Question

100 mL of 1.0 M aqueous NaOH solution was diluted to 1.0 L by adding water. Half of this solution was discarded. A new 100 mL of 0.5 M aqueous NaOH solution was added to the remaining solution. What is the concentration of the final aqueous NaOH solution?

• A. 0.17 M
• B. 0.10 M
• C. 0.50 M
• D. 0.33 M

Hint:

To avoid calculation errors, always keep track of the absolute number of moles of solute at each step.
Dilution changes the volume but keeps moles constant.
Discarding a fraction of a solution reduces both volume and moles by that same fraction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Concentration calculation involves tracking the total moles of solute and the final total volume of the mixture.
Key Formula or Approach:
\[ \text{Moles} (n) = \text{Molarity} (M) \times \text{Volume} (V, \text{ in Liters}) \]
\[ M_{final} = \frac{n_{total}}{V_{total}} \]

Step 2: Detailed Explanation:
1. Initial Solution: $100 \text{ mL of } 1.0 \text{ M NaOH}$.
Moles $n_{1} = 0.1 \text{ L} \times 1.0 \text{ mol/L} = 0.1 \text{ mol}$.
2. Dilution: Diluted to $1.0 \text{ L}$. Moles remain $0.1 \text{ mol}$.
3. Discard half: $500 \text{ mL}$ is discarded. The remaining $500 \text{ mL}$ ($0.5 \text{ L}$) contains half the moles.
Remaining moles $n_{rem} = \frac{0.1}{2} = 0.05 \text{ mol}$.
4. Addition: $100 \text{ mL of } 0.5 \text{ M NaOH}$ is added.
Added moles $n_{add} = 0.1 \text{ L} \times 0.5 \text{ mol/L} = 0.05 \text{ mol}$.
5. Final Totals:
Total moles $n_{total} = 0.05 + 0.05 = 0.10 \text{ mol}$.
Total volume $V_{total} = 500 \text{ mL} + 100 \text{ mL} = 600 \text{ mL} = 0.6 \text{ L}$.
6. Final Molarity:
\[ M_{final} = \frac{0.10 \text{ mol}}{0.6 \text{ L}} = 0.1666... \text{ M} \approx 0.17 \text{ M} \]

Step 3: Final Answer:
The final concentration is 0.17 M.
This matches option (A).