What volume (in mL) of 0.2400 M HBr is required to neutralize 40.00 mL of 0.8000 M NaOH?

Answer 1

HBr + NaOH ==> NaBr + H2O ... balanced equation

mols NaOH present = 40.00 ml x 1 L / 1000 ml x 0.8000 mol / L = 0.0320 mols NaOH

mols HBr needed = 0.0320 mols (since mol ratio is 1:1)

volume HBr needed:

(x L)(0.2400 mol/L) = 0.0320 mols

x = 0.1333 liters = 133.3 mls