Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.250 M pyridine, C5H5N(aq) with 0.250 M HBr(aq):?

(a)
Refer to: http://occonline.occ.cccd.edu/online/jmlaux/Kb%20Table%20App.%20D.pdf
Kb for C₅H₅N = 1.5 × 10⁻⁹
Then, Ka for C₅H₅NH⁺ = Kw / (Ka for C₅H₅N) = (1.0 × 10⁻¹⁴) / (1.5 × 10⁻⁹) = 6.67 × 10⁻⁶

____________ C₅H₅N(aq) __ + __ H₂O(l) __ ⇌ __ C₅H₅NH⁺(aq) __ + __ OH⁻(aq) ____ Kb = 1.5 × 10⁻⁹
Initial: _______ 0.250 M ______________________ 0 M _____________ 0 M
Change: ________ -y _________________________ +y M ____________ +y M
Equilibrium: _ (0.250 - y) M ___________________ y M ______________ y M

As Kb is very small, the dissociation of C₅H₅N is very small..
It is assumed that 0.250 ≫ y and thus [C₅H₅N] at equilibrium = (0.250 - y) M ≈ 0.250 M

At equilibrium:
Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]
1.5 × 10⁻⁹ = y² / 0.250
y = √{0.250 × (1.5 × 10⁻⁹)} = 1.94 × 10⁻⁵
[OH⁻] = 1.94 × 10⁻⁵ M

pOH = -log[OH⁻] = -log(1.94 × 10⁻⁵) = 4.7
PH = pKw - pOH = 14.0 - 4.7 = 9.3

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(b)
After addition of 12.5 mL HBr:
C₅H₅N(aq) + H⁺(aq) → C₅H₅NH⁺(aq)
Before dissociation of C₅H₅NH⁺:
[C₅H₅N] = (0.250 × 25.0 - 0.250 × 12.5) / (25.0 + 12.5) = 0.0833 M
[C₅H₅NH⁺] = (0.250 × 12.5) / (25.0 + 12.5) = 0.0933 M

Consider the dissociation of C₅H₅NH⁺:
C₅H₅NH⁺(aq) ⇌ C₅H₅N(aq) + H⁺(aq) …… Ka = 6.67 × 10⁻⁶

Henderson-Hasselbalch equation:
pH = pKa + log([C₅H₅N]/[C₅H₅NH⁺]) = -log(6.67 × 10⁻⁶) + log(0.0833/0.0833) = 5.2

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(c)
After addition of 21.0 mL HBr:
C₅H₅N(aq) + H⁺(aq) → C₅H₅NH⁺(aq)
Before dissociation of C₅H₅NH⁺:
[C₅H₅N] = (0.250 × 25.0 - 0.250 × 21.0) / (25.0 + 21.0) = 0.0217 M
[C₅H₅NH⁺] = (0.250 × 21.0) / (25.0 + 21.0) = 0.114 M

Consider the dissociation of C₅H₅NH⁺:
C₅H₅NH⁺(aq) ⇌ C₅H₅N(aq) + H⁺(aq) …… Ka = 6.67 × 10⁻⁶

Henderson-Hasselbalch equation:
pH = pKa + log([C₅H₅N]/[C₅H₅NH⁺]) = -log(6.67 × 10⁻⁶) + log(0.0217/0.114) = 4.5

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(d)
After addition of 25.0 mL HBr:
C₅H₅N(aq) + H⁺(aq) → C₅H₅NH⁺(aq)
Before dissociation of C₅H₅NH⁺:
[C₅H₅N] = (0.250 × 25.0 - 0.250 × 25.0) / (25.0 + 25.0) = 0 M
[C₅H₅NH⁺] = (0.250 × 25.0) / (25.0 + 25.0) = 0.125 M

Consider the dissociation of C₅H₅NH⁺:
______________ C₅H₅NH⁺(aq) __ ⇌ __ C₅H₅N(aq) __ + __ H⁺(aq) ____ Ka = 6.67 × 10⁻⁶
Initial: ________ 0.125 M __________ 0 M __________ 0 M
Change: ________ -z M ____________ +z M _________ +z M
Equilibrium: _ (0.125 - z) M _________ z M __________ z M

As Ka is very small, the dissociation of C₅H₅NH₄⁺ is very small..
It is assumed that 0.125 ≫ z and thus [C₅H₅NH₄⁺] at equilibrium = (0.125 - z) M ≈ 0.125 M

At equilibrium:
Ka = [C₅H₅N] [H⁺] / [C₅H₅NH⁺]
6.67 × 10⁻⁶ = z² / 0.125
z = √{0.125 × (6.67 × 10⁻⁶)} = 9.13 × 10⁻⁴
[H⁺] = 9.13 × 10⁻⁴ M

pH = -log[H⁺] = -log(9.13 × 10⁻⁴) = 3.0

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(e)
After addition of 37.0 mL HBr (in excess):
C₅H₅N(aq) + H⁺(aq) → C₅H₅NH⁺(aq)
[H⁺] = (0.250 × 37.0 - 0.250 × 25.0) / (25.0 + 37.0) = 0.0484 M

pH = -log[H⁺] = -log(0.0484) = 1.3

F