Final answer:
To neutralize 46.67 mL of 0.038 M nitric acid, 44.34 mL of 0.02 M calcium hydroxide solution is required, based on stoichiometry and the balanced chemical reaction between nitric acid and calcium hydroxide.
Step-by-step explanation:
The volume of 0.02 M calcium hydroxide (Ca(OH)₂) required to neutralize a given volume of 0.038 M nitric acid (HNO₃) can be determined using stoichiometry and the balanced chemical equation:
Ca(OH)₂ (aq) + 2HNO₃ (aq) → Ca(NO₃)₂ (aq) + 2H₂O (l).
First, we calculate the moles of nitric acid that you have:
# moles HNO₃ = 0.04667 L × 0.038 M = 0.00177346 mol HNO₃.
According to the balanced equation, it takes 1 mole of Ca(OH)₂ to neutralize 2 moles of HNO₃, hence we need half the moles of Ca(OH)₂:
0.00177346 mol HNO₃ × (1 mol Ca(OH)₂ / 2 mol HNO₃) = 0.00088673 mol Ca(OH)₂.
Now we can find the volume of Ca(OH)₂ needed:
Volume of Ca(OH)₂ = moles / concentration = 0.00088673 mol / 0.02 M = 0.0443365 liters or 44.34 mL.
Therefore, 44.34 mL of 0.02 M calcium hydroxide is required to neutralize 46.67 mL of 0.038 M nitric acid.