Question

The balanced equation for the neutralization reaction of aqueous H2SO4 with aqueous KOH is shown.

H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)

What volume of 0.130 M KOH is needed to react completely with 11.0 mL of 0.155 M H2SO4?

Answer 1

The balanced chemical equation for the neutralization of H₂SO₄ with KOH is given as

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

As per the stoichiometric equation,

1 mole H₂SO₄ = 2 moles KOH.

Millimoles H₂SO₄ = (volume of H₂SO₄ in mL)*(molarity of H₂SO₄)

= (11.0 mL)*(0.155 M) = 1.705 mmol.

Millimoles KOH needed to neutralize 1.705 mmol H₂SO₄

= (1.705 mmol H₂SO₄)*(2 moles KOH)/(1 mole H₂SO₄)

= 3.410 mmol.

Volume of 0.130 M KOH needed to neutralize 3.410 mmol H₂SO₄

= (millimoles H₂SO₄)/(molarity of H₂SO₄)

= (3.410 mmol)/(0.130 M)

= 26.23 mL ≈ 26.2 mL (ans, correct to 3 sig. figs).