Final answer:
To neutralize 275 mL of 0.500 M sulfuric acid, one would need 366.7 mL of 0.750 M sodium hydroxide according to stoichiometry and the balanced chemical equation for the reaction.
Step-by-step explanation:
The question asks for the volume of sodium hydroxide (NaOH) required to neutralize a given volume of sulfuric acid (H₂SO₄). We can solve this using stoichiometry and the balanced chemical equation which, for the reaction between sodium hydroxide and sulfuric acid, is:
H₂SO₄ (aq) + 2NaOH (aq) → Na₂SO₄ (aq) + 2H₂O (1)
The balanced equation shows that two moles of NaOH react with one mole of H₂SO₄. We are starting with 275 mL of 0.500 M H₂SO₄, so let's find out how many moles of H₂SO₄ we have:
- Moles of H₂SO₄ = Molarity of H₂SO₄ * Volume of H₂SO₄ in liters = 0.500 mol/L * 0.275 L = 0.1375 mol H₂SO₄.
- Now, remember the mole ratio from the balanced equation is 1:2, meaning you need twice as many moles of NaOH. Thus, we need 0.1375 mol H₂SO₄ * 2 = 0.275 mol NaOH.
- Finally, we can find the required volume of NaOH by using its molarity: Volume of NaOH = Moles of NaOH / Molarity of NaOH = 0.275 mol / 0.750 M = 0.3667 L, which is 366.7 mL.
Therefore, the volume of 0.750 M sodium hydroxide needed to neutralize 275 mL of 0.500 M sulfuric acid is 366.7 mL.