Final answer:
To find out the required volume of 0.558M HNO3 to react with 45.55 mL of 0.0515M Ba(OH)2, you calculate the moles of Ba(OH)2, determine the moles of HNO3 needed based on the stoichiometry of the reaction, and then divide by the molarity of HNO3 to get the volume. The result is 8.41 mL.
Step-by-step explanation:
To determine what volume of 0.558M HNO3 solution is required to react completely with 45.55 mL of 0.0515M Ba(OH)2, you need to apply stoichiometry. Given that the reaction between nitric acid and barium hydroxide is:
2 HNO3 (aq) + Ba(OH)2 (s) → Ba(NO3)2 (aq) + 2 H2O (l)
First, calculate the moles of Ba(OH)2 using the molarity and volume:
Moles of Ba(OH)2 = Molarity of Ba(OH)2 x Volume of Ba(OH)2 in liters
= 0.0515 M x 0.04555 L
= 0.002346825 mol
Since the reaction ratio is 2 moles of HNO3 per mole of Ba(OH)2, multiply the moles of Ba(OH)2 by 2 to find the moles of HNO3 required:
Moles of HNO3 needed = 2 x Moles of Ba(OH)2
= 2 x 0.002346825 mol
= 0.00469365 mol
Now calculate the volume of 0.558M HNO3 needed by dividing the moles of HNO3 by the molarity:
Volume of HNO3 = Moles of HNO3 / Molarity of HNO3
= 0.00469365 mol / 0.558M
= 0.008411791 L or 8.41 mL
Therefore, 8.41 mL of 0.558M HNO3 is required to completely react with 45.55 mL of 0.0515M Ba(OH)2.