What volume of 0.410 M KOH is needed to react completely with 23.3 mL of 0.280 M H2SO4 ?

Question

The balanced equation for the neutralization reaction of aqueous H2SO4H2⁢SO4 with aqueous KOHKOH is shown.

H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)

Answer 1

H₂SO₄(aq) + 2KOH(aq) ==> 2H₂O(l) + K₂SO₄(aq) ... balanced equation

moles H₂SO₄ present = 23.3 ml x 1 L / 1000 ml x 0.280 mol / L = 6.52x10^-3 moles H₂SO₄

From the balanced equation, it will take 2 mols KOH for each mol of H₂SO₄

moles KOH needed = 6.52x10^-3 mol H₂SO₄ x 2 mol KOH / mol H₂SO₄ = 0.013048 mol KOH needed

The KOH is 0.410 M meaning 0.410 moles / liter

Volume KOH needed = 0.013048 mol KOH x 1 L / 0.410 mol = 0.0318 L KOH = 31.8 ml KOH