35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO₃)₂. What is the concentration of nitrate ion in the final solution A) 0.481 M B) 0.296 M C) 0.854 M D) 1.10 M E) 0.0295 M

Question

Quizplus · Accepted Answer

The concentration of nitrate ion in the final solution can be calculated by considering the contributions from both nitric acid and magnesium nitrate. Nitric acid (HNO3) provides one nitrate ion per molecule, and magnesium nitrate [Mg(NO3)2] provides two nitrate ions per molecule.First, calculate the moles of nitrate from nitric acid:$$ \text{Moles of } \text{NO}_3^- \text{ from HNO}_3 = 0.255 \, \text{M} \times 0.035 \, \text{L} = 0.008925 \, \text{mol} $$Then, calculate the moles of nitrate from magnesium nitrate:$$ \text{Moles of } \text{NO}_3^- \text{ from Mg(NO}_3)_2 = 0.328 \, \text{M} \times 0.045 \, \text{L} \times 2 = 0.02952 \, \text{mol} $$Total moles of nitrate ions:$$ \text{Total moles of } \text{NO}_3^- = 0.008925 + 0.02952 = 0.038445 \, \text{mol} $$Total volume of the solution:$$ \text{Total volume} = 35.0 \, \text{mL} + 45.0 \, \text{mL} = 80.0 \, \text{mL} = 0.080 \, \text{L} $$Finally, calculate the concentration of nitrate ions:$$ \text{Concentration of } \text{NO}_3^- = \frac{0.038445 \, \text{mol}}{0.080 \, \text{L}} = 0.4805625 \, \text{M} $$Rounded to three significant figures, the concentration of nitrate ion in the final solution is 0.481 M.

350 ML of 0