chem help?

a)
Initial moles of C₃H₇CO₂H = (0.250 mol/L) × (20.00/1000 L) = 0.005 mol
Moles of OH⁻ added = (0.125 mol/L) × (20.00/1000 L) = 0.0025 mol

C₃H₇COOH(aq) + OH⁻(aq) → C₃H₇COO⁻(aq) + H₂O(ℓ)
After addition, [C₃H₇COO⁻]/[C₃H₇COOH]
= (Moles of C₃H₇COO⁻)/(Moles of C₃H₇COOH)
= 0.0025/(0.005 - 0.0025)
= 1

Consider the dissociation of C₃H₇COOH:
C₃H₇COOH(aq) + H₂O(ℓ) ⇌ C₃H₇COO⁻(aq) + H₃O⁺(aq) … Ka

Henderson-Hasselbalch equation:
pH = pKa + log([C₃H₇COO⁻]/[C₃H₇COOH])
pH = -log(1.3 × 10⁻⁵) + log(1)
pH = 4.89

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b)
Initial moles of C₃H₇CO₂H = (0.250 mol/L) × (20.00/1000 L) = 0.005 mol
Moles of OH⁻ added = (0.125 mol/L) × (40.00/1000 L) = 0.005 mol

C₃H₇COOH(aq) + OH⁻(aq) → C₃H₇COO⁻(aq) + H₂O(ℓ)
Both C₃H₇COOH and OH⁻ completely react.
Volume of the final solution = (20.00 + 40.00) mL = 60.00 mL = 0.06 L
[C₃H₇COO⁻] in the final solution = (0.005 mol) / (0.06 L) = 0.0833 M

Kb(C₃H₇COO⁻) = Kw / Ka(C₃H₇COOH) = (1.00 × 10⁻¹⁴) / (1.3 × 10⁻⁵) = 7.69 × 10⁻¹⁰

Consider the dissociation of C₃H₇COO⁻:
______ C₃H₇COO⁻(aq) + H₂O(ℓ) ⇌ C₃H₇COOH(aq) + OH⁻(aq) ___ Kb = 7.69 × 10⁻¹⁰
Initial: __ 0.0833 M _______________ 0 M ________ 0 M
Change: ___ -y M _______________ +y M _______ +y M
Eqm: _ (0.0833 - y) M _____________ y M ________ y M

As Kb is very small, the dissociation of C₃H₇COO⁻ is to a very small extent.
It can be assumed that0.0833 ≫ y
Equilibrium [C₃H₇COO⁻] = (0.0833 - y) M ≈ 0.0833 M

At equilibrium:
Kb = [C₃H₇COOH] [OH⁻] / [C₃H₇COO⁻]
7.69 × 10⁻¹⁰ = y² / 0.0833
y = √(0.0833 × 7.69 × 10⁻¹⁰)
y = 8.00 × 10⁻⁶

pOH = -log[OH⁻] = -log(8.00 × 10⁻⁶) = 5.10
pH = pKw - pOH = 14.00 - 5.10 = 8.90

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c)
Initial moles of C₃H₇CO₂H = (0.250 mol/L) × (20.00/1000 L) = 0.005 mol
Moles of OH⁻ added = (0.125 mol/L) × (50.00/1000 L) = 0.00625 mol

C₃H₇COOH(aq) + OH⁻(aq) → C₃H₇COO⁻(aq) + H₂O(ℓ)
Volume of the final solution = (20.00 + 50.00) mL = 70.00 mL = 0.07 L
[OH⁻] in the final solution = [(0.00625 - 0.005) mol] / (0.07 L) = 0.0179 M

pOH = -log[OH⁻] = -log(0.0179) = 1.75
pH = pKw - pOH = 14.00 - 1.75 = 12.25