What volume of 0.5720 M HBr(aq) is needed to fully neutralize 50.0 mL
of 0.147 M Ba(OH)2(aq)?

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Clarification of Question by lilgreeny-ga on 05 Mar 2006 15:15 PST

Could someone  please help me?

25.7 ml is required

You have 50 x 0.147 = 7.35 mmol of Ba(OH)2
You thus require 2 x 7.35/0.572 = 25.7 ml of HBr (aq)

The 2 is required because Ba(OH)2 is diacidic:-
Ba(OH)2 + 2 HBr -> BaBr2 + 2 H2O

Regards