What volume in milliliters of 0.0160 M Ba(OH)₂ is required to neutralize 50.0 mL of 0.0500 M HBr?

Answer 1

First, write the balanced equation for the reaction taking place:

2HBr + Ba(OH)₂ ==> 2H₂O + BaBr₂

Moles HBr present = 50.0 ml x 1 L / 1000 ml x 0.050 mols / L = 0.0025 mols HBr

Moles Ba(OH)₂ needed = 0.0025 mol HBr x 1 mol Ba(OH)2 / 2 mol HBr = 0.00125 mols Ba(OH)2

Volume Ba(OH)₂ needed = 0.00125 mols x 1 L / 0.016 mols = 0.0781 L = 78.1 mls

Answer 2

.05L HBr X .05 mole HBr/L = .0025 mole HBr

.0025 mole Ba(OH)2 X L Ba(OH)2/.016 mole Ba(OH)2 = .16L Ba(OH)2 = .00016mL Ba(OH)2