Final answer:
To neutralize 78.2 ml of 2.549 M hydrobromic acid, 99.65319 ml of 1 M barium hydroxide is needed based on the stoichiometry of the acid-base neutralization reaction.
Step-by-step explanation:
To find the volume of 1 M barium hydroxide (Ba(OH)2) needed to neutralize 78.2 ml of 2.549 M hydrobromic acid (HBr), we need to use the concept of titration from acid-base chemistry. The reaction between barium hydroxide and hydrobromic acid can be represented by the balanced chemical equation:
Ba(OH)2 + 2 HBr → BaBr2 + 2 H2O
For every mole of Ba(OH)2, two moles of HBr are neutralized based on the stoichiometry of the reaction. First, calculate the moles of HBr in the given volume:
moles HBr = 78.2 ml × 2.549 M = 0.0782 L × 2.549 mol/L = 0.19930638 mol HBr
Since the ratio of Ba(OH)2 to HBr is 1:2, we need half the moles of Ba(OH)2 to neutralize the HBr:
moles Ba(OH)2 needed = 0.19930638 mol HBr ÷ 2 = 0.09965319 mol Ba(OH)2
Now, use the molarity of Ba(OH)2 to find the volume needed:
volume Ba(OH)2 = moles Ba(OH)2 → M Ba(OH)2 = 0.09965319 mol → 1 M = 0.09965319 L
Simplifying the volume gives us:
volume Ba(OH)2 = 99.65319 ml