suppose that 20 mL of 2.50 x 10^-2 M aqueous H2SO4 is required to neutralize 10.0 mL of an aqueous solution of KOH. What is the molarity of the KOH solution?

Answer 1

...............H2SO4 + 2KOH ==> K2SO4 + 2H2O
moles H2SO4 = M x L = 0.025 M x 0.020 L = 0.0005
0.0005 moles H2SO4 x (2 mol KOH/1 mol H2SO4) = 00005 x 2 = 0.001
M KOH = mols KOH/L KOH = 0.001/0.01 L = 0.1 M
NOTE: In practice volumes are too small and the molarity of H2SO4 is low so the precision and accuracy will suffer unless you are using specialized measuring equipment.