What volume of 2.5 M nitric acid (HNO3) is required to prepare 500 mL of a 0.20 M HNO3 solution?

• A. 40 mL
• B. 25 mL
• C. 100 mL
• D. 250 mL

Answer: (A)

Verified step by step guidance

1. Identify the known values: the concentration of the stock solution \(C_1 = 2.5\,M\), the desired concentration of the diluted solution \(C_2 = 0.20\,M\), and the final volume of the diluted solution \(V_2 = 500\,mL\).
2. Recall the dilution equation which relates the concentrations and volumes before and after dilution: \(C_1 \times V_1 = C_2 \times V_2\), where \(V_1\) is the volume of the stock solution needed.
3. Rearrange the dilution equation to solve for \(V_1\): \(V_1 = \frac{C_2 \times V_2}{C_1}\).
4. Substitute the known values into the equation: \(V_1 = \frac{0.20\,M \times 500\,mL}{2.5\,M}\).
5. Calculate \(V_1\) to find the volume of 2.5 M nitric acid required to prepare 500 mL of 0.20 M solution.