Final answer:
To neutralize 150.0 ml of 0.280 M HNO₃, 181 ml of 0.116 M Ba(OH)₂ is required.
Step-by-step explanation:
The student's question involves determining the volume of 0.116 M Ba(OH)₂ required to completely neutralize 150.0 ml of 0.280 M HNO₃. To solve this, we can use the stoichiometry of the neutralization reaction, which is:
Ba(OH)₂(aq) + 2HNO₃(aq) → Ba(NO₃)₂(aq) + 2H₂O(l).
From the reaction, we know that one mole of Ba(OH)₂ reacts with two moles of HNO₃. We can calculate the moles of HNO₃ in the given solution and then determine how many moles of Ba(OH)₂ would be needed for complete neutralization.
moles of HNO₃ = 150.0 ml × 0.280 M = 0.0420 mol
Since the molar ratio is 1:2, we need half the amount of moles of Ba(OH)₂:
moles of Ba(OH)₂ required = 0.0420 mol / 2 = 0.0210 mol
The volume of Ba(OH)₂ needed can be found using its concentration:
Volume = moles / concentration
Volume of Ba(OH)₂ = 0.0210 mol / 0.116 M = 0.1810 L or 181 ml
Therefore, 181 ml of 0.116 M Ba(OH)₂ is needed to neutralize 150.0 ml of 0.280 M HNO₃.