First, calculate the moles of nitrate ions contributed by the nitric acid (\(\text{HNO}_3\)). Since nitric acid is a strong acid, it dissociates completely in water, providing one nitrate ion per molecule of \(\text{HNO}_3\).
\[
\text{Moles of } \text{HNO}_3 = \text{Volume (L)} \times \text{Concentration (M)} = 0.0350 \, \text{L} \times 0.255 \, \text{M} = 0.008925 \, \text{mol}
\]
Since each \(\text{HNO}_3\) provides one \(\text{NO}_3^-\), the moles of \(\text{NO}_3^-\) from \(\text{HNO}_3\) is also \(0.008925 \, \text{mol}\).
Next, calculate the moles of nitrate ions contributed by magnesium nitrate (\(\text{Mg(NO}_3\text{)}_2\)). Each \(\text{Mg(NO}_3\text{)}_2\) provides two nitrate ions.
\[
\text{Moles of } \text{Mg(NO}_3\text{)}_2 = \text{Volume (L)} \times \text{Concentration (M)} = 0.0450 \, \text{L} \times 0.328 \, \text{M} = 0.01476 \, \text{mol}
\]
The moles of \(\text{NO}_3^-\) from \(\text{Mg(NO}_3\text{)}_2\) is:
\[
2 \times 0.01476 \, \text{mol} = 0.02952 \, \text{mol}
\]
Add the moles of nitrate ions from both sources to find the total moles of nitrate ions in the solution.
\[
\text{Total moles of } \text{NO}_3^- = 0.008925 \, \text{mol} + 0.02952 \, \text{mol} = 0.038445 \, \text{mol}
\]
Calculate the total volume of the solution by adding the volumes of the two solutions.
\[
\text{Total volume} = 35.0 \, \text{mL} + 45.0 \, \text{mL} = 80.0 \, \text{mL} = 0.0800 \, \text{L}
\]
Finally, calculate the concentration of nitrate ions in the final solution.
\[
\text{Concentration of } \text{NO}_3^- = \frac{\text{Total moles of } \text{NO}_3^-}{\text{Total volume (L)}} = \frac{0.038445 \, \text{mol}}{0.0800 \, \text{L}} = 0.4806 \, \text{M}
\]
\[
\boxed{0.4806 \, \text{M}}
\]
Answered
