What will be the pH of the solution?

KOH(aq) + HI(aq) → KI(aq) + H₂O(l)
Mole ratio KOH : HI = 1 : 1

Initial number of moles of KOH = (0.3450 mol/L) × (30.00/1000 L) = 0.01035 mol
Initial number of moles of HI = (0.5000 mol/L) × (12.00/1000 L) = 0.00600 mol

Obviously, HI is the limiting reactant, and KOH is in excess.
Number of moles of HI reacted = 0.006000 mol
Number of moles of KOH reacted = 0.006000 mol
Number of moles of KOH left = (0.01035 - 0.006000) mol = 0.004350 mol

Volume of final solution = (30.00 + 12.00) mL = 42.00 mL = 0.04200 L
[KOH] = (0.004350 mol) / (0.04200 L) = 0.1036 M

KOH completely dissociates in water to give OH⁻ ions.
[OH⁻] = [KOH] = 0.1036 M
pOH = -log[OH⁻] = -log(0.1036) = 0.98
pH = 14 - pOH = 14 - 0.98 = 13.02

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