What volume of (dm$^3$) of water will be added to 10dm$^3$ of 2.0 mol/dm$^3$ HCL acid solution to give a final solution of 0.5 mol/dm$^3$?

Analysis

To solve this problem, we use the dilution formula. When you add water to a solution, the number of moles of the solute (HCl) remains constant, but the volume and concentration change.

Concept

The dilution formula is:

Where:

• $C_1$ = Initial concentration (2.0 mol/dm³)
• $V_1$ = Initial volume (10 dm³)
• $C_2$ = Final concentration (0.5 mol/dm³)
• $V_2$ = Final total volume (Initial volume + Volume of water added)

Solution

Step 1: Calculate the final total volume ($V_2$) Using the formula $C_1V_1 = C_2V_2$: $2.0 \times 10 = 0.5 \times V_2$ $20 = 0.5 \times V_2$

Divide both sides by 0.5: $V_2 = \frac{20}{0.5}$ $V_2 = 40 \text{ dm}^3$

Step 2: Calculate the volume of water added The total final volume ($V_2$) is the sum of the initial volume ($V_1$) and the volume of water added ($V_{added}$). $V_2 = V_1 + V_{added}$ $40 = 10 + V_{added}$ $V_{added} = 40 - 10$ $V_{added} = 30 \text{ dm}^3$

Conclusion

The volume of water to be added is 30 dm³. This matches option A.

Exam Tip

Always distinguish between the final volume ($V_2$) and the volume added. Many students stop at 40 dm³ and choose that as the answer, forgetting that the question asks for the volume of water added to the existing 10 dm³. Always read the question carefully to see if it asks for the final volume or the volume to be added.