A 1.000 g sample of Na₂CO₃ (MW: 105.99 g/mol) is dissolved in enough water to make 200.0 mL of solution. A 25.00 mL aliquot required 32.18 mL of HCl to completely neutralize it. What is the molar concentration of HCl? 
Na₂CO₃ (aq)  + 2 HCl (aq)  →  2 KCl (aq)  +  H₂O(l)  +  CO₂  (g)

• A. 0.5864 M
• B. 0.07330 M
• C. 0.01832 M
• D. 0.002359 M

Answer: (B)

Verified step by step guidance

1. Calculate the moles of Na2CO3 in the 1.000 g sample using its molar mass: \( \text{moles of Na}_2\text{CO}_3 = \frac{1.000 \text{ g}}{105.99 \text{ g/mol}} \).
2. Determine the concentration of Na2CO3 in the 200.0 mL solution by dividing the moles of Na2CO3 by the volume of the solution in liters: \( \text{Concentration of Na}_2\text{CO}_3 = \frac{\text{moles of Na}_2\text{CO}_3}{0.200 \text{ L}} \).
3. Calculate the moles of Na2CO3 in the 25.00 mL aliquot by multiplying the concentration of Na2CO3 by the volume of the aliquot in liters: \( \text{moles of Na}_2\text{CO}_3 = \text{Concentration of Na}_2\text{CO}_3 \times 0.02500 \text{ L} \).
4. Use the stoichiometry of the reaction to find the moles of HCl required to neutralize the moles of Na2CO3. According to the balanced equation, 2 moles of HCl are needed for every mole of Na2CO3: \( \text{moles of HCl} = 2 \times \text{moles of Na}_2\text{CO}_3 \).
5. Calculate the molar concentration of HCl by dividing the moles of HCl by the volume of HCl solution used in liters: \( \text{Concentration of HCl} = \frac{\text{moles of HCl}}{0.03218 \text{ L}} \).