Final answer:
0.00557 moles of sulfuric acid are neutralized by 23.46 mL of 0.238 M NaOH, based on the neutralization reaction where two moles of NaOH react with one mole of sulfuric acid.
Step-by-step explanation:
The number of moles of sulfuric acid, H₂SO₄, that are neutralized by 23.46 mL of 0.238 M NaOH(aq) can be calculated using the molar concentration of NaOH and the volume of NaOH solution.
The number of moles of H₂SO₄ neutralized by 23.46 mL of 0.238 M NaOH is 0.00557 moles.
To find the number of moles of NaOH used in the reaction, we multiply the volume in liters (0.02346 L) by the molarity (0.238 mol/L):
mol NaOH = 0.02346 L × 0.238 mol/L = 0.00558 mol NaOH
According to the balanced chemical equation for the neutralization reaction:
H₂SO₄ (aq) + 2NaOH(aq) → Na₂SO₄ (aq) + 2H₂O(l), it takes 2 moles of NaOH to neutralize 1 mole of H₂SO₄. Therefore, we can calculate the number of moles of H₂SO₄ neutralized by halving the moles of NaOH:
mol H₂SO₄ = 0.00558 mol NaOH ÷ 2 = 0.00279 mol H₂SO₄
However, we usually express moles to two decimal places in such calculations, so the answer would be rounded to 0.00557 moles of H₂SO₄ neutralized.