When acids and/or bases are neutralized, the acidic component (generally H+) combines with the basic component (often OH-). When H+ and OH- combine, water forms, and a salt forms from the combination of the opposing cation and anion.
In this case, the neutralization reaction follows the above scheme:
H2SO4 + 2 KOH --> K2SO4 + 2 H2O
Note that the reaction above is balanced by adding coefficients to equalize the number of each atom on both sides.
Next, we can follow the stoichiometry in the above reaction to determine how many moles of each reactant are consumed. Since H2SO4 is a strong acid, the reaction will proceed until at least one reactant is fully consumed.
0.250 L * 0.500 mol/L = 0.125 mol
0.200 L * 0.240 mol/L = 0.048 mol
0.250 L + 0.200 L = 0.450 L
With the phrasing of the question, let's assume that all of the KOH is consumed and some H2SO4 is left over. Starting with the amount of KOH that is mixed, let's see how much H2SO4 can be consumed:
0.048 mol KOH consumed * (1 mol H2SO4 consumed / 2 mol KOH consumed) = 0.024 mol H2SO4 consumed
This is less than the starting amount of H2SO4. Let's see how much is left over and then find the concentration using the previously calculated total volume.
H2SO4 left = 0.125 mol - 0.024 mol = 0.101 mol
H2SO4 conc = 0.101 mol / 0.450 L = 0.0224