Question

Exactly 59.1 mL of a perchloric acid solution of unknown concentration was titrated with 0.951 M sodium hydroxide. An endpoint was reached when 15.3 mL of the base was added. Calculate the molarity of the perchloric acid solution.

Answer 1

V_a(HClO_4) =59.1 m L
.c_b(NaOH)=0.951 M
V_b (NaOH)=15.3 m L
.c_a(HClO_4)=?
The reaction between the substances
HClO_4 +NaOH=NaClO_4 +H_2O
According to the titration rule V_a×c_a=V_b×c_b
.c_a =(V_b×c_b)/V_a =(0.951×15.3)/59.1=0.246 M