You have 75.0 mL of 0.125 M HNO3 to which you add 75.0 mL of 0.100 M KOH. What is the resulting pH of the solution?

HNO₃ + KOH → KNO₃ + H₂O
Mole ratio HNO₃ : KOH = 1 : 1

No. of moles of HNO₃ = (0.125 mol/L) × (75.0/1000 L) = 0.009375 mol
No. of moles of KOH = (0.100 mol/L) × (75.0/1000 L) = 0.0075 mol < No. of moles of HNO₃

No. of moles of excess KOH = (0.009375 - 0.0075) mol = 0.001875 mol

Each mole of KOH contains 1 mole of OH⁻ ions.
No. of moles of excess OH⁻ = 0.001875 mol
Volume of the final solution = (75.0 mL) × 2 = 150 mL = 0.150 L
[OH⁻] = (0.001875 mol) / (0.150 L) = 0.0125 M

Resulting pOH = -log[OH⁻] = -log(0.0125) = 1.9
Resulting pH = pKw - pOH = 14.0 - 1.9 = 12.1