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a)
_         HCOOH(aq) + H₂O(ℓ) ⇌ HCOO⁻(aq) + H₃O⁺(aq)   Ka = 1.8 × 10⁻⁴
Initial:    0.1225 M                       0 M                0 M
Change:    -y M                         +y M              +y M
Eqm:   (0.1225 - y) M                  y  M               y M

At equilibrium:
Ka = [HCOO⁻] [H₃O⁺] / [HCOOH]
1.8 × 10⁻⁴ = y² / (0.1225 - y)
y² + (1.8 × 10⁻⁴)y - 2.205 × 10⁻⁵ = 0
y = {-(1.8 × 10⁻⁴) + √[(1.8 × 10⁻⁴)² + 4 × (2.205 × 10⁻⁵)]} / 2
y = 4.61 × 10⁻³
pH = -log[H₃O⁺] = -log(4.61 × 10⁻³) = 2.34

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b)
Initial moles of HCOOH = (0.1225 mol/L) × (10.00/1000 L) = 0.001225 mol
Moles of OH⁻ added = (0.08220 mol/L) × (7.45/1000 L) = 0.0006124 mol

HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
After addition of NaOH, [HCOO⁻]/[HCOOH]
= (Moles of HCOO⁻)/(Moles of HCOOH)
= 0.0006124/(0.001225 - 0.0006124)
= 0.0006124/0.0006126
≈ 1

Consider the dissociation of HCOOH:
HCOOH(aq) + H₂O(ℓ) ⇌ HCOO⁻(aq) + H₃O⁺(aq) … Ka = 1.8 × 10⁻⁴

Henderson-Hasselbalch equation:
pH = pKa + log([HCOO⁻]/[HCOOH])
pH = -log(1.8 × 10⁻⁴) + log(1) = 3.74

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c)
Initial moles of HCOOH = 0.001225 mol
Moles of OH⁻ added = (0.08220 mol/L) × (18.50/1000 L) = 0.001521 mol

Volume of the final solution = (10.00 + 18.50) mL = 28.50 mL = 0.02850 L
[OH⁻] in the final solution = [(0.001521 - 0.001225) mol] / (0.0285 L) = 0.0104 M

pOH = -log[OH⁻] = -log(0.0104) = 1.98
pH = pKw - pOH = 14.00 - 1.98 = 12.02