925 ml of a 0.843M HNO3 solution is added to 1.5L of a 0.540M KOH solution. What is the pH of the mixture when the reaction is complete?

HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)
OR: Mole ratio HNO₃ : KOH = 1 : 1

No. of moles of HNO₃ = (0.843 mol/L) × (925/1000 L) = 0.7798 mol
No. of moles of KOH = (0.540 mol/L) × (1.5 L) = 0.8100 mol > 0.7798 mol
Hence, KOH is in excess, and HNO₃ completely reacts.

No. of moles of HNO₃ reacted = 0.7798 mol
No. of moles of KOH reacted = 0.7798 mol
No. of moles of KOH left in the final solution = (0.8100 - 0.7798) mol = 0.0302 mol
Volume of the final solution = [(925/1000) + 1.5] L = 2.425 L
Molarity of KOH in the solution = (0.0302 mol) / (2.425 L) = 0.0125 M

1 mole of KOH dissociates to give 1 mole of OH⁻ ions.
In the final solution, [OH⁻] = 0.0125 M
pOH = -log[OH⁻] = -log(0.0125) = 1.9
pH = 14 - pOH = 14 - 1.9 = 12.1