Question

What volume(l) of 7.48*10^-2 m perchloric acid can be neutralized with 115 ml?

Answer 1

HClO4 + NaOH + NaClO4 + H2O 

so equal molar amounts of acid and base. 

NaOH contains 0.244 * 115/1000 moles of OH so you need that number of moles of H+ from perchloric which is in x mL 
0.244 * 115/1000 = 0.0748 * x/1000 so x = 0.244 * 115/0.0748 = 375 mL