A 1.0 L buffer containing 0.118 mol L-1 HOCl and 0.103 mol L-1 OCl-. What is the pH of the solution after adding 7.7 x 10-3 mol of HCl?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka(HOCl) = 2.9 × 10⁻⁸
(The formula HOCl can be written as HClO.)
(Different Ka values taken may give slightly different answers.)

On addition of each mole of HCl, 1 mole of OCl⁻ is converted to 1 mole of HOCl.
Mole of OCl⁻ after reaction = (0.103 mol L⁻¹) × (1 L) - (7.7 × 10⁻³ mol) = 0.0953 mol
Moles of HOCl after reaction = (0.118 mol L⁻¹) × (1 L) + (7.7 × 10⁻³ mol) = 0.1257 mol
After reaction, [OCl⁻]/[HOCl] = (Moles of OCl⁻)/(Moles of HOCl) = 0.0953/0.1257

Consider the dissociation of HOCl:
HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + OCl⁻(aq) …… Ka = 2.9 × 10⁻⁸

Henderson-Hasselbalch equation: pH = pKa + log([OCl⁻]/[HOCl])
pH = -log(2.9 × 10⁻⁸) + log(0.0953/0.1257) = 7.4