Question

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

H₂SO₄(aq) + 2 KOH(aq) + 2H₂O(1) + K₂SO₄ (aq)
What volume of 0.470 M KOH is needed to react completely with 18.3 mL of 0.255 MH₂SO₄?
volume: ____ ml

Answer 1

Approximately 19.86 mL of a 0.470 M KOH solution is required to neutralize 18.3 mL of 0.255 M H₂SO₄ in this neutralization reaction.

Step-by-step explanation:

To calculate the volume of 0.470 M KOH needed to completely react with 18.3 mL of 0.255 M H₂SO₄, we first determine the moles of H₂SO₄ using its concentration and volume. Since the balanced chemical equation shows that 2 moles of KOH react with 1 mole of H₂SO₄, we can then find the required moles of KOH. Finally, using the concentration of KOH, we calculate the required volume.

Moles of H₂SO₄ = 18.3 mL * 0.255 mol/L = 0.0046665 mol

Moles of KOH required = 2 * Moles of H₂SO₄ = 2 * 0.0046665 mol = 0.009333 mol

Volume of KOH needed = Moles of KOH / Molarity of KOH = 0.009333 mol / 0.470 mol/L = 0.0198585 L = 19.86 mL

Therefore, approximately 19.86 mL of the 0.470 M KOH solution is required to neutralize 18.3 mL of 0.255 M H₂SO₄.