Question

42.0 mL of 0.130 M NaOH and 37.6 mL of 0.450 M NaOH. Assume that the volumes are additive.
Enter your answers, separated by a comma, using three significant figures.

Answer 1

To address the question, we need to find the total amount of NaOH in each solution and then calculate the total concentration when these solutions are mixed.

1. **Calculate the moles of NaOH in each solution:**

- For the first solution (42.0 mL of 0.130 M NaOH):
\[
\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.130 \, \text{mol/L} \times 0.0420 \, \text{L} = 0.00546 \, \text{mol}
\]

- For the second solution (37.6 mL of 0.450 M NaOH):
\[
\text{Moles of NaOH} = 0.450 \, \text{mol/L} \times 0.0376 \, \text{L} = 0.01692 \, \text{mol}
\]

2. **Add the moles of NaOH:**
\[
\text{Total moles of NaOH} = 0.00546 \, \text{mol} + 0.01692 \, \text{mol} = 0.02238 \, \text{mol}
\]

3. **Calculate the total volume:**
\[
\text{Total volume} = 42.0 \, \text{mL} + 37.6 \, \text{mL} = 79.6 \, \text{mL} = 0.0796 \, \text{L}
\]

4. **Calculate the new concentration after mixing:**
\[
\text{Concentration (C)} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.02238 \, \text{mol}}{0.0796 \, \text{L}} = 0.280 \, \text{M}
\]

5. **Final answers:**
- Total moles of NaOH: \(0.02238\) mol (to three significant figures: **0.0224 mol**)
- Concentration after mixing: \(0.280\) M

Thus, the results in three significant figures are:
**0.0224, 0.280**