Question

An acid-base titration was performed. It took 27.45 mL of the base, KOH, to titrate 3.115 g HBr, the acid. What was the Molarity of the KOH?

Answer 1

$$n(HBr)=3.115/81=0.0385 mol$$

According to the reaction n(HBr)=n(KOH)=0.038 mol

$$C(KOH)=n/V=0.0385/0.02745=1.4M$$