, pK_b = 14 – 8.04 = 5.96

Or, -log K_b = 5.96

Or, K_b = 10^(-5.96)

∴ K_b = 1.1 × 10^-6

The dissociation of Aziridine in water occurs as follows:

Az + H₂O → AzH⁺+ OH^-

(i) Case-I: Addition of 0.00 mL of HNO₃

There will no trace of HNO₃ and only Az will be present. K_b = 1.1 × 10^-6 and C = 0.0750 M

We know that [OH-] = √(K_b × C) = √{(1.1×10^-6)×(0.075)} = 2.87×10^-4

pOH = -log (2.87×10^-4) = 3.54 [∵ pH + pOH = 14]

∴ pH = 14 – 3.54 = 10.46

(ii) Case-II: Addition of 5.91 mL of HNO₃

Total conc. of HNO₃ = 5.91 × 0.0559 M = 0.33 M

Total conc. Az before addition of HNO₃ = 60 × 0.075 M = 4.5 M

Since conc. of Az is much higher than HNO₃ the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pOH = pK_b + log [salt]/[base]

[salt] = 0.33 / (60+5.91) = 0.33 / 65.91 = 5 × 10^-3 M

[base] = (4.5-0.33) / (60+5.91) = 4.17 / 65.91 = 0.063 M

pOH = 5.96 + log [5 × 10^-3] / [ 0.063] = 4.86

∴ pH = 14 – 4.86 = 9.14 [∵ pH + pOH = 14]

(iii) Case-III: 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,

∴ [salt] = [base]

pOH = pKb + log[salt]/[base] = 5.96 + log[1] = 5.96

pH = 14 – pOH = 14 – 5.96 = 8.04

(iv) Case-IV: Addition of 75.9 mL of HNO₃

Total conc. of HNO₃ = 75.9 × 0.0559 M = 4.24 M

Total conc. Az before addition of HNO₃ = 60 × 0.075 M = 4.5 M

Since conc. of Az is much higher than HNO₃ and the mixture will behave as a buffer solution.

pH of this mixture can be calculated from Henderson’s equation

pOH = pK_b + log [salt]/[base]

[salt] = 4.24 / (60+75.9) = 4.24 / 135.9 = 0.031 M

[base] = (4.5-4.24) / (60+75.9) = 0.26 / 135.9 = 0.0019 M

pOH = 5.96 + log [0.031] / [ 0.0019] = 7.17

∴ pH = 14 – 7.17 = 6.83 [∵ pH + pOH = 14]

(v) Case-V: Volume of HNO₃ equal to the equivalence point volume

At equivalence point, the concentration of the base is equal to the concentration of the acid.

∴ [acid] = [base]

Volume of HNO₃ needed for the equivalence point = (4.5 mmol) / (0.0559 mmol/ml) = 80.5 mL

At the equivalence point: [Az] = 0, [AzH⁺] = (4.5) / (60.0 + 80.5) = 4.5/140.5 = 0.032 M

As Ka is very small, the dissociation of AzH⁺ can be negligible.

Hence, [AzH⁺] ≈ 0.032 M.

[H⁺] = √(Ka.C) = √(10⁻⁸˙⁰⁴ × 0.032) = 1.73 × 10⁻⁵

pH = - log[H⁺] = - log(1.73×10⁻⁵) = 4.76

(vi) Case-VI: Addition of 88.43 mL of HNO₃

Total conc. of HNO₃ = 88.43 × 0.0559 M = 4.94 M

Total conc. Az before addition of HNO₃ = 60 × 0.075 M = 4.5 M

Since conc. of Az is much lower than HNO₃

There will be excess acid after addition of 88.43 mL of HNO₃

So, after addition [H⁺] = (4.94-4.50) / (88.43+60) = 0.44/148.43 = 0.003

∴ pH = -log [H⁺] = -log [0.003] = 2.52

Summary:

(i) Case-I: Addition of 0.00 mL of HNO3	pH = 10.46
(ii) Case-II: Addition of 5.91 mL of HNO3	pH = 9.14
(iii) Case-III: Volume of HNO₃ equal to half the equivalence point volume	pH = 8.04
(iv) Case-IV: Addition of 75.9 mL of HNO3	pH = 6.83
(v) Case-V: Volume of HNO₃ equal to the equivalence point volume	pH = 4.76
(vi) Case-VI: Addition of 88.43 mL of HNO3	pH = 2.52