Final answer:
To neutralize 2.001 g of NaOH, we need approximately 25.0 mL of 2.0 M HCl, which is option a) in the provided choices.
Step-by-step explanation:
The question involves determining the volume of HCl needed to neutralize a given amount of NaOH. First, we calculate the number of moles of NaOH using the formula moles = mass (g) / molar mass (g/mol). The molar mass of NaOH is 40.00 g/mol, so for 2.001 g of NaOH:
moles of NaOH = 2.001 g / 40.00 g/mol = 0.050025 mol
Since the reaction between HCl and NaOH is a 1:1 mole ratio (HCl + NaOH → NaCl + H2O), the number of moles needed to neutralize the NaOH is also 0.050025 mol.
To find the volume of 2.0 M HCl needed: Volume (L) = moles / molarity
Volume of HCl = 0.050025 mol / 2.0 mol/L = 0.0250125 L
Converting liters to milliliters (1 L = 1000 mL):
Volume of HCl = 0.0250125 L * 1000 mL/L = 25.01 mL
After rounding, the answer is approximately 25.0 mL of 2.0 M HCl, which corresponds to option a).