55.00 mL of 1.482 M H2SO4 solution and 10.00 mL of 0.269 M NaOH solution are combined, and then diluted with water until the total volume is 250.00 mL. What is the molarity of the final H2SO4 solution?

Volume of H₂SO₄ solution = 55.00 mL = 55/1000 = 0.055 L

Molarity of H₂SO₄ solution = 1.482 M

Molarity = moles / solution in Liter

Moles of H₂SO₄ = Molarity x solution in Liter

= 1.482 M x 0.055 L

= 0.08151

Volume of NaOH solution = 10.00 mL = 10/1000 = 0.01 L

Molarity of NaOH solution = 0.269 M

Moles of NaOH = 0.269 M x 0.01 L = 2.96 x 10⁻³

Now writing a balanced equation for the chemical reaction between H₂SO₄ and NaOH:

H₂SO₄ + 2NaOH - -> Na₂SO₄ + 2H₂O

So 2 moles of NaOH reacts with 1 mole of H₂SO₄. NaOH being least in the number of moles will be the limiting reactant.

So 2.96 x 10⁻³ moles of NaOH will neutralize 2.96 x 10⁻³ moles of H₂SO₄.

Unreacted moles of H₂SO₄ = 0.08151 - 2.96 x 10⁻³ = 0.07855

Total volume of the solution = 250 ml = 0.25 L

Thus, the molarity of the final H₂SO₄ solution = moles / Liter of solution

= 0.07855 moles / 0.25 L = 0.3142 M