titrating buffers ?

a)
Refer to https://depts.washington.edu/eooptic/links/acidstrength.html
Ka(CH₃COOH) = 1.8 × 10⁻⁵
(The above value may be slightly different from different sources.)

_       CH₃CO₂H(aq) + H₂O(ℓ) ⇌ CH₃CO₂⁻(aq) + H₃O⁺(aq)   Ka = 1.8 × 10⁻⁵
Initial:     0.20 M            0 M           0M
Change:    -y M            +y M        +y M
Eqm:    (0.20 - y) M        y M          y M
_           ≈ 0.20 M

At equilibrium:
Ka = [CH₃CO₂⁻] [H₃O⁺] / [CH₃CO₂H]
1.8 × 10⁻⁵ = y² / 0.20
y = √(0.20 × 1.8 × 10⁻⁵) = 1.90 × 10⁻³
pH = -log[H₃O⁺] = -log(1.90 × 10⁻³) = 2.72

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b)
Initial moles of CH₃CO₂H = (0.20 mol/L) × (50.0/1000 L) = 0.01 mol
Moles of OH⁻ added = (0.40 mol/L) × (12.5/1000 L) = 0.005 mol

CH₃CO₂H(aq) + OH⁻(aq) → CH₃CO₂⁻(aq) + H₂O(ℓ)
After addition of NaOH, [CH₃CO₂⁻]/[ CH₃CO₂H]
= (Moles of CH₃CO₂⁻)/(Moles of CH₃CO₂H)
= 0.005/(0.01 - 0.005)
= 1

Consider the dissociation of CH₃CO₂H:
CH₃CO₂H(aq) + H₂O(ℓ) ⇌ CH₃CO₂⁻(aq) + H₃O⁺(aq)   Ka = 1.8 × 10⁻⁵

Henderson-Hasselbalch equation:
pH = pKa + log([CH₃CO₂⁻]/[ CH₃CO₂H])
pH = -log(1.8 × 10⁻⁵) + log(1) = 4.74

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c)
Initial moles of CH₃CO₂H = 0.01 mol
Moles of OH⁻ added = (0.40 mol/L) × (40.0/1000 L) = 0.016 mol

CH₃CO₂H(aq) + OH⁻(aq) → CH₃CO₂⁻(aq) + H₂O(ℓ)
Volume of the final solution = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L
[OH⁻] in the final solution = [(0.016 - 0.01) mol] / (0.0900 L) = 0.0667 M

pOH = -log[OH⁻] = -log(0.0667) = 1.18
pH = pKw - pOH = 14.00 - 1.18 = 12.82