question is in the description

Question

Suppose 0.850 L of 0.400 M H2SO4 is mixed with 0.800 L of 0.250 M KOH. What concentration of sulfuric acid remains after neutralization

Answer 1

H₂SO₄ + 2KOH ==> K₂SO₄ + 2H₂O ... balanced equation

moles H₂SO₄ = 0.850 L x 0.400 mols / L = 0.340 mols H₂SO₄

moles KOH = 0.800 L x 0.250 mols / L = 0.200 mols KOH

moles H₂SO₄ reacted (used up) = 0.200 mols KOH x 1 mol H₂SO₄ / 2 mols KOH = 0.1 mols used

moles H₂SO₄ remaining after reaction = 0.340 mols - 0.100 mols = 0.240 mols H₂SO₄ left over

Final volume of solution = 0.850 L + 0.800 L = 1.65 L

Concentration H₂SO₄ after neutralization = 0.240 mols / 1.65 L = 0.145 M