Question

A titration neutralizes 25mL of an unknown acid using 32 mL of 6.0 M KOH. What is the molarity of the unknown acid in the original solution?

Answer 1

`[(Molarity Acid)(Volume Acid)] = [(Molarity Base)(Volume Base)]`

Explanation:

`(Molarity Acid)(25 ml) = (6.0M)(32ml)`

`Molarity Acid = [(6.0M)(32ml)]/[(25ml)] = 7.68M`