Calculate the pH after 0.15 mole of NaOH is added to 1.10 L of a solution...?

0.15 mole of NaOH :

HF + NaOH → NaF + H₂O
Addition of each mole of NaOH converts 1 mole of HF to 1 mole of F⁻.
Moles of HF after reaction = (0.56 mol/L) × (1.10 L) - (0.15 mol) = 0.466 mol
Moles of F⁻ after reaction = (1.18 mol/L) × (1.10 L) + (0.15 mol) = 1.448 mol
After reaction, [F⁻]/[HF] = (Moles of F⁻)/(Moles of HF) = 1.448/0.466

Consider the dissociation of HF:
HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq) …… Ka = 7.2 × 10⁻⁴

Henderson-Hasselbalch equation: pH = pKa + log([F⁻]/[HF])
pH = -log(7.2 × 10⁻⁴) + log(1.448/0.466) = 3.6

====
0.30 mole of HCl :

NaF + HCl → NaCl + HF
Addition of each mole of HCl converts 1 mole of F⁻ to 1 mole of HF.
Moles of F⁻ after reaction = (1.18 mol/L) × (1.10 L) - (0.30 mol) = 0.998 mol
Moles of HF after reaction = (0.56 mol/L) × (1.10 L) + (0.30 mol) = 0.916 mol
After reaction, [F⁻]/[HF] = (Moles of F⁻)/(Moles of HF) = 0.998/0.916

Consider the dissociation of HF:
HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq) …… Ka = 7.2 × 10⁻⁴

Henderson-Hasselbalch equation: pH = pKa + log([F⁻]/[HF])
pH = -log(7.2 × 10⁻⁴) + log(0.998/0.916) = 3.2