Answers of the description

Question

Given: 150.0 mL of 0.130 hydrobromic acid (HBr)

0.0200 barium hydroxide (Ba(OH)2)

Wanted; mL of 0.0200 Ba(OH)2 necessary to neutralize HBr?

Answer 1

2HBr + Ba(OH)₂ ==> 2H₂O + BaBr₂ ... balanced equation

moles HBr present = 150.0 mls x 1 L / 1000 mls x 0.130 mol/L = 0.0195 mols HBr

moles Ba(OH)₂ needed = 0.0195 mol HBr x 1 mol Ba(OH)₂ / 2 mol HBr = 0.00975 mol Ba(OH)₂

Volume Ba(OH)₂ needed = 0.00975 mol Ba(OH)₂ x 1 L / 0.0200 mol = 0.488 L = 488 mls (3 sig. figs.)