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?
Show 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.
Step 1: Understanding the Question:
This is a stoichiometry and solution chemistry problem. We need to calculate the molarity of a final sodium hydroxide ($\text{NaOH}$) solution obtained through a sequence of dilution, volume reduction, and mixing steps. Step 2: Key Formula or Approach:
Use the fundamental definitions of molarity ($M$) and moles ($n$):
\[ n = M \times V \]
Where $V$ is the volume in liters.
To find the final concentration after mixing:
\[ M_{\text{final}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{n_{\text{remaining}} + n_{\text{added}}}{V_{\text{remaining}} + V_{\text{added}}} \]
Step 3: Detailed Explanation:
Let's calculate the moles of $\text{NaOH}$ and volumes at each individual stage:
• Stage 1: Initial solution:
- Volume $V_1 = 100\text{ mL} = 0.1\text{ L}$
- Molarity $M_1 = 1.0\text{ M}$
- Moles of $\text{NaOH}$ initially:
\[ n_1 = M_1 \times V_1 = 1.0\text{ mol/L} \times 0.1\text{ L} = 0.10\text{ mol} \]
• Stage 2: Dilution to 1.0 L:
- The solution volume is increased to $V_2 = 1.0\text{ L}$ by adding water.
- The total moles of solute remains unchanged: $n_2 = 0.10\text{ mol}$.
• Stage 3: Discarding half of the solution:
- Since half of the homogeneous solution is discarded, the volume and moles are both halved:
- Remaining volume:
\[ V_{\text{remaining}} = \frac{1.0\text{ L}}{2} = 0.5\text{ L} \]
- Remaining moles of $\text{NaOH}$:
\[ n_{\text{remaining}} = \frac{0.10\text{ mol}}{2} = 0.05\text{ mol} \]
• Stage 4: Addition of a new NaOH solution:
- Added volume $V_{\text{added}} = 100\text{ mL} = 0.1\text{ L}$
- Added molarity $M_{\text{added}} = 0.5\text{ M}$
- Moles of $\text{NaOH}$ added:
\[ n_{\text{added}} = M_{\text{added}} \times V_{\text{added}} = 0.5\text{ mol/L} \times 0.1\text{ L} = 0.05\text{ mol} \]
• Stage 5: Calculating final concentration:
- Total moles of $\text{NaOH}$ in final mixture:
\[ n_{\text{total}} = n_{\text{remaining}} + n_{\text{added}} = 0.05\text{ mol} + 0.05\text{ mol} = 0.10\text{ mol} \]
- Total volume of the final mixture:
\[ V_{\text{total}} = V_{\text{remaining}} + V_{\text{added}} = 0.5\text{ L} + 0.1\text{ L} = 0.6\text{ L} \]
- Final concentration ($M_{\text{final}}$):
\[ M_{\text{final}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{0.10\text{ mol}}{0.6\text{ L}} = 0.167\text{ M} \approx 0.17\text{ M} \]
Step 4: Final Answer:
The final concentration of the aqueous $\text{NaOH}$ solution is $0.17\text{ M}$, which matches option (A).
