Final answer:
To neutralize 25.00 mL of 0.450 m potassium hydroxide, 22.5 milliliters of 0.250m sulfuric acid are needed.
Step-by-step explanation:
To determine how many milliliters of 0.250m sulfuric acid are required to neutralize 25.00 ml of 0.450 m potassium hydroxide, we need to use the concept of titration and stoichiometry from acid-base reactions. The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is:
H2SO4 (aq) + 2KOH (aq) → K2SO4 (aq) + 2H2O (l)
This equation tells us that one mole of sulfuric acid reacts with two moles of potassium hydroxide. First, we calculate the number of moles of KOH:
Number of moles of KOH = Volume × Molarity = 25.00 mL × 0.450 m = 0.01125 mol
Because the ratio of moles of H2SO4 to KOH is 1:2, the number of moles of sulfuric acid needed is half the number of moles of KOH:
Number of moles of H2SO4 required = 0.01125 mol × (1 mol H2SO4/2 mol KOH) = 0.005625 mol
Finally, we find the volume of 0.250 M H2SO4 needed:
Volume of H2SO4 = Moles of H2SO4 × (1 L of solution/0.250 mol H2SO4) × 1000 mL/L = 0.005625 mol × 4000 mL/mol = 22.5 mL
Therefore, 22.5 milliliters of 0.250m sulfuric acid are required to neutralize 25.00 ml of 0.450 m potassium hydroxide.