
David J. answered 03/11/21
Highly experienced chem/org chem tutor. Free initial consultation!
First, what is the balanced rxn?
H2SO4 + 2KOH ------> K2SO4 + 2H2O
Then, what are the mol quantities of the reactants? (V)(M) = #mol
(0.250 L)(0.500M) = 0.125 mol H2SO4
(0.200L)(0.240 M) = 0.048 mol KOH
0.048 mol KOH can only neutralize 0.024 mol H2SO4, from the balanced rxn above.
0.125 mol H2SO4 - 0.024 mol H2SO4 = 0.101 mol H2SO4 remaining
0.101 mol /(0.250 + 0.200L) = 0.224 M H2SO4 remains