To find the volume of the phosphoric acid solution required to neutralize the barium hydroxide solution, we can use the stoichiometry of the balanced chemical equation between barium hydroxide (Ba(OH)2) and phosphoric acid (H3PO4). The balanced equation is:
3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O
From the balanced equation, we can see that the ratio of barium hydroxide to phosphoric acid is 3:2. This means that for every 3 moles of barium hydroxide, we need 2 moles of phosphoric acid.
Given:
Mass of barium hydroxide (Ba(OH)2) = 13.75 g
Molar mass of Ba(OH)2 = 171.34 g/mol
First, let's calculate the number of moles of barium hydroxide:
Moles of Ba(OH)2 = mass / molar mass
Moles of Ba(OH)2 = 13.75 g / 171.34 g/mol
Moles of Ba(OH)2 = 0.0802 mol
Using the stoichiometry of the balanced equation, we can determine the number of moles of phosphoric acid needed:
Moles of H3PO4 = (Moles of Ba(OH)2) × (2 moles of H3PO4 / 3 moles of Ba(OH)2)
Moles of H3PO4 = 0.0802 mol × (2/3)
Moles of H3PO4 = 0.0535 mol
Now, let's calculate the volume of the phosphoric acid solution required:
Volume (L) of H3PO4 = Moles of H3PO4 / Concentration of H3PO4
Volume (L) of H3PO4 = 0.0535 mol / 0.150 mol/L
Volume (L) of H3PO4 = 0.3567 L
Therefore, the volume of the phosphoric acid solution required to neutralize the barium hydroxide solution is 0.3567 liters (or 356.7 mL).