Final answer:
To neutralize 50.0 mL of 0.0500 M HBr, 69.4 mL of 0.0180 M Ba(OH)2 is required, as calculated through stoichiometry using the mole ratio from the balanced chemical reaction between Ba(OH)2 and HBr.
Step-by-step explanation:
The student is asking about the volume of 0.0180 M barium hydroxide (Ba(OH)2) required to neutralize a given volume of hydrobromic acid (HBr) in a titration. To solve this, we'll use the concept of stoichiometry and the molarity (concentration) of the solutions. The reaction between Ba(OH)2 and HBr is:
Ba(OH)2 + 2HBr → BaBr2 + 2H2O
From the balanced chemical equation, we see that one mole of Ba(OH)2 reacts with two moles of HBr. Therefore, the mole ratio of Ba(OH)2 to HBr is 1:2. Using this information, we can calculate the volume of Ba(OH)2 needed to neutralize the HBr.
- First, calculate the moles of HBr:
0.0500 M × 0.050 L = 0.0025 moles HBr - Using the stoichiometry of the reaction, calculate the moles of Ba(OH)2 needed:
0.0025 moles HBr × (1 mole Ba(OH)2/2 moles HBr) = 0.00125 moles Ba(OH)2 - Calculate the volume of 0.0180 M Ba(OH)2 that contains 0.00125 moles:
0.00125 moles / 0.0180 M = 0.0694 L, or 69.4 mL
Therefore, 69.4 mL of 0.0180 M Ba(OH)2 is required to neutralize 50.0 mL of 0.0500 M HBr.