What volume of 0.0748 M perchloric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide?

0.375 L

Explanation:

We know that at neutralization, the number of mol of acid must equal the number of equivalents of base.

This is a reaction 1:1 acid to base:

HClO₄ + NaOH ⇒ NaClO₄ + H₂O

We re given the moles of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.

Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L = 0.028 mol

Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:

Molarity = # moles / V ⇒ V = # moles / M

V = 0.028 mol / 0.0748 mol/L = 0.375 L

Note that this problem can be solved in just one step since

M (HClO₄) x V (HClO₄) = M (NaOH) x V (NaOH) ⇒

V (HClO₄) = M (NaOH) x V (NaOH) / M (HClO₄)