Titrations, pH curves, and Indicators? 10 Points!?

a)
Refer to: http://occonline.occ.cccd.edu/online/jmlaux/Kb%20Table%20App.%20D.pdf
Kb for CH₃NH₂ = 4.2 × 10⁻⁴

Consider the dissociation of CH₃NH₂:
____________ CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq) ___ Kb = 4.2 × 10⁻⁴
Initial: _______ 0.175 M _____________ 0 M _______ 0 M
Change: ______ -y M _______________ +y M _______ +y M
At eqm: ____ (0.175 - y) M ___________ y M ________ y M

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]
4.2 × 10⁻⁴ = y² / (0.175 - y)
(7.35 × 10⁻⁵) - (4.2 × 10⁻⁴)y = y²
y² + (4.2 × 10⁻⁴)y - (7.35 × 10⁻⁵) = 0
y = {-(4.2 × 10⁻⁴) ± √[(4.2 × 10⁻⁴)² + 4(7.35 × 10⁻⁵)]} / 2
y = 0.00837 or y = -0.00879 (rejected)

pOH = -log[OH⁻] = -log(0.00837) = 2.08
pH = Kw - pOH = 14.00 - 2.08 = 11.92


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b)
HBr is a strong acid which completely dissociates to give H⁺ ion.

CH₃NH₂(aq) + H⁺(aq) → CH₃NH₃⁺(aq)
Mole ratio CH₃NH₂ : H⁺ = 1 : 1

No. of milli-moles of CH₃NH₂ = (0.175 mmol/mL) × (25.0 mL) = 4.375 mmol
No. of milli-moles of H⁺ required = 4.375 mmol
Volume of the acid required = (4.375 mmol) / (0.150 mmol/mL) = 29.2 mL


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c)
When 5 mL of acid is added, total volume of the solution = (25 + 5) mL = 30 mL
[CH₃NH₂] = {0.175 × (25/30)} - {0.150 × (5/30)} M = 0.121 M
[CH₃NH₃⁺] = {0.150 × (5/30)} M = 0.025 M

Consider the dissociation of CH₃CH₂:

CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq) ___ Kb = 4.2 × 10⁻⁴
Henderson-Hasselbalch equation:
pOH = pKb + log([CH₃NH₃⁺]/[ CH₃NH₂])
pOH = -log(4.2 × 10⁻⁴) + log(0.025/0.121) = 2.69
pH = Kw - pOH = 14.00 - 2.69 = 11.31


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d)
At 1/2 of the equivalence point: [CH₃NH₃⁺] = [ CH₃NH₂]

Henderson-Hasselbalch equation:
pOH = pKb + log([CH₃NH₃⁺]/[ CH₃NH₂])
pOH = -log(4.2 × 10⁻⁴) + log(1) = 3.38
pH = Kw - pOH = 14.00 - 3.38 = 10.62


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e)
At the equivalent point, CH₃NH₂ is completely converted to CH₃NH₃⁺.
[CH₃NH₃⁺]ₒ = 0.175 × {25.0/(25.0 + 29.2)} M = 0.0807 M

Consider the dissociation of CH₃NH₃⁺:
____________ CH₃NH₃⁺(aq) ⇌ CH₃NH₂(aq) + H⁺(aq) ___ Ka = (1.0 × 10⁻¹⁴) / (4.2 × 10⁻⁴)
Initial: _______ 0.0807 M ______ 0 M _____ 0 M
Change: _______ -z M ________ +z M _____ +z M
At eqm: ____ (0.0807 - z) M ____ z M ______ z M

Ka is very small, and thus the dissociation of CH₃NH₃⁺ ion is to a very small extent.
It is assumed that 0.0807≫ z, and thus [CH₃NH₃⁺] = (0.0807 - z) M ≈ 0.0807 M

Ka = [CH₃NH₂] [H⁺] / [CH₃NH₃⁺]
(1.0 × 10⁻¹⁴) / (4.2 × 10⁻⁴) = z² / 0.0807
z = √{0.0807 × (1.0 × 10⁻¹⁴) / (4.2 × 10⁻⁴)} = 1.39 × 10⁻⁶ M
pH = -log[H⁺] = -log(1.39 × 10⁻⁶) = 5.86


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f)
After added 5mL of acid beyond equivalence point:
Total volume = (25.0 + 29.2 + 5) mL = 59.2 mL
[H⁺] = 0.150 × (29.2/59.2) M = 0.0740 M
pH = -log[H⁺] = -log(0.0740) = 1.13