0.150 L of 0.470 M H2SO4 is mixed with 0.100 L of 0.200 M KOH. What concentration of sulfuric acid remains after neutralization?

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
OR: Mole ratio H₂SO₄ : KOH = 1 : 2

Initial number of moles of H₂SO₄ = (0.470 mol/L) × (0.150 L) = 0.0705 mol
Initial number of moles of KOH = (0.200 mol/L) × (0.100 L) = 0.0200 mol

When KOH completely reacts :
Number of moles of H₂SO₄ reacted = (0.0200 mol) × (1/2) = 0.0100 mol
Number of moles of H₂SO₄ remains = (0.0705 - 0.0100) mol = 0.0605 mol
Volume of the final solution = (0.150 + 0.100) L = 0.250 L
Concentration of H₂SO₄ remains = (0.0605 mol) / (0.250 L) = 0.242 M