10.0 dm3 of water was added to 2.0 mol $dm^{-3}$ of 2.5dm3 solution of HCl. What is the concentration of the final solution in mol $dm^{-3}$?

Analysis

This is a dilution problem. You are taking a concentrated solution of Hydrochloric acid (HCl) and adding more water to it. When you add water (the solvent), the number of moles of the solute (HCl) remains the same, but the total volume increases, which causes the concentration to decrease.

Concept

To solve this, we use the dilution formula:

Where:

• $C_1$ = Initial concentration
• $V_1$ = Initial volume
• $C_2$ = Final concentration
• $V_2$ = Final volume

Step-by-Step Solution

Step 1: Identify the given values.

• $C_1 = 2.0 \text{ mol dm}^{-3}$
• $V_1 = 2.5 \text{ dm}^3$
• Volume of water added = $10.0 \text{ dm}^3$

Step 2: Calculate the final volume ($V_2$). The final volume is the sum of the initial volume and the volume of water added:

Step 3: Plug the values into the dilution formula.

Step 4: Solve for $C_2$.

Conclusion

The concentration of the final solution is 0.4 mol dm⁻³.

Exam Tip

Always remember to add the volume of the water added to the initial volume to get the total final volume. A common mistake students make is using the volume of water added ($10.0 \text{ dm}^3$) as the final volume instead of the sum ($12.5 \text{ dm}^3$).