Final answer:
To calculate the molarity of a Ba(OH)₂ solution, you can use the equation Moles of solute = molarity x volume. In this case, if 26.1 mL of 0.505 M HNO₃ titrates 1950 mL of the Ba(OH)₂ solution, the molarity of Ba(OH)₂ is 0.003369 M.
Step-by-step explanation:
To calculate the molarity of a Ba(OH)₂ solution, we can use the equation:
Moles of solute = molarity × volume of solution (in liters)
Given that 26.1 mL of 0.505 M HNO₃ completely titrates 1950 mL of the Ba(OH)₂ solution, we can calculate the moles of HNO3:
Moles of HNO₃ = 0.505 M × 0.0261 L = 0.0131605 mol
The balanced chemical equation between HNO₃ and Ba(OH)₂ is:
2HNO₃ + Ba(OH)₂ -> Ba(NO₃)₂ + 2H₂O
From the equation, we can see that 2 moles of HNO₃ react with 1 mole of Ba(OH)₂.
Therefore, the moles of Ba(OH)₂ in the solution is half the moles of HNO₃:
Moles of Ba(OH)₂ = (1/2) × 0.0131605 mol = 0.00658025 mol
Lastly, we can calculate the molarity of the Ba(OH)₂ solution:
Molarity of Ba(OH)₂ = Moles of Ba(OH)₂ / Volume of Ba(OH)₂ solution (in liters)
Molarity of Ba(OH)₂ = 0.00658025 mol / 1.95 L
= 0.003369 M