Two solutions were mixed together. One solution was 300 mL of 0.45 M sodium chloride. The other solution was 200 mL of 1.5 M sodium nitride. What was the final concentration of sodium ions in the solution? Question 4 options: 1.95 M 0.435 M 2.07 M 0.975 M 0.87 M

To find the final concentration of sodium ions in the mixed solution, we need to follow these steps:

1. Calculate the moles of sodium ions from each solution:

   • For the sodium chloride (NaCl) solution: $\text{Moles of NaCl} = \text{Volume (L)} \times \text{Molarity (M)}$ $\text{Moles of NaCl} = 0.300 \, \text{L} \times 0.45 \, \text{M} = 0.135 \, \text{moles}$ Since each NaCl molecule provides one Na$^+$ ion: $\text{Moles of Na}^+ \text{ from NaCl} = 0.135 \, \text{moles}$

   • For the sodium nitride (Na$_3$N) solution: $\text{Moles of Na}_3\text{N} = \text{Volume (L)} \times \text{Molarity (M)}$ $\text{Moles of Na}_3\text{N} = 0.200 \, \text{L} \times 1.5 \, \text{M} = 0.300 \, \text{moles}$ Since each Na$_3$N molecule provides three Na$^+$ ions: $\text{Moles of Na}^+ \text{ from Na}_3\text{N} = 0.300 \, \text{moles} \times 3 = 0.900 \, \text{moles}$

2. Add the moles of sodium ions from both solutions: $\text{Total moles of Na}^+ = 0.135 \, \text{moles} + 0.900 \, \text{moles} = 1.035 \, \text{moles}$

3. Calculate the total volume of the mixed solution: $\text{Total volume} = 300 \, \text{mL} + 200 \, \text{mL} = 500 \, \text{mL} = 0.500 \, \text{L}$

4. Calculate the final concentration of sodium ions: $\text{Final concentration of Na}^+ = \frac{\text{Total moles of Na}^+}{\text{Total volume (L)}}$ $\text{Final concentration of Na}^+ = \frac{1.035 \, \text{moles}}{0.500 \, \text{L}} = 2.07 \, \text{M}$

Therefore, the final concentration of sodium ions in the solution is $2.07 \, \text{M}$.