If you have 500mls of 10N NaOH, how much 2.5N Sulfuric Acid is needed for neutralization?

To determine the amount of 2.5N sulfuric acid needed for neutralization of 500 mL of 10N sodium hydroxide, you can use the concept of normality and the stoichiometry of the neutralization reaction.

Sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) react in a 2:1 molar ratio because sulfuric acid can donate two protons (H⁺ ions) per molecule, while sodium hydroxide provides one hydroxide ion (OH⁻) per molecule. This means that for every 2 equivalents of NaOH, you will need 1 equivalent of H₂SO₄.

First, calculate the equivalents of NaOH in 500 mL of 10N solution:

Equivalents of NaOH = Volume (L) × Normality (N) = 0.5 L × 10 N = 5 equivalents of NaOH

Since sulfuric acid donates two protons, you will need half the number of equivalents of H₂SO₄ for neutralization:

Equivalents of H₂SO₄ needed = 5 equivalents of NaOH ÷ 2 = 2.5 equivalents