50 mL of 2M H2SO4 are combined with 100 mL of 1M KOH. What is the pOH of the resulting solution?

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
1 mole of H₂SO₄ reacts with 2 moles of KOH.

Initial number of moles of H₂SO₄ = 2 × (50/1000) = 0.1 mol
Initial number of moles of KOH = 1 × (100/1000) = 0.1 mol

0.1 mole of H₂SO₄ needs 0.1 × 2 = 0.2 mole of KOH for complete reaction.
Thus, KOH is the limiting reactant (completely reacts).

No. of moles of KOH reacted = 0.1 mol
No. of moles of H₂SO4 reacted = 0.1 × (1/2) = 0.05 mol
No. of moles of H₂SO₄ unreacted = 0.1 - 0.05 = 0.05 mol
H⁺ ions left in the solution = 0.05 × 2 = 0.1 mol
Volume of the final solution = (50 + 100)/1000 L = 0.15 L

In the final solution :
[H⁺] = 0.1/0.15 = 0.667 M
pH = -log(0.667) = 0.18
OH = 14 - 0.18 = 13.82

50 mL x 2 M = 100 mmol of H2SO4 and 200mmol of H^+1
100 mL x 1 M = 100 mmol of KOH and 100 mmol of OH^-1
After reaction of H^+1 and OH^-1 to form H2O, 100 mmol of H^+1 remains in 150 mL of soln.
100 mmol / 150 mL = 0.667 M H^+1
pH = 0.18
pOH = 13.82