0.280 gram of koh will just neutralize what volume of 0.200 m h2so4?

The neutralization reaction is expressed as:

2KOH + H2SO4 = K2SO4 + 2H2O

We are given the amount of the KOH that is used in the reaction. We use this and the relations from the reaction to determine the required acid.

0.280 g ( 1 mol / 56.11 g ) ( 1 mol H2SO4 / 2 mol KOH ) = 0.0025 mol H2SO4

Volume needed = 0.0025 mol / ( 0.200 mol / L ) = 0.0125 L or 12.5 mL of the H2SO4 is needed.