a.
Initial moles of (CH₃CH₂)₃N = (0.1000 mol/L) × (20.00/1000 L) = 0.002 mol
Moles of HCl added = (0.1000 mol/L) × (13.00/1000 L) = 0.0013 mol
Addition of 1 mole of HCl reacts with 1 mole of (CH₃CH₂)₃N to 1 mole of (CH₃CH₂)₃NH⁺.
After addition of 13.00 mL of HCl:
Moles of (CH₃CH₂)₃N left unreacted = (0.002 - 0.0013) mol = 0.0007 mol
Moles of (CH₃CH₂)₃NH⁺ produced = 0.0013 mol
Hence, [(CH₃CH₂)₃NH⁺]/[(CH₃CH₂)₃N] = 0.0013/0.0007 = 13/7
Consider the dissociation of (CH₃CH₂)₃N:
(CH₃CH₂)₃N(aq) + H₂O(l) ⇌ (CH₃CH₂)₃NH⁺(aq) + OH⁻(aq) …… Kb = 5.2 × 10⁻⁴
Henderson-Hasselbalch equation:
pOH = pKb + log([(CH₃CH₂)₃NH⁺]/[(CH₃CH₂)₃N])
pOH = -log(5.2 × 10⁻⁴) + log(13/7) = 3.55
pH = pKw - pOH = 14.00 - 3.55 = 10.45
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b.
Initial moles of (CH₃CH₂)₃N = 0.002 mol
Moles of HCl added = (0.1000 mol/L) × (20.80/1000 L) = 0.00208 mol
HCl is in excess.
Moles of HCl left unreacted = (0.00208 - 0.002) mol = 0.00008 mol
Volume of the final solution = (20.00 + 20.80) mL = 40.8 mL = 0.0408 L
In the final solution:
[H₃O⁺] = [HCl] = (0.00008 mol) / (0.0408 L) = 0.00196
pH = -log[H₃O⁺] = -log(0.00196) = 2.71
====
c.
Initial moles of (CH₃CH₂)₃N = 0.002 mol
Moles of HCl added = (0.1000 mol/L) × (29.00/1000 L) = 0.0029 mol
HCl is in excess.
Moles of HCl left unreacted = (0.0029 - 0.002) mol = 0.0009 mol
Volume of the final solution = (20.00 + 29.00) mL = 49 mL = 0.049 L
In the final solution:
[H₃O⁺] = [HCl] = (0.0009 mol) / (0.049 L) = 0.0184
pH = -log[H₃O⁺] = -log(0.0184) = 1.74
