Final answer:
To neutralize the solution from 2.00 g of PCl3 with 0.200 M NaOH, a volume of approximately 72.85 mL of NaOH is required, following the molar ratio of 1:1 for H3PO3 to NaOH, taking into account that only one proton reacts. The provided options do not include this answer,
Step-by-step explanation:
To find the volume of 0.200 M NaOH needed to neutralize the solution produced by dissolving 2.00 g of PCl3 in an excess of water, we first determine the number of moles of PCl3. The molar mass of PCl3 is approximately 137.33 g/mol, which makes the moles of PCl3 dissolved:
moles of PCl3 = mass (g) / molar mass (g/mol) = 2.00 g / 137.33 g/mol ≈ 0.01457 mol
When PCl3 dissolves in water, it reacts to form H3PO3 and HCl. However, it is noted that only one proton from H3PO3 reacts. Hence, the molar ratio of H3PO3 to NaOH is 1:1. Therefore, we need the same number of moles of NaOH to neutralize the acid:
moles of NaOH needed = moles of PCl3
Next, we calculate the volume of 0.200 M NaOH needed:
volume of NaOH (L) = moles of NaOH / concentration (M) = 0.01457 mol / 0.200 M ≈ 0.07285 L or 72.85 mL
The closest possible answer to 72.85 mL is 73.0 mL, which is not listed among the options provided. The question possibly contains an error or requires an updated set of multiple-choice answers.