chem help?

b)
Initial moles of HCOOH = (0.15 mol/L) × (25.0/1000 L) = 0.00375 mol

HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
Moles of OH⁻ added = (0.20 mol/L) × (13.00/1000 L) = 0.0026 mol
After addition of 13.00 mL NaOH:
Moles of HCOOH = (0.00375 - 0.0026) mol = 0.00115 mol
Moles of HCOO⁻ = 0.0026 mol
[HCOO⁻]/[HCOOH] = (Moles of HCOO⁻)/(Moles of HCOOH) = 0.0026/0.00115

Consider the dissociation of HCOOH in the final solution:
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(0.0026/0.00115)
pH = 4.10

====
c)
Initial moles of HCOOH = 0.00375 mol

HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
Moles of OH⁻ added = (0.20 mol/L) × (18.75/1000 L) = 0.00375 mol
Both HCOOH and OH⁻ completely react to give 0.00375 mol of HCOO⁻.
[HCOO⁻] = (0.00375 mol) / [(25.0 + 18.75)/1000 L] = 0.0857 M

Kb for HCOO⁻ = Kw/Ka(HCOOH) = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁴) = 5.56 × 10⁻¹¹

Consider the dissociation of HCOO⁻ in the final solution:
_________ HCOO⁻(aq) + H₂O(l) ⇌ HCOOH(aq) + OH⁻(aq) … Kb = 5.56 × 10⁻¹¹
Initial: ____ 0.0857 M ___________ 0 M _______ 0M
Change: ____ -w M ____________ +w M ______ +w M
Eqm: ___ (0.0857 - w) M _________ w M ______ w M
________ ≈ 0.0857 M

At equilibrium:
Kb = [HCOOH] [OH⁻] / [HCOO⁻]
5.56 × 10⁻¹¹ = w² / 0.0857
w = √(0.0857 × 5.56 × 10⁻¹¹) = 2.18 × 10⁻⁶ M

pOH = -log[OH⁻] = -log(2.18 × 10⁻⁶) = 5.66
pH = pKw - pOH = 14.00 - 5.66 = 8.34

====
c)
Initial moles of HCOOH = 0.00375 mol

HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
Moles of OH⁻ added = (0.20 mol/L) × (25.00/1000 L) = 0.005 mol > 0.00375 mol
Hence, OH⁻ is in excess.
[OH⁻] in the final solution = [(0.005 - 0.00375) mol] / [(25.0 + 25.00)/1000 L] = 0.025 M

pOH = -log[OH⁻] = -log(0.025) = 1.60
pH = pKw - pOH = 14.00 - 1.60 = 12.40