Final answer:
To calculate the volume of a 0.175 M HBr solution required to neutralize 40.95 mL of 0.300 M Ba(OH)₂ solution, use stoichiometry. The volume of HBr solution needed is 0.07 L or 70 mL.
Step-by-step explanation:
To calculate the volume of a 0.175 M HBr solution required to neutralize 40.95 mL of 0.300 M Ba(OH)₂ solution, we can use the concept of stoichiometry.
1. First, we need to write the balanced chemical equation for the neutralization reaction between HBr and Ba(OH)₂: HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
2. We can see from the equation that 1 mole of HBr reacts with 1 mole of Ba(OH)₂.
3. Therefore, the moles of HBr required to neutralize the Ba(OH)₂ solution can be calculated as Moles of HBr = Molarity of Ba(OH)₂ × Volume of Ba(OH)₂ Moles of HBr = 0.300 M × 0.04095 L
4. Substituting the values: Moles of HBr = 0.012285 mol
5. Since the molar ratio between HBr and Ba(OH)₂ is 1:1, the moles of Ba(OH)₂ are also 0.012285 mol.
6. To calculate the volume of the HBr solution, we can use the equation: Volume of HBr = Moles of HBr / Molarity of HBr Volume of HBr = 0.012285 mol / 0.175 M
7. Substituting the values: Volume of HBr = 0.07 L or 70 mL