What was the concentration of the unknown HClO4 solution? The neutralization reaction is: HClO4(aq)+NaOH(aq)→H2O(l)+NaClO4(aq)?

Question

A 35.0 mL sample of an unknown HClO4 solution requires 42.3 mL of 0.101 M NaOH for complete neutralization.

What was the concentration of the unknown HClO4 solution? The neutralization reaction is:
HClO4(aq)+NaOH(aq)→H2O(l)+NaClO4(aq)?

Answer 1

This is a titration problem.  At the point of neutralization, moles of acid = moles of base since the balanced equation shows that 1 mole of HClO₄ reacts with 1 mole of NaOH.  

(1)  moles NaOH used = 0.101 moles/L x 0.0423 L = 0.0042723 moles NaOH

(2)  moles HClO4 present = 0.0042723 moles NaOH x 1 moles HClO4/mole NaOH =  0.0042723 moles HClO₄

(3)  concentration of HClO4 = moles/liter =  0.0042723 moles/0.035 L = 0.122 mole/L = 0.122 M (3 sig. figs.)