Final answer:
To neutralize 225 mL of 3.0 M sulfuric acid, 1125 mL of 1.20 M sodium hydroxide is required.
Step-by-step explanation:
Neutralization Calculation
To determine how much 1.20 M NaOH is needed to neutralize 225 mL of 3.0 M H₂SO₄, we use the balanced chemical reaction:
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
First, we calculate the number of moles of H₂SO₄:
- Moles of H₂SO₄ = Molarity (M) × Volume (L) = 3.0 M × 0.225 L = 0.675 moles
Since each mole of H₂SO₄ requires 2 moles of NaOH for neutralization, we have:
- Moles of NaOH needed = 0.675 moles of H₂SO₄ × 2 = 1.35 moles
Then, we calculate the volume of 1.20 M NaOH required:
- Volume of NaOH = Moles of NaOH ÷ Molarity of NaOH = 1.35 moles ÷ 1.20 M = 1.125 L
Therefore, to neutralize 225 mL of 3.0 M H₂SO₄, 1.125 liters or 1125 mL of 1.20 M NaOH is needed.