0.450 L of 0.440 M H2SO4 is mixed with 0.400 L of 0.230 M KOH. What concentration of sulfuric acid remains after neutralization?

H₂SO₄(aq)+ 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
OR: Mole ratio H₂SO₄ : KOH = 1 : 2

Initial number of moles of H₂SO₄ = (0.440 mol/L) × (0.450 L) = 0.198 mol
Initial number of moles of KOH = (0.230 mol/L) × (0.400 L) = 0.092 mol

When KOH is completely reacted :
Number of moles of KOH reacted = 0.092 mol
Number of moles of H₂SO₄ reacted = (0.092 mol) × (1/2) = 0.046 mol
Number of moles of H₂SO₄ unreacted = (0.198 - 0.046) = 0.152 mol
Volume of the final solution = (0.450 + 0.400 L) = 0.850 L
Concentration of H₂SO₄ remains after neutralization = (0.152 mol) / (0.850 L) = 0.179 M

H2SO4(aq)+ 2 KOH(aq) ----> K2SO4(aq) + 2 H2O(l).

Reaction stoichiometry H2SO4 1 : 2 KOH.

0.400 L of 0.230 mol KOH/L = 0.092 moles KOH.

0.092 moles * 1/2 = 0.046 moles H2SO4 reacted.

Total H2SO4 = 0.450 L * 0.440 mol/L = 0.198 moles.

Unreacted H2SO4 = 0.198 moles - 0.046 moles = 0.152 moles.

Total volume solution = 0.850 L.

Concentration H2SO4 = 0.152 mol / 0.850 L = 0.179 M.