We can start by writing a balanced chemical equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2):
2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of Ca(OH)2. Therefore, we need to determine the number of moles of Ca(OH)2 in 750.0 mL of 0.400 M solution, and then use the mole ratio to calculate the number of moles of HNO3 required.
First, we need to convert the volume of Ca(OH)2 solution into moles:
750.0 mL × (1 L / 1000 mL) × 0.400 mol/L = 0.300 mol Ca(OH)2
Next, we can use the mole ratio from the balanced equation to determine the amount of HNO3 required:
2 mol HNO3 / 1 mol Ca(OH)2 × 0.300 mol Ca(OH)2 = 0.600 mol HNO3
Finally, we can use the definition of molarity to calculate the volume of 0.650 M HNO3 required to provide 0.600 moles of HNO3:
0.600 mol HNO3 × (1 L / 0.650 mol) ×(1000 mL / 1 L) = 923.1 mL
Therefore, approximately 923.1 mL (or 0.9231 L) of 0.650 M HNO3 is required to react with 750.0 mL of 0.400 M Ca(OH)2. Note that we assumed the volumes are additive and that the reaction goes to completion.