Final answer:
To neutralize 38.00 mL of 0.0500 M nitric acid, 17.27 mL of 0.0550 M calcium hydroxide is needed, based on the stoichiometry of the neutralization reaction.
Step-by-step explanation:
The student is asking about the volume of a 0.0550 M calcium hydroxide (Ca(OH)2) solution that is needed to neutralize a given volume and concentration of nitric acid (HNO3). To solve this, we can use the stoichiometry of the reaction between Ca(OH)2 and HNO3, which shows that 1 mole of Ca(OH)2 reacts with 2 moles of HNO3. First, calculate the number of moles of HNO3:
- moles HNO3 = volume x concentration = 0.038 L x 0.0500 M = 0.0019 moles
Using the stoichiometry of the reaction (1 Ca(OH)2 : 2 HNO3), we need half the number of moles of Ca(OH)2, which is 0.0019 moles / 2 = 0.00095 moles of Ca(OH)2. Now calculate the required volume:
- Volume = moles / concentration = 0.00095 moles / 0.0550 M = 0.01727 L = 17.27 mL
The volume of 0.0550 M calcium hydroxide required to neutralize 38.00 mL of 0.0500 M nitric acid is therefore 17.27 mL.