Would the following working help?
Balanced equation:
H2SO4(l) + 2 KOH - - - > K2SO4(aq) + 2 H2O(l)
Available H2SO4 = 20 mL x 0.08 mol/L = 1.6 mmol of H2SO4
Available KOH = 25 mL x 0.35 mol/L = 8.75 mmol of KOH
Since the available H2SO4 (1.6 mmol) < the available KOH (8.75mol), so, H2SO4 acts as the LIMITING REAGENT.
From the above equation, mole ratio of H2SO4 and KOH = 1 : 2.
Here, 0.16 mmol of H2SO4 will stoichiometrically react with 2 x 0.16 mmol of KOH = 0.32 mmol of KOH. Thus, the remaining KOH = (8.75 mmol — 0.32 mmol) = 5.55 mmol of KOH or 5.55 x 10-³ mol.
Since, the volume of total solution = 20 mL + 25 mL = 45 mL, so the concentration of KOH = 5.55 x 10-³ mol/45 mL solution or (1000 mL/45 mL) x 5.55 x 10-³ mol/L = 0.1233 mol/L.
KOH is a strong acid, so in water, it will completely dissociate to form OH- ion and K+ ion.
KOH(aq) - - - > K+(aq) + OH-(aq)
0.1233 mol of KOH ~ 0.1233 mol of OH- ion - - - > [OH-] = 0.1233 mol/L.
pOH = — log [OH-]
pOH = — log 0.1233
Solving for pOH, we get,
pOH = 0.91
pH + pOH = 14
pH = 13.09
Thus, the pH of the resulting solution = 13.09, i.e. (C).
