What volume (in mL) of 0.2800 M HBr is required to neutralize 50.00 mL of 0.7000 M KOH?

Answer 1

Since the HBr and KOH are a strong acid and strong base respectively, they dissociate into their ion forms completely. The Br- anion has a charge of -1 and the K+ cation has a charge of +1, so they neutralize each other at a ratio of 1 mole to 1 mole. Therefore we can solve this problem using the formula M1V1 = M2V2, with solution 1 being 0.28M HBr and solution 2 being 0.7M KOH. Plugging in for the respective concentrations and volumes, we get (0.2800M HBr)(V1) = (50.00mL)(0.7000M KOH). This solves to V1 = 125.0mL of HBr.