Questions: 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?

Solution Steps

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.

$$\text{Concentration of } \text{NO}_3^- = \frac{\text{Total moles of } \text{NO}_3^-}{\text{Total volume (L)}} = \frac{0.038445 \, \text{mol}}{0.0800 \, \text{L}} = 0.4806 \, \text{M}$$

Final Answer